Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12.1 जयपूर फूटचे जनक

Balbharti Maharashtra State Board Marathi Yuvakbharati 12th Digest Chapter 12.1 जयपूर फूटचे जनक Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board 12th Marathi Yuvakbharati Solutions Chapter 12.1 जयपूर फूटचे जनक

12th Marathi Guide Chapter 12.1 जयपूर फूटचे जनक Textbook Questions and Answers

नमुना कृती

1. परिणाम लिहा :

प्रश्न 1.

घटना परिणाम
अ. अपघातामध्ये सुधाला एक पाय गमवावा लागला. …………………..
आ. परदेशातून आयात केलेला कृत्रिम पाय बसवलेल्या रुग्णांना नीट चालता येत नव्हते. ………………….

उत्तर :

घटना परिणाम
अ. अपघातामध्ये सुधाला एक पाय गमवावा लागला. तिचे नृत्य कायमचेच बंद पडण्याच्या मार्गावर होते.
आ. परदेशातून आयात केलेला कृत्रिम पाय बसवलेल्या रुग्णांना नीट चालता येत नव्हते. पंडितजींच्या कल्पकतेला येथे आव्हान मिळाले.

Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12.1 जयपूर फूटचे जनक

2. अभिव्यक्ती.

प्रश्न 1.
‘कृत्रिम पायाच्या मदतीने दि व्यां गत्वाव र मात करता येते’, हे सोदाहरण स्पष्ट करा.
उत्तर :
जयपूर फूट आता जगप्रसिद्ध झाला आहे. आजपर्यंत हजारो विकलांग लोकांनी हा कृत्रिम पाय बसवून घेतला आहे. ही मुलेमाणसे आता सर्वसाधारण आयुष्य जगत आहेत. यांच्या कथा प्रेरणादायक आहेत.

एक कथा आहे नायरा नावाच्या तीन वर्षांच्या मुलीची. तिचे वडील स्वतः फार्मासिस्ट आहे. नायरा जन्मतः च कमकुवत होती. तीन वर्षांपर्यंत तिला उभे राहता येत नव्हते, चालता येत नव्हते. हा जयपूर फूट बसवल्यावर मात्र नायरा सोबतच्या मुलीबरोबर खेळू लागली; बागडू लागली; धावू लागली.

कॉर्पोरेट जगताने आता यात लक्ष घालायला सुरुवात केली आहे. कॉर्पोरेट कंपन्यांपैकी एक आहे – लेमन ट्री हॉटेल. एक २३ वर्षांचा तरुण पायाने अधू होता. त्याला एक दानशूर व्यक्तीने जयपूर फूट बसवून दिला. तो आता सहायक म्हणून या हॉटेलमध्ये काम करतो. हॉटेलमध्ये आलेल्या पाहुण्यांची देखभाल करण्याचे काम तो करतो. तो आता हॉटेलचा मॅनेजर होण्याचे स्वप्न पाहत आहे. त्या कंपनीत आतापर्यंत ४०० विकलांगांची भरती केली गेली आहे.
एकंदरीत, विकलांगत्वावर मात करून सर्वसाधारण माणसाचे आयुष्य नक्की जगता येऊ शकते, हे आता सगळ्यांनाच मान्य झालेले आहे.

Marathi Yuvakbharati 12th Digest Chapter 12.1 जयपूर फूटचे जनक Additional Important Questions and Answers

कारणे लिहा :

प्रश्न 1.
पंडितजींच्या कल्पकतेला आव्हान मिळाले; कारण –
उत्तर :
पंडितजींच्या कल्पकतेला आव्हान मिळाले; कारण उपलब्ध कृत्रिम पाय खूप महागडे होते आणि ज्यांना ते परवडणारे होते त्यांची चाल सुलभ होताना दिसत नव्हती.

प्रश्न 2.
त्या कृत्रिम पायाचे नाव ‘जयपूर फूट’ ठेवण्यात आले; कारण –
उत्तर :
त्या कृत्रिम पायाचे नाव ‘जयपूर फूट’ ठेवण्यात आले; कारण जयपूरमधल्या एका रुग्णालयात तो प्रथम विकसित झाला होता.

Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12.1 जयपूर फूटचे जनक

चौकटी पूर्ण करा :

प्रश्न 1.

  1. सुधाचंद्राच्या आयुष्यावरील चित्रपटाचे नाव – [ ]
  2. जयपूरमध्ये मिळणाऱ्या कृत्रिम पायाचे नाव – [ ]
  3. विकलांगांवर उपचार करणाऱ्या डॉक्टरांचे नाव – [ ]
  4. कृत्रिम पाय तयार करणारे – [ ]
  5. सुरुवातीला पंडितर्जीनी पाय तयार करण्यासाठी वापरलेला पदार्थ – [ ]

उत्तर :

  1. सुधाचंद्राच्या आयुष्यावरील चित्रपटाचे नाव : नाचे मयूरी
  2. जयपूरमध्ये मिळणाऱ्या कृत्रिम पायाचे नाव : जयपूर फूट
  3. विकलांगांवर उपचार करणाऱ्या डॉक्टरांचे नाव : डॉ. प्रमोद किरण सेठी
  4. कृत्रिम पाय तयार करणारे : पंडित रामचरण शर्मा
  5. सुरुवातीला पंडितजींनी पाय तयार करण्यासाठी वापरलेला पदार्थ : बांबू

Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12.1 जयपूर फूटचे जनक

जयपूर फूटचे जनक Summary in Marathi

जयपूर फूटचे जनक

‘नाचे मयूरी’ हा चित्रपट अनेकांनी पाहिला असेल. सुप्रसिद्ध नर्तिका सुधा चंद्रन हिच्या आयुष्यावर तो आधारित होता. एका अपघातामध्ये सुधाला एक पाय गमवावा लागला होता. साहजिकच तिचं नृत्य कायमचंच बंद पडण्याच्या मार्गावर होतं; पण सुधा जयपूरला गेली आणि तिथं तयार करण्यात आलेला कृत्रिम पाय आपल्या गमावलेल्या पायाच्या जागी बसवला, नृत्याची कारकीर्द तिनं नव्यानं सुरू केली आणि त्या कृत्रिम पायाच्या आधारानं तिनं भरघोस यश मिळवलं. सुधानं बसवून घेतलेल्या त्या कृत्रिम पायाचंच नाव आहे ‘जयपूर फूट’, जयपूरमधल्या एका रुग्णालयात तो प्रथम विकसित केला गेला म्हणून त्याला ते नाव मिळालं.

जयपूरच्या रुग्णालयात डॉ. प्रमोद किरण सेठी अनेक विकलांगांवर उपचार करीत होते. पोलिओची बाधा झाल्यामुळे दिव्यांगत्व आलेल्या मुलांना पाहून त्यांना एक कल्पना सुचली. पंडित राम चरण शर्मा या कलाकाराला विविध प्रकारची विलक्षण साधनं तयार करताना त्यांनी पाहिलं होतं. त्यांनी पंडित ना सणालयात येण्याचं आमंत्रण दिलं.

पंडितींनी सणालयात, ज्यांचे पाय काही कारणांनी गमावले आहेत अशांना परदेशातून आयात केलेले, महागडे कृत्रिम पाय बसवताना पाहिलेलं होतं, ते परवडणारे नव्हते आणि ज्यांना ते परवडणारे होते त्यांचीही चाल काही सुलभ होत असताना त्यांना दिसली नव्हती. ते पाहून त्यांच्या कल्पकतेला आव्हान मिळालं. त्यांनी व्हल्कनाईज्ड रबर आणि लाकूड या सहजगत्या उपलब्ध असलेल्या कच्च्या मालापासून हालचाल करण्यास सुलभ असा पाय तयार केला.

डॉ. सेठी यांनी तो आपल्या एका रुग्णाला बसवून पाहिला. त्यासाठी शस्त्रक्रियेची नवी पद्धत विकसित केली, त्या रुपणाला त्याचा फायदा झाल्याचं पाहून त्यांनी पंडितजींना आणखी तसेच पाय तयार करायला सांगितलं. आता परदेशातून कृत्रिम पाय आयात न करता हे लाकडी पाय बसवण्याचाच सिलसिला सुरू झाला. सुरुवातीला तर पंडितजींनी बांबूचाच वापर केला होता; पण हळूहळू इतरही पदार्थाचा वापर करायला त्यांनी सुरुवात केली.

आता जगभर त्याचं रोपण केलं जातं. अदययावत प्लास्टिक व अॅल्युमिनियम यांचा वापरही आता करण्यात येतो. पण मूळ कल्पना मात्र पंडितर्जीचीच राहिली आहे.
– डॉ. बाळ फोंडके

Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा

Balbharti Maharashtra State Board Marathi Yuvakbharati 12th Digest Chapter 12 रंगरेषा व्यंगरेषा Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board 12th Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा

12th Marathi Guide Chapter 12 रंगरेषा व्यंगरेषा Textbook Questions and Answers

कृती

1. अ. लेखकाने खालील गोष्टी कळण्यासाठी व्यंगचित्रात वापरलेली प्रतीके लिहा.

प्रश्न 1.
Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा 1
उत्तर :

पाठातीलगोष्टी प्रतीके
1. चिमुरड्या मुलीचं डोकं – नारळ
2. आई हे नातं – ईश्वराचा अंश
3. भरपावसातली छत्री – बाबा

आ. वैशिष्ट्ये लिहा.

प्रश्न 1.
Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा 3
उत्तर :
Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा 5

प्रश्न 2.
Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा 4
उत्तर :
Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा 6

इ. योग्य जोड्या लावा.

प्रश्न 1.

‘अ’ गट
लेखकाचीव्यंगचित्रे
‘ब’ गट
व्यंगचित्रांचीकार्ये
1. लेखकाचे स्त्रीभ्रूणहत्येचे पोस्टर अ. भाषेइतकी संवादी बनून प्रेक्षकांशी संवाद साधतात.
2. लेखकाच्या मते व्यंगचित्रे ही आ. स्वकल्पनाशक्तीने चित्र समजून घेऊन इतरांचे उद्बोधन केल
3. शेतकऱ्याने व्यंगचित्राचा अर्थ इतरांना सांगताना इ. लिहिता वाचता न येणाऱ्यांना संदेश देते.

उत्तर :

‘अ’ गट
लेखकाचीव्यंगचित्रे
‘ब’ गट
व्यंगचित्रांचीकार्ये
1. लेखकाचे स्त्रीभ्रूणहत्येचे पोस्टर इ. लिहिता वाचता न येणाऱ्यांना संदेश देते.
2. लेखकाच्या मते व्यंगचित्रे ही अ. भाषेइतकी संवादी बनून प्रेक्षकांशी संवाद साधतात.
3. शेतकऱ्याने व्यंगचित्राचा अर्थ इतरांना सांगताना आ. स्वकल्पनाशक्तीने चित्र समजून घेऊन इतरांचे उद्बोधन केल

Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा

ई. लेखकाला लागू पडण्याचा व्यक्तिवैशिष्ट्यांसमोर (✓) अशीखूणकरा.

प्रश्न 1.

  1. लेखकामध्ये जबरदस्त निरीक्षणशक्ती होती.
  2. लेखकाच्या व्यंगचित्रांना सहजासहजी प्रसिद्धी मिळाली.
  3. लेखकामध्ये प्रयोगशीलता पुरेपूर भरलेली होती.
  4. अपेक्षित उत्तर मिळेपर्यंत ते विचारांचा पाठलाग करत.
  5. प्राप्त प्रसंगांतून आणि भेटलेल्या व्यक्तींकडून शिकत राहण्याची वृत्ती होती.
  6. नवनिर्मितीक्षमता हा त्यांचा गुण होता.
  7. इतरांच्या आधाराने पुढे जाण्याची त्यांची वृत्ती नव्हती.

उत्तर :

  1. लेखकामध्ये जबरदस्त निरीक्षणशक्ती होती.(✓)
  2. लेखकाच्या व्यंगचित्रांना सहजासहजी प्रसिद्धी मिळाली. (✗)
  3. लेखकामध्ये प्रयोगशीलता पुरेपूर भरलेली होती. (✓)
  4. अपेक्षित उत्तर मिळेपर्यंत ते विचारांचा पाठलाग करत. (✓)
  5. प्राप्त प्रसंगांतून आणि भेटलेल्या व्यक्तींकडून शिकत राहण्याची वृत्ती होती. (✓)
  6. नवनिर्मितीक्षमता हा त्यांचा गुण होता. (✓)
  7. इतरांच्या आधाराने पुढे जाण्याची त्यांची वृत्ती नव्हती. (✓)

2. वर्णनकरा.

प्रश्न अ.
वाई येथील प्रदर्शनाला भेट देणारा शेतकरी कुटुंबप्रमुख.
उत्तर :
चित्रकलेच्या प्रदर्शनाला साधारणपणे सुशिक्षित व उच्चभ्रू वर्गातील लोक जास्त असतात. ग्रामीण भागातील लोक तर अशा प्रदर्शनांकडे सहसा फिरकत नाहीत. मात्र वाई येथे लेखकांनी भरवलेल्या स्वत:च्या व्यंगचित्रांच्या प्रदर्शनाला शेतकरी कुटुंबातील थोडीथोडकी नव्हेत, तर चक्क वीस-बावीस माणसे भेट देण्यासाठी आली होती. त्यांचा कुटुंबप्रमुख त्यांना हे प्रदर्शन दाखवण्यासाठी घेऊन आला होता.

त्या शेतकरी कुटुंबप्रमुखाचे वय सत्तरीच्या आसपास होते. पांढऱ्या मिशा, रंग काळाकभिन्न, डोक्याला खूप मोठे बांधलेले मुंडासे असा नखशिखान्त शेतकरी पण अंगावर वागवीत होता. असा हा कुटुंबप्रमुख सर्वांना व्यंगचित्र समजावून सांगत होता. तो एकेका चित्रासमोर उभा राही आणि त्याला समजलेला चित्राचा अर्थ स्वत:च्या माणसांना समजावून सांगे. मुलानातवंडापासून लहानथोर सोबत आलेले ते कुटुंबीय आपल्या प्रमुखाच्या मार्गदर्शनाखाली चित्रांचा आस्वाद घेत होती. हे दृश्यच विलक्षण व दुर्मीळ होते. लेखकांच्या मनातली चित्र काढण्यामागील कल्पना आणि त्या कुटुंबप्रमुखाला जाणवलेला अर्थ यांतली तफावत लेखक समजावून घेत होते. फार मोठे अनौपचारिक शिक्षण लेखकांना या प्रसंगातून मिळत होते.

Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा

प्रश्न आ.
स्त्रीभ्रूणहत्येबद्दलचे लेखकाने तयार केलेले पोस्टर
उत्तर :
लेखक एका कार्यक्रमाला गेले होते. पाहुणे म्हणून त्यांना पुष्पगुच्छ व नारळ दिला गेला. तो नारळ टेबलावर ठेवला. लेखक त्या नारळाचे निरीक्षण करीत बसले. त्या नारळात त्यांना चिमुरड्या मुलीचे डोके भासले. त्यांना नारळावरून, देवळासमोर दगडावर आपटून नारळ फोडतात. हा प्रसंग आठवला. एवढ्या तपशिलाच्या आधारे त्यांनी स्त्रीभ्रूणहत्येविरुद्धचे पोस्टर तयार केले. एक पुरुषी हात. त्या हातात नारळ, नारळात मुलीचे रूप भासावे म्हणून बारीकसा कानातला डूल दाखवला. तो हात वरून खाली या दिशेने येत दगडावर नारळ फोडणार होता. तेवढ्यात एका तरुण हाताने तो पुरुषी हात अडवला. चिमुरड्या मुलींची, भ्रूणाची हत्या होऊ न देण्याचा निर्धार त्या चित्रातून व्यक्त झाला.

प्रश्न इ.
लेखकांनी रेखाटलेले आईचे काव्यात्म चित्र.
उत्तर :
लेखकांनी आईचे, आईच्या प्रेममय हृदयाचे अत्यंत हृदय चित्र रेखाटले आहे. चित्रातल्या परिसरात वैशाखातला वणवा पेटला आहे. त्यात एक सुकलेले झाड आहे. त्या झाडावर एकही पान नाही. अत्यंत भकास वातावरण आहे. तरीही त्या झाडावर एका पक्ष्याचे घरटे बांधले आहे. त्या घरट्यात चोच वासून आकाशाकडे व्याकुळपणे बघणारी तीनचार पिल्ले आहेत. अत्यंत हृदयद्रावक असे हे दृश्य आहे. त्या चित्रात वर दूरवर ठिपक्यासारखी दिसणारी पक्षीण पाण्याच्या ढगाला चोचीत धरून जिवाच्या आकांताने ओढीत घरट्याकडे नेत आहे. एवढ्या या एका कृतीतून त्या पक्षिणीची आपल्या पिल्लांना वाचवण्याची चाललेली जिवापाड धडपड प्रभावीपणे व्यक्त होते. आईची अपार माया या चित्रातून दिसून येते.

3. व्याकरण.

अ. खालील वाक्यांतील प्रयोग ओळखा.

प्रश्न 1.
या चित्रांचे स्रोत मला सापडतात.
उत्तर :
कर्तरी प्रयोग.

प्रश्न 2.
हा संदेश मला पोहोचवता आला.
उत्तर :
कर्मणी प्रयोग.

Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा

प्रश्न 3.
त्यांनी ती सातआठ चित्रं पुन्हा चितारुन दाखवली.
उत्तर :
कर्मणी प्रयोग.

प्रश्न 4.
मार्गदर्शन संपवून चहा मागवला.
उत्तर :
कर्मणी प्रयोग.

आ. खालील वाक्यांचा प्रकार ओळखून सूचनेप्रमाणे तक्ता पूर्ण करा.

प्रश्न 1.
Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा 7
उत्तर :

वाक्य वाक्यप्रकार बदलासाठी सूचना
1. अशी माणसं क्वचितच सापडतात. विधानार्थी – होकारार्थी वाक्य नकारार्थी → अशी माणसे बहुतेक सापडत नाहीत.
2. ती जुनी कौलारू वास्तू होती. विधानार्थी वाक्य उद्गारार्थी → किती जुनी कौलारू वास्तू होती ती!
3. तुम्ही मला बोलू देणार आहात की नाही? प्रश्नार्थी वाक्य आज्ञार्थी → तुम्ही मला बोलू दया.

Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा

इ. समासाचे नाव व सामासिक शब्द यांच्या जोड्या जुळवून लिहा

प्रश्न 1.
Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा 8
उत्तर :

समासाचे नाव सामासिक शब्द
1. तत्पुरुष समास आ. महात्मा, पंचधातू
2. अव्ययीभाव समास इ. प्रतिवर्ष, आजन्म
3. बहुव्रीही समास ई. लक्ष्मीकांत, निर्धन
4. द्वंद्व समास अ. स्त्रीपुरुष, गुण दोष.

ई. कंसात समासांची नावे दिलेली आहेत. पुढे दिलेल्या सामासिक शब्दांसमोर त्यापैकी योग्य समासाचे नाव लिहा :
(विभक्ती तत्पुरुष समास, कर्मधारय समास, द्‌विगू समास, इतरेतर द्वंद्‌व समास, वैकल्पिक द्वंद्‌व समा, समाहार द्‌वंद्‌व समास)

प्रश्न 1.

  1. चहापाणी – …………………. .
  2. सद्गुर – …………………. .
  3. सुईदोरा – …………………. .
  4. चौघडी – …………………. .
  5. कमीअधिक – …………………. .
  6. जलदुर्ग – …………………. .

उत्तर :

  1. चहापाणी – समाहार द्वंद्व समास
  2. सद्गुरू – कर्मधारय समास
  3. सुईदोरा – इतरेतर द्वंद्व समास
  4. चौघडी – द्विगु समास
  5. कमीअधिक – वैकल्पिक द्वंद्व समास
  6. जलदुर्ग – विभक्ती तत्पुरुष समास.

4. स्वमत.

प्रश्न अ.
‘एखादे व्यंगचित्र हे प्रत्यक्ष भाषेपेक्षा संवादाचे प्रभावी माध्यम असू शकते’, या विधानाशी तुम्ही सहमत वा असहमत अाहात ते सकारण स्पष्ट करा.
उत्तर :
भाषा आणि व्यंगचित्र ही दोन भिन्न माध्यमे आहेत. त्यांचे सादरीकरण भिन्न असते आणि त्यांचे परिणामही भिन्न असतात. आपण घर हा शब्द लिहितो, तेव्हा घर या शब्दाचा आकार पाहून किंवा घर या शब्दाच्या उच्चारातून जो ध्वनी निर्माण होतो, तो ध्वनी ऐकून घराचा बोध होऊ शकणार नाही. फक्त मराठी भाषा समजणाऱ्यालाच घर या शब्दाचा बोध होऊ शकतो. घर शब्दाच्या आकाराशी व उच्चाराशी घर ही वस्तू जोडलेली आहे.

ज्याला हा संकेत माहीत आहे आणि ज्याने तो संकेत लक्षात ठेवला आहे, त्यालाच घर या शब्दाचा बोध होऊ शकतो. याउलट, घराचे चित्र दाखवल्यावर ते जगातल्या कोणत्याही व्यक्तीला त्या चित्रातून व्यक्त होणाऱ्या वस्तूचा बोध होतो. तिथे भाषेची आडकाठी येत नाही. देशाच्या सीमा आड येत नाहीत. सामाजिक, सांस्कृतिक दर्जाचा संबंध येत नाही. चित्रातून आशयाचे थेट आकलन होते. म्हणून भाषेपेक्षा चित्र आशयाला प्रेक्षकांपर्यंत पटकन व थेट पोहोचवते.

चित्र व व्यंगचित्र यांत काहीएक फरक आहे. व्यंगचित्रात काही व्यक्ती दाखवलेल्या असतात. व्यंगचित्रकार कधीही विषयवस्तूचे फक्त वर्णन करीत नाही. त्याला माणसाच्या वृत्ती-प्रवृत्तींवर काहीएक भाष्य करायचे आहे. त्या वृत्ती-प्रवृत्ती ठळकपणे लक्षात याव्यात म्हणून माणसांचे चेहेरे, त्यांवरचे हावभाव, त्यांच्या हालचाली यांतील काही रेषा मुद्दाम ठळकपणे चितारतो. त्यामुळे चित्र हे व्यंगचित्र बनते. व्यंगचित्रातून माणसाच्या वृत्ती-प्रवृत्तींवर केलेली टीका कोणालाही पटकन कळू शकते. तोच भाव समजावून सांगण्यासाठी खूप शब्द वापरावे लागतात. खूप शब्द वापरूनही सर्व आशय नेमकेपणाने व्यक्त होतोच असे नाही. व्यंगचित्राच्या बाबतीत असे घडत नाही. तेथे आशय स्पष्टपणे लक्षात येतो.

व्यंगचित्रात चालू घडामोडींवर भाष्य असते. व्यंगचित्र पाहणारा प्रेक्षक चालू घडामोडींचा साक्षीदारही असतो. म्हणून त्याला व्यंगचित्र पटकन कळू शकते.

Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा

प्रश्न आ.
‘वाहत्या आयुष्यामध्ये सावधगिरीनं उभं राहिलं तर व्यंगचित्राची कल्पना अगदी जवळून जाते,’ या विधानाचा तुम्हांला समजलेला अर्थ लिहा.
उत्तर :
कलावंतांविषयी, कलेविषयी सर्वच समाजात विलक्षण कुतूहल असते. व्यंगचित्रकार हाही एक कलावंतच असतो. व्यंगचित्रकाराची व्यंगचित्र दररोज प्राधान्याने वर्तमानपत्रांतून लोकांसमोर येत असतात. दररोज त्याची लोकांशी गाठभेट होत असते. वर्तमानपत्र वाचणारा प्रत्येक वाचक व्यंगचित्र पाहतोच पाहतो. व्यंगचित्र पाहताच त्या वाचकाला त्याचे मर्म खाडकन जाणवते. ते मर्म क्षणार्धात त्याच्या मनात शिरते. चेहेऱ्यावर स्मित तरळते. त्याच क्षणी तो वाचक व्यंगचित्रकाराला मनोमन दाद देतो. किती सुंदर कल्पना आहे ही! कशी सुचली असेल या व्यंगचित्रकाराला? आपल्यासारखाच हा माणूस. हातपाय, नाक, कान, डोळे हे सर्व अवयव आपल्यासारखेच. यांना कशी काय सुचते हे व्यंगचित्र?

लोकांच्या मनातील या प्रश्नालाच मंगेश तेंडुलकर यांनी प्रस्तुत पाठात दोन वाक्यात उत्तर दिले आहे. ‘वाहत्या आयुष्यात सावधगिरीने उभे राहिले, तर व्यंगचित्राची कल्पना आपल्या जवळूनच जाताना नजरेस पडेल. तिथून ती उचलायची आणि कागदावर उतरवायची.’ त्यांनी अत्यंत सोप्या शब्दांत उत्तर दिले आहे. व्यंगचित्रकार कमीत कमी रेषांमध्ये आपला आशय व्यक्त करतो. तसेच इथे लेखकांनी कमीत कमी शब्दांत आशय व्यक्त केला आहे. वाहत्या आयुष्यात म्हणजे दैनंदिन जीवन जगत असतानाच. त्याच जीवनाचे निरीक्षण केले असता, आपण जगत असलेल्या प्रसंगातच व्यंगचित्राची कल्पना सापडते. या कल्पनेसाठी रानावनात जाऊन वेगळी तपश्चर्या करावी लागत नाही.

मग काय करावे लागते? तर आपल्या जगण्याचेच तटस्थपणाने, त्रयस्थासारखे निरीक्षण करावे लागते. लेखकांनी यासाठीच ‘सावधगिरीने’ हा शब्द वापरला आहे. सावधगिरीने याचा अर्थ समजून घ्यायला हवा. आपण आपल्या दैनंदिन जीवनात साधारणपणे आपल्या अनुभवांकडे दुर्लक्ष करतो किंवा त्यांत पूर्णपणे बुडून जातो. त्यामुळे आपल्या अनुभवाचे मर्म आपल्या ध्यानात येत नाही. म्हणून काही क्षण तरी आपल्या अनुभवांकडे रेंगाळून पाहिले पाहिजे.

जो विचार आपल्याला उत्कटपणे सांगावासा वाटतो, त्याला साजेसा प्रसंग आपल्याला दिसतो, असे लेखकांना सुचवायचे आहे. एकदा कल्पना सुचली की चित्र काढणे सोपे असते. खरे सर्वच कलांच्या बाबतीत हे पूर्णपणे खरे आहे. लेखकांनी खूप गूढ अशा प्रश्नाला साध्या, सोप्या शब्दांत उत्तर दिले आहे.

प्रश्न इ.
लेखकांनी व्यंगचित्रांतून वडिलांना वाहिलेली श्रद्धांजली तुमच्या शब्दांत लिहा.
उत्तर :
लेखक मंगेश तेंडुलकर हे अत्यंत मनस्वी वृत्तीचे होते. ते जे काही करायला घेत त्यावर ते उत्कटपणे प्रेम करीत. ते अत्यंत करारी होते. ते ऐहिक सुखसमृद्धीच्या मागे धावले नाहीत. कोणासमोर हात पसरले नाहीत. या सगळ्या गुणांची देणगी लेखकांना मिळाली. आयुष्यात अनेक संकटे आली, वावटळी आल्या. पण त्यांना तोंड देऊन लेखक भक्कमपणे स्वत:च्या पायावर उभे राहिले. हे ते करू शकले कारण त्यांना वडिलांकडून मिळालेला नैतिक वारसा. त्या बळावर आयुष्यात तग धरून राहिले. कोलमडले नाहीत. आपल्याला मिळालेल्या या वारशाबद्दल स्वत:च्या मनात लेखकांना अपार कृतज्ञता वाटत होती. ही कृतज्ञता व्यक्त करण्यासाठी त्यांनी स्वतःच्या वडिलांना व्यंगचित्र मधून खूप मनापासून श्रद्धांजली वाहिली.

ही श्रद्धांजली वाहण्यासाठी त्यांनी दोन चित्रे काढली. एका चित्रात लेखकांचे बालरूप आहे. बालरूपातले लेखक भर पावसात उभे आहेत. त्यांच्या डोक्यावर ‘बाबा’ ही दोन अक्षरे आहेत. ‘बाबा’ या शब्दाच्या दोन्ही बाजूंनी पाऊस पडत आहे. खाली इवलासा मुलगा पूर्णपणे सुरक्षित आहे. बाबा या शब्दांनी, म्हणजे बाबांनी त्यांचे रक्षण केले.

दुसऱ्या चित्रात लेखक सरत्या वयातले, वयाच्या पंच्याहत्तराव्या वर्षातले, उभे आहेत. भर पावसात उभे आहेत. डोक्यावर बाबा हा शब्द नाही. तरीही पाऊस लेखकांना न भिजवता त्यांच्या बाजूने पडत आहे. ते आता पंच्याहत्तराव्या वर्षीही सुरक्षित आहेत. बाबांकडून मिळालेला नैतिक वारसा त्यांच्या मृत्यूनंतर लेखकांना खूप मोठा आधार, आश्रय देत आला आहे. वडिलांबद्दलची ही कृतज्ञता लेखक या व्यंगचित्रातून व्यक्त करू पाहतात. स्वत:च्या पित्याला इतकी उत्कट श्रद्धांजली क्वचितच कोणीतरी वाहिली असेल.

Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा

5. अभिव्यक्ती.

प्रश्न अ.
‘स्त्रीभ्रूणहत्या एक अपराध’ याविषयी तुमचे विचार स्पष्ट करा.
उत्तर :
आपला देश अंधश्रद्धेच्या मगरमिठीत पूर्णपणे अडकलेला आहे. मागासलेल्या विचारांच्या दलदलीत बुडालेला आहे. यामुळे समाजात सामाजिक – सांस्कृतिक दुर्गुण निर्माण झाले आहेत. अत्यंत दुष्ट, अन्यायकारक रूढी-परंपरा निर्माण झाल्या आहेत. या रूढींच्या बरोबरीने समाजाच्या मनात घातक व अन्यायकारक विचार रुजत चालले आहे. त्यांपैकी एक आहे – स्त्री पुरुष समानता. समाजमनात स्त्री ही कनिष्ठ व पुरुष हा श्रेष्ठ अशी धारणा निर्माण झाली आहे.

मुलगा हा कुलदीपक व मुलगी परक्याची धनसंपदा, अशी समजूत. मुलगी घरात जन्मणे हे अशुभ. आपल्या हातून पाप घडले असेल तरच आपल्या पोटी मुलगी जन्माला येते, यामुळे घरात मुलगा जन्माला यावा यासाठी लोक धडपड करतात. उपासतापास करतात. लोक मुलगी होणार असेल, तर तिला जन्म होण्याच्या आधीच मारतात. हे सर्रास होत होते. अनेक डॉक्टर यात सामील होते. याविरुद्ध आता कडक कायदे झालेले आहेत. तरीही अधूनमधून हे कृत्य घडताना दिसते.

खरे तर मुलगी जन्मण्याच्या आधीच तिला मारणे, हा खूनच होय. हा एका व्यक्तीचा खून या अर्थाने ही भीषणच घटना आहे. पण हे व्यक्तीच्या मृत्यूपुरते थांबत नाही. यामुळे संपूर्ण मानव जातच धोक्यात येऊ शकते. लोकसंख्येत पुरुषांचे प्रमाण जास्त आणि स्त्रियांचे प्रमाण कमी या वास्तवामुळे फार मोठे सामाजिक प्रश्न निर्माण होणार आहेत. या प्रश्नांच्या आगीत संपूर्ण समाज होरपळून जाण्याची शक्यता आहे.

या वस्तुस्थितीकडे जरा नीट पाहिले, तिच्यातली विपरीतता स्पष्ट होईल. स्त्रियांची संख्या समाजात सामान्यत: निम्मी असते. आपला अर्धा समाज अन्यायग्रस्त राहिला तर त्याची प्रगती होणार तरी कशी?

स्त्रीभ्रूणहत्येच्या समस्येवर ‘डॉक्टर तुम्हीसुद्धा?’ या नावाचे नाटकही येऊन गेले आहे. समाजाचा एक भाग या समस्येची भीषणता ओळखून आहे. पण ज्याला कळलेच नाही, असा समाजाचा जो भाग आहे तो खूप मोठा आहे. हा समाजगट कितीही मोठा असला, तरी सुज्ञ लोकांनी याविरुद्ध लढले पाहिजे. स्त्रीभ्रूणहत्या हा खूनच होय. आणि अशी कृत्ये करणारी माणसे खुनी होत, असेच समाजाने मानले पाहिजे. तरच या भीषण रूढीला आळा बसेल.

Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा

प्रश्न आ.
‘आईचं नातं सगळ्या जगातलं एकमेव खरं आणि सुंदर नातं आहे,’ या वाक्यातील आशयसौंदर्य उलगडून दाखवा.
उत्तर :
रवींद्रनाथांची एक कथा आहे. एका आईचे आपल्या मुलावर अतोनात प्रेम होते. तिने त्याला कष्टपूर्वक वाढवले. एकदा त्याला देवाचा साक्षात्कार झाला. देवाने त्याला वर दिला, “बाळा काय हवे ते माग.” त्याने देवाकडे अमरत्व मागितले. देवाने ही मागणी मान्य केली, पण त्याने एक अट घातली. “मला तू तुझ्या आईचे हृदय आणून दे.”

बाळाला खात्री होती की आई आपल्याला तिचे हृदय देणारच. तिचे त्याच्यावर अपार प्रेम होते. मुलाने आईकडे येऊन हृदयाची मागणी केली. आईचे हृदय घेतले. आई मरून पडली. मुलगा धावत धावत देवाकडे निघाला. वाटेत त्याला ठेच लागली. त्याच्या हातातले हृदय जमिनीवर पडले. मुलगा हृदय उचलायला धावला तेवढ्यात त्याच्या कानावर शब्द आले, “बाळा, तुला काही लागलं नाही ना?”

ही कथा काल्पनिक आहे, यात शंका नाही. पण या कथेतून आईचे अपार ममत्व, त्यातील उदात्तता, त्यातील शुद्धता, मुलाबाबतची ओढ हे सारे विलक्षण नजाकतीने व्यक्त झाले आहे. आई आपल्या बाळाला नऊ महिने आपल्या कुशीत सांभाळते. तो निव्वळ गोळा असतो, तेव्हा ती स्वत:चे रक्त देऊन त्याचे पालनपोषण करते. खरे तर आपल्याला जसे हात, पाय इत्यादी आपले अवयव असतात, तसाच कोणताही बाळ त्याच्या आईला स्वत:च्या देहाचाच भाग वाटत असतो.

त्यामुळे तिच्या आयुष्यातली सगळी शुद्धता, सगळे पावित्र्य त्या नात्यात एकवटलेले असते. बाळाची भूक प्रथम आईला लागते. बाळाला जरा कुठे काही लागले, तर त्याची कळ आईच्या हृदयात प्रथम उमटते. नीट बारकाईने निरीक्षण केले तर लक्षात येईल की, बाळाच्या वाटचालीकडे, त्याच्या शिक्षणाकडे, त्याच्या प्रकृतीकडे आईचे पूर्ण लक्ष असते. त्याच्या विकासाबाबत ती आत्यंतिक संवेदनशील असते. आईचे प्रेम शुद्ध, पवित्र असते, याचे कारण ती आपल्या बाळासाठी जे काही करते, त्याबदल्यात बाळाकडून कोणतीही अपेक्षा ठेवत नाही. म्हणून जगभरात, सर्व मानवी समाजात आई-मूल हे नाते सर्वश्रेष्ठ मानले गेले आहे.

उपक्रम :

अ. तुमच्या शाळेतील/महाविद्यालयातील स्नेहसंमेलनात ‘व्यंगचित्रांतून सामाजिक प्रबोधन’ या विषयावर चित्रप्रदर्शनाचे आयोजन करा.

आ. ‘बेटी बचाओ, बेटी पढाओ।’ यासारखी स्त्री शिक्षणाशी संबंधित पाच घोषवाक्ये तयार करा.

Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा

खाली दिलेल्या मंगेश तेंडुलकर यांच्या व्यंगचित्राचे निरीक्षण करा. या चित्रातून व्यंगचित्रकाराला काय सुचवायचे असेल असे तुम्हांला वाटते. ते तुमच्या शब्दांत लिहा.

प्रश्न 1.
Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा 9
उत्तर :
कुल्हाड हे दहशत, दडपशाही, हिंसा यांचे प्रतीक आहे. मुलगी हे शांतताप्रेमींचे प्रतीक आहे. पाणी शिंपणे हे शांततेच्या प्रयत्नांचे प्रतीक आहे. कु-हाडीवरील रोपटे हे शांततेचे, प्रेमाचे प्रतीक आहे. हिंसेला हिंसेने उत्तर देऊन हिंसा कधीच संपत नाही. हिंसाचारी माणसांचे मन प्रेमानेच बदलता येते. शांतता, प्रेम निर्माण करण्याचा तोच एक मार्ग आहे.

Marathi Yuvakbharati 12th Digest Chapter 12 रंगरेषा व्यंगरेषा Additional Important Questions and Answers

लेखकांनी पुढील गोष्टी कळण्यासाठी व्यंगचित्रात वापरलेली प्रतीके लिहा :

प्रश्न 1.
पाठातील गोष्टी – प्रतीके
1. नारळ घेतलेला हात – [ ]
2. जवळून जाणारी व्यंगचित्राची कल्पना – [ ]
उत्तर :
1. नारळ घेतलेला हात : रूढीग्रस्त पुरुष
2. जवळून जाणारी व्यंगचित्राची कल्पना : जवळून जाणारे मासे

जोड्या लावा :

प्रश्न 1.

‘अ’ गट ‘ब’ गट
1. तो नारळ नसून ते चिमुरड्या मुलीचे डोके आहे, हे सूचित व्हावे म्हणून 1. आणि ती सतत परत यायची.
2. मासिके, नियतकालिके यांना मी सतत चित्र पाठवायचो 2. म्हणून ते दीनानाथांशी खोटे बोलले.
3. लेखकांना स्वत:च्या बळावर स्वत:ची पात्रता सिद्ध करायची होती 3. त्या नारळाला बारीकसा कानातला डूल दाखवला.

उत्तर :

‘अ’ गट ‘ब’ गट
1. तो नारळ नसून ते चिमुरड्या मुलीचे डोके आहे, हे सूचित व्हावे म्हणून 3. त्या नारळाला बारीकसा कानातला डूल दाखवला.
2. मासिके, नियतकालिके यांना मी सतत चित्र पाठवायचो 1. आणि ती सतत परत यायची.
3. लेखकांना स्वत:च्या बळावर स्वत:ची पात्रता सिद्ध करायची होती 2. म्हणून ते दीनानाथांशी खोटे बोलले.

Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा

कृती करा :

प्रश्न 1.
Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा 10
उत्तर :
Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा 11

लेखकांनी नोंदवलेल्या व्यंगचित्र सुचण्याच्या दोन प्रक्रिया सांगा.

प्रश्न 1.
लेखकांनी नोंदवलेल्या व्यंगचित्र सुचण्याच्या दोन प्रक्रिया सांगा.
उत्तर :
1. जवळून जात असलेली एखादी व्यंगचित्राची कल्पना तिथून उचलली जाते आणि कागदावर उतरवली जाते.
2. आधी एखादी कल्पना मनात निश्चित करून चित्र रेखाटता रेखाटता कधीतरी अचानकपणे आपली मनातील सुप्त इच्छा त्या चित्रात स्वतंत्र रूप घेऊन येते.

लेखकांना लागू पडणाऱ्या व्यक्तिवैशिष्ट्यांसमोर (✓) अशी खूण करा आणि लागू न पडणाऱ्या व्यक्तिवैशिष्ट्यांसमोर (✗) अशी खूण करा :

प्रश्न 1.
वाईच्या व्यंगचित्र प्रदर्शनात लेखकांना मोलाचे शिक्षण मिळाले. ( )
उत्तर :
वाईच्या व्यंगचित्र प्रदर्शनात लेखकांना मोलाचे शिक्षण मिळाले. (✓)

Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा

कारणे लिहा :

प्रश्न 1.

  1. ‘दीपावली’त स्वत:ची चित्रे प्रसिद्ध व्हावीत असे लेखकांना वाटत होते; कारण –
  2. लेखक खोटे बोलले, हे दीनानाथांना कळले होते, पण ते त्या वेळी काहीही बोलले नाहीत; कारण –
  3. दीनानाथांना लेखकांचे स्पष्टीकरण आवडले; कारण –

उत्तर :

  1. ‘दीपावली त स्वत:ची चित्रे प्रसिद्ध व्हावीत असे लेखकांना वाटत होते; कारण दीपावली हे मासिक दर्जेदार असल्याने त्यात आपली चित्रे प्रसिद्ध झाली; तर आपल्याला मान्यता मिळेल, असे लेखकांना वाटत होते.
  2. लेखक खोटे बोलले, हे दीनानाथांना कळले होते, पण ते त्या वेळी काहीही बोलले नाहीत; कारण लेखक कधी कबूल करतात, हे त्यांना पाहायचे होते.
  3. दीनानाथांना लेखकांचे स्पष्टीकरण आवडले; कारण लेखकांसारखी माणसे क्वचितच आढळतात.

रंगरेषा व्यंगरेषा Summary in Marathi

पाठ परिचय :

  • व्यंगचित्रांसाठी विषय मिळण्यात लेखकांना कधीही अडचण आली नाही. त्यांना खोलवर विचार करण्याची सवय होती. या विचार करण्याच्या वाटेवर त्यांना व्यंगचित्रांचे विषय मिळत असत.
  • लेखक अनेक कार्यक्रमांना आवर्जून उपस्थित राहत. अशा कार्यक्रमांतील घडामोडींमधून व्यंगचित्रांसाठी त्यांना विषय मिळत. उदा., स्त्रीभ्रूणहत्येबद्दलचे पोस्टर.
  • व्यंगचित्रांचे विषय आपल्या अवतीभवतीच असतात. जवळून जात-येत असतात. त्यांनाच अलगद उचलायचे आणि कागदावर चितारायचे.
  • व्यंगचित्र ही नि:शब्द भाषा असते. अंत:करणातील खोलवरचे भाव व्यंगचित्रातून उत्तम रितीने व्यक्त होऊ शकतात, ही त्यांची ठाम धारणा होती, आपले हे मत त्यांनी आपल्या चित्रांतून समर्थपणे व्यक्त केले. उदा., आईवरील चित्र आणि बाबांना श्रद्धांजली वाहणारे चित्र.
  • देशविदेशात 80 हून अधिक व्यंगचित्र प्रदर्शने भरवली.
  • व्यंगचित्र म्हणजे एक छान भाषाच होय, हा संदेश देण्यासाठी त्यांनी प्रदर्शने भरवली.
  • वाईच्या प्रदर्शनात एक प्रेक्षक शेतकरी आपल्या कुटुंबीयांना चित्रे समजावून देत होता. चित्र निर्माण करण्याचा आपला हेतू आणि
  • चित्राचा प्रेक्षकांवर होणारा परिणाम यातून आपल्या चित्रांचे यशापयश समजून घेण्यातून चित्रकलेचे फार मोठे शिक्षण लेखकांना मिळाले.
  • प्रारंभीच्या काळात प्रसिद्धीसाठी धडपड करावी लागली. दीनानाथ दलालांनी संधी दिली. त्यांनी लेखकांची मुलाखत घेतली. लेखकांना व्यंगचित्रांचा नवीन ट्रेंड समजावून सांगितला. या भेटीतून खूप मोठा लाभ लेखकांना झाला. त्यांची चित्रेही दलालांनी स्वीकारली.
  • दलालांनी चित्र काढण्याचे प्रात्यक्षिक दाखवले. व्यंगचित्र कलेतील नवा प्रवाह समजावून सांगितला. चित्राचा ‘परस्पेक्टिव्ह’ समजावून सांगितला.
  • दलालांशी झालेल्या बोलण्यात लेखकांनी विजय तेंडुलकर हे आपले बंधू आहेत, हे प्रथम नाकारले, नंतर स्वीकारले. विजय तेंडुलकरांच्या ओळखीमुळे आपली चित्रे स्वीकारली गेली, असे होऊ नये; आपल्या चित्रांच्या दर्जामुळे ती स्वीकारली जावीत, असा लेखकांचा आग्रह होता.

Maharashtra Board Class 12 Marathi Yuvakbharati Solutions Chapter 12 रंगरेषा व्यंगरेषा

शब्दार्थ :

  1. स्रोत – उगमस्थान.
  2. प्रत्यंतर – अनुभव, पुरावा, खात्री.
  3. स्वारस्य – मनापासून असलेली, उत्कट इच्छा, नैसर्गिक आवड.
  4. कणसदृश्य – कणाएवढी.
  5. अपार – अमर्याद (पार = काठ, बांध. बांध नसलेला, अमर्याद पसरलेला.)
  6. इझेल – लिहिण्याचा फळा अडकवण्याचे साधन, घोडा.
  7. परस्पेक्टिव्ह – विशिष्ट कोनातून पाहिल्यास नजरेस पडणारे दृश्यरूप.

वाक्प्रचार व त्याचा अर्थ

हात देणे – मदत करणे.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

1. Choose the correct option from the given alternatives.

Question i.
Nylon fibers are …………………………………..
A. Semisynthetic fibres
B. Polyamide fibres
C. Polyester fibres
D. Cellulose fibres
Answer:
B. polyamide fibres

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question ii.
Which of the following is naturally occurring polymer?
A. Telfon
B. Polyethylene
C. PVC
D. Protein
Answer:
D. Protein

Question iii.
Silk is a kind of …………………………………. fibre
A. Semisynthetic
B. Synthetic
C. Animal
D. Vegetable
Answer:
C. an animal

Question iv.
Dacron is another name of …………………………………. .
A. Nylon 6
B. Orlon
C. Novolac
D. Terylene
Answer:
D. Terylene

Question v.
Which of the following is made up of polyamides?
A. Dacron
B. Rayon
C. Nylon
D. Jute
Answer:
C. Nylon

Question vi.
The number of carbon atoms present in the ring of ε – caprolactam is
A. Five
B. Two
C. Seven
D. Six
Answer:
D. Six

Question vii.
Terylene is …………………………………. .
A. Polyamide fibre
B. Polyester fibre
C. Vegetable fibre
D. Protein fibre
Answer:
B. Polyester fibre

Question viii.
PET is formed by …………………………………. .
A. Addition
B. Condensation
C. Alkylation
D. Hydration
Answer:
D. Hydration

Question ix.
Chemically pure cotton is …………………………………. .
A. Acetate rayon
B. Viscose rayon
C. Cellulose nitrate
D. Cellulose
Answer:
D. Cellulose

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question x.
Teflon is chemically inert, due to presence of …………………………………. .
A. C-H bond
B. C-F bond
C. H- bond
D. C=C bond
Answer:
A. C-H bond

2. Answer the following in one sentence each.

Question i.
Identify ‘A’ and ‘B’ in the following reaction …………………………………. .
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 1
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 70

Question ii.
Complete the following statements
a. Caprolactam is used to prepare …………………………………. .
b. Novolak is a copolymer of …………………………………. and …………………………………. .
c. Terylene is ………………………………….. polymer of terephthalic acid and ethylene glycol.
d. Benzoyl peroxide used in addtion polymerisation acts as …………………………………. .
e. Polyethene consists of polymerised …………………………………. .
Answer:
a. Nylon-6
b. Phenol, formaldehyde
c. polyester
d. initiator (catalyst)
e. linear or branched-chain

Question iii.
Draw the flow chart diagram to show the classification of polymers based on type of polymerisation.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 71

Question iv.
Write examples of Addition polymers and condensation polymers.
Answer:
Addition polymers : Polyvinyl chloride, polythene
Condensation polymers : Bakelite, terylene, Nylon-66

Question v.
Name some chain-growth polymers.
Answer:
Chain growth polymers : Polythene, polyacrylonitrile and polyvinyl chloride.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question vi.
Define the terms :
1) Monomer
2) Vulcanisation
3) Synthetic fibres
Answer:
1. Monomer is a small and simple molecule and has a capacity to form two chemical bonds with other monomers. Examples : Ethene, Propylene.
2. The process by which a network of cross-links is introduced into an elastomer is called vulcanisation or it can also be defined as the process of heating natural rubber with sulphur to increase the tensile strength, toughness and elasticity of natural rubber is known as vulcanization of rubber.
3. The man-made fibres prepared by polymerization of one monomer or copolymerization of two or more monomers are called synthetic fibres.

Question vii.
What type of intermolecular force leads to high-density polymer?
Answer:
High density polymers have low degree of branching along the hydrocarbon chain. The molecules are closely packed together during crystallization. This closer packing means that the van der Waals attraction between the chains are greater and so the plastic (high density polymer) is stronger and has a melting point.

Question viii.
Give one example each of copolymer and homopolymer.
Answer:
Homopolymer : PVC, Nylon-6
Copolymer : Terylene, Buna-S

Question ix.
Identify Thermoplastic and Thermosetting Plastics from the following …………………………………. .
1. PET
2. Urea-formaldehyde resin
3. Polythene
4. Phenol formaldehyde
Answer:
Thermoplastic plastics : PET, Polythene
Thermosetting plastics : Urea formaldehyde resin, Phenol formaldehyde

3. Answer the following.

Question i.
Write the names of classes of polymers formed according to intermolecular forces and describe briefly their structural characteristics.
Answer:
Molecular forces bind the polymer chains either by hydrogen bonds or Vander Waal’s forces. These forces are called intermolecular forces. On the basis of magnitude of intcrmolccular forces, polymers are further classified as ebstomers, fibres, thermoplastic polymers. thermosetting polymers.

(1) Elastomers: Weak van der Waals type of intermolecular forces of attraction between the polymer chains are observed in cbstomcrs. When polymer is stretched, the polymer chain stretches and when the strain is relieved the chain returns to its odginal position, Thus, polymer shows elasticity and is called elastomers. Elastomers, the elastic polymers, have weak van der Waals type of intermolecular forces which permit the polymer to be stretched. Lilastorners are soft and stretchy and used in making rubber bands. E.g.. neoprene, vulcanized rubber, buna.S, buna-N.

(2) Fibres : It consists of strong intermolecular forces of’ attraction due to hydrogen bonding and strong dipole-dipole forces. These polymers possess high tensile strength. Due to these strong intermolecular forces the fibres are crystalline in nature. They are used in textile industries, strung tyres. etc.. e.g., nylon, terylene.

(3) Thermoplastic polymers: These polymer possess moderately strong intermolecular forces of attraction between those of elastomers and fibres. These polymers arc called thermoplastic because they become soft on heating and hard on cooling. They are either linear or branched chain polymers. They can be remoulded and recycled. E.g. polyethenc, PVC, polystyrene.

(4) Thermosetting polymers: These polymers are cross linked or branched molecules and are rigid polymers. During their formation they have property of being shaped on heating. but they get hardened while hot. Once hardened these become infusible, cannot be softened by heating and therefore, cannot be remoulded and recycled.
This shows extensive cross linking by covalent bonds formed in the moulds during hardening/setting process while hot. E.g. Bakelite, urea formaldehyde resin.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question ii.
Write reactions of formation of :
a. Nylon 6
b. Terylene
Answer:
Terylene is polyester fibre formed by the polymerization of terephthalic acid and ethylene glycol.

Terylene is obtained by condensation polymerization of ethylene glycol and terephthalic acid in presence of catalyst zinc acetate and antimony trioxide at high temperature.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 26

Properties :

  • Terylene has relatively high melting point (265 °C)
  • It is resistant to chemicals and water.

Uses :

  • It is used for making wrinkle free fabrics by blending with cotton (terycot) and wool (terywool), and also as glass reinforcing materials in safety helmets.
  • PET is the most common thermoplastic which is another trade name of the polyester polyethylenetereph- thalate.
  • It is used for making many articles like bottles, jams, packaging containers.

Question iii.
Write the structure of natural rubber and neoprene rubber along with the name and structure of thier monomers.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 27

Question iv.
Name the polymer type in which the following linkage is present.
Answer:
The Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 74 linkage is present in terylene or dacron polymer.

Question v.
Write the structural formula of the following synthetic rubbers :
a. SBR rubber
b. Buna-N rubber
c. Neoprene rubber
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 41

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question vi.
Match the following pairs :
Name of polymer – Monomer
1. Teflon – a. CH2 = CH2
2. PVC – b. CF2 = CF2
3. Polyester – c. CH2 = CHCl
4. Polythene – d. C6H5OH and HCHO
5. Bakelite – e. Dicarboxylic acid and polyhydoxyglycol
Answer:

  1. Teflon – CF2 = CF2
  2. PVC – CH2 = CHCI
  3. Polyester-Dicarboxylic acid and polyhydoxyglycol
  4. Polythene – CH2 = CH2
  5. Bakelite – C6H5OH and HCHO

Question vii.
Draw the structures of polymers formed from the following monomers
1. Adipic acid + Hexamethylenediamine
2. e – Aminocaproic acid + Glycine
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 32

Question viii.
Name and draw the structure of the repeating unit in natural rubber.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 14Repeating unit of natural rubber (Basic unit : isoprene)

Question ix.
Classify the following polymers as natural and synthetic polymers
a. Cellulose
b. Polystyrene
c. Terylene
d. Starch
e. Protein
f. Silicones
g. Orlon (Polyacrylonitrile)
h. Phenol-formaldehyde resins
Answer:

Natural Polymers 1. Cellulose 4. Starch 5. Protein
Synthetic Polymers 2. Polystyrene 3. Terylene 6. Silicones 7. Orion (Polyacrylonitrile) 8. phenol-formaldehyde resin

Question x.
What are synthetic resins? Name some natural and synthetic resins.
Answer:
Synthetic resins are artificially synthesised high molecular weight polymers. They are the basic raw material of plastic. The main properties of plastic depend on the synthetic resin it is made from.

Examples of natural resins : Rosin, Damar, Copal, Sandarac, Amber, Manila
Examples of synthetic resins : Polyester resin, Phenolic resin, Alkyl resin, Polycarbonate resin, Polyamide resin, Polyurethane resin, silicone resin, Epoxy resin, Acrylic resin.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question xi.
Distinguish between thermosetting and thermoplastic resins. Write example of both the classes.
Answer:

Thermosetting resin Thermoplastic resin
(1) They harden when heated. Once hardened it no longer melts. (1) They soften when heated and harden again when cooled.
(2) They cannot be re-shaped. (2) They can be reshaped
(3) They are strong, hard. (3) They are weak, soft.
(4) Thermosetting resin show cross-linking.
Examples : Melamine resin Epoxy resins, Bake-lite.
(4) Thermoplastic molecules do not cross link, hence are flexible.
Examples : Polythene, polypropylene, nylon, polyester.

Question xii.
Write name and formula of raw material from which bakelite is made.
Answer:
The raw material or monomers used to prepare bakelite are o-hydroxymethyl phenol Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 35 and formaldehyde (HCHO)

4. Attempt the following :

Question i.
Identify condensation polymers and addition polymers from the following.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 2
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 68

Question ii.
Write the chemical reactions involved in the manufacture of Nylon 6, 6
Answer:
Nylon-6, 6 is a linear polyamide polymer formed by the condensation polymerisation reaction. The monomers used in the preparation of Nylon-6, 6 are :
(1) Adipic acid : HOOC-(CH2)4-COOH
(2) Hexamethylene diamine : H2N-(CH2)6-NH2

When equimolar aqueous solutions of adipic acid and hexamethylene diamine are mixed and heated, there is neutralization to form a nylon salt. During polymerisation at 553 k nylon salt loses a water molecule to form nylon 6, 6 polymer. Both monomers (hexamethylene diamine and adipic acid) contain six carbon atoms each, hence the polymer is termed as Nylon-6,6.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 24

Properties and uses : Nylon 6,6 is high molecular mass (12000 – 50000 u) linear condensation polymer. It possesses high tensile strength. It does not soak in water. It is used for making sheets, bristles for brushes, surgical sutures, textile fabrics, etc.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question iii.
Explain the vulcanisation of rubber. Which vulcanizing agents are used for the following synthetic rubber.
a. Neoprene
b. Buna-N
Answer:
The process by which a network of cross links is introduced into an elastomer is called vulcanization.

Vulcanization enhances the properties of natural rubber like tensile strength, stiffness, elasticity, toughness etc. Sulphur forms cross links between polyisoprene chains which results in improved properties of rubber.

  • For neoprene vulcanizing agent is MgO.
  • For Buna-N vulcanizing agent is sulphur.

Question iv.
Write reactions involved in the formation of …………………………………. .
1) Teflon
2) Bakelite
Answer:
The monomers phenol and formaldehyde undergo polymerisation in the presence of alkali or acid as catalyst.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 33
Phenol reacts with formaldehyde to form ortho or p-hydroxy methyl phenols, which further reacts with phenol to form a linear polymer called Novolac. It is used in paints.

In the third stage, various articles are shaped from novolac by putting it in appropriate moulds and heating at high temperature (138 °C to 176 °C) and at high pressure forms rigid polymeric material called bakelite. Bakelite is insoluble and infusible and has high tensile strength.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 34
Bakelite is used in making articles like telephone instrument, kitchenware, electric insulators frying pans, etc.

2. Teflon is polytetrafluoroethylene. The monomer used in preparation of teflon is tetrafluoroethylene, (CF2 = CF2) which is a gas at room temperature. Tetrafluoroethylene is polymerized by using free radical initiators such as hydrogen peroxide or ammonium persulphate at high pressure.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 22

Properties:

  • Telflon is tough, chemically inert and resistant to heat and attack by corrosive reagents.
  • C – F bond is very difficult to break and remains unaffected by corrosive alkali, organic solvents.
    Uses : Telflon is used in making non-stick cookware, oil seals, gaskets, etc.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question v.
What is meant by LDP and HDP? Mention the basic difference between the same with suitable examples.
Answer:

  • LDP is low density polyethylene and HDP is high density polyethylene.
  • LDP is a branched polymer with low density due to chains are loosely held and HDP is a linear polymer with density due to close packing.
  • HDP is much stiffer than LDP and has high tensile strength and hardness.

LDP is mainly used in preparation of pipes for agriculture, irrigation and domestic water line connections. HDP is used in manufacture of toys and other household articles like bucket, bottles, etc.

Question vi.
Write preparation, properties and uses of Teflon.
Answer:
Teflon is polytetrafluoroethylene. The monomer used in preparation of teflon is tetrafluoroethylene, (CF2 = CF2) which is a gas at room temperature. Tetrafluoroethylene is polymerized by using free radical initiators such as hydrogen peroxide or ammonium persulphate at high pressure.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 22

Properties:

  • Telflon is tough, chemically inert and resistant to heat and attack by corrosive reagents.
  • C – F bond is very difficult to break and remains unaffected by corrosive alkali, organic solvents.
    Uses : Telflon is used in making non-stick cookware, oil seals, gaskets, etc.

Question vii.
Classify the following polymers as straight-chain, branched-chain and cross-linked polymers.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 3
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 8

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

5. Answer the following.

Question i.
How is polythene manufactured? Give their properties and uses.
Answer:
LDP means low density polyethylene. LDP is obtained by polymerization of ethylene under high pressure (1000 – 2000 atm) and temperature (350 – 570 K) in presence of traces of O2 or peroxide as initiator.

The mechanism of this reaction involves free radical addition and H-atom abstraction. The latter results in branching. As a result the chains are loosely held and the polymer has low density.

Properties of LDP :

  • LDP films are extremely flexible, but tough chemically inert and moisture resistant.
  • It is poor conductor of electricity with melting point 110 °C.

Uses of LDP :

  • LDP is mainly used in preparation of pipes for agriculture, irrigation, domestic water line connections as well as insulation to electric cables.
  • It is also used in submarine cable insulation.
  • It is used in producing extruded films, sheets, mainly for packaging and household uses like in preparation of squeeze bottles, attractive containers, etc.

HDP means high density polyethylene. It is a linear polymer with high density due to close packing.

HDP is obtained by polymerization of ethene in presence of Zieglar-Natta catalyst which is a combination of triethyl aluminium with titanium tetrachloride at a temperature of 333K to 343K and a pressure of 6-7 atm.

Properties of HDP :

  • HDP is crystalline, melting point in the range of 144 – 150 °C.
  • It is much stiffer than LDP and has high tensile strength and hardness.
  • It is more resistant to chemicals than LDP.

Uses of HDP :

  • HDP is used in manufacture of toys and other household articles like buckets, dustbins, bottles, pipes, etc.
  • It is used to prepare laboratory wares and other objects where high tensile strength and stiffness is required.

Question ii.
Is synthetic rubber better than natural rubber? If so, in what respect?
Answer:
Yes. Synthetic rubber is more resistant to abrasion than natural rubber and is also superior in resistance to heat and the effects of aging (lasts longer). Many types of synthetic rubber are flame-resistant, so they can be used in insulation for electrical devices.

It also remains flexible at low temperatures and is resistant to grease and oil. It is resistant to heat, light and certain chemicals.

Question iii.
Write main specialities of Buna-S, Neoprene rubber?
Answer:
Buna-S is an elastomer and it is copolymer of styrene with butadiene. Its trade name is SBR. Buna-S is superior to natural rubber, because of its mechanical strength and abrasion resistance. It is used in tyre industry. It is vulcanized with sulphur. Neoprene is a synthetic rubber and it is a condensation polymer of chloroprene (2-chloro-l, 3-butadiene). Vulcanization of neoprene takes place in presence of MgO. It is resistant to petroleum, vegetable oils. Neoprene is used in making hose pipes for transport of gasoline and making gaskets.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Question iv.
Write the structure of isoprene and the polymer obtained from it.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 82

Question v.
Explain in detail the free radical mechanism involved during the preparation of the addition polymer.
Answer:
Polymerisation of ethylene is carried out at high temperature and at high pressure in presence of small amount of acetyl peroxide as initiator.

(1) Formation of free radicals : The first step involves clevage of acetyl peroxide to form two carboxy radicals. These carboxy radicals immediately undergo decarboxylation to give methyl initiator free radicals.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 15

(2) Chain initiating step : The methyl radical thus formed adds to ethylene to form a new larger free radical.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 16

(3) Chain propagation step : The larger free radical formed in the chain initiating step reacts with another molecule of ethene to form another big size free radicals and chain grows. This is called chain propagation step.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 17

The chain reaction continues till thousands of ethylene molecules are added.

(4) Chain terminating step : The continuous chain reaction can be terminated by the combination of free radicals to form polyethene.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 18

Activity :
i. Collect the information of the process like extrusion and moulding in Textile Industries.
ii. Make a list of polymers used to make the following articles
a. Photographic film
b. Frames of spectacles
c. Fountain pens
d. Moulded plastic chains
e. Terywool or Terycot fabric
iii. Prepare a report on factors responsible for degradation of polymers giving suitable example.
iv. Search and make a chart/note on silicones with reference to monomers, structure, properties and uses.
v. Collect the information and data about Rubber industry, plastic industry and synthetic fibre (rayon) industries running in India.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

12th Chemistry Digest Chapter 15 Introduction to Polymer Chemistry Intext Questions and Answers

Use your brain power! (Textbook Page No 323)

Question 1.
Differentiate between natural and synthetic polymers.
Answer:

Natural polymers Synthetic polymers
(1) The polymers are obtained either from plants or animals. (1) The man made fibres prepared by polymerization of monomer or copolymerization of two or more monomers.
(2) They are further divided into two types :
(i) plant polymers
(ii) Animal polymers.Examples: Cotton, linen, latex
(2) They are further divided into three subtypes :
(i) fibres
(ii) synthetic rubbers
(iii) plastics.Examples : Nylon, terylene Buna-S

Use your brain power! (Textbook Page No 325)

Question 1.
What is the type of polymerization in the following examples?
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 11
Answer:
(i) Addition polymerization
(ii) Condensation polymerization

Problem 15.1 : (Textbook Page No 326)

Question 1.
Refer to the following table listing for different polymers formed from respective monomers. Identify from the list whether it is copolymer or homopolymer.

Monomer Polymers
Ethylene Polyethene
Vinyl chloride Polyvinyl chloride
Isobutylene Polyisobutylene
Acrylonitrile Polyacrylonitrile
Caprolactam Nylon 6
Hexamethylene diammonium adipate Nylon 6, 6
Butadiene + styrene Buna-S

Solution :
In each of first five cases, there is only one monomer which gives corresponding homopolymer. In the sixth case hexamethylene diamine reacts with adipic acid to form the salt hexamethylene diammonium adipate which undergoes condensation to form Nylon 6, 6. Hence nylon 6, 6 is homopolymer. The polymer Buna-S is formed by polymerization of the monomers butadiene and styrene in presence of each other. The repeating units corresponding to the monomers butadiene and styrene are randomly arranged in the polymer. Hence Buna-S is copolymer.

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Use your brain power! (Textbook Page No 328)

Question 1.
(1) From the cis-polyisoprene structure of natural rubber explain the low strength of van der Waals forces in it.
(2) Explain how the vulcanization of natural rubber improves its elasticity. (Hint : consider the intermolecular links.)
Answer:
(1) (i) Natural rubber is cis-polyisoprene. It is obtained by polymerization of isoprene units at 1, 4 positions. In rubber molecule, double bonds are located between C2 and C3 of each isoprene unit. These cis-double bonds do not allow the polymer chain to come closer. Therefore, only weak vander Waals’ forces are present. Since the chains are not linear, they can be stretched just like springs and exhibit elastic properties.

(ii) Cis-1, 4 polyisoprene (Natural rubber), due to this cis configuration about the double bonds, has the adjacent chain that do not fit together well (there is no close packing of adjacent chains). The only force that interact is the weak or low strength of van der Waals’ forces.

(iii) Cis-polyisoprene has a coiled structure in which the various polymer chains are held together by weak van der Waals’ forces.

(2) (i) Vulcanization of rubber is a process of improvement of the rubber elasticity and strength by heating it in the presence of sulphur, which results in three dimensional cross-linking of the chain rubber molecules (polyisoprene) bonded to each other by sulphur atoms.

(ii) Vulcanisation makes rubber more elastic and more stiffer. On vulcanisation, sulphur forms cross links at the reactive sites of double bonds and thus rubber get stiffened.

(iii) The improved properties of vulcanised rubber are (i) high elasticity (ii) low water absorption tendency,

(iii) resistance to oxidation.

Use your brain power! (Textbook Page No 334)

Question 1.
Write structural formulae of styrene and polybutadiene.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 43

(1) Classify the following polymers as addition or condensation.
(i) PVC (ii) Polyamides
(iii) Polystyrene
(iv) Polycarbonates
(v) Novolac
Answer:
Addition polymers: PVC, Polystyrene
(ondensatlon polymers: Polyamides. Polycarbonates, Novolac

Question 2.
Completed the following table :
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 44

Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

Use your brain power! (Textbook Page No 335)

(1) Represent the copolymerization reaction between glycine and e aminocaproic acid to form the copolymer nylon 2-nylon 6.
(2) What is the origin of the numbers 2 and 6 in the name of this polymer?
Answer:
(1) It is a copolymer and has polyamide linkages. The monomers glycine and e-amino caproic acid undergo condensation polymerisation to form nylon-2-nylon-6.
Maharashtra Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry 46
Nylon-2-nylon-6 is used in orthopaedic devices and implants.

(2) Monomer glycine contains two carbon atoms and e amino caproic acid contains six carbon atoms, hence the polymer is termed as nylon-2-nylon-6.

Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 14 Biomolecules Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 14 Biomolecules

1. Select the most correct choice.

Question i.
CH2OH-CO-(CHOH)4-CH2OH is an example of
a. Aldohexose
b. Aldoheptose
c. Ketotetrose
d. Ketoheptose
Answer:
(d) Ketoheptose

Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules

Question ii.
The open chain formula of glucose does not contain
a. Formyl group
b. Anomeric hydroxyl group
c. Primary hydroxyl group
d. Secondary hydroxyl group
Answer:
(b) Anomeric hydroxyl group

Question iii.
Which of the following does not apply to CH2NH2 – COOH
a. Neutral amino acid
b. L – amino acid
c. Exists as zwitterion
d. Natural amino acid
Answer:
(d) Natural amino acid

Question iv.
Tryptophan is called essential amino acid because
a. It contains an aromatic nucleus.
b. It is present in all the human proteins.
c. It cannot be synthesized by the human body.
d. It is an essential constituent of enzymes.
Answer:
(c) It cannot be synthesised by human body.

Question v.
A disulfide link gives rise to the following structure of protein.
a. Primary
b. Secondary
c. Tertiary
d. Quaternary
Answer:
(c) Tertiary

Question vi.
RNA has
a. A – U base pairing
b. P – S – P – S backbone
c. double helix
d. G – C base pairing
Answer:
(a) A – U base pairing

2. Give scientific reasons :

Question i.
The disaccharide sucrose gives negative Tollens test while the disaccharide maltose gives positive Tollens test.
Answer:
(1) In disaccharide sucrose, the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a nonreducing sugar. As there is no free aldehyde group, it does not reduce Tollen’s reagent to metallic silver. Hence, sucrose gives negative Tollen’s test.

(2) While the disaccharide maltose is a reducing sugar because a free aldehyde group can be produced at C1 of second sugar molecule. It is a reducing sugar. It reduces Tollen’s reagent to shining silver mirror. Hence, Maltose gives positive Tollen’s test.

Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules

Question ii.
On complete hydrolysis DNA gives equimolar quantities of adenine and thymine.
Answer:
On complete hydrolysis DNA yields 2-deoxy-D-ribose, adenine, thymine, guanine, cystosine and phosphoric acid. Since adenine always forms two hydrogen bonds with thymine, the hydrolysis of DNA gives equimolar quantities of adenine and thymine.

Question iii.
α – Amino acids have high melting points compared to the corresponding amines or carboxylic acids of comparable molecular mass.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 102
α-Amino acids have high melting points compared to the corresponding amines or carboxylic acids of comparable molecular mass due to the presence of both acidic (carboxylic group) and basic (amino group) groups in the same molecule. In aqueous solution, proton transfer from acidic group to amino (basic) group of amino acid forms a salt, which is a dipolar ion called zwitter ion.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 103

Question iv.
Hydrolysis of sucrose is called inversion.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 104
Sucrose is dextro rotatory. On hydrolysis it gives equimolar mixture of D – ( + ) glucose and D – ( -) fructose. Since the laevorotation of fructose (- 92.4°) is more than dextrorotation of glucose ( + 52.7°), the hydrolysis product has net laevorotation. Thus, hydrolysis of sucrose brings about a change in the sign of rotation, from dextro ( + ) to laevo (-) and the product is called as invert sugar and so the hydrolysis of sucrose is called inversion.

Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules

Question v.
On boiling, egg albumin becomes opaque white.
Answer:
Upon boiling the egg, denaturation αcurs. During denaturation, secondary and tertiary structures are destroyed, but primary structure remains intact. Egg contains soluble globular proteins, which forms insoluble fibrous proteins (opque) on boiling egg.

3. Answer the following

Question i.
Some of the following statements apply to DNA only, some to RNA only and some to both. Lable them accordingly.
a. The polynucleotide is double stranded. ( …………… )
b. The polynucleotide contains uracil. ( …………… )
c. The polynucleotide contains D-ribose ( …………… ).
d. The polynucleotide contains Guanine ( …………… ).
Answer:
(1) The polynucleotide is double stranded. (DNA)
(2) The polynucleotide contains uracil. (RNA)
(3) The polynucleotide contain D-ribose (RNA)
(4) Thc polynucleotide contains Guanine (DNA, RNA)

Question ii.
Write the sequence of the complementary strand for the following segments of a DNA molecule.
a. 5′ – CGTTTAAG – 3′
b. 5′ – CCGGTTAATACGGC – 3′
Answer:
(1) DNA molecule : 5′ – CGTTTAAG – 3′
The complementary strand runs in opposite direction from the 3′ end to the 5′ end. It has the base sequence decided by complementary base pairs A – T and C – G.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 100
(2) DNA molecule : 5′ – CCGGTTAATACGGC – 3′
The complementary strand runs in opposite direction from the 3′ end to the 5′ end. It has the base sequence decided by complementary base pairs A – T and C – G.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 101

Question iii.
Write the names and schematic representations of all the possible dipeptides formed from alanine, glycine and tyrosine.
Answer:
(1) Dipeptide formed from alanine and glycine
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 67
(2) Dipeptide formed from alanine and tyrosine
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 68
(3) Dipeptide formed from glycine and tyrosine.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 69

Question iv.
Give two pieces of evidence for the presence of the formyl group in glucose.
Answer:
(1) Glucose reacts with hydroxyl amine in an aqueous solution to form glucose oxime. This indicates the presence of – CHO (formyl group) in glucose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 15
(2) Glucose on oxidation with mild oxidising agent like bromine water gives gluconic acid which shows carbonyl group in glucose is aldehyde (formyl group) group.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 16

Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules

4. Draw a neat diagram for the following:

Question i.
Haworth formula of glucopyranose
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 30

Question ii.
Zwitter ion
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 61

Question iii.
Haworth formula of maltose
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 38

Question iv.
Secondary structure of the protein

Answer:
The structure of proteins can be studied at four different levels i.e. primary, secondary, tertiary and quaternary levels. Each level is more complex than the previous one.
(1) Primary structure of proteins :
(a) Representation by structural formula
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 76

(b) Representation with amino acid symbols
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 77

Primary structure of proteins is the sequence of constituent a-amino acid residues linked by peptide bonds. Any change in the sequence of amino acid residue creates different protein molecule. Primary structure of proteins is represented by writing the three letter symbols of amino acid residues as per their sequence in the concerned protein. The symbols are separated by dashes. According to the convention, the N-terminal amino acid residue as written at the left end and the C-terminal amino acid residue at the right end.

Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules

(2) Secondary structure of proteins : The three-dimensional arrangement of lαalized regions of a long polypeptide chain is called the secondary structure of protein. Hydrogen bonding between N-H proton of one amide linkage and C = O oxygen of another gives rise to the secondary structure. There are two different types of secondary structures i.e. α-helix and β-pleated sheet.

α-Helix : In a-helix structure, a polypeptide chain gets coiled by twisting into a right handed or clαkwise spiral known as a-helixn. The characteristic features of α-helical structure of protein are :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 78
(1) Each turn of the helix has 3.6 amino acids.
(2) A C = O group of one amino acid is hydrogen bonded to N – H group of the fourth amino acid along the chain.
(3) Hydrogen bonds are parallel to the axis of helix while R groups extend outward from the helix core.
Myosin in muscle and a-keratin in hair are proteins with almost entire a-helical secondary structure.

β-Pleated sheet : In secondary structure, when two or more polypeptide chains (strands) line up side-by-side is called β-pleated sheets. The β-picate sheet structure of protein consists of extended strands of polypeptide chains held together by intermolecular hydrogen bonding. The characteristics of β-pleated sheet structure are :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 79

  • The C = O and N – H bonds lie in the planes of the sheet.
  • Hydrogen bonding occurs between the N – H and C = O groups of nearby amino acid residues in the neighbouring chains.
  • The R groups are oriented above and below the plane of the sheet.

The β-pleated sheet arrangement is favoured by amino acids with small R groups.

(3) Tertiary structure of proteins :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 80
The three-dimensional shape acquired by the entire polypeptide chain of a protein is called its tertiary structure. The structure is stabilized and has attractive interaction with the aqueous environment of the cell due to the folding of the chain in a particular manner. Tertiary structure gives rise to two major molecular shapes i.e. globular and fibrous proteins. The main forces which stabilize a particular tertiary structure include hydrogen bonding, dipole-dipole attraction (due to polar bonds in the side chains), electrostatic attraction (due to the ionic groups like -COO, \(\mathrm{NH}_{3}^{+}\) in the side chain) and also London dispersion forces. Finally, disulfide bonds formed by oxidation of nearby – SH groups (in cysteine residues) are the covalent bonds which stabilize the tertiary structure.

Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules

(4) Quaternary structure of proteins The two or more polypeptide chains with folded tertiary structures forms complex protein. The spatial arrangements of these polypeptide chains with respect to each other is known as quaternary structure. Each individual polypeptide chain is called a subunit of the overall protein. For example: Haemoglobin consists of four subunits called haeme held together by intermolecular forces in a compact three dimensional shape. Haemoglobin can do its function of oxygen transport only when all the four subunits are together.

Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 81

Question v.
AMP
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 105

Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules

Question vi.
dAMP
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 106

Question vii.
One purine base from nucleic acid
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 88

Question viii.
Enzyme catalysis
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 85

Activity :

  • Draw the structure of a segment of DNA comprising at least ten nucleotides on a big chart paper.
  • Make a model of DNA double stranded structure as group activity.

12th Chemistry Digest Chapter 14 Biomolecules Intext Questions and Answers

Try ….. this (Textbook Page No 298)

Question 1.
Observe the following structural formulae carefully and answer the questions.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 2
(1) How many OH groups are present in glucose, fructose and ribose respectively?
(2) Which other functional groups are present in these three compounds?
Answer:
(1) Glucose contains five hydroxyl (- OH) groups.
Fructose contains five hydroxyl ( – OH) groups.
Ribose contains four hydroxyl ( – OH) groups.

(2) Glucose contains aldehyde ( – CHO) as other functional group.
Fructose contains ketonic group Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 3 as other functional group.
Ribose contains aldehyde ( – CHO) as other functional group.

Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules

Use your brain power! (Textbook Page No 299)

Question 1.
Give IUPAC names to the following monosaccharides.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 4
Answer:
(1) Aldotriose
(2) Aldopentose
(3) Ketoheptose

Problem 14.1 : (Textbook Page No 300)

Question 1.
An alcoholic compound was found to have molecular mass of 90 u. It was acetylated. Molecular mass of the acetyl derivative was found to be 174 u. How many alcoholic (- OH) groups must be present in the original compound?
Solution :
In acetylation reaction H atom of an (- OH) group is replaced by an acetyl group (- COH3).

This results in an increase in molecular mass by [(12 + 16 + 12 + 3 x 1) – 1] that has, 42 u. In the given alcohol, increase in molecular mass = 174 u – 90 u = 84 u
∴ Number of – OH groups \(=\frac{84 \mathrm{u}}{42 \mathrm{u}}=2\)

Use your brain power! (Textbook Page No 301)

(1) Write structural formula of glucose showing all the bonds in the molecule.
(2) Number all the carbons in the molecules giving number 1 to the ( – CHO) carbon.
(3) Mark the chiral carbons in the molecule with asterisk (*).
(4) How many chiral carbons are present in glucose?
Answer:
Refer structural formula of glucose for (1) (2) and (3).
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 14
(4) There are 4 chiral carbon atoms present in glucose.

Use your brain power! (Textbook Page No 306)

Question 1.
(1) Is galactose an aldohexose or a ketohexose?
(2) Which carbon in galactose has different configuration compared to glucose?
(3) Draw Haworth formulae of α-D-galactose and β-D-galactose.
(4) Which disaccharides among sucrose, maltose and lactose is/are expected to give positive Fehling test?
(5) What are the expected products of hydrolysis of lactose?
Answer:

  1. Galactose is an aldohexose.
  2. Fourth carbon in galactose has different configuration compared to glucose.
  3. Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 42
  4. Maltose and lactose are expected to give positive Fehling solution test.
  5. The expected products of hydrolysis of lactose are D – ( +) glucose and D – ( +) galactose.

Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules

Can you think? (Textbook Page No 307)

Question 1.
When you chew plain bread, chapati or bhaakari for long time, it tastes sweet. What could be the reason?
Answer:
When chapati, bread or bhakari are chewed for long time the pulp mixes with saliva and carbohydrate component in them diseminates and gives the sweet taste.

Use your brain power! (Textbook Page No 309)

Question 1.
Tryptophan and histidine have the structures (I) and (II) respectively. Classify them into neutral? acidic/basic &amino acids and justify your answer. (Hint: Consider învolvement of lone pair in resonance).
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 56
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 57
In tryptophan, nitrogen atom present in cyclic structure cannot donate pair of electrons as it is stabilized by resonance. The other amino group and carboxylic group present in the side chain neutralize each other. Tryptophan has equal number of amino and carboxylic groups. Hence, tryptophan is a neutral amino acid.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 58
In histidine, amino groups are more in number than carboxyl groups therefore histidine ¡s basic in nature.

Can you think? (Textbook Page No 309)

Question 1.
Compare the molecular masses of the following compounds and explain the observed melting points.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 59
Answer:
Above compounds have same molecular masses but they have different melting points, a-amino acids have higher melting points compared to the corresponding amines or carboxylic acids of comparable masses. This property is due to the presence of both carboxylic group (acidic) and amino group (basic) in the molecule. In aqueous solution, protons transfer from acidic group to amino (basic) group of amino acid forms a salt, which is a dipolar ion called – Zwitter ion.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 60

Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules

Use your brain power! (Textbook Page No 310)

Question 1.
(1) Write the structural formula of dipeptide formed by combination of carboxyl group of alanine and amino group of glycine.
(2) Name the resulting dipeptide.
(3) Is this dipeptide same as glycyalanine or its structural isomer?
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 66
(2) ala-glycine. OR ala-gly
(3) It is a structural isomer.

Question 54.
Write the names and schematic representations of all the possible dipeptides formed from alanine, glycine and tyrosine.

Problem 14.3 : (Textbook Page No 311)

Question 1.
Chymotrypsin is a digestive enzyme that hydrolyzes those amide bonds for which the carbonyl group comes from phenylalanine, tyrosine or tryptophan. Write the symbols of the amino acids and peptides smaller than pentapeptide formed by hydrolysis of the following hexapeptide with chymotrypsin. Gly-Tyr-Gly-Ala-Phe-Val
Solution :
In the given hexapeptide hydroylsis by chymotripsin can take place at two points, namely, Phe and Tyr. The carbonyl group of these residues is towards the right side, that is, toward the C-terminal. Therefore the hydrolysis products in required range will be :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 70

Problem 14.4 : (Textbook Page No 311)

Question 1.
Write down the structures of amino acids constituting the following peptide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 71
Solution :
The given peptide has two amide bonds linking three amino acids. The structures of these amino acids are obtained by adding one H2O molecule across the amide bond as follows :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 72

Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules

Use your brain power! (Textbook Page No 313)

A protein chain has the following amino acid residues. Show and label the interactions that can be present in various pairs from these giving rise to tertiary level structure of protein.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 82
Answer:
Tertiary level structure from amino residues.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 83

Can you tell? (Textbook Page No 313)

Question 1.
What is the physical change observed when (a) egg is boiled, (b) milk gets curdled on adding lemon juice?
Answer:
(a) When egg is boiled, coagulation of eggwhite (insoluble fibrous proteins) takes place. This is a common example of denaturation.
(b) When lemon juice is added to milk, it gets curdled due to the formation of lactic acid. This is another example of denaturation.

Can you tell? (Textbook Page No 315)

Question 1.
What is the single term that answers all the following questions?
(1) What decides whether you are blue eyed or brown eyed?
(2) Why does wheat grain germinate to produce wheat plant and not rice plant?
(3) Which acid molecules are present in nuclei of living cells?
Answer:
(1) Nucleic acid (DNA)
(2) Nucleic acid (DNA)
(3) Nucleic acid (DNA + RNA)

Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules

Use your brain power! (Textbook Page No 317)

Question 1.
Draw structural formulae of nucleosides formed from the following sugars and bases.
(1) D – ribose and guanine
(2) D – 2 – deoxyribose and thymine
Answer:
(1) D-ribose and guanine
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 93
(2) D – 2 – deoxyribose and thymine
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 94

Problem 14.5 (Textbook Page No 318)

Queston 1.
Draw a schematic representaion of trinucicotide segment ACT of a DNA molecule.
Solution :
In DNA molecule sugar is deoxyribose. The base ‘A’ in the given segment is at 5 end while the base T at the 3’ end. I-fence the schematic representation of the given segment of DNA is
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 97

Problem 14.6 : (Textbook Page No 320)

Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules

Question 1.
Write the sequence of the complementary strand of the following portion of a DNA molecule : 5 -ACGTAC-3
Solution :
The complementary strand runs in opposite direction from the 3′ end to the 5′ end. It has the base sequence decided by complementary base pairs A – T and C – G.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 99

Problem 14.2 : (Text Page No 303)

Question 1.
Assign D/L configuration to the following monosaccharides.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 20
Solution :
D/L configuration is assigned to Fischer projection formula of monosaccharide on the basis of the lowest chiral carbon.

Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 21
Threose has two chiral carbons C-2 and C-3. The given Fischer projection formula of threose has OH groups at the lowest C -3 chiral carbon on the right side.
∴ It is D-threose.

Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 22
Ribose has three chiral carbons C – 2, C – 3 and C -4.
The given Fischer projection formula of ribose has – OH group at the lowest C -4 chiral carbon on the left side.
∴ It is L-ribose

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 13 Amines Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 13 Amines

1. Choose the most correct option.

Question i.
The hybridization of nitrogen in primary amine is ………………………..
a. sp
b. sp2
c. sp3
d. sp3d
Answer:
c. sp3

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines

Question ii.
Isobutylamine is an example of ………………………..
a. 2° amine
b. 3° amine
c. 1° amine
d. quaternary ammonium salt.
Answer:
a. 2° amine

Question iii.
Which one of the following compounds has the highest boiling point?
a. n-Butylamine
b. sec-Butylamine
c. isobutylamine
d. tert-Butylamine
Answer:
a. n-Butylamine

Question iv.
Which of the following has the highest basic strength?
a. Trimethylamine
b. Methylamine
c. Ammonia
d. Dimethylamine
Answer:
d. Dimethylamine

Question v.
Which type of amine does produce N2 when treated with HNO2?
a. Primary amine
b. Secondary amine
c. Tertiary amine
d. Both primary and secondary amines
Answer:
a. Primary amine

Question vi.
Carbylamine test is given by
a. Primary amine
b. Secondary amine
c. Tertiary amine
d. Both secondary and tertiary amines
Answer:
a. Primary amine

Question vii.
Which one of the following compounds does not react with acetyl chloride?
a. CH3-CH2-NH2
b. (CH3-CH2)2NH
c. (CH3-CH2)3N
d. C6H5-NH2
Answer:
Answer:
c. (CH3 – CH2)3N

Question viii.
Which of the following compounds will dissolve in aqueous NaOH after undergoing reaction with Hinsberg reagent?
a. Ethylamine
b. Triethylamine
c. Trimethylamine
d. Diethylamine
Answer:
a. Ethyl amine

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines

Question ix.
Identify ‘B’ in the following reactions
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 239
Answer:
d. CH3-CH2-OH

Question x.
Which one of the following compounds contains azo linkage?
a. Hydrazine
b. p-Hydroxyazobenzene
c. N-Nitrosodiethylamine
d. Ethylenediamine
Answer:
b. p-Hydroxyazobenzene

2. Answer in one sentence.

Question i.
Write reaction of p-toluenesulfonyl chloride with diethylamine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 223

Question ii.
How many moles of methylbromide are required to convert ethanamine to N, N-dimethyl ethanamine?
Answer:
2 moles of methylbromide are required to convert ethanamine to N, N-dimethyl ethanamine.

Question iii.
Which amide does produce ethanamine by Hofmann bromamide degradation reaction?
Answer:
Propanamide (CH3 – CH2 – CONH2) produces ethanamine by Hofmann bromamide degradation reaction.

Question iv.
Write the order of basicity of aliphatic alkylamine in gaseous phase.
Answer:
The order of basicity of aliphatic alkyl amines in the gaseous follows the order : tertiary amine > secondary amine > primary amine > NH3.

Question v.
Why are primary aliphatic amines stronger bases than ammonia?
Answer:
The alkyl group tends to increase the electron density on the nitrogen atom. As a result, amines can donate the lone f pair of electrons on nitrogen more easily than ammonia. Hence, aliphatic amines are stronger bases than ammonia.

Question vi.
Predict the product of the following reaction. Nitrobenzene Sn/Conc. HCl?
Answer:
The product is aniline/Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 224

Question vii.
Write the IUPAC name of benzylamine.
Answer:
The IUPAC name is Phenylmethanamine.

Question viii.
Arrange the following amines in an increasing order of boiling points. n-propylamine, ethylmethyl amine, trimethylamine.
Answer:
Amines in an increasing order of boiling points : trimethyl amine, ethyl methyl amine, n-propyl amine

Question ix.
Write the balanced chemical equations for the action of dil H2SO4 on diethylamine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 225

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines

Question x.
Arrange the following amines in the increasing order of their pKb values. Aniline, Cyclohexylamine, 4-Nitroaniline
Answer:
Cyclohexyl amine (pKA 3.34), aniline (pKA 9.13) 4-nitroaniline (pKA 12.99)

3. Answer the following

Question i.
Identify A and B in the following reactions.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 240
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 77

Question ii.
Explain the basic nature of amines with suitable example.
Answer:
The basic strength of amines is expressed in terms of Kb or pKb value. According to Lowry-Bron-sted theory the basic nature of amines is explained by the following equilibrium equation.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 80

In this equilibrium amine accepts H+, hence an amine is a Lowry-Bronsted base.

According to Lewis theory, the species which donates a pair of electrons is called a base.

The nitrogen atom in amiqes has a lone pair of electrons, which can be donated to suitable acceptor like proton H+.

The aqueous solutions of amines are basic in nature due to release of free OH ions in solutions. Hence amines are Lewis bases. There exists an equilibrium in their aqueous solutions as follows :

R – NH2 + H2O ⇌ RNH3 + OH

Since OH is a stronger base, equilibrium shifts towards left-hand side giving less concentration of OH.

Here, Kb value is smaller and pKb value is larger.

Hence amines are weak bases.

Question iii.
What is diazotisation? Write diazotisation reaction of aniline.
Answer:
Aryl amines react with nitrous acid in cold condition (273 – 278 K) forms arene diazonium salts. The conversion of primary aromatic amine into diazonium salts is called diazotisation.

Diazotisation of aniline :
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 127

Question iv.
Write reaction to convert acetic acid into methylamine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 72

Question v.
Write a short note on coupling reactions.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 138
Reactions involving retention of diazo group : (Coupling reactions) :

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines

Question vi.
Explain Gabriel phthalimide synthesis.
Answer:
Phthalimide is reacted with alcoholic KOH to form potassium phthalimide. Further potassium phthalimide is treated with an ethyl iodide. The product N-ethylphthalimide is hydrolysed with aq NaOH to form ethyl amine. This reaction is known Gabriel phthalimide synthesis.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 52

Question vii.
Explain carbylamine reaction with suitable examples.
Answer:
Aliphatic or aromatic primary amines on heating with chloroform and alcoholic potassium hydroxide solution form carbyl amines or isocyanides with extremely unpleasant smell. This reaction is a test for primary amines.

Secondary and tertiary amines do not give this test.

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 120
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 121

Question viii.
Write reaction to convert
(i) methanamine into ethanamine
(ii) Aniline into p-bromoaniline.
Answer:
(1) Methanamine into ethanamine
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 195
(2) Aniline into p-bromo aniline
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 196

Question ix.
Complete the following reactions :
a. C6H5N2 Cl + C2H5OH →
b. C6H5NH2 + Br2(aq) → ?
Answer:
(a)
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 211

(b)
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 213

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines

Question x.
Explain Ammonolysis of alkyl halides.
Answer:
When an alkyl halide is heated with alcoholic ammonia in a sealed tube under pressure at 373 K, a mixture of primary, secondary, tertiary amines and a quaternary ammonium salt is obtained. In this reaction, breaking of C – X bond by ammonia is called ammonolysis of alkyl halides. The reaction is also known as alkylation. For example, when methyl bromide is heated with alcoholic ammonia at 373 K, it gives a mixture of methylamine (a primary amine), dimethylamine (a secondary amine), trimethyl amine (a tertiary amine) and tetramethylam- monium bromide (a quaternary ammonium salt).
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 30

The order of reactivity of alkyl halides with ammonia is R – I > R – Br > R – Cl.

Question xi.
Write reaction to convert ethylamine into methylamine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 72

4. Answer the following.

Question i.
Write the IUPAC names of the following amines :
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 241
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 20

Question ii.
What are amines? How are they classified?
Answer:
Amines are classified on the basis of the number of hydrogen atoms of ammonia that are replaced by alkyl group. Amines are classified as primary (1°), secondary (2°) and tertiary (3°).

(1) Primary amines (1° amines) : The amines in which only one hydrogen atom of ammonia is replaced by an alkyl group or aryl group are called primary (1°) amines.

Examples :
(i) CH3 – NH2 methylamine
(ii) CH3 – CH2 – NH2 ethylamine
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 2

(2) Secondary amines (2° amines) : The amines in which two hydrogen atoms of ammonia are replaced by two, same or different alkyl or aryl groups are called secondary (2°) amines.

Examples :
(i) C2H5 – NH – CH3 ethylmethylamine
(ii) CH3 – NH – CH3 dimethylamine
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 3

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines

(3) Tertiary amines (3° amines) : The amines in which all the three hydrogen atoms of ammonia are replaced by three same or different alkyl or aryl groups are called tertiary (3°) amines.

Examples :
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 4

Secondary and tertiary amines are further classified as (1) Simple or symmetrical amines (2) Mixed or unsymmetrical amines.

(i) Simple or symmetrical amines : In simple amines same alkyl groups are attached to the nitrogen e.g.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 5
(ii) Mixed or unsymmetrical amines : In mixed amines different alkyl groups are attached to the nitrogen.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 6

Question iii.
Write IUPAC names of the following amines.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 242
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 21

Question iv.
Write reactions to prepare ethanamine from
a. Acetonitrile
b. Nitroethane
c. Propionamide
Answer:
a. Ethanamine from acetonitrile :
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 73
b. Ethanamine from nitroethane :
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 74
c. Ethanamine from Propionamide :
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 75

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines

Question v.
What is the action of acetic anhydride on ethylamine, diethylamine and triethylamine?
Answer:
Acetylation of amines : The reaction in which the H atom attached to nitrogen in amine is replaced by acetyl group Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 112 is called acetylation of amines.

(1) Ethylamine on reaction with acetic anhydride forms monoacetyl derivative, N-acetylethylamine.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 113
(2) Diethylamine (a secondary amine) on reaction with acetic anhydride forms a monoacetyl derivative, N-acetyldiethyl amine (or N,N-diethyl acetamide).
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 114
(3) Triethylamine does not react with acetic anhydride as it does not have any H atom attached nitrogen atom of amin e
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 115

Question vii.
Distinguish between ethylamine, diethylamine and triethylamine by using Hinsberg’s reagent?
Answer:
This reaction is useful for the distinction of primary, secondary and tertiary amines.

(i) Primary amine (like ethyl amine) is treated with Hinsberg’s reagent (benzene sulphonyl chloride) forms N-alkyl benzene sulphonamide which dissolve in aqueous KOH solution to form a clear solution of potassium salt and upon acidification gives insoluble N-alkyl benzene sulphonamide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 141
(ii) Secondary amine like diethyl amine is treated with benzene sulphonyl chloride forms N,N-diethyl benzene which sulphonyl amide remains insoluble in aqueous KOH and does not dissolve in acid.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 142
(iii) Tertiary amine like triethyl amine does not react with benzene sulphonyl chloride and remains insoluble in KOH, however it dissolves in dil. HCl to give a clear solution due to formation of ammonium salt.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 143

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines

Question viii.
Write reactions to bring about the following conversions :
a. Aniline into p-nitroaniline
b. Aniline into sulphanilic acid?
Answer:
(1) Aniline into p-nitroaniline
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 189
(2) Aniline into sulphanilic acid.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 190

Activity :

  • Prepare a chart of azodyes, colours and its application.
  • Prepare a list of names and structures of N-containing ingredients of diet.

12th Chemistry Digest Chapter 13 Amines Intext Questions and Answers

Use your brain power! (Textbook Page No 282)

Question 1.
Classify the following amines as simple/mixed; 1°, 2°, 3° and aliphatic or aromatic. (C2H5)2NH, (CH3)3N, C2H5 – NH – CH3,
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 11
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 12

(A) Common Names : Rules

  1. According to common naming system, the amines are named as alkylamines.
  2. The common name of a primary amine is obtained by writing the name of the alkyl group followed by the word ‘amine’.
    Example : CH3 – NH2 : methyl-amine
  3. The simple {symmetrical) secondary and tertiary amines are written by adding prefix ‘di- (forpresence of two alkyl groups) and ‘tri’- (for presence of three alkyl groups) respectively to the name of alkyl groups.
    Examples: (i) CH3 – NH – CH3 dimethylamine, (ii) (C2H5)3 N triethylamine
  4. The mixed (or unsymmetrical) secondary and tertiary amines are given names by writing the names of alkyl groups in alphabetical order, followed by the word ‘amine’.
    Example : CH3 – CH2 – NH – CH3 ethyhnethylamine

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines

(B) IUPAC names : Rules

  1. According to IUPAC system of nomenclature of amines, aliphatic amines are named as alkanarnines.
  2. The name of the amine is obtained by replacing the suffix ‘e’ from parent alkane’s name by ‘amine’.
  3. The position of the amino group is indicated by the lowest possible locant.
    Example :
    Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 13
  4. In case of secondary and tertiary amines, the largest alkyl group is considered to be the parent alkane and other alkyl groups are written as N-substituents.
    Example : ClH5NH – CH3 N – Methylethanamine
  5. A complete name of amine is written as one word.

Try this….. (Textbook Page No 283)

Question 1.
Draw possible structures of all the isomers of C4H11N. Write their common as well as IUPAC names.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 18
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 19

Use your brain power! (Textbook Page No 283)

Question 1.
Write chemical equations for

(i) reaction of alc. NH with C2H5I.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 31

(ii) Amonolysis of benzyl chloride followed by the reaction with 2 moles of CH3I.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 32

(2) Ammonolysis of alkyl halides is not suitable method to prepare primary amines.
Answer:
In the laboratory, ammonolysis of alkyl halides is not a suitable method to prepare primary amines as it gives a mixture of primary, secondary, tertiary amines and quaternary ammonium salts. (Refer to the reaction in answer to Question 16). The separation of primary amine becomes difficult.

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines

Problem 13.1 : (Textbook Page No 285)

Question 1.
Write reaction to convert methyl bromide into ethyl amine? Also, comment on the number of carbon atoms in the starting compound and the product.
Solution :
Methyl bromide can be converted into ethyl amine in two stage reaction sequence as shown below.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 47
The starting compound methyl bromide contains one carbon atom while the product ethylamine contains two carbon atoms. A reaction in which number of carbons increases involves a step up reaction. The overall conversion of methyl bromide into ethyl amine is a step up conversion.

Use your brain power! (Textbook Page No 285)

Identify ‘A’ and ‘B’ in the following conversions.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 48
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 49

Use your brain power! (Textbook Page No 286)

Question 1.
Write the chemical equations for the following conversions :
(1) Methyl chloride to ethylamine.
(2) Benzamide to aniline.
(3) 1, 4-Dichlorobutane to hexane-1, 6-diamine.
(4) Benzamide to benzylamine.
Answer:
(1) Methyl chloride to ethylamine
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 56
(2) Benzamide to aniline
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 57
(3) 1, 4-Dichlorobutane to hexane-1, 6-diamine
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 58
(4) Benzamide to benzylamine
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 59

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines

Use your brain power! (Textbook Page No 287)

Question 1.
Arrange the following :
(1) In decreasing order of the boiling point C2H5 – OH, C2H5 – NH2, (CH3)2 NH
(2) In increasing order of solubility in water: C2H5 – NH2, C3H7 – NH2, C6H5 – NH2
Answer:
(1) Decreasing order of the boiling point : C2H5 — OH, C2H5 — NH2, (CH3)2 NH
(2) Increasing order of solubility in water : C6H5NH2, C3H7 — NH2, C2H5 — NH2

Use your brain power! (Textbook Page No 288)

Question 1.
Refer to pKb values and answer which compound from the following pairs is the stronger base?
(1) CH3 – NH2 and (CH3)2 NH
(2) (C2H5)2 NH and (C2H5)3 N
(3) NH3 and (CH3)2 CH – NH2
Answer:
(1) CH3 -NH2 and (CH3)2 NH
(CH3)2 NH is a stronger base

(2) (C2H5)2 NH and (C2H5)3 N
(C2H5)2 NH is a stronger base

(3) NH3 and (CH3)2 CHNH2
(CH3)2 CHNH2 is a stronger base

Use your brain power! (Textbook Page No 290)

Question 1.
Arrange the following amines in decreasing order of their basic strength :
NH3, CH3 – NH2, (CH3)2 NH, C6H5NH2
Answer:
Decreasing order of basic strength :
(CH3)2NH, CH3 -NH2, NH3, C6H5NH2

Use your brain power! (Textbook Page No 291)

Question 1.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 94
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 95
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 96

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines

Use your brain power! (Textbook Page No 291)

Question 1.
Complete the following reaction :
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 100
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 101

Use your brain power! (Textbook Page No 292)

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 118
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 119

Use your brain power! (Textbook Page No 292)

Question 1.
Write the carbylamine reaction by using aniline as starting material.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 122

Can you tell? (Textbook Page No 292)

(1) What is the formula of nitrous acid ?
(2) Can nitrous acid be stored in bottle ?
Answer:
(1) Formula of nitrous acid : H – O – N = O
(2) Nitrous acid cannot be stored in bottle.

Use your brain power! (Textbook Page No 294)

Question 1.
How will you distinguish between methyl amine, dimethylamine and trimethylamine by Hinsberg’s test?
Answer:
(1) Methyl amine (primary amine) reacts with benzene sulphonyl chloride to form N-methylbenzene sulphona- mide
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 147
(2) Dimethyl amine reacts with benzene sulphonyl chloride to give N, N – dimethylbenzene sulphonamide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 148
(3) Trimethyl amine does not react with benzene sulphonyl chloride and remains insoluble in KOH
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 149

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines

Problem 13.1 : (Textbook Page No 295)

Question 1.
Write the scheme for preparation of p-bromoaniline from aniline. Justify your answer.
Solution :
NH2 – group in aniline is highly ring activating and o – /p – directing due to involvement of the lone pair of electrons on ‘N’ in resonance with the ring. As a result, on reaction with Br2 it gives 2,4,6-tribromoaniline. To get a monobromo product, it is necessary to decrease the ring activating effect of – NH2 group. This is done by acetylation of aniline. The lone pair of ‘N’ in acetanilide is also involved in resonance in the acetyl group. To that extent, ring activation decreases.

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 167

Hence, acetanilide on bromination gives a monobromo product p-bromoacetanilide. After monobromination the original – NH2 group is regenerated. The protection of – NH2 group in the form of acetyl group is removed by acid catalyzed hydrolysis to get p-bromoaniline, as shown in the following scheme.

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 168

Use your brain power! (Textbook Page No 296)

Question 1.
(1) Can aniline react with a Lewis acid?
(2) Why aniline does not undergo Frledel – Craft’s reaction using aluminium chloride?
Answer:
(1) Aniline reacts with a Lewis acid, forms salt.
(2) Aniline does not undergo Friedcl-Crafr’s reaction (alkylation and acetylation) due to salt formation with aluminium chloride (Lewis acid), which is used as catalyst. Due to this, nitrogen of anime acquires + ve charge and hence acts as strong deactivating effect on the ring and makes it difficult for electrophilic attack.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 214

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines

Can you tell? (Textbook Page No 294)

(1) Do tertiary amines have ‘H’ bonded to ‘N?
(2) Why do tertiary amines not react with benzene sulfonyl chloride?
Answer:
(1) Tertiary amines Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 146 do not have ‘H’ bonded to ‘N’.
(2) Tertiary amine does not undergo reaction with benzene sulphonyl chloride as it does not have any H atom attached to nitrogen atom of amine.

Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

1. Choose the most correct option.

Question i.
In the following resonating structures A and B, the number of unshared electrons in valence shell present on oxygen respectively are
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 1
Answer:
c. 4, 6

Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question ii.
In the Wolf -Kishner reduction, alkyl aryl ketones are reduced to alkylbenzenes. During this change, ketones are first converted into
a. acids
b. alcohols
c. hydrazones
d. alkenes
Answer:
c. hydrazones

Question iii.
Aldol condensation is
a. electrophilic substitution reaction
b. nucleophilic substitution reaction
c. elimination reaction
d. addition – elimination reaction
Answer:
d. addition-elimination reaction

Question iv.
Which one of the following has the lowest acidity?
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 2
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 3
Answer:
(c)

Question v.
Diborane reduces
a. ester group
b. nitro group
c. halo group
d. acid group
Answer:
d. acid group

Question vi.
Benzaldehyde does NOT show positive test with
a. Schiff reagent
b. Tollens’ ragent
c. Sodium bisulphite solution
d. Fehling solution
Answer:
d. Fehling solution

2. Answer the following in one sentence

Question i.
What are aromatic ketones?
Answer:
The compounds in which Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 23 group is attached to either two aryl groups or one aryl and one alkyl group are called aromatic ketones.

For example :
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 24

Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question ii.
Is phenylacetic acid an aromatic carboxylic acid?
Answer:
Phenylacetic acid is not an aromatic carboxylic acid.

Question iii.
Write reaction showing conversion of ethanenitrile into ethanol.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 311

Question iv.
Predict the product of the following reaction:
\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{COOCH}_{3} \frac{\mathrm{i} \cdot \mathrm{AlH}(\mathrm{i}-\mathrm{Bu})_{2}}{\text { ii. } \mathrm{H}_{3} \mathrm{O}^{\oplus}}{\longrightarrow} ?\)
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 313

Question v.
Name the product obtained by reacting toluene with carbon monoxide and hydrogen chloride in presence of anhydrous aluminium chloride.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 314

Question vi.
Write reaction showing conversion of Benzonitrile into benzoic acid.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 315

Question vii.
Name the product obtained by the oxidation of 1,2,3,4-tetrahydronaphthalene with acidified potassium permanganate.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 316

Question viii.
What is formalin?
Answer:
The aqueous solution of formaldehyde (40%) is known as formalin.

Question ix.
Arrange the following compounds in the increasing order of their boiling points : Formaldehyde, ethane, methyl alcohol.
Answer:
Ethane, formaldehyde, methyl alcohol.

Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question x.
Acetic acid is prepared from methyl magnesium bromide and dry ice in presence of dry ether. Name the compound which serves not only reagent but also as cooling agent in the reaction.
Answer:
The cooling agent used in the above reaction is dry ice (O = C = O).

3. Answer in brief.

Question i.
Observe the following equation of reaction of Tollens’ reagent with aldehyde. How do we know that a redox reaction has taken place. Explain.
\(\begin{array}{r}
\mathrm{R}-\mathrm{CHO}+2 \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}+\mathrm{OH}^{-} \stackrel{\Delta}{\longrightarrow} \\
\mathrm{R}-\mathrm{COO}^{-}+2 \mathrm{Ag} \downarrow+4 \mathrm{NH}_{3}+2 \mathrm{H}_{2} \mathrm{O}
\end{array}\)
Answer:
Tollen’s reagent oxidises acetaldehyde to acetic acid (carboxylate ion) and Ag in Tollen’s reagent complex are reduced to silver. In this reaction, oxidation and reduction takes place simultaneously hence, it is a redox reaction.

Question ii.
Formic acid is stronger than acetic acid. Explain.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 146
In acetic acid, the methyl group is an electron-donating group. The acetate ion formed gets destabilized due to the electron releasing effect of methyl group ( +1 effect) which is higher than that of H-atom in the corresponding formic acid. As a result, acetic acid dissociates to a lesser extent. Thus decreasing the acidity of acetic acid.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 147
Formic acid having lower pKa value than acetic acid. Hence, formic acid is a stronger acid than acetic acid.

Question iii.
What is the action of hydrazine on cyclopentanone in presence of —. KOH in ethylene glycol?
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 308

Question iv.
Write reaction showing conversion of Acetaldehyde into acetaldehyde dimethyl acetal.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 184

Question v.
Aldehydes are more reactive toward nucleophilic addition reactions than ketones. Explain.
Answer:
The reactivity of aldehydes and keones is due to the polarity of carbonyl group which results in electrophilicity of carbon. The reactivity is further explained on the basis of electronic effect and steric effects.

(1) Influence of electronic effects: A ketone has two electron-donating aJ.kyl groups ( + I effect) bonded to carbonyl carbon which are responsible for decreasing its positive polarity and electrophilicity. In contrast. aldehydes have only electron-donating group bonded to carbonyl carbon. This shows aldehydes are more electrophilic than ketones.

(2) Steric effects : Two bulky alkyl groups in ketone come in the way of the incoming nucleophile. This is called steric hindrance to nucleophilic attack.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 160
On the other hand. nucleophile can easily attack the carbonyl carbon in aldehyde because has one alkyl group and is less crowded or sterically less hindered.

Hence aldehydes are more reactive and can easily be attacked by nucleophiles.

Question vi.
Write reaction showing the action of the following reagent on propane nitrile
a. Dilute NaOH
b. Dilute HCl ?
Answer:
(1) Action of dil NaOH:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 115
(2) Action of dil HCl:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 116

Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question vi.
Arrange the following carboxylic acids with increasing order of their acidic strength and justify your answer.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 4
Answer:
The increasing order of acidity will be 1 <3 <2. Acidity depends on mainly two factors : (1) ease of proton release (2) stability of conjugate base formed. In example (3) the ether O exerts a I effect and is closer to COOH group than in 2 (1 effect diminishes). Also the conjugate base formed will be stabilized by the same – I effect by delocalization of charge.

4. Answer the following

Question i.
Write a note on
a. Cannizaro reaction
b. Stephen reaction.
Answer:
(1) The carbon atom adjacent to carbonyl carbon atom is called a-carbon atom (α – C) and the hydrogen atom attached to a-carbon atom is called α-hydrogen atom (α – H).
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 226
(2) The α-hydrogen of aldehydes and ketones is acidic in nature due to (i) the strong-I effect of carbonyl group (ii) resonance stabilization of the carbanion.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 227
(3) Aldol condensation reaction is characteristic reaction of aldehydes and ketones containing active α-hydrogen atom.

(4) When aldehydes or ketones containing α – H atoms are warmed with a dilute base or dilute acid, two molecules of them undergo self condensation to give β-hydroxy aldehyde (aldol) or β-hydroxy ketone (ketol) respectively. The reaction is known as Aldol addition Reaction.

(5) In aldol condensation, the product is formed by the nucleophilic addition of α-carbon atom of a second molecule which gets attached to carbonyl carbon atom of the first molecule and α-hydrogen atom of the second molecule gets attached to carbonyl oxygen atom of the first molecule forming (- OH) group to give β-hydroxy aldehyde or ketone.

(6) This is a reversible reaction, establishing an equilibrium favouring aldol formation to a greater extent than ketol formation.

(7) For aldehyde :
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 228
Acetaldol on heating undergoes subsequent elimination of water giving rise to α, β unsaturated aldehyde.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 229
The overall reaction is called aldol condensation. It is a nucleophilic addition-elimination reaction. For ketone :
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 230
Diacetone alcohol on dehydration by heating forms α, β unsaturated ketone.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 231

Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question ii.
What is the action of the following reagents on toluene ?
a. Alkaline KMnO4, dil. HCl and heat
b. CrO2Cl2 in CS2
c. Acetyl chloride in presence of anhydrous AlCl3.
Answer:
(1) Action of alkaline KMnO4 : When toluene is heated with alkaline KMnO4. (methyl group gets oxidised to earboxy lic group) benzoic acid is obtained
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 304

(2) Action of CrO2Cl2 in C2:
Answer:
When Loluenc is ircated with soluion of chromyl chloride (CrO2Cl2) in Cs2, brown chromium complex is obtained, which on acid hydrolysis gives benzaldehyde.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 305

(3) Action of acetyl chloride in presence of anhyd. AlCl3.
Answer:
When toluene is treated with acetyl chloride in presence of anhydrous AlCl3 4-methyl acetophenone is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 306

Question iii.
Write the IUPAC names of the following structures :
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 5

Question iv.
Write reaction showing conversion of p- bromoisopropyl benzene into p-Isopropyl benzoic acid (3 steps).
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 117

Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question v.
Write reaction showing aldol condensation of cyclohexanone.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 247

Activity :
Draw and complete the following reaction scheme which starts with acetaldehyde. In each empty box, write the structural formula of the organic compound that would be formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 323

12th Chemistry Digest Chapter 12 Aldehydes, Ketones and Carboxylic Acids Intext Questions and Answers

Use your brain power! (Textbook Page No 254)

Question 1.
Classify the following as aliphatic and aromatic aldehydes.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 18
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 19

Use your brain power! (Textbook Page No 255)

Question 1.
Classify the following as simple and mixed ketones. Benzoplienone, acetoneq hutanoneq acetophenone.

Compoun
Benzophenone ……………………………………………..
Acetone ……………………………………………..
Butanone ……………………………………………..
Acetophenone ……………………………………………..

Answer:

Compound
Benzophenone Simple ketone
Acetone Simple ketone
Butanone Mixed ketone
Acetophenone Mixed ketone

Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Use your brain power! (Textbook Page No 264)

Write IUPAC names for the following compounds.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 49

Try this ….. (Textbook Page No 260)

Question 1.
Draw structures for the following :
(1) 2-Methylpentanal
(2) Hexan-2-one
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 50

Can you tell? (Textbook Page No 260)

Question 1.
Which is the reagent that oxidizes primary alcohols to only aldehydes and does not oxidize aldehydes further into carboxylic acid ?
Answer:
The reagent that is used to make only aldehydes is-heated Cu at 573 K.

Use your brain power! (Textbook Page No 261)

Question 1.
Write the structure of the product formed on Rosenmund reduction of ethanoyl chloride and benzoyl chloride.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 84

Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Can you think? (Textbook Page No 261)

Question 1.
What is the alcohol formed when benzoyl chloride is reduced with pure palladium as the catalyst ?
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 85

Use your brain power! (Textbook Page No 262)

Question 1.
Name the compounds which are used for the preparation of benzophenone by Friedel-Crafts acylation reaction. Draw their structures.
Answer:
The compounds which are used in preparation of benzophenone by Friedel – Crafts reaction are :
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 106

Use your brain power! (Textbook Page No 263)

Identify the reagents necessary to achieve each of the following transformations:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 108
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 109

Use your brain power! (TextBook Page No 264)

Predict the products (name and structure) in the following reactions.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 133
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 134

Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Problem 12.1 : (Textbook Page No 276)

Question 1.
Alcohols (R – OH), phenols (Ar – OH) and carboxylic acids (R – COOH) can undergo ionization of O – H bond to give away proton H+; yet they have different pKa values, which are 16, 10 and 4.5 respectively. Explain.
Solution :
pKa value is indicative of acid strength. Lower of pKa value stronger the acid. Alcohols, phenols and carboxylic acids, all involve ionization of an O – H bond. But their different pKa values indicate that their acid strengths are different. This is because the resulting conjugate bases are stabilized to different extents.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 150
As the conjugate base of carboxylic acid is best stabilized, among the three, carboxylic acids are strongest and have the lowest pKa value. As conjugate base of alcohols is destabilized, alcohols are weakest acids and have highest pIQ value. As conjugate base of phenols is moderately stabilized, phenols are moderately acidic and have intermediate pBQ value.

Try this….. (Textbook Page No 277)

Question 1.
Compare the following two conjugate bases and answer.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 151
(1) Indicate the inductive effects of CH3 – group in (a) and Cl-group in (b) by putting arrowheads in the middle of appropriate covalent bonds.
(2) Which species is stabilized by inductive effect, (a) or (b) ?
(3) Which species is destabilized by inductive effect, (a) or (b) ?
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 152
(2) The monochloroacetate ion formed gets stabilised due to electron-withdrawing of Cl atom (- I effect).
(3) Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 153 The acetate ion formed gets destabilised due to electron releasing effect of methyl group

Use your brain power! (Textbook Page No 277)

Question 1.
(1) Compare the pKa values and arrange the following in an increasing order of acid strength. CI3CCOOH, ClCH2COOH, CH3COOH, Cl2CHCOOH
(2) Draw structures of conjugate bases of monochloroacetic acid and dichloroacetic acid. Which one is more stabilized by – 1 effect?
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 154

Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 155 The dichloroacetate ion formed gets stabilised due electron-withdrawing effect of two chlorine atoms. (- 1 effect)

Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Try this….. (Textbook Page No 277)

Question 1.
Arrange the following acids in order of their decreasing acidity.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 156
Answer:
Acidity in the decreasing order

Try this ….. (Textbook Page No 267)

Question 1.
Draw the structure of propanone and indicate its polarity.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 161

Can you tell? (Textbook Page No 268)

Question 1.
Simple hydrocarbons, ethers, ketones and alcohols do not get oxidized by Tollen’s reagent. Explain, Why?
Answer:
(1) Due to the presence of hydrogen atom in aldehyde group Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 167, an aldehyde is oxidised to carboxylic acid Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 168 which is not possible in case of ethers, ketones, alcohols and hydrocarbons.
(2) In ketones, carbonyl atom is attached to C-atom, hence C – C bond in Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 169 can’t be broken easily.
(3) H atom attached to carbonyl carbon can be oxidised to – OH group giving carboxylic group Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 168 Therefore, aldehyde reduces Tollen’s reagent, whereas simple hydrocarbons, ethers, ketones and alcohols do not reduce Tollen’s reagent.

Use your brain power! (Textbook Page No 269)

Question 1.
Sodium bisulfite is sodium salt of sulfurous acid, write down its detailed bond structure.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 179
Bond structure of sodium bisulfite

Use your brain power! (Textbook Page No 270)

Predict the product of the following reaction:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 185
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 186

Use your brain power! (Textbook Page No 271)

Question 1.
Draw the structures of
(1) The semicarbazone of cyclohexanone
(2) The imine formed in the reaction between 2-methylhexanal and ethyl amine
(3) 2, 4-dinitrophenyl hydrazone of acetaldehyde.
Answer:
(1) The semi carbazone of cyclohexanone.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 217
(2) The imine formed between 2-methyl hexanal and ethyl amine.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 218
(3) 2, 4-dinitrophenylhydrazone of acetaldehyde.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 219

Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Try this….. (Textbook Page No 272)

Question 1.
Write chemical reactions taking place when propan-2-ol is treated with iodine and sodium hydroxide.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 224

Question 2.
When acetaldehyde Is treated with dilute NaOH, the following reaction is observed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 225
(1) What are the functional groups in the product?
(2) Can another product be formed during the same reaction? (Deduce the answer by doing atomic audit of reactant and product)
(3) Is this an addition reaction or condensation reaction?
Answer:
(1) There arc two functiona’ groups in the product: -CRO and -OH
(2) No other product can be formed in the same reaction.
(3) This is an addition reaction.

Use your brain power! (Textbook Page No 273)

Question 1.
Observe the following reaction.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 234

Question 2.
Will this reaction give a mixture of products like a cross aldol reaction ?
Answer:
No, since benzaldehyde, Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 235 does not have a-hydrogen atom, it will not undergo self aldol condensation.

Use your brain power! (Textbook Page No 274)

Question 1.
Can isobutyraldehyde undergo a Cannizzaro reaction? Explain.
Answer:
Since isobutyraldehydeMaharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 252contains a-carbon atom, it cannot undergo Cannizzaro reaction.

Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Can you tell? (Textbook Page No 279)

What is the term used for elimination of water molecule ?
Answer:
Dehydration.

Use your brain power! (Textbook Page No 278)

Question 1.
Fill in the blanks and rewrite the balanced equations.
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 302
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 303

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 11 Alcohols, Phenols and Ethers Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

1. Choose the correct option.

Question i.
Which of the following represents the increasing order of boiling points of (1), (2), and (3)?
(1) CH3 – CH2 – CH2 – CH2 – OH
(2) (CH3)2 CH – O – CH3
(3) (CH3)3COH
A. (1) < (2) < (3)
B. (2) < (1) < (3)
C. (3) < (2) < (1)
D. (2) < (3) < (1)
Answer:
(a) (1) < (2) < (3)

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Question ii.
Which is the best reagent for carrying out following conversion ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 272
A. LiAlH4
B. Conc. H2SO4, H2O
C. H2/Pd
D. B2H6, H2O2 – NaOH
Answer:
B. Conc. H2SO4, H2O

Question iii.
Which of the following reaction will give ionic organic product on reaction ?
A. CH3 – CH2 – OH + Na
B. CH3 – CH2 – OH + SOCl2
C. CH3 – CH2 – OH + PCl5
D. CH3 – CH2 – OH + H2SO4
Answer:
C. CH3 – CH2 – OH + PCl5

Question iv.
Which is the most resistant alcohol towards oxidation reaction among the follwoing ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 273
Answer:
(c)

Question v.
Resorcinol on distillation with zinc dust gives
A. Cyclohexane
B. Benzene
C. Toluene
D. Benzene-1, 3-diol
Answer:
(b) Benzene

Question vi.
Anisole on heating with concerntrated HI gives
A. Iodobenzene
B. Phenol + Methanol
C. Phenol + Iodomethane
D. Iodobenzene + methanol
Answer:
B. Phenol + Methanol

Question vii.
Which of the following is the least acidic compound ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 274
Answer:
(b)

Question viii.
The compound incapable of hydrogen bonding with water is ……
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 275
Answer:
(b)

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Question ix.
Ethers are kept in air tight brown bottles because
A. Ethers absorb moisture
B. Ethers evaporate readily
C. Ethers oxidise to explosive peroxide
D. Ethers are inert
Answer:
C. Ethers oxidise to explosive peroxide

Question x.
Ethers reacts with cold and concentrated H2SO4 to form
A. oxonium salt
B. alkene
C. alkoxides
D. alcohols
Answer:
A. oxonium salt

2. Answer in one sentence/ word.

Question i.
Hydroboration-oxidation of propene gives…..
Answer:
n-propyl alcohol (CH3 – CH2 – CH2 – OH)

Question ii.
Write the IUPAC name of alcohol having molecular formula C4H10O which is resistant towards oxidation.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 244

Question iii.
Write the structure of optically active alcohol having molecular formula C4H10O
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 245

Question iv.
Write name of the electrophile used in Kolbe’s Reaction.
Answer:
Electrophile : Carbon dioxide (O = C = O)

3. Answer in brief.

Question i.
Why phenol is more acidic than ethyl alcohol ?
Answer:
(1) In ethyl alcohol, the -OH group is attached to sp3 – hybridised carbon while in phenols, it is attached to sp2 – hybridised carbon.

(2) Due to higher electronegativity of sp2 – hybridised carbon, electron density on oxygen decreases. This increases the polarity of O-H bond and results in more ionization of phenol than that of alcohols.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 117

(3) Electron donating inductive effect (+1 effect) of the alkyl group destabilizes alkoxide ion. As a result alcohol does not ionize much in water, therefore alcohol is neutral compound in aqueous medium.

(4) In alkoxide ion, the negative charge is localized on oxygen, while in phenoxide ion the negative charge is delocalized. The delocalization of the negative charge (structure I to V) makes phenoxide ion more stable than that of phenol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 118

The delocalization of charge in phenol (structures VI to X), the resonating structures have charge separation (where oxygen atom of OH group to be positive and delocalization of negative charge over the ortho and para positions of aromatic ring) due to which phenol molecule is less stable than phenoxide ion. This favours ionization of phenol. Thus phenols are more acidic than ethyl alcohol.

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Question ii.
Why p-nitrophenol is a stronger acid than phenol ?
Answer:
(1) In p-nitrophenol, nitro group (NO2) is an electron withdrawing group present at para position which enhances the acidic strength (-1 effect). The O-H bond is under strain and release of proton (H+) becomes easy. Further p-nitrophenoxide ion is more stabilised due to resonance.

(2) Since the absence of electron withdrawing group (like – NO2) in phenol at ortho and para position, the acidic strength of phenol is less than that of p-nitrophenol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 119

Question iii.
Write two points of difference between properties of phenol and ethyl alcohol.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 122

Question iv.
Give the reagents and conditions necessary to prepare phenol from
a. Chlorobenzene
b. Benzene sulfonic acid.
Answer:
(1) From chlorobenzene : Reagents required : NaOH and dil. HC1 Temperature : 623 K, Pressure : 150 atm
(2) From Benzene sulphonic acid : Reagents required : aq NaOH, caustic soda, dil. HC1 Temperature : 573 K

Question v.
Give the equations of the reactions for the preparation of phenol from isopropyl benzene.
Answer:
Preparation of phenol from cumene (isopropylbenzene) : This is the commercial method of preparation of phenol. When a stream of air is passed through cumene (isopropylbenzene) suspended in aqueous Na2CO3 solution in the presence of cobalt naphthenate catalyst, isopropyl benzene hydroperoxide or cumene hydroperoxide is formed. Isopropylbenzene hydroperoxide on warming with dil. H2SO4 gives phenol and acetone. Acetone is an important by-product of the reaction and is separated by distillation. The reaction is called auto oxidation.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 111

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Question vi.
Give a simple chemical test to distinguish between ethanol and ethyl bromide.
Answer:
When ethyl bromide is heated with aq NaOH; ethyl alcohol is formed whereas ethanol does not react with aq NaOH
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 123

4. An ether (A), C5H12O, when heated with excess of hot HI produce two alkyl halides which on hydrolysis form compound (B)and (C), oxidation of (B) gave and acid (D), whereas oxidation of (C) gave a ketone (E). Deduce the structural formula of (A), (B), (C), (D) and (E).
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 243

5. Write structural formulae for

a. 3-Methoxyhexane
b. Methyl vinyl ether
c. 1-Ethylcyclohexanol
d. Pentane-1,4-diol
e. Cyclohex-2-en-1-ol
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 35

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

6. Write IUPAC names of the following

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 276
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 36
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 37

Activity :
• Collect information about production of ethanol as byproduct in sugar industry and its importance in fuel economy.
• Collect information about phenols used as antiseptics and polyphenols having antioxidant activity.

12th Chemistry Digest Chapter 11 Alcohols, Phenols and Ethers Intext Questions and Answers

Use your brain power! (Textbook Page No 235)

Question 1.
Classify the following alcohols as l0/2°/3° and allylic/benzylic
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 7
Answer:
(1) Ally lie alcohol (primary)
(2) Allylic alcohol (secondary)
(3) Allylic alcohol (tertiary)
(4) Benzylic alcohol (primary)
(5) Benzylic alcohol (secondary)

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Use your brain power ….. (Textbook Page No 236)

Question 1.
Name t-butyl alcohol using carbinol system of nomenclature.
Answer:
Trimethyl carbinol.

Problem 11.1 (Textbook Page No 238)

Question 1.
Draw structures of following compounds:
(i) 2,5-DiethIphenoI
(ii) Prop-2-en-I-oI
(iii) 2-methoxypropane
(iv) Phenylmethanol
Solution :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 32

Try this ….. (Textbook Page No 238)

Write IUPAC names ol (he following compounds.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 33
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 34

Do you know (Textbook Page No 238)

Question 1.
The mechanism of hydration of ethylcnc to ethyl alcohol.
Answer:
The mechanism of hydration of ethylene involves three steps:

Step 1: Ethylene gets protonated to form carbocation by electrophilic attack of H3O (Formation of carbocation intermediate).
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 59
Step 2 : Nucleophilic attack of water on carbocation
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 60
Step 3 : Deprotonation to form an alcohol
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 61

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Problem 11.2 : (Textbook Page No 239)

Question 1.
Predict the products for the following reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 76
Solution:
The substrate (A) contains an isolated Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 77 and an aldehyde group. H2/Ni can reduce both these functional groups while LiAlH4 can reduce only – CHO of the two, Hence
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 78

Try this ….. (Textbook page 240)

Question 1.
Arrange O – H, C – H and N – H bonds in increasing order of their bond polarity.
Answer:
Increasing order of polarity :C – H, N – H, O – H

Problem 11.3 : (Textbook Page No 241)

Question 1.
The boiling point of n-butyl alcohol, isobutyl alcohol, sec-butyl alcohol and tert-butyl alcohol are 118 °C, 108 °C. 99 °C and 82 °C respectively. Explain.
Solution:
As branching increases, intermolecular van der Waal’s force become weaker and the boiling point decreases. Therefore, n-butyl alcohol has highest boiling point 118 °C and tert-butyl alcohol has lowest boiling point 83 °C. Isobutyl alcohol is a primary alcohol and hence its boiling point is higher than that of sec-butyl alcohol.

Problem 11.4 : (Textbook Page No 242)

The solubility of o-nitrophenol and p-nitrophenol is 0.2 g and 1.7 g/100 g of H2O respectively. Explain the difference.
Solution :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 115
p-Nitrophenol has strong intermolecular hydrogen bonding with solvent water. On the other hand, o-nitrophenol has strong intramolecular hydrogen bonding and therefore the intermolecular attraction towards solvent water is weak. The stronger the intermolecular attraction between solute and solvent higher is the solubility. Hence p-nitrophenol has higher solubility in water than that of o-nitrophenol.

Problem 11.5 : (Textbook Page No 243 & 244)

Question 1.
Arrange the following compounds in decreasing order of acid strength and justify.
(1) CH3 – CH2 – OH
(2) (CH3)3 C – OH
(3) C6H5 – OH
(4) p-NO2 – C6H4 – OH
Solution :
Compounds (3) and (4) are phenols and therefore are more acidic than the alcohols (1) and (2). The acidic strengths of compounds depend upon stabilization of the corresponding conjugate bases. Hence let us compare electronic effects in the conjugate bases of these compounds :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 133

The conjugate base of the alcohol (1) is destabilized by + 1 effect of one alkyl group, whereas conjugate base of the alcohol (2) is destabilized by +1 effect of three alkyl groups. Hence (2) is weaker acid than (1)
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 134

Phenols : The conjugate base of p-nitrophenol (4) is better resonance stabilized due to six resonance structures compared to the five resonance structure of conjugate base of phenol (3). The resonance structure VI has – ve charge on only electronegative oxygens. Hence the phenol (4) is stronger acid than (3). Thus the decreasing order of acid strength is (4), (3), (1), (2).

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Use your brain power (Textbook Page No 244)

Question 1.
What are the electronic effects exerted by – OCH3 and – Cl? Predict the acid strength of
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 135
Answer:
The electronic effects exerted by – Cl and – O CH3 are as follows :
(1) Cl being more electronegative atom it pulls the bonding electrons towards itself. This is known as negative inductive effect (- I).

(2) – OCH3 is less electronegative group which repels the bonding electrons away from it. This is known as positive inductive effect ( + I).

(3) The relative to parent phenol, Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 136 is more acidic than Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 137.

Problem 11.6 : (Textbook Page No 245)

Question 1.
Mechanism of acid catalyzed dehydration of ethanol to give ethene.
Answer:
The mechanism of dehydration of ethanol involves the following order :
Step 1 : Formation of protonated alcohols : Initially ethyl alcohol gets protonated to form ethyl oxonium ion.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 164
Step 2 : Formation of carbocation : It is the slowest step and hence, the rate determining step of the reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 165
Steps 3: Formation of ethene: Removal of a proton (H+) from carbocation.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 166

The acidused in step I is released in step 3, the equilibrium is shifted to the right, ethene is removed as it is formed.

Problem 11.6 : (Textbook Page No 245)

Question 1.
Write the reaction showing major and minor products formed on heating butan-2-ol with concentrrated sulphuric acid.
Solution :
In the reaction described butan-2-ol undergoes dehydration to give but-2-ene (major) and but-l-ene (minor) in accordance with Saytzeff rule.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 167

Problem 11.7 : (Textbook Page No 246)

Question 1.
Write and explain reactions to convert propan-l-ol into propan-2-ol.
Solution :
The dehydration of propane-l-ol to propene is the first step. Markownikoff hydration of propene is the second step to get the product propan-2-ol. This is brought about by reaction with concemtrated H2SO4 followed by hydrolysis.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 168

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Problem 11.8 : (Textbook Page No 246)

Question 1.
An organic compound gives hydrogen on reaction with sodium metal. It forms an aldehyde having molecular formula C2H4O on oxidation with pyridinium chlorochromate. Name the compounds and give equations of these reactions.
Solution :
The given molecular formula C2H4O of aldehyde is written as CH3 – CHO. Hence the formula of alcohol from which this is obtained by oxidation must be CH3 – CH2 – OH. The two reactions can, therefore, be represented as follows :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 178

(Do you know? Textbook Page No 248)

Question 90.
Write the mechanism of dehydration of alcohol to give ether.
Answer:
Dehydration of alcohols to form ether is SN2 reaction. The mechanism of dehydration of ethanol involves the following steps.

Step 1 (Protonation) : Initially ethyl alcohol gets protonated in the presence of acid to form ethyl oxonium ion.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 191
Step 2 (SN2 mechanism) : Protonated alcohol species undergoes a backside attack by second molecule of alcohol is a slow step.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 192

Step 3 (Deprotonation) : Formation of diethyl ether by elimination of proton
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 193

Problem 11.9 : (Textbook Page No 249)

Question 1.
Ethyl isopropyl ether does not form on reaction of sodium ethoxide and isopropyl chloride.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 202
(i) What would be the main product of this reaction?
(ii) Write another reaction suitable for the preparation of ethyl isopropyl ether.
Solution :
(i) Isopropyl chloride is a secondary chloride. On treating with sodium ethoxide it gives elimination reaction to form propene as the main product.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 203
(ii) Ethyl isopropyl ether can be prepared as follows using ethyl chloride (10 chloride) as substrate.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 204

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Do you know? (Textbook Page No 250)

Question 1.
The mechanism of the reaction of HI with methoxy ethane.
Answer:
The reaction mechanism takes place as follows :
Step 1 : Protonation of ether Initially the ether molecule (methoxy ethane) protonated by cone. HI to form oxonium ion.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 229

Step 2 : Iodide is a good nucleophile. It attacks the least substituted carbon of the oxonium ion formed in step 1 and displaces an alcohol molecule by SN2 mechanism.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 230

For example :
• Use of excess HI converts the alcohol into alkyl iodide.
• In case of ether having one tertiary alkyl group the reaction with hot HI follows SN1 mechanism, and tertiary iodide is formed rather than tertiary alcohol.

Step 1 :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 231
Step 2 :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 232

Maharashtra Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 8 Transition and Inner Transition Elements Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements

1. Choose the most correct option.

Question i.
Which one of the following is diamagnetic
a. Cr3⊕
b. Fe3⊕
c. Cu2⊕
d. Sc3⊕
Answer:
d. Sc3⊕

Maharashtra Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements

Question ii.
Most stable oxidation state of Titanium is
a. +2
b. +3
c. +4
d. +5
Answer:
c. +4

Question iii.
Components of Nichrome alloy are
a. Ni, Cr, Fe
b. Ni, Cr, Fe, C
c. Ni, Cr
d. Cu, Fe
Answer:
(c) Ni, Cr

Question iv.
Most stable oxidation state of Ruthenium is
a. +2
b. +4
c. +8
d. +6
Answer:
(b) +4

Question v.
Stable oxidation states for chromium are
a. +2, +3
b. +3, +4
c. +4, +5
d. +3, +6
Answer:
d. +3, +6

Question vi.
Electronic configuration of Cu and Cu+1
a. 3d10, 4s0; 3d9, 4s0
b. 3d9, 4s1; 3d94s0
c. 3d10, 4s1; 3d10, 4s0
d. 3d8, 4s1; 3d10, 4s0
Answer:
c. 3d10, 4s1; 3d10, 4s°

Question vii.
Which of the following have d0s0 configuration
a. Sc3⊕
b. Ti4⊕
c. V5⊕
d. all of the above
Answer:
d. All of the above

Question viii.
Magnetic moment of a metal complex is 5.9 B.M. Number of unpaired electrons in the complex is
a. 2
b. 3
c. 4
d. 5
Answer:
d. 5

Question ix.
In which of the following series all the elements are radioactive in nature
a. Lanthanoids
b. Actinoids
c. d-block elements
d. s-block elements
Answer:
b. Actinides

Question x.
Which of the following sets of ions contain only paramagnetic ions
a. Sm3⊕, Ho3⊕, Lu3⊕
b. La3⊕, Ce3⊕, Sm3⊕
c. La3⊕, Eu3⊕, Gd3⊕
d. Ce3⊕, Eu3⊕, Yb3⊕
Answer:
d. Ce3⊕, Eu3⊕, Yb3⊕

Maharashtra Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements

Question xi.
Which actinoid, other than uranium, occur in a significant amount naturally?
a. Thorium
b. Actinium
c. Protactinium
d. Plutonium
Answer:
a. Thorium

Question xii.
The flux added during extraction of Iron from hematite are its?
a. Silica
b. Calcium carbonate
c. Sodium carbonate
d. Alumina
Answer:
b. Calcium carbonate

2. Answer the following

Question i.
What is the oxidation state of Manganese in
\(\text { (i) } \mathrm{MnO}_{4}^{2-}(\mathrm{ii}) \mathrm{MnO}_{4}^{-} \text {? }\)
Answer:
Oxidation state of Manganese in
\((i) \mathrm{MnO}_{4}^{2-} is +6
(ii) \mathrm{MnO}_{4}^{-}is +7\)

*Question ii.
Give uses of KMnO4

Question iii.
Why salts of Sc3⊕, Ti4⊕, V5⊕ are colorless?
Answer:
(i) Sc3+ salts are colourless :

  • The electronic configuration of 21Sc [Ar| 3d1 4s2 and Sc3+ [Ar] d°.
  • Since there are no unpaired electrons in 3d subshell, d → d transition is not possible.
  • Therefore, Sc3+ ions do not absorb the radiations in the visible region. Hence salts of Sc3+ are colourless (or white).

(ii) Ti4+ salts are colourless :

  • The electronic configuration of 22Ti [Ar] 3d24s2 and Ti4+ : [Ar] d°
  • Since there are no unpaired electrons in 3d subshell, d-*d transition is not possible.
  • Therefore, Ti3+ ions do not absorb the radiation in visible region. Hence salts of Ti3+ are colourless.

(iii) Vs5+ salts are eolourless :

  • The electronic configuration of 23V : [Ar] 3d34s2 and V5+ : [Ar] 3d°
  • Since there are no unpaired electrons in 3d-subshell, d – d transition is not possible.
  • Therefore, V5+ ions do not absorb the radiations in the visible region. Hence, V5+ salts are colourless, a

Question iv.
Which steps are involved in the manufacture of potassium dichromate from chromite ore?
Answer:
Steps in the manufacture of potassium dichromate from chromite ore are :

  • Concentration of chromite ore.
  • Conversion of chromite ore into sodium chromate (Na2CrO4).
  • Conversion of Na2CrO4 into sodium dichromate (Na2Cr2O7).
  • Conversion of Na2Cr2O7 into K2Cr2O7.

Question v.
Balance the following equation
(i) KMnO4 + H2C2O4 + H2SO4 → MnSO4 + K2SO4 + H2O + O2
(ii) K2Cr2O7 + KI + H2SO4 → K2SO4 + Cr2(SO4)3 + 7H2O + 3I2
Answer:
(i) 2KMnO4 + 3H2SO4 + 5H2C2O4 → K2SO4 + 2MnSO4 + 8H2O + 10CO2
(ii) Acidified potassium dichromate oxidises potassium iodide (KI) to iodine (I2). Potassium dichromate is reduced to chromic sulphate. Liberated iodine turns the solution brown K2Cr2O7 + 6KI + 7H2SO4 → 4K2SO4 + Cr2(SO4)3 + 7H2O + 3I2 [Oxidation state of iodine increases from – 1 to zero]

Maharashtra Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements

Question vi.
What are the stable oxidation states of plutonium, cerium, manganese, Europium?
Answer:
Stable oxidation states :
Plutonium + 3 to + 7
Cerium + 3, + 4
Manganese + 2, + 4, + 6, + 7
Europium +2, +3

Question vii.
Write the electronic configuration of chromium and copper.
Answer:
Chromium (24Cr) has electronic configuration,
24Cr (Expected) : Is2 2s2 2p6 3s2 3p6 3d4 4s2
(Observed) : Is2 2s2 2p6 3s2 3p6 3d5 4s1

Explanation :

  • The energy difference between 3d- and 45-orbitals is very low.
  • d-orbitals being degenerate, they acquire more stability when they are half-filled (3d5).
  • Therefore, there arises a transfer of one electron from 45-orbital to 3d-orbital in Cr giving more stable half-filled orbital. Hence, the configuration of Cr is [Ar] 3d5 4s1 and not [Ar] 3d4 4s2.

Copper (29CU) has electronic configuration,
29Cu (Expected) : Is2 2s3 2p6 3s3 3p6 3d9 4s2
(Observed) : Is2 2s2 2p6 3s2 3p6 3d10 4s1

Explanation :

  • The energy difference between 3d- and 45-orbitals is very low.
  • d-orbitals being degenerate, they acquire more stability when they are completely filled.
  • Therefore, there arises a transfer of one electron from 45-orbital to 3d-orbital in Cu giving completely filled more stable d-orbital.

Hence, the configuration of Cu is [Ar] 3d10 4s1 and not [Ar] 3d9 4s2.

Question viii.
Why nobelium is the only actinoid with +2 oxidation state?
Answer:

  • Nobelium has the electronic configuration 102NO : [Rn] 5f146d°7s2
  • No2+ : [Rn] 5f146d°
  • Since the 4f subshell is completely filled and 6d° empty, + 2 oxidation state is the stable oxidation state.
  • Other actinoids in + 2 oxidation state are not as stable due to incomplete 4f subshell.

Question ix.
Explain with the help of balanced chemical equation, why the solution of Ce(IV) is prepared in acidic medium.
Answer:
Ce4+ undergoes hydrolysis as, Ce4++ 2H2O → Ce(OH)4 + 4H+.
Due to the presence of H+ in the solution, the solution is acidic.

Question x.
What is meant by ‘shielding of electrons’ in an atom?
Answer:
The inner shell electrons in an atom screen or shield the outermost electron from the nuclear attraction. This effect is called the shielding effect.

The magnitude of the shielding effect depends upon the number of inner electrons.

Question xi.
The atomic number of an element is 90. Is this element diamagnetic or paramagnetic?
Answer:
The 90th element thorium has an electronic configuration, [Rn] 6d27s2. Since it has 2 unpaired electrons it is paramagnetic.

3. Answer the following

Question i.
Explain the trends in atomic radii of d-block elements
Answer:

  1. The atomic or ionic radii of 3-d series transition elements are smaller than those of representative elements, with the same oxidation states.
  2. For the same oxidation state, there is an increase in nuclear charge and a gradual decrease in ionic radii of 3d-series elements is observed. Thus ionic radii of ions with oxidation state + 2 decreases with increase in atomic number.
  3. There is slight increase is observed in Zn2+ ions. With the higher oxidation states, effective nuclear charge increases. Therefore ionic radii decrease with increase in oxidation state of the same element. For example, Fe2+ ion has ionic radius 77 pm whereas Fe3+ has 65 pm.

Maharashtra Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements

Question ii.
Name different zones in the Blast furnace. Write the reactions taking place in them.
Answer:
(i) Zone of combustion : The hot air oxidises coke to CO which is an exothermic reaction, due to which the temperature of furnace rises.
C + 1/2 O2 → CO ΔH= – 220kJ
Some part of CO dissociates to give finely divided carbon and O2.
2CO → 2C + O2
The hot gases with CO rise up in the furnace and heats the charge coming down. CO acts as a fuel as well as a reducing agent.

(ii) Zone of reduction : At about 900 °C, CO reduces Fe2O3 to spongy (or porous) iron.
Fe2O3 + 3CO → 2Fe + 3CO2
Carbon also reduces partially Fe203 to Fe.
Fe2O3 + 3C → 2Fe + 3CO

(iii) Zone of slag formation : At 1200 K limestone, CaCO3 in the charge, decomposes and forms a basic flux CaO which further reacts at 1500 K with gangue (SiO2, Al2O3) and forms a slag of CaSiO3 and Ca3AlO3.
CaCO3 + CaO + CO2.
CaO + SiO2 → CaSiO3
12CaO + 2Al2O3 → 4Ca3AlO3 + 3O2

The slag is removed from the bottom of the furnace through an outlet.

(iv) Zone of fusion : The impurities in ore like MnO2 and Ca3(PO4)2 are reduced to Mn and P while SiO2 is reduced in Si. The spongy iron moving down in the furnace melts in the fusion zone and dissolves the impurities like C, Si, Mn, phosphorus and sulphure. The molten iron collects at the bottom of furnace. The lighter slag floats on the molten iron and prevents its oxidation.

The molten iron is removed and cooled in moulds. It is called pig iron or cast iron. It contains about 4% carbon.

Question iii.
What are the differences between cast iron, wrought iron and steel?
Answer:

Cast iron Wrought iron Steel
(1) Hard and brittle
(2) Contains 4% carbon.
(3) Used for making pipes, manu­facturing automotive parts, pots, pans, utensils
(1) Very soft
(2) Contains less than 0.2% carbon.
(3) Used for making pipes, bars for stay bolts, engine bolts and rivets.
(1) Neither too hard nor too soft.
(2) Contains 0.2 to 2% carbon
(3) Used in buildings infrastruc­ture, tools, ships, automobiles, weapons etc.

Question iv.
Iron exhibits +2 and +3 oxidation states. Write their electronic configuration. Which will be more stable? Why?
Answer:
The electronic configuration of Fe2 + and Fe3+ :
Fe2+ : Is2 2s2 2p6 3s2 3p6 3d6
Fe3+ : Is2 2s2 2p6 3s2 3p6 3d5

Due to loss of two electrons from the 4.v-orbital and one electron from the 3d-orbital, iron attains 3+ oxidation state. Since in Fe3+, the 3d-orbital is half-filled, it gets extra stability, hence Fe3+ is more stable than Fe2+.

Question v.
Give the similarities and differences in elements of 3d, 4d and 5d series.
Answer:
Similarity :

  • They are placed between .s-block and p-block of the periodic table.
  • All elements are metals showing metallic characters.
  • Some are paramagnetic.
  • Most of them give coloured compounds.
  • They have catalytic properties.
  • They form complexes.
  • They have variable oxidation states.

Differences :

  • In 4d and 5d series lanthanide and actinoid contraction is observed. In 3d series atomic size changes are less marked.
  • 4d and 5d elements have high coordination numbers compared to 3d elements.
  • 4d and 5d series have similar properties whereas 3d series have different properties.

Question vi.
Explain trends in ionisation enthalpies of d-block elements.
Answer:

  1. The ionisation enthalpies of transition elements are quite high and lie between those of 5-block and p-block elements. This is because the nuclear charge and atomic radii of transition elements lie between those of 5-block and p-block elements.
    Maharashtra Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements 12
  2. As atomic number of transition elements increases along the period and along the group, first ionisation enthalpy increases even though the increase is not regular.
  3. If IE1; IE2 and IE3 are the first, second and third ionisation enthalpies of the transition elements, then IE1 < IE2 < IE3.
  4. In the transition elements, the added last differentiating electron enters into (n – 1) d-orbital and shields the valence electrons from the nuclear attraction. This gives rise to the screening effect of (n – 1) d-electrons.
  5. Due to this screening effect of (n – 1) d electrons, the ionisation enthalpy increases slowly and the increase is not very regular.

Maharashtra Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements

Question vii.
What is meant by diamagnetic and paramagnetic metal? Give one example of diamagnetic and paramagnetic transition metal and lanthanoid metal.
Answer:

  1. Paramagnetic substances : When a magnetic field is applied, substances which are attracted towards the applied magnetic field are called paramagnetic substances. Example : Ni2+, Pr4+
  2. Diamagnetic substances : When a magnetic field is applied, substances which are repelled by the magnetic fields are called diamagnetic substances. Example : Zn2+, La3+
  3. Ferromagnetic substances : When a magnetic field is applied, substances which are attracted very strongly are called ferromagnetic substances. These substances can be magnetised. For example, Fe, Co, Ni are ferromagnetic.

Question viii.
Why the ground-state electronic configurations of gadolinium and lawrencium are different than expected?

Question ix.
Write steps involved in the metallurgical process
Answer:
The various steps and principles involved in the extraction of pure metals from their ores are as follows.:

  • Concentration of ores in which impurities (gangue) are removed.
  • Conversion of ores into oxides or other reducible compounds of metals.
  • Reduction of ores to obtain crude metals.
  • Refining of metals giving pure metals.

Question x.
Cerium and Terbium behaves as good oxidising agents in +4 oxidation state. Explain.
Answer:

  • The most stable oxidation state of lanthanoids is +3.
  • Hence, Ce4+ (cerium) and Tb4+ (terbium) tend to get + 3 oxidation state which is more stable.
  • Since they get reduced by accepting electron, they are good oxidising agents in + 4 oxidation state.

Question xi.
Europium and Ytterbium behave as good reducing agents in +2 oxidation state explain.
Answer:

  • The most stable oxidation state of lanthanoids is + 3.
  • Hence, Eu2+ and Yb2+ tend to get + 3 oxidation states by losing one electron.
  • Since they get oxidised, they are good reducing agents in + 2 oxidation state.

Activity :
Make groups and each group prepare a PowerPoint presentation on the properties and applications of one element. You can use your imagination to create some innovative ways of presenting data.

You can use pictures, images, flow charts, etc. to make the presentation easier to understand. Don’t forget to cite the reference(s) from where data for the presentation is collected (including figures and charts). Have fun!

12th Chemistry Digest Chapter 8 Transition and Inner Transition Elements Intext Questions and Answers

Do you know? (Textbook Page No 165)

Question 1.
In which block of the modern periodic table are the transition and inner transition elements placed?
Answer:
The transition elements are placed in d-block and inner transition elements are placed in f-block of the modern periodic table.

Use your brain power! (Textbook Page No 167)

Question 1.
Fill in the blanks with correct outer electronic configurations.
Answer:
Answers are given in bold.
Maharashtra Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements 6

Maharashtra Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements

Try this….. (Textbook Page No 168)

Question 1.
Write the electronic configuration of Cr and Cu.
Answer:
24Cr : [Ar] 3d54s1 30Cu : [Ar] 3d104s1

Can you tell? (Textbook Page No 168)

Question 1.
Which of the first transition series element shows the maximum number of oxidation states and why?
Answer:

  • 25Mn shows the maximum number of oxidation states, + 2 to +7.
  • 25Mn : [Ar] 3d54s3
  • Mn has incompletely filled J-subshell.
  • Due to small difference in energy between 3d and 4s -orbitals, Mn can lose (or share) electrons from both the orbitals.
  • Hence Mn shows oxidation states from + 2 to +7.

Question 2.
Which elements in the 4d and 5d-series will show maximum number of oxidation states?
Answer:
In 4d-series maximum number of oxidation states are for Ruthenium Ru ( + 2, +3, + 4„ +6, +7, + 8). In 5d-series, maximum number of oxidation states are for Osmium, Os ( + 2 to + 8).

Try this ….. (Textbook Page No 168)

Question 1.
Write the electronic configuration of Mn6+, Mn4+, Fe4+, Co5+, Ni2+.
Answer:

Ions Electronic configuration of valence shell
Mn6+ [Ar] 3d1
Mn4+ [Ar] 3d3
Fe4+ [Ar] 3d4
Co5+ [Ar] 3d4
Ni2+ [Ar] 3d8

Try this ….. (Textbook Page No 171)

Question 1.
Pick up the paramagnetic species from the following : Cu1+, Fe3+, Ni2+, Zn2+, Cd2+, Pd2+.
Answer:
The following ions are paramagnetic : Fe3+, Ni2+, Pd2+

Try this ….. (Textbook Page No 171)

Question 1.
What will be the magnetic moment of transition metal having 3 unpaired electrons?
(a) equal to 1.73 B.M.?
(b) less than 1.73 BM.
(c) more than 1.73 B.M.?
Answer:
By spin-only formula, \(\mu=\sqrt{n(n+2)}\) where n is number of unpaired electrons.
\(\mu=\sqrt{3(3+2)}=\sqrt{3(5)}=3.87 \mathrm{~B} . \mathrm{M}\)
Thus the value is more than 1.73 B.M.

Use your brain power! (Textbook Page No 171)

Question 1.
A metal ion from the first transition series has two unpaired electrons. Calculate the magnetic moment.
Answer:
\(\)\begin{aligned}
\mu &=\sqrt{n(n+2)} \\
&=\sqrt{2(2+2)} \\
&=\sqrt{8} \\
&=2.84 \text { B.M. }
\end{aligned}\(\)

Maharashtra Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements

Problem (Textbook Page No 172)

Question 1.
Calculate the spin-only magnetic moment of divalent cation of a transition metal with atomic number 25.
Answer:
For element with atomic number 25. electronic configuration of its divalent cation will be : [Ar] 3d5.
Maharashtra Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements 16

Try this….. (Textbook Page No 172)

Question 1.
Calculate the spin-only magnetic moment of a divalent cation of element Slaving atomic number 27.
Answer:
Electronic configuration of divalent ion of an element with atomic number 27 : [Ar] 3d7;
Maharashtra Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements 17

Can you tell? (Textbook Page No 172)

Question 1.
Compounds of s and p-block elements are almost white. What could be the absorbed radiation? (uv or visible)?
Answer:
The white colour of a compound indicates the absorption of uv radiation.

Can you tell? (Textbook Page No 181)

Question 1.
Why f-block elements are called inner transition metals?
Answer:
f-block elements are called inner transition elements since f-orbital lies much inside the f-orbital in relation to the transition metals, These elements have 1 to 14 electrons in their f-orbital.

Question 2.
Are there an similarities between transition and inner transition metals?
Answer:
There are some properties similarity between transition and inner transition metals.

  • They are placed between s and p-block elements.
  • They are metals with filling of inner suhshells in their electronic configuration.
  • They show variable oxidation slates.
  •  They show magnetism.
  • They form coloured compounds.
  • They have catalytic property.

Problem (Textbook Page No 184)

Question 1.
Which of the following will have highest fourth ionisation enthalpy, La4+, Gd4+, Lu4+.
Answer:
La : 4f°5d16s2
Gd : 4f15d16s2
Lu : 4f145d16s2
Lu will have the highest fourth ionisation enthalpy since Lu3+ has the most stable configuration of 4f14.

Use your brain power! (Textbook Page No 185)

Question 1.
Do you think that lanthanoid complex would show magnetism?
Answer:
Lanthanoid complexes may show magnetism.

Question 2.
Can you calculate the spin only magnetic moment of lanthanoid complexes using the same formula that you used for transition metal complexes?
Answer:
You cannot calculate magnetic moment of lanthanoid complexes using spin only formula as you have to consider orbital momentum also.

Maharashtra Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements

Question 3.
Calculate the spin only magnetic moment of La3+. Compare the value with that given in the table.
Answer:
La3+ ion has no unpaired electron.
Maharashtra Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements 56
La3+ ion has zero value of magnetic moment same as given in the table.

Maharashtra Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements 51

Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 7 Elements of Groups 16, 17 and 18 Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

1. Select appropriate answers for the following.

Question i.
Which of the following has the highest electron gain enthalpy?
A. Fluorine
B. Chlorine
C. Bromine
D. Iodine
Answer:
B. Chlorine

Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

Question ii.
Hydrides of group 16 are weakly acidic. The correct order of acidity is
A. H2O > H2S > H2Se > H2Te
B. H2Te > H2O > H2S > H2Se
C. H2Te > H2Se > H2S > H2O
D. H2Te > H2Se > H2O > H2S
Answer:
C. H2Te > H2Se > H2S > H2O

Question iii.
Which of the following element does not show oxidation state of +4 ?
A. O
B. S
C. Se
D. Te
Answer:
A. O

Question iv.
HI acid when heated with conc. H2SO4 forms
A. HIO3
B. KIO3
C. I2
D. KI
Answer:
C. I2

Question v.
Ozone layer is depleted by
A. NO
B. NO2
C. NO3
D. N2O5
Answer:
A. NO

Question vi.
Which of the following occurs in liquid state at room temperature?
A. HIO3
B. HBr
C. HCl
D. HF
Answer:
D. HF

Question vii.
In pyrosulfurous acid oxidation state of sulfur is
A. Only +2
B. Only +4
C. +2 and +6
D. Only +6
Answer:
B. Only + 4

Question viii.
Stability of interhalogen compounds follows the order
A. BrF > IBr > ICl > ClF > BrCl
B. IBr > BeF > ICl > ClF > BrCl
C. ClF > ICl > IBr > BrCl > BrF
D. ICl > ClF > BrCl > IBr > BrF
Answer:
C. ClF > ICl > IBr > BrCl > BrF

Question ix.
BrCl reacts with water to form
A. HBr
B. Br2 + Cl2
C. HOBr
D. HOBr + HCl
Answer:
D. HOBr + HCl

Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

Question x.
Chlorine reacts with excess of fluorine to form.
A. ClF
B. ClF3
C. ClF2
D. Cl2F3
Answer:
B. ClF3

Question xi.
In interhalogen compounds, which of the following halogens is never the central atom.
A. I
B. Cl
C. Br
D. F
Answer:
D. F

Question xii.
Which of the following has one lone pair of electrons?
A. IF3
B. ICl
C. IF5
D. ClF3
Answer:
C. IF5

Question xiii.
In which of the following pairs, molecules are paired with their correct shapes?
A. [I3] : bent
B. BrF5 : trigonal bipyramid
C. ClF3 : trigonal planar
D. [BrF4] : square planar
Answer:
A. [I3] : bent

Question xiv.
Among the known interhalogen compounds, the maximum number of atoms is
A. 3
B. 6
C. 7
D. 8
Answer:
D. 8

2. Answer the following.

Question i.
Write the order of the thermal stability of the hydrides of group 16 elements.
Answer:
The thermal stability of the hydrides of group 16 elements decreases in the order of H2O > H2S > H2Se > H2Te.

Question ii.
What is the oxidation state of Te in TeO2?
Answer:
The oxidation state of Te in TeO2 is + 4.

Question iii.
Name two gases which deplete ozone layer.
Answer:
Nitrogen oxide (NO) released from exhaust systems of car or supersonic jet aeroplanes and chlorofluorocarbons (Freons) used in aerosol sprays and refrigerators deplete ozone layer.

Question iv.
Give two uses of ClO2
Answer:
(i) ClO2 is used as a bleaching agent for paper pulp and textiles.
(ii) It is also used in water treatment.

Question v.
What is the action of bromine on magnesium metal?
Answer:
Bromine reacts instantly with magnesium metal to give magnesium bromide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 27

Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

Question vi.
Write the names of allotropic forms of selenium.
Answer:
Selenium has two allotropic forms as follows :
(i) Red (non-metallic) form
(ii) Grey (metallic) form

Question vii.
What is the oxidation state of S in H2SO4.
Answer:
The oxidation state of S in H2SO4 is + 6.
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 29

Question viii.
The pKa values of HCl is -7.0 and that of HI is -10.0. Which is the stronger acid?
Answer:
For HCl, pKa = -7.0, hence its dissoClation constant is, Ka = 1 x 10-7.
For HI pKa = – 10.0, hence its dissoClation constant is Ka = 1 x 10-7. Hence HCl dissoClates more than HI.
Therefore HCl is a stronger acid than HI.

Question ix.
Give one example showing reducing property of ozone.
Answer:
Ozone decomposes to liberate nascent oxygen, hence it is a powerful oxidising agent. O3(g) → O2(g) + O

For example :
(i) It oxidises lead sulphide (PbS) to lead sulphate (PbSO4).
pbS(s) + 4O3(g) → PbSO(s) + 4O2(g)
(ii) Potassium iodide, KI is oxidised to iodine, I2 in the solution.
2KI(aq) + H2O(1) + O3(g) → 2KOH(aq) + I2(s) + O2(g)

Question x.
Write the reaction of conc. H2SO4 with sugar.
Answer:
Concentrated sulphuric acid when added to sugar, it is dehydrated giving carbon.
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 70
The carbon that is left behind is called sugar charcoal and the process is called char.

Question xi.
Give two uses of chlorine.
Answer:
Chlorine is used for :

  • for sterilization of drinking water.
  • bleaching wood pulp required for the manufacture of paper and rayon, cotton and textiles are also bleached using chlorine.
  • in the manufacture of organic compounds like CHCl3, CCl4, DDT, dyes and drugs.
  • in the extraction of metals like gold and platinum.
  • in the manufacture of refrigerant like Freon (i.e., CCl2F2).
  • in the manufacture of several poisonous gases like mustard gas (Cl-C2H4-S-C2H4-Cl), phosgene (COCl2) used in warfare.
  • in the manufacture of tear gas (CCl3NO2).

Question xii.
Complete the following.
1. ICl3 + H2O …….. + …….. + ICl
2. I2 + KClO3 ……. + KIO2
3. BrCl + H2O ……. + HCl
4. Cl2 + ClF3 ……..
5. H2C = CH2 + ICl …….
6. XeF4 + SiO2 ……. + SiF4
7. XeF6 + 6H2O …….. + HF
8. XeOF4 + H2O ……. + HF
Answer:
1. 2ICI3 + 3H2O → 5HCl + HlO3 + ICl
2. I2 + KCIO3 → ICl + KIO3
3. BrCl + H2O → HOBr + HCl
4. Cl2 + C1F3 → 3ClF
5. CH2 = CH2 + ICl → CH2I – CH2Cl
6. 2XeF6 + SiO2 → 2XeOF4 + SiF4
7. XeF6 + 3H2O → XeO3 + 6HF
8. XeOF4 + H2O→  XeO2F2 + 2HF

Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

Question xiii.
Match the following
A – B
XeOF2 – Xenon trioxydifluoride
XeO2F2 – Xenon monooxydifluoride
XeO3F2 – Xenon dioxytetrafluoride
XeO2F4 – Xenon dioxydifluoride
Answer:
XeOF2 – Xenon monooxydifluoride
XeO2F2 – Xenon dioxydifluoride
XeO3F2 – Xenon trioxydifluoride
XeO2F4 – Xenon dioxytetrafluoride

Question xiv.
What is the oxidation state of xenon in the following compounds?
XeOF4, XeO3, XeF5, XeF4, XeF2.
Answer:

Compound Oxidation state of Xe
XeOF4 + 6
XeO3 + 6
XeF6 + 6
XeF4 + 4
XeF2 + 2

3. Answer the following.

Question i.
The first ionisation enthalpies of S, Cl and Ar are 1000, 1256 and 1520 kJ/mol-1, respectively. Explain the observed trend.
Answer:
(i) The atomic number increases as, 16S < 17Cl < 18Ar1.
(ii) Due to decrease in atomic size and increase in effective nuclear charge, Cl binds valence electrons strongly.
(iii) Hence ionisation enthalpy of Cl (1256 kJ mol-1) is higher than that of S(1000 kJ mol-1)
(iv) Ar has electronic configuration 3s23p6. Since all electrons are paired and the octet is complete, it has the highest ionisation enthalpy, (1520 kJ mol-1)

Question ii.
“Acidic character of hydrides of group 16 elements increases from H2O to H2Te” Explain.
Answer:
(i) The thermal stability of the hydrides of group 16 elements decreases from H2O to H2Te. This is because the bond dissociation enthalpy of the H-E bond decreases down the group.
(ii) Thus, the acidic character increases from H2O to H2Te.

Question iii.
How is dioxygen prepared in laboratory from KClO3?
Answer:
By heating chlorates, nitrates and permanganates.
Potassium chlorate in the presence of manganese dioxide on heating decomposes to form potassium chloride and oxygen.
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 39

Question iv.
What happens when
a. Lead sulfide reacts with ozone (O3).
b. Nitric oxide reacts with ozone.
Answer:
(i) It oxidises lead sulphide (PbS) to lead sulphate (PbSO4) changing the oxidation state of S from – 2 to +6.
PbS(s) + 4O3(g) → PbSO(s) + 4O2(g)

(ii) Ozone oxidises nitrogen oxide to nitrogen dioxide.
NO(g) + O3(g) → NO2(g) + O2(g)

Question v.
Give two chemical reactions to explain oxidizing property of concentrated H2SO4.
Answer:
Hot and concentrated H2SO4 acts as an oxidising agent, since it gives nascent oxygen on heating.
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 68

Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

Question vi.
Discuss the structure of sulfur dioxide.
Answer:
(i) SO2 molecule has a bent V shaped structure with S-O-S bond angle 119.5° and bond dissoClation enthalpy is 297 kJ mol-1.
(ii) Sulphur in SO2 is sp2 hybridised forming three hybrid orbitals. Due to lone pair electrons, bond angle is reduced from 120° to 119.5°.
(iii) In SO2, each oxygen atom is bonded to sulphur by σ and a π bond.
(iv) a bond between S and O are formed by sp2-p overlapping.
(v) One of π bonds is formed by pπ – pπ overlapping while other n bond is formed by pπ – dπ overlap.
(vi) Due to resonance both the bonds are identical having observed bond length 143 pm due to resonance,
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 63

Question vii.
Fluorine shows only -1 oxidation state while other halogens show -1, +1, +3, +5 and +7 oxidation states. Explain.
Answer:

  • Halogens have outer electronic configuration ns2 np5.
  • Halogens have tendency to gain or share one electron to attain the stable configuration of nearest inert element with configuration ns2np6.
  • Hence they are monovalent and show oxidation state – 1.
  • Since fluorine does not have vacant d-orbital, it shows only one oxidation state of – 1 while all other halogens show variable oxidation states from – 1 to +7.
  • These oxidation states are, – 1, +1, + 3, +5 and + 7. Cl and Br also show oxidation states + 4 and + 6 in their oxides and oxyaClds.

Question viii.
What is the action of chlorine on the following
a. Fe
b. Excess of NH3
Answer:
(a) Chlorine reacts with Fe to give ferric chloride.
2Fe + 3Cl2 → 2FeCl3

(b) Chlorine reacts with the excess of ammonia to form ammonium chloride, NH4Cl and nitrogen.
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 111

Question ix.
How is hydrogen chloride prepared from sodium chloride?
Answer:

  1. In the laboratory, hydrogen chloride, HCl is prepared by heating a mixture of NaCl and concentrated H2SO4.
    Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 88
  2. Hydrogen chloride gas, is dried by passing it through a dehydrating agent like concentrated H2SO4 and then collected by upward displacement of air.

Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

Question x.
Draw structures of XeF6, XeO3, XeOF4, XeF2.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 105
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 106

Question xi.
What are interhalogen compounds? Give two examples.
Answer:
Interhalogen compounds : Compounds formed by the combination of atoms of two different halogens are called interhalogen compounds. In an interhalogen compound, of the two halogen atoms, one atom is more electropositive than the other. The interhalogen compound is regarded as the halide of the more electropositive halogen.
For example ClF, BrF3, ICl

Question xii.
What is the action of hydrochloric acid on the following?
a. NH3
b. Na2CO3
Answer:
a. Hydrochloric acid reacts with ammonia to give white fumes of ammonium chloride.
NH3 + HCl → NH4Cl

b. Hydrochloric acid reacts with sodium carbonate to give sodium chloride, water with the liberation of carbon dioxide gas.
Na2CO3 + 2HCl → 2NaCl + H2O + CO2

Question xiii.
Give two uses of HCl.
Answer:
Hydrogen chloride (OR hydrochloric acid) is used :

  • in the manufacture of chlorine and ammonium chloride,
  • to manufacture glucose from com, starch
  • to manufacture dye
  • in mediClne and galvanising
  • as an important reagent in the laboratory
  • to extract glue from bones and for the purification of bone black.
  • for dissolving metals, Fe + 2HCl(aq) → FeCl2 + H2(g)

Question xiv.
Write the names and structural formulae of oxoacids of chlorine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 37
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 38

Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

Question xv.
What happens when
a. Cl2 reacts with F2 in equal volume at 437 K.
b. Br2 reacts with excess of F2.
Answer:
(a) Cl2 reacts with F2 in equal volumes at 437 K to give chlorine monofluoride ClF.
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 97

(b) Br2 reacts with excess of F2 to give bromine trifluoride BF3.
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 98

Question xvi.
How are xenon fluorides XeF2, XeF4 and XeF6 obtained ? Give suitable reactions.
Answer:
Xenon fluorides are generally prepared by the direct reaction of xenon and fluorine in different ratios and under appropriate experimental conditions, such as temperature, in the presence of an electric discharge and by a photochemical reaction.
(i) Preparation of XeF2 :
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 102
(ii) Preparation of XeF4 :
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 103
(iii) Preparation of XeF6 :
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 104

Question xvii.
How are XeO3 and XeOF4 prepared?
Answer:
Preparation of XeO3 : Xenon trioxide (XeO3) is prepared by the hydrolysis of XeF4 or XeF6.

  • By hydrolysis of XeF4 :
    3XeF4 + 6H20 → 2Xe + XeO3 + 12 HF + \(1 \frac{1}{2} \mathrm{O}_{2}\)
  • By hydrolysis of XeF6 :
    XeF6 + 3H2O → XeO3 + 6HF
  • Preparation of XeOF4 :
    Xenon oxytetrafluoride (XeOF4) is prepared by the partial hydrolysis of XeF6.
    XeF6 + H2O → XeOF4 + 2HF

Question xviii.
Give two uses of neon and argon.
Answer:
Uses of neon (Ne) :

  • Neon is used in the production of neon discharge lamps and signs by filling Ne in glass discharge tubes.
  • Neon signs are visible from a long distance and also have high penetrating power in mist or fog.
  • A mixture of neon and helium is used in voltage stabilizers and current rectifiers.
  • Neon is also used in the production of lasers and fluorescent tubes.

Uses of argon (Ar) :

  • Argon is used to fill fluorescent tubes and radio valves.
  • It is used to provide inert atmosphere for welding and production of steel.
  • It is used along with neon in neon sign lamps to obtain different colours.
  • A mixture of 85% Ar and 15% N2 is used in electric bulbs to enhance the life of the filament.

Question xix.
Describe the structure of Ozone. Give two uses of ozone.
Answer:
(A)

  • Ozone has molecular formula O3.
  • The lewis dot and dash structures for O3 are :
    Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 55
  • Infrared and electron diffraction spectra show that O3 molecule is angular with 0-0-0 bond angle 117°.
  • Both 0-0 bonds are identical having bond length 128 pm which is intermediate between single and double bonds.
  • This is explained by considering resonating structures and resonance hybrid.
    Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 56

(B) Uses of Ozone :

  • Ozone sterilises drinking water by oxidising germs and bacteria present in it.
  • It is used as a bleaching agent for ivory, oils, starch, wax and delicate fabrics like silk.
  • Ozone is used to purify the air in crowded places like Clnema halls, railways, tunnels, etc.
  • In industry, ozone is used in the manufacture of synthetic camphor, potassium permanganate, etc.

Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

Question xx.
Explain the trend in following atomic properties of group 16 elements.
i. Atomic radii
ii. Ionisation enthalpy
iii. Electronegativity.
Answer:
(1) Atomic and ionic radii :

  1. As compared to group 15 elements, the atomic and ionic radii of group 16 elements are smaller due to higher nuclear charge.
  2. The atomic and ionic radii increase down the group from oxygen to polonium. This is due to the addition of a new shell at each successive elements on moving down the group. The atomic radii increases in the order O < S < Se < Te < Po

(2) Ionisation enthalpy :

  • The ionisation enthalpy of group 16 elements has quite high values.
  • Ionisation enthalpy decreases down the group from oxygen to polonium. This is due to the increase in atomic volume down the group.
  • The first ionisation enthalpy of the lighter elements of group 16 (O, S, Se) have lower values than those of group 15 elements in the corresponding periods. This is due to difference in their electronic configurations.

Group 15 : (valence shell) ns2 npx1 npy1 npz1
Group 16 : (valence shell) ns2 npx2 npy1 npz1

Group 15 elements have extra stability of half-filled and more symmetrical orbitals, while group 16 elements acquire extra stability by losing one of paired electrons from npx- orbital forming half-filled p-orbitals.

Hence group 16 elements have lower first ionisation enthalpy than group 15 elements.

(3) Electronegativity :

  • The electronegativity values of group 16 elements have higher values than corresponding group 15 elements in the same periods.
  • Oxygen is the second most electronegative elements after fluorine. (O = 3.5, F = 4)
  • On moving down the group electronegativity decreases from oxygen to polonium.
  • On moving down the group atomic size increases, hence nuclear attraction decreases, therefore electro-negativity decreases.
Elements O S Se Te Po
Electronegativity 3.5 2.44 2.48 2.01 1.76

4. Answer the following.

Question i.
Distinguish between rhombic sulfur and monoclinic sulfur.
Answer:

Rhombic sulphur Monoclinic sulphur
1. It is pale yellow. 1. It is bright yellow.
2. Orthorhombic crystals 2. Needle-shaped monoclinic crystals
3. Melting point, 385.8 K 3. Melting point, 393 K
4. Density, 2.069 g/cm3 4. Density: 1.989 g/cm3
5. Insoluble in water, but soluble in CS2 5. Soluble in CS2
6. It is stable below 369 K and transforms to α-sulphur above this temperature. 6. It is stable above 369 K and transforms into β-sulphur below this temperature.
7. It exists as S8 molecules with a structure of a puckered ring. 7. It exists as S8 molecules with a structure of a puckered ring.
8. It is obtained by the evaporation of roll sulphur in CS2 8. It is prepared by melting rhombic sulphur and cooling it till a crust is formed. Two holes are pierced in the crust and the remaining liquid is poured to obtain needle-shaped crystals of monoclinic sulphur (β-sulphur).

Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

Question ii.
Give two reactions showing oxidizing property of concentrated H2SO4.
Answer:
Hot and concentrated H2SO4 acts as an oxidising agent, since it gives nascent oxygen on heating.
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 68

Question iii.
How is SO2 prepared in the laboratory from sodium sulfite? Give two physical properties of SO2.
Answer:
(A) Laboratory method (From sulphite) :

  • Sodium sulphite on treating with dilute H2SO4 forms SO2.
    Na2SO3 + H2SO4(aq) → Na2SO4 + H2O(1) + SO2(g)
  • Sodium sulphite, Na2SO3 on reaction with dilute hydrochloric acid solution forms SO2.
    Na2SO3(aq) + 2HCl(aq) → 2NaCl9aq0 + H2O(1) + SO2(g)

(B) Physical properties of SO2

  • It is a colourless gas with a pungent smell.
  • It is highly soluble in water and forms sulphurous acid, H2SO3 SO2(g) + H2O(1) → H2SO3(aq)
  • It is poisonous in nature.
  • At room temperature, it liquefies at 2 atmospheres. It has boiling point 263K.

Question iv.
Describe the manufacturing of H2SO4 by contact process.
Answer:
Contact process of the manufacture of sulphuric acid involves following steps :

(1) Preparation of SO2 : Sulphur or pyrite ores like iron pyrites, FeS2 on burning in excess of air, form SO2.
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 64
(2) Oxidation of SO2 to SO3 : SO2 is oxidised to SO3 in the presence of a heterogeneous catalyst V2O5 and atmospheric oxygen. This oxidation reaction is reversible.
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 65
To avoid the poisoning of a costly catalyst, it is necessary to make SO2 free from the impurities like dust, moisture, As2O3 poison, etc.

The forward reaction is exothermic and favoured by increase in pressure. The reaction is carried out at high pressure (2 bar) and 720 K temperature. The reacting gases, SO2 and O2 are taken in the ratio 2:3.

(3) Dissolution of SO3 : SO3 obtained from catalytic converter is absorbed in 98%. H2SO4 to obtain H2S2O7, oleum or fuming sulphuric acid.
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 66
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 67
Flow diagram for the manufacture of sulphuric acid

Question 7.1 (Textbook Page No 141)

12th Chemistry Digest Chapter 7 Elements of Groups 16, 17 and 18 Intext Questions and Answers

Question 1.
Elements of group 16 generally show lower values of first ionisation enthalpy compared to the elements of corresponding period of group 15. Why?
Answer:
Group 15 elements have extra stable, half filled p-orbitals with electronic configuration (ns2np3). Therefore more amount of energy is required to remove an electron compared to that of the partially filled orbitals (ns2np4) of group 16 elements of the corresponding period.

Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

Question 7.2 (Textbook Page No 141)

Question 1.
The values of first ionisation enthalpy of S and Cl are 1000 and 1256 kJ mol-1, respectively. Explain the observed trend.
Answer :
The elements S and Cl belong to second period of the periodic table.
Across a period effective nuclear charge increases and atomic size decreases with increase in atomic number. Therefore the energy required for the removal of electron from the valence shell (I.E.) increases in the order S < Cl.

Question 7.4 (Textbook Page No 141)

Question 1.
Fluorine has less negative electron gain affinity than chlorine. Why?
Answer :
The size of fluorine atom is smaller than chlorine atom. As a result, there are strong inter electronic repulsions in the relatively small 2p orbitals of fluorine and therefore, the incoming electron does not experience much attraction. Thus fluorine has less negative electron gain affinity than chlorine.

Try this… (Textbook Page No 140)

Question 1.
Explain the trend in the following properties of group 17 elements.

(1) Atomic size,
(2) Ionisation enthalpy,
(3) Electronegativity,
(4) Electron gain enthalpy.
Answer:
(1) Atomic size :

  • Atomic and ionic radii increase down the group as atomic number increases due to the addition of new electronic valence shell to each succeeding element.
  • The atomic radii increase in the order F < Cl < Br < 1
  • Halogens possess the smallest atomic and ionic radii in their respective periods since the effective nuclear charge experienced by valence electrons in halogen atoms is the highest.

(2) Ionisation enthalpy :

  • The ionisation enthalpies of halogens are very high due to their small size and large nuclear attraction.
  • The ionisation ethalpies decrease down the group since the atomic size increases.
  • The ionisation enthalpy decreases in the order F > Cl > Br > I.
  • Among halogens fluorine has the highest ionisation enthalpy due to its smallest size.
Element F Cl Br I
Ionisation enthalpy kJ/mol 1680 1256 1142 1008

(3) Electronegativity :

  • Halogens have the highest values for electronegativity due to their small atomic radii and high effective nuclear charge.
  • Each halogen is the most electronegative element of its period.
  • Fluorine has the highest electronegativity as compared to any element in the periodic table.
  • The electronegativity decreases as,
    F > Cl > Br > I
    4.0 3.2 3.0 2.7 (electronegativity)

(4) Electron gain enthalpy (ΔegH) :

  • The halogens have the highest negative values for electron gain enthalpy.
  • Electron gain enthalpies of halogens are negative indicating release of energy.
  • Halogens liberate maximum heat by gain of electron as compared to other elements.
  • Since halogens have outer valence electronic configuration, ns2 np5, they have strong tendency to accept an electron to complete an octet and acquire electronic configuration of the nearest inert elements.
  • In case of fluorine due to small size of 2 p-orbitals and high electron density, F has less negative electron gain enthalpy than Cl.
    F(g) + e → F(g) ΔegH = – 333 klmol-1
    Cl(g) + e → Cl(g) ΔegH = – 349 kJ mol-1
  • The variation in electron gain enthalpy is in the order of, Cl > F > Br > I.

Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

Question 2.
Oxygen has less negative electron gain enthalpy than sulphur. Why?
Answer:

  • Oxygen has a smaller atomic size than sulphur.
  • It is more electronegative than sulphur.
  • It has a larger electron density.
  • Due to high electron density, oxygen does not accept the incoming electron easily and therefore has less electron gain enthalpy than sulphur.

Question 7.3 (Textbook Page No 141)

Question 1.
Why is there a large difference between the melting and boiling points of oxygen and sulphur?
Answer :
Oxygen exists as diatomic molecule (O2) whereas sulphur exists as polyatomic molecule (S8). The van der Waals forces of attraction between O2 molecules are relatively weak owing to their much smaller size. The large van der Waals attractive forces in the S8 molecules are due to large molecular size. Therefore oxygen has low m.p. and b.p. as compared to sulphur.

Question 7.5 (Textbook Page No 141)

Question 1.
Bond dissoClation enthalpy of F2 (158.8 kj mol-1) is lower than that of Cl2 (242.6 kj mol-1) Why?
Answer :
Fluorine has small atomic size than chlorine. The lone pairs on each F atom in F2 molecule are so close together that they strongly repel each other, and make the F – F bond weak. Thus, it requires less amount of energy to break the F – F bond. In Cl2 molecule the lone pairs on each Cl atom are at a larger distance and the repulsion is less.

Thus Cl – Cl bond is comparatively stronger. Therefore bond dissoClation enthalpy of F2 is lower than that of Cl2.
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 8

Question 7.6 (Textbook Page No 142)

Question 1.
Noble gases have very low melting and boiling points. Why?
Answer :
Noble gases are monoatomic, the only type of interatomic interactions which exist between them are weak van der Waals forces. Therefore, they can be liquefied at very low temperatures and have very low melting or boiling points.

Can you tell? (Textbook Page No 142)

Question 1.
The first member of the a group usually differs in properties from the rest of the members of the group. Why?
Answer:
The first member of a group usually differs in properties from the rest of the members of the group for the following reasons :

  • Its small size
  • High electronegativity
  • Absence of vacant d-orbitals in its valence shell.

Use your brain power! (Textbook Page No 142)

Question 1.
Oxygen forms only OF2 with fluorine while sulphur forms SF6. Explain. Why?
Answer:

  • Oxygen combines with the most electronegative element fluorine to form OF2 and exhibits positive oxidation state (+ 2). Since, oxygen does not have vacant J-orbitals it cannot exhibit higher oxidation states.
  • Sulphur has vacant d-orbitals and hence can exhibit + 6 oxidation state to form SF6.

Question 2.
Which of the following possesses hydrogen bonding? H2S, H2O, H2Se, H2Te
Answer:

  • Oxygen being more electronegative, is capable of forming hydrogen bonding in the compound H2O.
  • The other elements S, Se and Te of Group 16, being less electronegative do not form hydrogen bonds.
  • Thus, hydrogen bonding is not present in the other hydrides H2S, H2Se and H2Te.

Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

Question 3.
Show hydrogen bonding in the above molecule with the help of a diagram.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 9

Try this….. (Textbook Page No 143)

Question 1.
Complete the following tables :
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 108
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 109

Can you tell? (Textbook Page No 146)

Question 1.
What is allotropy?
Answer:
The property of some elements to exist in two or more different forms in the same physical state is called allotropy.

Question 2.
What is the difference between allotropy and polymorphism?
Answer:

  • Allotropy is the existence of an element in more than one physical form. It means that under different conditions of temperature and pressure an element can exist in more than one physical forms.
  • Coal, graphite and diamond etc., are different allotropic forms of carbon.
  • Polymorphism is the existence of a substance in more than one crystalline form.
  • It means that under different conditions of temperature and pressure, a substance can form more than one type of crystal. For example, mercuric iodide exists in the orthorhombic and trigonal form.

Question 7.7 (Textbook Page No 146)

Which form of sulphur shows paramagnetic behaviour?
Answer :
In the vapour state, sulphur partly exists as S2 molecule, which has two unpaired electrons in the antibonding π* orbitals like O2. Hence it exhibits paramagnetism.

Try this….. (Textbook Page No 149)

Question 1.
Why water in a fish pot needs to be changed from time to time?
Answer:
A fish pot is an artificial ecosystem and the fish in it are selective and maintained in a restricted environment.

In a fish pot, the unwanted food and waste generated by the fish mix with the water and remain untreated due to lack of decomposers.

Accumulation of waste material will decrease the levels of dissolved oxygen in the water pot.

Hence, it is necessary to change the water from time to time.

Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

Question 7.8 (Textbook Page No 149)

Dioxygen is paramagnetic in spite of having an even number of electrons. Explain.
Answer :
Dioxygen is a covalently bonded molecule.
The paramagnetic behaviour of O2 can be explained with the help of molecular orbital theory.
Electronic configuration O2
KK σ(2s)2 σ(2s)2 σ*(2pz)2 π(2px)2 π(2px)2 π(2py)2 π*(2px)1 π*(2py)1. Presence of two unpaired electrons in antibonding orbitals explains paramagnetic nature of dioxygen.

Question 7.9 (Textbook Page No 150)

High concentration of ozone can be dangerously explosive. Explain.
Answer :
Thermal stability : Ozone is thermodynamically unstable than oxygen and decomposes into O2. The decomposition is exothermic and results in the liberation of heat (ΔH is – ve) and an increase in entropy (ΔS is positive). This results in large negative Gibbs energy change (ΔG). Therefore high concentration of ozone can be dangerously explosive. Eq O3 → O2 + O

Try this…… (Textbook Page No 151)

(a) Ozone is used as a bleaching agent. Explain.
Answer:

  • Ozone due to its oxidising property can act as a bleaching agent. O3(g) → O2(g) + O
  • It bleaches coloured matter. coloured matter + O → colourless matter
  • Ozone bleaches in the absence of moisture, so it is also known as dry bleach.
  • Ozone can bleach ivory and delicate fabrics like silk.

(b) Why does ozone act as a powerful oxidising agent?
Answer:
Ozone decomposes to liberate nascent oxygen, hence it is a powerful oxidising agent. O3(g) → O2(g) + O
For example :

  • It oxidises lead sulphide (PbS) to lead sulphate (PbSO4).
    pbS(s) + 4O3(g) → PbSO(s) + 4O2(g)
  • Potassium iodide, KI is oxidised to iodine, I2 in the solution.
    2KI(aq) + H2O(1) + O3(g) → 2KOH(aq) + I2(s) + O2(g)

Question 7.10 : (Textbook Page No 154)

What is the action of concentrated H2SO4 on (a) HBr (b) HI
Answer :
Concentrated sulphuric acid oxidises hydrobromic acid to bromine.

2HBr + H2SO4 → Br2 + SO2 + 2H2O
It oxidises hydroiodic acid to iodine.
2HI + H2SO4 → I2 + SO2 + 2H2O

Try this….. (Textbook Page No 156)

Question 1.
Give the reasons for the bleaching action of chlorine.
Answer:

  • Chlorine acts as a powerful bleaching agent due to its oxidising nature.
  • In moist conditions or in the presence of water it forms unstable hypochlorous acid, HOCl which decomposes giving nascent oxygen which oxidises the vegetable colouring matter of green leaves, flowers, litmus, indigo, etc.
    Cl2 + H2O → HCl + HOCl
    HOCl → HCl + [O]
    Vegetable coloured matter + [O] → colourless matter.

Question 2.
Name two gases used in war.
Answer:
Phosgene : COCl2
Mustard gas: Cl – CH2 – CH2 – S – CH2 – CH2 – Cl

Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

Use your brain power! (Textbook Page No 157)

Question 1.
Chlorine and fluorine combine to form interhalogen compounds. The halide ion will be of chlorine or fluorine?
Answer:
Among the- two halogens, chlorine is more electropositive than fluorine (Electronegativity values: F = 4.0, Cl = 3.2)

The interhalogen compound is regarded as the halide of the more electropositive halogen. Hence, the interhalogen compound is the fluoride of chlorine, i.e. chlorine monofluoride, CiF.

Question 2.
Why does fluorine combine with other halogens to form maximum number of fluorides?
Answer:
Since fluorine is the most electronegative element and has the smallest atomic radius compared to other halogen fluorine forms maximum number of fluorides.

Use your brain power! (Textbook Page No 158)

Question 1.
What will be the names of the following compounds: ICl, BrF?
Answer:
ICl : Iodine monochloride
BrF : Bromine monofluoride

Question 2.
Which halogen (X) will have maximum number of other halogen (X ) attached?
Answer:
The halogen Iodine (I) will have the maximum number of other halogens attached.

Question 3.
Which halogen has tendency to form more interhalogen compounds?
Answer:
The halogen fluorine (F) has the maximum tendency to form more interhalogen compounds as it has a small size and more electronegativity.

Question 4.
Which will be more reactive?
(a) ClF3 or ClF,
(b) BrF5 or BrF
Answer:
ClF3 is more reactive than ClF, while BrF5 is more reactive than BrF. Both ClF3 and BrF5 are unstable compared to ClF and BrF respectively due to steric hindrance hence are more reactive.

Question 5.
Complete the table :

Formula Name
ClF Chlorine monofluoride
ClF3 …………………………………
………………………………… Chlorine pentachloride
BrF …………………………………
………………………………… Bromine pentafluoride
ICl …………………………………
ICl3 …………………………………

Answer:

Formula Name
ClF Chlorine monofluoride
ClF3 Chlorine trifluoride
CIF5 Chlorine pentafluoride
BrF Bromine monofluoride
BrF5 Bromine pentafluoride
ICl Iodine monochloride
ICl3 Iodine trichloride

Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

Use your brain power! (Textbook Page No 159)

Question 1.
In the special reaction for ICl, identify the oxidant and the reductant? Denote oxidation states of the species.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18 99
Potassium chlorate, KClO3 is the oxidising agent or oxidant and iodine is the reducing agent or reductant.

Use your brain power! (Textbook Page No 162)

Question 1.
What are missing entries?

Formula Name
XeOF2
……………
XeO3F2
XeO2F4
Xenon monooxyfluoride
Xenon dioxydifluoride
……………………………………..
……………………………………..

Answer:

Formula Name
XeOF2
XeO2F2
XeO3F2
XeO2F4
Xenon monooxydifluoride
Xenon dioxydifluoride
Xenon trioxydifluoride
Xenon dioxytetrafluoride

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 6 Chemical Kinetics Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

1. Choose the most correct option.

Question i.
The rate law for the reaction aA + bB → P is rate = k[A] [B]. The rate of reaction doubles if
a. concentrations of A and B are both doubled.
b. [A] is doubled and [B] is kept constant
c. [B] is doubled and [A] is halved
d. [A] is kept constant and [B] is halved.
Answer:
b. [A] is doubled and [B] is kept constant

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Question ii.
The order of the reaction for which the units of rate constant are mol dm-3 s-1 is
a. 1
b. 3
c. 0
d. 2
Answer:
c. 0

Question iii.
The rate constant for the reaction 2N2O5(g) → 2N2O4(g) + O2(g) is 4.98 × 10-4 s-1. The order of reaction is
a. 2
b. 1
c. 0
d. 3
Answer:
b. 1

Question iv.
Time required for 90 % completion of a certain first order reaction is t. The time required for 99.9 % completion will be
a. t
b. 2t
c. t/2
d. 3t
Answer:
d. 3t

Question v.
Slope of the graph ln[A]t versus t for first order reaction is
a. -k
b. k
c. k/2. 303
d. -k/2. 303
Answer:
a. -k

Question vi.
What is the half life of a first order reaction if time required to decrease concentration of reactant from 0.8 M to 0.2 M is 12 h?
a. 12 h
b. 3 h
c. 1.5 h
d. 6 h
Answer:
d. 6 h

Question vii.
The reaction, 3ClO ClO3Θ + 2 ClΘ occurs in two steps,
(i) 2 ClO → ClO2Θ
(ii) ClO2Θ + ClOΘ → ClO3Θ + ClΘ

The reaction intermediate is
a. ClΘ
b. ClO2Θ
c. ClO3Θ
d. ClOΘ
Answer:
b. ClO2Θ

Question viii.
The elementary reaction O2(g) + O(g) → 2O2(g) is
a. unimolecular and second order
b. bimolecular and first order
c. bimolecular and second order
d. unimolecular and first order
Answer:
c. bimolecular and second order

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Question ix.
Rate law for the reaction, 2NO + Cl2 → 2 NOCl is rate = k[NO2]2[Cl2]. Thus k would increase with
a. increase of temperature
b. increase of concentration of NO
c. increase of concentration of Cl2
d. increase of concentrations of both Cl2 and NO
Answer:
a. increase of temperature

Question x.
For an endothermic reaction, X ⇌ Y. If E f is activation energy of the forward reaction and Er that for reverse reaction, which of the following is correct?
a. Ef = Er
b. Ef < Er
c. Ef > Er
d. ∆H = Ef – Er is negative
Answer:
(c) Ef → Er

2. Answer the following in one or two sentences.

Question i.
For the reaction,
N2(g) + 3 H2(g) → 2NH3(g), what is the relationship among \(\frac{\mathrm{d}\left[\mathrm{N}_{2}\right]}{\mathrm{dt}}\)\(\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}} \text { and } \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}} ?\)
Answer:
N2(g) + 3H2(g) → 2NH3(g)
From the above reaction, when 1 mole of N2 reacts, 3 moles of H2 are consumed and 2 moles of NH3 are formed.

If the instantaneous rate R of the reaction is represented in terms of rate of the consumption of N2 then, \(R=-\frac{d\left[\mathrm{~N}_{2}\right]}{d t}\)
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 10
Hence the rate of reaction in terms of concentration changes in N2, H2 and NH3 may be represented as,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 11

Question ii.
For the reaction,
CH3Br(aq) + OH-(aq) → CH3OHΘ (aq) +BrΘ (aq), rate law is rate = k[CH3Br][OHΘ]
a. How does reaction rate changes if [OHΘ] is decreased by a factor of 5?
b. What is change in rate if concentrations of both reactants are doubled?
Solution :
Given :
(a) Rate = R = k [CH3Br] x [OH]
If R1 and R2 are initial and final rates of reaction then,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 84
Rate will be increased 4 time.

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Question iii.
What is the relationship between coeffients of reactants in a balanced equation for an overall reaction and exponents in rate law. In what case the coeffients are the exponents?
Answer:
Explanation : Consider the following reaction, aA + bB → products

If the rate of the reaction depends on the concentrations of the reactants A and B, then, by rate law,
R α [A]a [B]b
∴ R = k [A]a [Bb
where [A] = concentration of A and
[B] = concentration of B

The proportionality constant k is called the velocity constant, rate constant or specific rate of the reaction.

a and b are the exponents or the powers of the concentrations of the reactants A and B respectively when observed experimentally.

The exponents or powers may not be necessarily a and b but may be different x and y depending on experimental observations. Then the rate R will be,
R = k [A]x [B]y
For example, if x = 1 and y = 2, then,
R = k [A] x [B]2

Question iv.
Why all collisions between reactant molecules do not lead to a chemical reaction?
Answer:
(i) Collisions of reactant molecules : The basic re-quirements of a reaction is that the reacting species (atoms, ions or molecules) must come together and collide for a reaction to occur. Therefore the rate of the reaction shall depend on the rate and frequency of collisions which increase with the amount of reacting species and temperature.

However it is observed that the rate of reaction is very low as compared to the rate of collisions between reacting species in gaseous phase or liquid phase. This suggests that all the collisions are not fruitful leading to a reaction. Hence it is necessary to consider another factor like energy of colliding species along with collision frequency.

(ii) Energy requirement (Activation energy) : The colliding molecules must possess a certain mini-mum energy called activation energy required far breaking and making bonds resulting in the reaction. This implies that the colliding molecules must have energy equal to or greater than the activation energy. The colliding molecules with less energy do not react.

(iii) Orientation of reactant molecules : The concept of activation energy is satisfactory in case of simple molecules or ions but not in case of complex or higher polyatomic molecules. It is observed that the rates of reaction are less as compared to the rates of collisions between activated molecules with activation energy.

This suggests that in addition, the colliding molecules must have proper orientations relative to each other during collisions. For example, consider the reaction, A – B-l-C → A + B – C. For the reaction to occur, C must collide with B while collisions with A will not be fruitful. Since B has to bond with C.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 118

Question v.
What is the activation energy of a reaction?
Answer:
Activation energy : The energy required to form activated complex or transition state from the reactant molecules is called activation energy.
OR
The height of energy barrier in the energy profile diagram is called activation energy.

Question vi.
What are the units for rate constants for zero order and second order reactions if time is expressed in seconds and concentration of reactants in mol/L?
Answer:
(a) For a zero order reaction, the rate constant has units, molL-1s-1.
(b) For second order reaction,
Rate = k x [Reactant]2

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 176

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Question vii.
Write Arrhenius equation and explain the terms involved in it.
Answer:
Arrhenius equation is represented as k = A x e-Ea/RT
where
k = Rate constant at absolute temperature T
Ea = Energy of activation R = Gas constant
A = Frequency factor or pre-exponential factor.

Question viii.
What is the rate determining step?
Answer:
Many chemical reactions take place in a series of elementary steps. Among many steps of the reaction, one of the steps is the slowest step compared to other steps.

Rate determining step : The slowest step in the reaction mechanism which involves many steps is called the rate-determining step.

Question ix.
Write the relationships between rate constant and half life of fist order and zeroth order reactions.
Answer:
(a) For first order reaction, half-life period t1/2 is, \(t_{1 / 2}=\frac{0.693}{k}\) where k is the rate constant.
(b) For zeroth-order reaction, half half period (t1/2) is, \(t_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 k}\) where k is the rate constant and [A]0 is initial concentration of the reactant.

Question x.
How do half lives of the fist order and zero order reactions change with initial concentration of reactants?
Answer:
(A) For the first order reaction, half life, t1/2 is given by, \(t_{1 / 2}=\frac{0.693}{k}\) where k is rate constant. Hence it is independent of initial concentration of the reactant.

(B) Zero order reaction,
\(t_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 k}\) where [A]0 is initial concentration of the reactant.

Hence, half life period increases with the increase in concentration of the reactant.

3. Answer the following in brief.

Question i.
How instantaneous rate of reaction is determined?
Answer:
(1) The instantaneous rate is expressed as an infinite¬simal change in concentration (- dc) of the reactant with the infinitesimal change in time (dt).
For a reaction, A → B, let an infinitesimal change in A be – dc in time dt, then Rate \(=\frac{d[\mathrm{~A}]}{d t}\).

Hence, it is represented as,
∴ Instantaneous rate \(=-\frac{d[\mathrm{~A}]}{d t}\)

The negative sign indicates a decrease in the concentration of A.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 8
It is obtained by drawing a tangent to the curve obtained by plotting the concentration against the time. Hence, the slope at a given point represents the instantaneous rate of the reaction.

(2) The instantaneous rate can also be expressed as an infinitesimal change (or increase) in the concentration of the product with the infinitesimal change in time (dt).

Let dB be an infinitesimal change in the concentration of product B in time dt, then Rate \(=\frac{d[\mathrm{~B}]}{d t}=\frac{d x}{d t}\).

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Hence,
Instantaneous rate \(=\frac{d x}{d t}\)
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 9
It is obtained from the slope of the curve obtained by plotting the concentration of the product against time.

The instantaneous rate is more useful in obtaining the rate law integrated equations.

Question ii.
Distinguish between order and molecularity of a reaction.
Answer:

Order Molecularity
1. It is the sum of the exponents to which the concentration terms in rate law expression are raised. 1. ¡t is the number of molecules (or atoms or ions) of the reaCtants taking part in the elementary reaction.
2. It is experimentally determined and indicates the dependence of the reaction rate on the concentration of particular reactants. 2. It is the oretical property and indicátes the number of molecules of reactant in each step of the reaction.
3. It may have values that are integer, fractional, or zero. 3. It is always an integer.
4. Its value depends upon experimental conditions. 4. Its value does not depend upon experimental conditions.
5. It is the property of elementary and complex reactions. 5. It is the property of elementary reactions only.
6. Rate law expression describes the order of the reaction. 6. Rate law does not describe molecularity.

Question iii.
A reaction takes place in two steps,
1. NO(g) + Cl2(g) NOCl2(g)
2. NOCl2(g) + NO(g) → 2NOCl(g)
a. Write the overall reaction.
b. Identify reaction intermediate.
c. What is the molecularity of each step?
Solution :
Given :
(1) NO(g) + Cl2(g) → NOCl2(g)
(2) NOCl2(g) + NO(g) → 2NOCl(g)

(a) Overall reaction is obtained by adding both the reactions
2NO(g) + Cl2(g) → 2NOCl2(g)
(b) The reaction intermediate is NOCl2, since it is formed in first step and consumed in the second step.
(c) Since the first step is a slow and rate determin­ing step, the molecularity is two.

Since the second step is a fast step its molecularity is not considered.

Question iv.
Obtain the relationship between the rate constant and half-life of a fist order reaction.
Answer:
Consider the following reaction,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 175
If [A]0 and [A]t are the concentrations of A at start and after time t, then [A]0 = a and [A]t = a – x.

The velocity constant or the specific rate constant k for the first order reaction can be represented as,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 57

where, a is the initial concentration of the reactant A, x is the concentration of the product B after time t, so that (a – x) is the concentration of the reactant A after time t.

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Half-life of a reaction : The time required to reduce the concentration of the reactant to half of its initial value is called the half-life period or the half-life of the reaction.

If t1/2 is the half-life of a reaction, then at t = t1/2, x = a/2, hence a – x = a – a/2 = a/2
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 58
Hence, for a first order reaction, the half-life of the reaction is independent of the initial concentration of the reactant.

Question v.
How will you represent zeroth-order reaction graphically?
Answer:
(1) A graph of concentration against time : In case of a zero-order reaction, the rate of reaction is independent of the concentration of the reactant. The concentration [A]t of the reactant at a time t is given by
[A]t = – kt + [A]0 (y = – mx + c)
where [A]0 is the initial concentration of the reactant and k is a rate constant.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 76

Hence in case of zero order reaction, when the concentration of the reactant is plotted against time, a straight line with the slope equal to – k is obtained. The concentration of the reactants de-crease with time. The intercept on the concentration axis gives the initial concentration, [A]0.

(2) A graph of rate of a reaction against the concen-tration of the reactant: Rate of a zero order reaction is independent of the concentration of the reactant.

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Rate, R = k [A]0 = k

Hence even if the concentration of the reactant decreases, the rate of the reaction remains constant.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 77

Therefore if the rate of a zero order reaction is plotted against concentration, then a straight line with zero slope is obtained indicating, no change in the rate of the reaction with a change in the concentration of the reactants.

(3) A graph of half-life period against concentration : The half-life period of a zero-order reaction is given by, \(t_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 k}\) where [A]0 is initial con-centration of the reactant and k is the rate constant. Hence the half-life period is directly proportional to the concentration.

When a graph of t1/2 is plotted against concentration, a straight line passing through origin is obtained, and the slope gives \(\frac{1}{2 k}\), where k is the rate constant.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 78

Question vi.
What are pseudo-fist order reactions? Give one example and explain why it is pseudo-fist order.
Answer:
Pseudo-first-order reaction : A reaction which has higher-order true rate law but is experimentally found to behave as first order is called pseudo first order reaction.
Explanation : Consider an acid hydrolysis reaction of an ester like methyl acetate.
CH3COOCH3(aq) + H2O(1) \(\stackrel{\mathrm{H}_{(\mathrm{aq})}^{+}}{\longrightarrow}\) CH3COOH(aq) + CH3OH(aq)
Since the reaction involves two substances, ester and water, it is a bimolecular reaction and the true rate law should be, Rate = k’ [CH3COOCH3] x [H2O]

Hence the reaction is expected to follow second order kinetics. However experimentally it is found that the reaction follows first order kinetics.

This is because solvent water being in a large excess, its concentration remains constant. Hence, [H2O] = constant = k”
Rate = k [CH3COOCH3] x [H2O]
= k [CH3COOCH3] x k”
= k’ x k” x [CH3COOCH3]
If k’ x k” = k, then Rate = k [CH3COOCH3],

This indicates that second-order true rate law is forced into first order rate law. Therefore this bimolecular reaction which appears of second order is called pseudo first order reaction.

Question vii.
What are the requirements for the colliding reactant molecules to lead to products?
Answer:
Collisions of reactant molecules : The basic re­quirements of a reaction is that the reacting species (atoms, ions or molecules) must come together and collide for a reaction to occur. Therefore the rate of the reaction shall depend on the rate and frequency of collisions which increase with the amount of reacting species and temperature.

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

However it is observed that the rate of reaction is very low as compared to the rate of collisions between reacting species in gaseous phase or liquid phase. This suggests that all the collisions are not fruitful leading to a reaction. Hence it is necessary to consider another factor like energy of colliding species along with collision frequency.

Energy requirement (Activation energy) : The colliding molecules must possess a certain mini­mum energy called activation energy required far breaking and making bonds resulting in the reac­tion. This implies that the colliding molecules must have energy equal to or greater than the activation energy. The colliding molecules with less energy do not react.

Orientation of reactant molecules : The concept of activation energy is satisfactory in case of simple molecules or ions but not in case of complex or higher polyatomic molecules. It is observed that the rates of reaction are less as compared to the rates of collisions between activated molecules with activa­tion energy.

This suggests that in addition, the colliding mole­cules must have proper orientations relative to each other during collisions. For example, consider the reaction, A – B + C → A + B – C. For the reaction to occur, C must collide with B while collisions with A will not be fruitful. Since B has to bond with C.

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 118

Question viii.
How catalyst increases the rate of reaction? Explain with the help of a potential energy diagram for catalyzed and uncatalyzed reactions.
Answer:
(i) A catalyst is a substance, when added to the reactants, increases the rate of the reaction without being consumed. For example, the decomposition of KClO3 in the presence of small amount of MnO2 is very fast but very slow in the absence of MnO2.

2KClO3(s) \(\frac{\mathrm{MnO}_{2}}{\Delta}\) 2KCl(s) + 3O2(g)

(ii) The phenomenon of catalysed reaction is called catalysis and depends on nature of the catalyst. In heterogeneous catalysis, the reactant molecules are adsorbed on the solid catalyst surface while in case of homogeneous catalysis, the catalyst reacts with one of the reactants, forms intermediate and decomposes reforming original catalyst and the products.

(iii) The catalyst provides alternative and lower energy path or mechanism for the reaction.

(iv) In the presence of the catalyst, the activation energy of the reaction is lowered. The height of activation energy barrier is less than that in the uncatalysed reaction.

(v) Due to lowering of energy of activation, (Ea) the number of molecules possessing Ea increases, hence the rate of the reaction increases.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 134

(vi) The rate constant = k = A x e-Ea/RT where A is a frequency factor and hence the rates of the catalysed reaction are higher than those of un-catalysed reactions.

(vii) The catalyst does not change the extent of the reaction but hastens the reaction.

(viii) The catalyst enters the reaction but does not appear in the balanced equation since it is consumed in one step and regenerated in the another.

Question ix.
Explain with the help of the Arrhenius equation, how does the rate of reaction changes with (a) temperature and (b) activation energy.
Answer:
(a) By Arrhenius equation, k = Ax e-Ea/RT where k is rate constant, A is a frequency factor and Ed is energy of activation at temperature T. As Ea increases, the rate constant and rate of the reaction decreases.

(b) As temperature increases Ea/RT decreases but due to negative sign, k and rate increase with the increase in temperature.

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Question x.
Derive the integrated rate law for first order reaction.
Answer:
Consider following gas phase reaction,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 68

Let initial pressure of A be P0 at t = 0. If after time t the pressure of a A decreases by jc then the partial pressures of the substances will be, PA = PQ – x; PB = x and Pc = x

Total pressure will be,

PT + P0 – x + x + x = Po + x
∴ x = PT – Pn

The partial pressures at time t will be,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 69

Question xi.
How will you represent first-order reactions graphically.
Answer:
(1) A graph of rate of a reaction and concentra­tion : The differential rate law for first-order reac­tion, A → Products is represented as, Rate = [/latex]-\frac{d[\mathrm{~A}]}{d t}=k[\mathrm{~A}]\(

∴ Rate = k x [A]t (y = mx). When the rate of a first order reaction is plotted against concentration, [A]t, a straight line graph is obtained.

With the increase in the concentration [A]t, rate R, increases. The slope of the line gives the value of rate constant k.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 59

(2) A graph of concentration against time : When the concentration of the reactant is plotted against time t, a curve is obtained. The concentration [A], of the reactant decreases exponentially with time. The variation in the concentration can be represented as,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 60
where [A]0 and [A]t are initial and final concentra­tions the reactant and k is the rate constant. The time required to complete the first order reaction is infinity.

(3) A graph of log10 (a – x) against time t :
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 61
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 62

When log10(a – x) is plotted against time t, a straight line with negative slope is obtained, from which the velocity constant k can be calculated.

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

(4) A graph of half-life period and concentration : The half-life period, t1/2 of a first order reaction is given by, where k is the rate constant.

For the given reaction at a constant temperature, t1/2 is constant and independent of the concentration of the reactant.

Hence when a graph of t1/2 is plotted against concentration, a straight line parallel to the concen­tration axis (slope = zero) is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 63

(5) A graph of log10 [latex]\left(\frac{a}{a-x}\right)\) against time : The rate constant, for a first order reaction is represented as, Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 64where [A0] and [A]t are the respective initial and final concentrations of the reactant after time t.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 65
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 66
When \(\log _{10}\left(\frac{a}{a-x}\right)\) is plotted against time t, a straight line graph passing through the origin is obtained and the slope gives the value of k/2.303. From this slope, the rate constant can be calculated.

Question xii.
Derive the integrated rate law for the first order reaction, A(g) → B(g) + C(g) in terms of pressure.
Answer:
Consider following gas phase reaction,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 68

Let initial pressure of A be P0 at t = 0. If after time t the pressure of a A decreases by jc then the partial pressures of the substances will be, PA = PQ – x; PB = x and Pc = x

Total pressure will be,

PT + P0 – x + x + x = Po + x
∴ x = PT – Pn

The partial pressures at time t will be,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 69

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Question xiii.
What is zeroth-order reaction? Derive its integrated rate law. What are the units of rate constant?
Answer:
Definition : Zero order reaction : A reaction in which the rate of the reaction does not depend on the concentration of any reactant taking part in the reaction is called zero order reaction.

Consider a zero order reaction, A → Products
The rate of the reaction is, Rate \(=\frac{-d[\mathrm{~A}]}{d t}\)

By rate law,
Rate = k x [A]0 = k
∴ – d[A] = k x dt

If [A]0 is the initial concentration of the reactant A at t = 0 and [A]t is the concentration of A present after time t, then by integrating above equation,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 73
This is the integrated rate law expression for rate constant for zero order reaction.
∴ k x t = [A]0 – [A]t
∴ [A]t = – kt + A0

For a zero order reaction :
The rate of reaction is R = k [A]0 = k
Hence, the velocity constant k has the unit of the rate of the reaction, i.e., mol dm-3 s-1.

Question xiv.
How will you determine activation energy: (a) graphically using Arrhenius equation (b) from rate constants at two different temperatures?
Answer:
(a) By Arrhenius equation,
Rate constant = = A x e-Ea/RT where A is a fre-quency factor.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 130
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 131

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

When log10k is plotted against 1/T a straight line with negative slope is obtained. From the slope of the graph, energy of activation Ea, is obtained as follows :
Slope = \(\frac{E_{\mathrm{a}}}{2.303 R}\)
∴ Ea = 2303R x sloPe

(b) For the given reaction, rate constants k1 and k2 are measured at two different temperatures T1 and T2 respectively. Then \(\log _{10} \frac{k_{2}}{k_{1}}=\frac{E_{\mathrm{a}}\left(T_{2}-T_{1}\right)}{2.303 R \times T_{1} \times T_{2}}\) where Ea is the energy of activation.

Hence by substituting appropriate values, energy of activation Ea for the reaction is determined.

Question xv.
Explain graphically the effect of temperature on the rate of reaction.
Answer:
(i) It has been observed that the rates of chemical reactions increase with the increase in temperature.
(ii) The kinetic energy of the molecules increases with the increase in temperature. The fraction of molecules possessing minimum energy barrier,
i. e. activation energy Ea increases with increase in temperature.
(iii) Hence the fraction of colliding molecules that possess kinetic energy (Ea) also increases, hence the rate of the reaction increases with increase in temperature.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 132
(iv) The above figure shows that the area that represents the fraction of molecules with kinetic energy exceeding Ea is greater at higher temperature T2 than at lower temperature T1. This explains that the rate of the reaction increases at higher temperature.
(v) The shaded area to the right of activation energy Ea represents fraction of collisions of activated molecules having energy Ea or greater.

Question xvi.
Explain graphically the effect of catalyst on the rate of reaction.
Answer:
(i) The phenomenon of catalysed reaction is called catalysis and depends on nature of the catalyst. In heterogeneous catalysis, the reactant molecules are adsorbed on the solid catalyst surface while in case of homogeneous catalysis, the catalyst reacts with one of the reactants, forms intermediate and decomposes reforming original catalyst and the products.
(ii) The catalyst provides alternative and lower energy path or mechanism for the reaction.
(iii) In the presence of the catalyst, the activation energy of the reaction is lowered. The height of activation energy barrier is less than that in the uncatalysed reaction.

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Question xvii.
For the reaction 2A + B → products, find the rate law from the following data.

[A]/M [A]/M rate/M s-1
0.3 0.05 0.15
0.6 0.05 0.30
0.6 0.2 1.20

Solution :
Given : 2A + B → Products
Rates : R1 = 0.15 Ms-1 R2 = 0.3 Ms-1
[A]1 = 0.3 M [A]2 = 0.6 M
[B]1 = 0.05 M [B]2 = 0.05 M
(i) If order of the reaction in A is x and in B is y then, by rate law,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 104
∴ y = 1. Hence the reaction has order one in B.
The order of overall reaction = n = nA + nB = 1 + 1 = 2
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 105
Answer:
(i) Rate law : Rate = fc [A] x [B]
Rate constant = k = 10M-1s-1
Order of the reaction = 2

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

4. Solve

Question i.
In a first order reaction, the concentration of reactant decreases from 20 mmol dm-3 to 8 mmol dm-3 in 38 minutes. What is the half life of reaction? (28.7 min)
Solution :
Given: [A]0 =20 mmol dm-3;
[A]t=8 mmol dm-3; t=38 mm;
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 85
Answer:
Half life period = 28.74 min

Question ii.
The half life of a first order reaction is 1.7 hours. How long will it take for 20% of the reactant to react? (32.9 min)
Solution :
Given : t1/2 = 1.7 hr; [A]0 = 100;
[A]t = 100 – 20 = 80; t =?
\(t_{1 / 2}=\frac{0.693}{k}\)
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 87
Answer:
Time required = t = 32.86 min

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Question iii.
The energy of activation for a first order reaction is 104 kJ/mol. The rate constant at 25 0C is 3.7 × 10-5 s-1. What is the rate constant at 300C? (R = 8.314 J/K mol) (7.4 × 10-5)
Solution :
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 137
Answer:
k2 = 7.382 x 10-5 s-1

Question iv.
What is the energy of activation of a reaction whose rate constant doubles when the temperature changes from 303 K to 313 K? (54.66 kJ/mol)
Solution :
Given : k2 = 2kt, T1 = 303 K; T2 = 313 K; Ea = ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 138
Answer:
Energy of activation = Ea = 54.66 kJ

Question v.
The rate constant of a reaction at 5000C is 1.6 × 103 M-1 s-1. What is the frequency factor of the reaction if its activation energy is 56 kJ/mol. (9.72 × 106 M-1 s-1)
Solution :
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 139
Answer:
Frequency factor = A = 9.727 x 106 M-1s-1

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Question vi.
Show that time required for 99.9% completion of a first order reaction is three times the time required for 90% completion.
Solution :
Given : For 99.9 % completion, if [A]0 = 100,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 89
If t1 and t2 are the times required for 99.9 % and 90 % completion of reaction respectively, then
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 90
Answer:
Time required for 99.9 % completion of a first order reaction is three time the time required for 90 % completion of the reaction.

Question vii.
A first order reaction takes 40 minutes for 30% decomposition. Calculate its half life. (77.66 min)
Solution :
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 88
Answer:
Half life period = 77.70 min.

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Question viii.
The rate constant for the first order reaction is given by log10 k = 14.34 – 1.25 × 104 T. Calculate activation energy of the reaction. (239.3 kJ/mol)
Solution :
Given : log10 k = 14.34 – \(\frac{1.25 \times 10^{4}}{T}\) ……………………. (1)
From Arrhenius equation we can write,
\(\log _{10} k=\log _{10} A-\frac{E_{\mathrm{a}}}{2.303 R \times T}\) ……………………. (2)
By comparing equations (1) and (2),
\(\frac{E_{\mathrm{a}}}{2.303 \times R}\) = 1.25 x 104
∴ Ea = 1.25 x 104 x 2.303 x R
= 1.25 x 104 x 2.303 x 8.314
= 23.93 x 104 = 239.3 kJ mol-1

[Note : Frequency factor A may also be calculated as follows : log10 A = 14.34
∴ A = Antilog 14.34 = 2.188 x 104
Answer:
Energy of activation = Ea = 239.3 kJ mol-1.

Question ix.
What fraction of molecules in a gas at 300 K collide with an energy equal to activation energy of 50 kJ/mol? (2 × 10-9)
Solution :
Given : T = 300 K; Ea = 50 kJ mol-1
= 50 x 103 mol-1
The fraction of molecules undergoing fruitful collisions is
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 140
Answer:
Fraction of molecules undergoing collision = 2 x 10-9

Activity :
1. If you wish to determine the reaction order and rate constant for the reaction, 2AB2 → A2 + 2B2.
a) What data would you collect?
b) How would you use these data to determine whether the reaction is zeroth or first order?

2. The activation energy for two reactions are Ea and E’a with Ea > E’a. If the temperature of reacting system increases from T1 to T2, predict which of the following is correct?
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 1
k values are rate constants at lower temperatures and k values at higher temperatures.

12th Chemistry Digest Chapter 6 Chemical Kinetics Intext Questions and Answers

(Textbook Page No 121)

Question 1.
Write the expressions for rates of reaction for :
2N2O5(g) → 4NO2(g) + O2(g)?
Answer:
For the given reaction, Rate of reaction =
\(=R=-\frac{1}{2} \frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}\)
\(\begin{aligned}
&=+\frac{1}{4} \frac{d\left[\mathrm{NO}_{2}\right]}{d t} \\
&=+\frac{d\left[\mathrm{O}_{2}\right]}{d t}
\end{aligned}\)

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Problem 6.1: (Textbook Page No 121)

Question 1.
For the reaction,
\(\mathbf{3 I}_{(a q)}^{-}+\mathbf{S}_{2} \mathbf{O}_{8(a q)}^{2-} \longrightarrow \mathbf{I}_{3(\text { (aq) }}^{-}+2 \mathbf{S O}_{4(\mathrm{aq})}^{2-}\)
Calculate (a) the rate of formation of I3,
(b) the rates of consumption of 1 and S2O and (c) the overall rate of reaction if the rate of formation of \(\mathrm{SO}_{4}^{2-}\) is 0.O22 moles dm-3 sec-1.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 19
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 20
∴ (a) Rate of formation of \(\mathrm{I}_{3}^{-}\) = 0.011 mol dm-3 s-1
(b) Rate of consumption of I = 0.033 mol dm-3 s-1
(c) Rate of consumption of \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) = 0.011 mol dm-3 s-1
(d) Overall rate of reaction = Rate of consumption of reactant = Rate of formation of product

Try this….. (Textbook Page No 122)

Question 1.
For the reaction :
NO2(g) + CO(g) → NO(g) + CO2(g), the rate of reaction is experimentally found to be proportional to the square of the concentration of NO2 and independent that of CO. Write the rate law.
Answer:
Since the rate of the reaction is proportional to [NO2]2 and [CO]0, the rate law is R = k[NO2]2 x [CO]0
∴ R = k[NO2]2.

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Try this….. (Textbook Page No 124)

Question 1.
The reaction,
CHCl3(g) + Cl2(g) → CCl4(g) + HCl(g) is first order in CHCl3 and 1/2 order in Cl2. Write the rate law and overall order of reaction.
Answer:
Since the reaction is first order in CHCl3 and 1/2 order in Cl2, the rate law for the reaction will be, Rate = k[CHCl3] X [Cl2]1/2
The overall order (n) of the reaction will be, n = l + = \(\frac{1}{2}=\frac{3}{2}\)

Use your brain power! (Textbook Page No 124)

Question 1.
The rate of the reaction 2A + B → 2C + D is 6 x 10-4 mol dm-3 s-1 when [A] =[B] = O.3 mol dm-3 If the reaction is of first order in A and zeroth order in B, what is the rate constant?
Answer:
For the reaction,
2A + B → 2C + D,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 51

(Problem 6.7) (Textbook Page No 126)

Question 1.
A reaction occurs in the following steps :
(i) NO2(g) + F2(g) → NO2F(g) + F(g) (slow)
(ii) F(g) + NO2(g) → NO2F(g) (fast)
(a) Write the equation of overall reaction.
(b) Write down rate law.
(c) Identify the reaction intermediate.
Solution :
(a) The addition of two steps gives the overall reaction as
2NO2(g) + F2(g) → 2NO2 F(g)
(b) Step (i) is slow. The rate law of the reaction is predicted from its stoichiometry. Thus, rate = k [NO2] [F2]
(c) F is produced in step (i) and consumed in step (ii) hence F is the reaction intermediate.

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Try this….. (Textbook Page No 126)

Question 1.
A complex reaction takes place in two steps :
(i) NO(g) + O3(g) → NO3(g) + O(g)
(ii) NO3(g) + O(g) → NO2(g) + O2(g)
The predicted rate law is rate = k [NO] [O3]. Identify the rate-determining step. Write the overall reaction. Which is the reaction inter-mediate? Why?
Answer:
(i) NO(g) + O3(g) → NO3(g) + O(g)
(ii) NO3(g) + O(g) NO2(g) + O2(g)
(a) The first step is slow and rate determining step since the rate depends on concentrations of NO(g) and O3(g). (Given : Rate = k [NO] x [O])
(b) The overall reaction is the combination of two steps.
NO(g) + O3(g) → NO2(g) + O2(g)
(c) NO3(g) and O(g) are reaction intermediates. They are formed in first step (i) and removed in the second step (ii).

Try this….. (Textbook Page No 129)

Question 1.
The half-life of a first-order reaction is 0.5 min. Calculate (a) time needed for the reactant to reduce to 20% and (b) the amount decomposed in 55 s.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 70

Try this….. (Textbook Page No 123)

Question 1.
For the reaction 2A + 2B → 2C + D, if concentration of A is doubled at constant [B] the rate increases by a factor of 4. If the concentration of B is doubled with [A] being constant the rate is doubled. Write the rate law of the reaction.
Answer:
Rate = R1 = k[A]x [B]y
When concentration of A = [2A] and
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 71
Hence order with respect to A is 2 and with respect to B is 1. By rate law,
Rate = A: [A]2 [B]

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

Question 2.
The rate law for the reaction A + B → C is found to be rate = k [A]2 x [B]. The rate constant of the reaction at 25 °C is 6.25 M-2 S-1. What is the rate of reaction when [A] = 1.0 mol dm-3 and [B] = 0.2 mol dm-3?
Answer:
Rate = k x [A]2 x [B]
= 6.25 x 12 x 0.2
Rate = 1.25 x 102 mol dm-3 s-1