Class 11

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 6 Redox Reactions Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 6 Redox Reactions

1. Choose the most correct option

Question A.
Oxidation numbers of Cl atoms marked as Cl^a and Cl^b in CaOCl₂ (bleaching powder) are

a. zero in each
b. -1 in Cl^a and +1 in Cl^b
c. +1 in Cl^a and -1 in Cl^b
d. 1 in each
Answer:
b. -1 in Cl^a and +1 in Cl^b

Question B.
Which of the following is not an example of redox reacton ?
a. CuO + H₂ → Cu + H₂O
b. Fe₂O₃ + 3CO₂ → 2Fe + 3CO₂
c. 2K + F₂ → 2KF
d. BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl
Answer:
d. BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

Question C.
A compound contains atoms of three elements A, B and C. If the oxidation state of A is +2, B is +5 and that of C is -2, the compound is possibly represented by
a. A₂(BC₃)₂
b. A₃(BC₄)₂
c. A₃(B₄C)₂
d. ABC₂
Answer:
b. A₃(BC₄)₂

Question D.
The coefficients p, q, r, s in the reaction
\(\mathrm{pCr}_{2} \mathrm{O}_{7}^{2-}\) + q Fe^2⊕ → r Cr^3⊕ + s Fe^3⊕ + H₂O respectively are :
a. 1, 2, 6, 6
b. 6, 1, 2, 4
c. 1, 6, 2, 6
d. 1, 2, 4, 6
Answer:
c. 1, 6, 2, 6

Question E.
For the following redox reactions, find the correct statement.
Sn^2⊕ + 2Fe^3⊕ → Sn^4⊕ + 2Fe^2⊕
a. Sn^2⊕ is undergoing oxidation
b. Fe^3⊕ is undergoing oxidation
c. It is not a redox reaction
d. Both Sn^2⊕ and Fe^3⊕ are oxidised
Answer:
a. Sn^2⊕ is undergoing oxidation

Question F.
Oxidation number of carbon in H₂CO₃ is
a. +1
b. +2
c. +3
d. +4
Answer:
d. +4

Question G.
Which is the correct stock notation for magenese dioxide ?
a. Mn(I)O₂
b. Mn(II)O₂
c. Mn(III)O₂
d. Mn(IV)O₂
Answer:
d. Mn(IV)O₂

Question I.
Oxidation number of oxygen in superoxide is
a. -2
b. -1
c. –\(\frac {1}{2}\)
d. 0
Answer:
c. –\(\frac {1}{2}\)

Question J.
Which of the following halogens does always show oxidation state -1 ?
a. F
b. Cl
c. Br
d. I
Answer:
a. F

Question K.
The process SO₂ → S₂Cl₂ is
a. Reduction
b. Oxidation
c. Neither oxidation nor reduction
d. Oxidation and reduction.
Answer:
a. Reduction

2. Write the formula for the following compounds :
A. Mercury(II) chloride
B. Thallium(I) sulphate
C. Tin(IV) oxide
D. Chromium(III) oxide
Answer:
i. HgCl₂
ii. Tl₂SO₄
iii. SnO₂
iv. Cr₂O₃

3. Answer the following questions

Question A.
In which chemical reaction does carbon exibit variation of oxidation state from -4 to +4 ? Write balanced chemical reaction.
Answer:
In combustion of methane, carbon exhibits variation from -4 to +4. The reaction is as follows:
CH₄ + 2O₂ → CO₂ + 2H₂O
In CH₄, the oxidation state of carbon is -4 while in CO₂, the oxidation state of carbon is +4.

Question B.
In which reaction does nitrogen exhibit variation of oxidation state from -3 to +5 ?
Answer:

C. Calculate the oxidation number of underlined atoms.
a. H₂SO₄
b. HNO₃
c. H₃PO₃
d. K₂C₂O₄
e. H₂S₄O₆
f. Cr₂O₇^2-
g. NaH₂PO₄
Answer:
i. H₂SO₄
Oxidation number of H = +1
Oxidation number of O = -2
H₂SO₄ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms of H₂SO₄ = 0
∴ 2 × (Oxidation number of H) + (Oxidation number of S) + 4 × (Oxidation number of O) = 0
∴ 2 × (+1) + (Oxidation number of S) + 4 × (-2) = 0
∴ Oxidation number of S + 2 – 8 = 0
∴ Oxidation number of S in H₂SO₄ = +6

ii. HNO₃
Oxidation number of H = +1
Oxidation number of O = -2
HNO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms of HNO₃ = 0
∴ (Oxidation number of H) + (Oxidation number of N) + 3 × (Oxidation number of O) = 0
∴ (+1) + (Oxidation number of N) + 3 × (-2) = 0
∴ Oxidation number of N + 1 – 6 = 0
∴ Oxidation number of N in HNO₃ = +5

iii. H₃PO₃
Oxidation number of O = -2
Oxidation number of H = +1
H₃PO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 3 × (Oxidation number of H) + (Oxidation number of P) + 3 × (Oxidation number of O) = 0
∴ 3 × (+1) + (Oxidation number of P) + 3 × (-2) = 0
∴ Oxidation number of P + 3 – 6 = 0
Oxidation number of P is H₃PO₃ = +3

iv. K₂C₂O₄
Oxidation number of K = +1
Oxidation number of O = -2
K₂C₂O₄ is a neutral molecule.
∴ Sum of the oxidation number of all atoms = 0
∴ 2 × (Oxidation number of K) + 2 × (Oxidation number of C) + 4 × (Oxidation number of O) = 0
∴ 2 × (+1) + 2 × (Oxidation number of C) + 4 × (-2) = 0
∴ 2 × (Oxidation number of C) + 2 – 8 = 0
∴ 2 × (Oxidation number of C) = + 6
∴ Oxidation number of C = +\(\frac {6}{2}\)
∴ Oxidation number of C in K₂C₂O₄ = +3

v. H₂S₄O₆
Oxidation number of H = +1
Oxidation number of O = -2
H₂S₄O₆ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of H) + 4 × (Oxidation number of S) + 6 × (Oxidation number of O) = 0
∴ 2 × (+1) + 4 × (Oxidation number of S) + 6 × (-2) = 0
∴ 4 × (Oxidation number of S) + 2 – 12 = 0
∴ 4 × (Oxidation number of S) = + 10
∴ Oxidation number of S = +\(\frac {10}{4}\)
∴ Oxidation number of S in H₂S₄O₆ = +2.5

vi. Cr₂O₇^2-
Oxidation of O = -2
Cr₂O₇^2- is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 2
∴ 2 × (Oxidation number of Cr) + 7 × (Oxidation number of O) = -2
∴ 2 × (Oxidation number of Cr) + 7 × (-2) = – 2
∴ 2 × (Oxidation number of Cr) – 14 = – 2
∴ 2 × (Oxidation number of Cr) = – 2 + 14
∴ Oxidation number of Cr = +\(\frac {12}{2}\)
∴ Oxidation number of Cr in Cr₂O₇^2- = +6

vii. NaH₂PO₄
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = -2
NaH₂PO₄ is a neutral molecule
Sum of the oxidation numbers of all atoms = 0
(Oxidation number of Na) + 2 × (Oxidation number of H) + (Oxidation number of P) + 4 × (Oxidation number of O) = 0
(+1) + 2 × (+1) + (Oxidation number of P) + 4 × (-2) = 0
(Oxidation number of P) + 3 – 8 = 0
Oxidation number of P in NaH₂PO₄ = +5

Question D.
Justify that the following reactions are redox reaction; identify the species oxidized/reduced, which acts as an oxidant and which act as a reductant.
a. 2Cu₂O_(s) + Cu₂S_(s) → 6Cu_(s) + SO_2(g)
b. HF_(aq) + OH^–_(aq) → H₂O_(l) + F^–_(aq)
c. I_2(aq) + 2 S₂O₃^2-_(aq) → S₄O₆^2-_(aq) + 2I^–_(aq)
Answer:
i. 2Cu₂O_(s) + Cu₂S_(s) → 6Cu_(s) + SO_2(g)
a. Write oxidation number of all the atoms of reactants and products.

b. Identify the species that undergoes change in oxidation number.

c. The oxidation number of S increases from -2 to +4 and that of Cu decreases from +1 to 0. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and it itself is oxidised. On the other hand, the oxidation number of Cu decreases by gain of electrons and therefore, Cu is an oxidising agent and itself is reduced.

Result:

1. The given reaction is a redox reaction.
2. Oxidant/oxidising agents (Reduced species): Cu₂O/ Cu₂S
3. Reductant/reducing agent (Oxidised species): Cu₂S

[Note: Cu in both Cu₂O and Cu₂S undergoes reduction. Hence, both Cu₂O and Cu₂S can be termed as oxidising agents in the given reaction.]

ii. HF_(aq) + OH^–_(aq) → H₂O_(l) + F^–_(aq)
a. Write oxidation number of all the atoms of reactants and products.

b. Since, the oxidation numbers of all the species remain same, this is NOT a redox reaction. Result:
The given reaction is NOT a redox reaction.

iii. I_2(aq) + 2 S₂O₃^2-_(aq) → S₄O₆^2-_(aq) + 2I^–_(aq)
a. Write oxidation number of all the atoms of reactants and products.

b. Identify the species that undergoes change in oxidation number.

c. The oxidation number of S increases from +2 to +2.5 and that of I decreases from 0 to -1. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and itself is oxidised. On the other hand, the oxidation number of I decreases by gain of electrons and therefore, I is an oxidising agent and itself is reduced.

Result:

1. The given reaction is a redox reaction.
2. Oxidant/oxidising agent (Reduced species): I₂
3. Reductant/reducing agent (Oxidised species): S₂O₃^2-

Question E.
What is oxidation? Which one of the following pairs of species is in its oxidized state ?
a. Mg / Mg²⁺
b. Cu / Cu²⁺
c. O₂ / O^2-
d. Cl₂ / Cl^–
Answer:
a. Mg / Mg²⁺
Here, Mg loses two electrons to form Mg²⁺ ion.
\(\mathrm{Mg}_{(\mathrm{s})} \longrightarrow \mathrm{Mg}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-}\)
Hence, Mg / Mg²⁺ is an oxidized state.

b. Cu/Cu²⁺
Here, Cu loses two electrons to form Cu²⁺ ion.
\(\mathrm{Cu}_{(\mathrm{s})} \longrightarrow \mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-}\)
Hence, Cu/Cu²⁺ is in an oxidized state.

c. O₂ / O^2-
Here, each O gains two electrons to form O^2- ion.
\(\mathrm{O}_{2(\mathrm{~g})}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}_{(\mathrm{aq})}^{2-}\)
Hence, O₂ / O^2- is in a reduced state.

d. Cl₂ / Cl^–
Here, each Cl gains one electron to form Cl^– ion.
\(\mathrm{Cl}_{2(\mathrm{~g})}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cl}_{(\mathrm{aq})}^{-}\)
Hence, Cl₂ / Cl^– is in a reduced state.

Question F.
Justify the following reaction as redox reaction.
2 Na_2(s) + S_(s) → Na₂S_(s)
Find out the oxidizing and reducing agents.
Answer:
i. Redox reaction can be described as electron transfer as shown below:
2Na_(s) + S_(s) → 2Na⁺ + S^2-
ii. Charge development suggests that each sodium atom loses one electron to form Na⁺ and sulphur atom gains two electrons to form S^2-. This can be represented as follows:

iii. When Na is oxidised to Na₂S, the neutral Na atom loses electrons to form Na⁺ in Na₂S while the elemental sulphur gains electrons and forms S^2- in Na₂S.
iv. Each of the above steps represents a half reaction which involves electron transfer (loss or gain).
v. Sum of these two half reactions or the overall reaction is a redox reaction.
vi. Oxidising agent is an electron acceptor and hence, S is an oxidising agent. Reducing agent is an electron donor and hence, Na is a reducing agent.

Question G.
Provide the stock notation for the following compounds : HAuCl₄, Tl₂O, FeO, Fe₂O₃, MnO and CuO.
Answer:

Question H.
Assign oxidation number to each atom in the following species.
a. Cr(OH)₄^–
b. Na₂S₂O₃
c. H₃BO₃
Answer:
i. Cr(OH)₄^–
Oxidation number of O = -2
Oxidation number of H = +1
Cr(OH)₄^– is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 1
∴ Oxidation number of Cr + 4 × (Oxidation number of O) + 4 × (Oxidation number of H) = – 1
∴ Oxidation number of Cr + 4 × (-2) + 4 × (+1) = – 1
∴ Oxidation number of Cr – 8 + 4 = – 1
∴ Oxidation number of Cr – 4 = – 1 –
∴ Oxidation number of Cr = – 1 + 4
∴ Oxidation number of Cr in Cr(OH)₄^– = +3

ii. Na₂S₂O₃
Oxidation number of Na = +1
Oxidation number of O = -2
Na₂S₂O₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of Na) + 2 × (Oxidation number of S) + 3 × (Oxidation number of O) = 0
∴ 2 × (+1) + 2 × (Oxidation number of S) + 3 × (-2) = 0
∴ 2 × (Oxidation number of S) + 2 – 6 = 0
∴ 2 × (Oxidation number of S) = + 4
∴ Oxidation number of S = +\(\frac {4}{2}\)
∴ Oxidation number of S in Na₂S₂O₃ = +2

iii. H₃BO₃
Oxidation number of H = +1
Oxidation number of O = -2
H₃BO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 3 × (Oxidation number of H) + (Oxidation number of B) + 3 × (Oxidation number of O) = 0
∴ 3 × (+1) + (Oxidation number of B) + 3 × (-2) = 0
∴ Oxidation number of B + 3 – 6 = 0
∴ Oxidation number of B in H₃BO₃ = +3

Question I.
Which of the following redox couple is stronger oxidizing agent ?
a. Cl₂ (E⁰ = 1.36 V) and Br₂ (E⁰ = 1.09 V)
b. \(\mathrm{MnO}_{4}^{\Theta}\) (E0 = 1.51 V) and \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\) (E⁰ = 1.33 V)
Answer:
a. Cl₂ has a larger positive value of E⁰ than Br₂. Thus, Cl₂ is a stronger oxidizing agent than Br₂.
b. \(\mathrm{MnO}_{4}^{\Theta}\) has larger positive value of E⁰ than \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\). Thus, \(\mathrm{MnO}_{4}^{\Theta}\) is stronger oxidizing agent than \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\)

Question J.
Which of the following redox couple is stronger reducing agent ?
a. Li (E⁰ = – 3.05 V) and Mg(E⁰ = – 2.36 V)
b. Zn(E⁰ = – 0.76 V) and Fe(E⁰ = – 0.44 V)
Answer:
a. Li has a larger negative value of E⁰ than Mg. Thus, Li is a stronger reducing agent than Mg.
b. Zn has a larger negative value of E⁰ than Fe. Thus, Zn is a stronger reducing agent than Fe.

4. Balance the reactions/equations :

Question A.
Balance the following reactions by oxidation number method

Answer:
i. \(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+\mathrm{SO}_{3(\mathrm{aq})}^{2-} \longrightarrow \mathrm{Cr}_{(\mathrm{aq})}^{3+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-} \quad(\text { acidic })\)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(a q)}^{2-}+\mathrm{SO}_{3(a)}^{2-} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-}\)
Step 2: Assign oxidation number to Cr and S. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 3 atoms of S and 2 atoms of Cr. (There are already 2 Cr atoms.)
Step 3: Balance ‘O’ atoms by adding 4H₂O to the right-hand side.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+3 \mathrm{SO}_{3(\mathrm{aq})}^{2-} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+3 \mathrm{SO}_{4(\mathrm{aq})}^{2-}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add 8H on the left-hand side.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+3 \mathrm{SO}_{3(\mathrm{aq})}^{2-}+8 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+3 \mathrm{SO}_{4(\mathrm{aq})}^{2-}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation:

ii. \(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3}^{-}{(a q)} \quad \text { (basic) }\)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3}^{-}{ }_{(\mathrm{aq})}\)
Step 2: Assign oxidation number to Mn and Br. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 2 atoms of Mn.
\(2 \mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow 2 \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3(\mathrm{aq})}^{-}\)
Step 3: Balance ‘O’ atoms by adding H₂O to the right-hand side.
\(2 \mathrm{MnO}_{4(a q)}^{-}+\mathrm{Br}_{(2 q)}^{-} \longrightarrow 2 \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3 \text { (aq) }}^{-}+\mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 4: The medium is basic. To make the charges and hydrogen atoms on the two sides equal, add 2H⁺ on the left-hand side.

iii. H₂SO_4(aq) + C_(s) → CO_2(g) + SO_2(g) + H₂O_(l) (acidic)
Step 1: Write skeletal equation and balance the elements other than O and H.
H₂SO_4(aq) + C_(s) → CO_2(g) + SO_2(g) + H₂O_(l)
Step 2: Assign oxidation number to S and C. Calculate the increase and decrease in the oxidation number and make them equal.


To make the net increase and decrease equal, we must take 2 atoms of S.
2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l)
Step 3: Balance ‘O’ atoms by adding H2O to the right-hand side.
2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l) + H₂O_(l)
Step 4: The medium is acidic. There is no charge on either side. Hydrogen atoms are equal on both side.
2H₂SO_4(aq) + C_(s) → CO₂ + 2SO_2(g) + H₂O_(l)
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation: 2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l)

iv. \(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{Sn}(\mathrm{OH})_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{Bi}_{(\mathrm{s})}+\mathrm{Sn}(\mathrm{OH})_{6(\mathrm{aq})}^{2-}\) (basic)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{Sn}(\mathrm{OH})_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{Bi}_{(\mathrm{s})}+\mathrm{Sn}(\mathrm{OH})_{6(\mathrm{aq})}^{2-}\)
Step 2: Assign oxidation numbers to Bi and Sn. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 3 atoms of Sn and 2 atoms of Bi.

Step 4: The medium is basic. To make hydrogen atoms on the two sides equal, add 3W on the right-hand side.

Question B.
Balance the following redox equation by half reaction method

Answer:
i. H₂C₂O_4(aq) + \(\mathrm{MnO}_{4(a q)}^{-}\) → CO_2(g) + \(\mathrm{Mn}_{(\mathrm{aq})}^{2+}\)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.

Step 2: Balance the atoms except O and H in each half equation. Balance half equation for O atoms by adding 4H₂O to the right side of reduction half equation.

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence, add 2H⁺ ions to the right side of oxidation half equation and 8H⁺ ions to the left side of reduction half equation.

Step 4: Now add 2 electrons to the right side of oxidation half equation and 5 electrons to the left side of reduction half equation to balance the charges.

Step 5: Multiply oxidation half equation by 5 and reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.

ii. \(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{SnO}_{2(\mathrm{aq})}^{2-} \longrightarrow \mathrm{SnO}_{3(\mathrm{aq})}^{2-}+\mathrm{Bi}_{(\mathrm{s})}\)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.

Step 2: Balance half equations for O atoms by adding H₂O to the side with less O atoms. Add 1H₂O to left side of oxidation half equation and 3H₂O to the right side of reduction half equation.

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence, add 2H⁺ ions to the right side of oxidation half equation and 3H⁺ ions to the left side of reduction half equation.

Step 4: Now add 2 electrons to the right side of oxidation half equation and 3 electrons to the left side of reduction half equation to balance the charges.

Step 5: Multiply oxidation half equation by 3 reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.

Reaction occurs in basic medium. However, H⁺ ions cancel out and the reaction is balanced. Hence, no need to add OH^– ions. The equation is balanced in terms of number of atoms and the charges.
Hence, balanced equation:

5. Complete the following table :

Assign oxidation number to the underlined species and write Stock notation of compound

Compound	Oxidation number	Stock notation
AuCl3	……………..	……………..
SnCl2	……………..	……………..
\(\underline{\mathrm{V}}_{2} \mathrm{O}_{7}^{4-}\)	……………..	……………..
\(\underline{\mathrm{Pt}} \mathrm{Cl}_{6}^{2-}\)	……………..	……………..
H3AsO3	……………..	……………..

Answer:

Compound	Oxidation number	Stock notation
AuCl3	+3	Au(III)Cl3
SnCl2	+2	Sn(II)Cl2
\(\underline{\mathrm{V}}_{2} \mathrm{O}_{7}^{4-}\)	+5	V2(V)\(\mathrm{O}_{7}^{4-}\)
\(\underline{\mathrm{Pt}} \mathrm{Cl}_{6}^{2-}\)	+4	Pt(IV)\(\mathrm{Cl}_{6}^{2-}\)
H3AsO3	+3	H3As(III)O3

11th Chemistry Digest Chapter 6 Redox Reactions Intext Questions and Answers

Can you tell? (Textbook Page No. 81)

Question i.
Why does cut apple turn brown when exposed to air?
Answer:
Cut apple turns brown when exposed to air because polyphenols are released. These polyphenols undergo oxidation in the presence of air and impart brown colour.

Question ii.
Why does old car bumper change colour?
Answer:
Car bumper is made of iron which undergoes rusting over a period of time. Hence, old car bumper changes colour.

Question iii.
Why do new batteries become useless after some days?
Answer:
Batteries generate electricity by redox reactions. Once the chemicals taking part in redox reaction are used up, the battery cannot generate power. Hence, new batteries become useless after some days.

Can you recall? (Textbook Page No. 81)

Question i.
What is combustion reaction?
Answer:
Combustion is a process in which a substance combines with oxygen.

Question ii.
Write an equation for combustion of methane.
Answer:
Combustion of methane: CH₄ + 2O₂ → CO₂ + 2H₂O + Heat + Light

Question iii.
What is the driving force behind reactions of elements?
Answer:
The ability of element to combine with other element or the ability of element to replace other element in compound is the driving force behind the reactions. This may involve formation of precipitates, formation of water, release of gas, etc.

Try this. (Textbook Page No. 82)

Question 1.
Complete the following table of displacement reactions. Identify oxidising and reducing agents involved.

Reactants	Products
Zn(s) + ————(aq)	————-(aq) + Cu(s)
Cu(s) + 2Ag+(aq)	—————– + ————–
———– + ————-	\( \mathrm{Co}_{(\mathrm{aq})}^{2+}\) + Ni(s)

Answer:

Try this (Textbook Page No. 88)

Question 1.
Classify the following unbalanced half equations as oxidation and reduction.

Answer:

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 5 Chemical Bonding Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 5 Chemical Bonding

1. Select and write the most appropriate alternatives from the given choices.

Question A.
Which molecule is linear?
a. SO₃
b. CO₂
c. H₂S
d. Cl₂O
Answer:
b. CO₂

Question B.
When the following bond types are listed in decreasing order of strength (strongest first). Which is the correct order?
a. covalent > hydrogen > van der waals
b. covalent > vander waal’s > hydrogen
c. hydrogen > covalent > vander waal’s
d. vander waal’s > hydrogen > covalent.
Answer:
a. covalent > hydrogen > van der waals

Question C.
Valence Shell Electron Pair repulsion (VSEPR) theory is used to predict which of the following :
a. energy levels in an atom
b. the shapes of molecules and ions.
c. the electron negetivities of elements.
d. the type of bonding in compounds.
Answer:
b. the shapes of molecules and ions.

Question D.
Which of the following is true for CO₂?

	C=O bond	CO2 molecule
A	polar	non-polar
B	non-polar	polar
C	polar	polar
D	non-polar	non-polar

Answer:

	C=O bond	CO2 molecule
A	polar	non-polar

Question E.
Which O₂ molecule is pargmagnetic. It is explained on the basis of :
a. Hybridisation
b. VBT
c. MOT
d. VSEPR
Answer:
c. MOT

Question F.
The angle between two covalent bonds is minimum in:
a CH₄
b. C₂H₂
c. NH₃
d. H₂O
Answer:
d. H₂O

2. Draw

Question A.
Lewis dot diagrams for the folowing
a. Hydrogen (H₂)
b. Water (H₂O)
c. Carbon dioxide (CO₂)
d. Methane (CH₄)
e. Lithium Fluoride (LiF)
Answer:

[Note: H atom in H₂ and Li atom in LiF attain the configuration of helium (a duplet of electrons).]

Question B.
Diagram for bonding in ethene with sp² Hybridisation.
Answer:

Question C.
Lewis electron dot structures of
a. HF
b. C₂H₆
c. C₂H₄
d. CF₃Cl
e. SO₂
Answer:

Question D.
Draw orbital diagrams of
a. Fluorine molecule
b. Hydrogen fluoride molecule
Answer:
a.

b.

3. Answer the following questions

Question A.
Distinguish between sigma and pi bond.
Answer:

σ (sigma) bond	π (pi) bond
1. It is formed when atomic orbitals overlap along internuclear axis.	1. It is formed when atomic orbitals overlap side-ways (laterally).
2. Electron density is high along the axis of the molecule (i.e., internuclear axis).	2. Electron density is zero along the axis of the molecule (i.e., internuclear axis).
3. In the formation of sigma bond, the extent of overlap is greater, hence, more energy is released.	3. In the formation of pi bond, the extent of overlap is less, hence, less energy is released.
4. It is a strong bond.	4. It is a weak bond.
5. Formation of sigma bonds involves s-s, s-p, p-p overlap and overlap between hybrid orbitals.	5. Formation of pi bonds involves p-p or d-d overlap. The overlap between hybrid orbitals is not involved.

Question B.
Display electron distribution around the oxygen atom in water molecule and state shape of the molecule, also write H-O-H bond angle.
Answer:
Electron distribution around oxygen atom in water molecule:
Shape of water molecule: Angular or V shaped H-O-H bond angle = 104°35′

Question C.
State octet rule. Explain its inadequecies with respect to
a. Incomplete octet
b. Expanded octet
Answer:
Statement: During the formation of chemical bond, atom loses, gains or shares electrons so that its outermost orbit (valence shell) contains eight electrons. Therefore, the atom attains the nearest inert gas electronic configuration.

a. Molecules with incomplete octet: e.g. BF₃, BeCl₂, LiCl
In these covalent molecules, the atoms B, Be and Li have less than eight electrons in their valence shell but these molecules are stable.
Li in LiCl has only two electrons, Be in BeCl₂ has four electrons while B in BF₃ has six electrons in the valence shell.

b. Molecules with expanded octet: Some molecules like SF₆, PCl₅, H₂SO₄ have more than eight electrons around the central atom.

Question D.
Explain in brief with one example:
a. Ionic bond
b. covalent bond
c. co-ordinate bond
Answer:
a. Formation of calcium chloride (CaCl₂):
i. The electronic configurations of calcium and chlorine are:
Na (Z = 11): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² or (2, 8, 8, 2)
Cl (Z = 17): 1s² 2s² 2p⁶ 3s² 3p⁵ or (2, 8, 7)
ii. Calcium has two electrons in its valence shell. It has tendency to lose two electrons to acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iii. Chlorine has seven electrons in its valence shell. It has tendency to gain one electron and thereby acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iv. During the combination of calcium and chlorine atoms, the calcium atom transfers its valence electrons to two chlorine atoms.
v. Calcium atom changes into Ca²⁺ ion while the two chlorine atoms change into two Cl^– ions. These ions are held together by strong electrostatic force of attraction.
vi. The formation of ionic bond(s) between Ca and Cl can be shown as follows:

b. Formation of Cl₂ molecule:
i. The electronic configuration of Cl atom is [Ne] 3s² 3p⁵.
ii. It needs one more electron to complete its valence shell.
iii. When two chlorine atoms approach each other at a certain internuclear distance, they share their valence electrons. In the process, both the atoms attain the valence shell of octet of nearest noble gas, argon.
iv. The shared pair of electrons belongs equally to both the chlorine atoms. The two atoms are said to be linked by a single covalent bond and a Cl₂ molecule is formed.

c. co-ordinate bond:
i. A coordinate bond is a type of covalent bond where both of the electrons that form the bond originate from the same atom
ii. An atom with a lone pair of electrons (non-bonding pair of electrons) is capable of forming a coordinate bond.
iii. For example, reaction of ammonia with boron trifluoride: Before the reaction, nitrogen (N) in ammonia has eight valence electrons, including a lone pair of electrons. Boron (B) in boron trifluoride has only six valence electrons, so it is two electrons short of an octet. The two unpaired electrons form a bond between nitrogen and boron, resulting in complete octets for both atoms. A coordinate bond is represented by an arrow. The direction of the arrow indicates that the electrons are moving from nitrogen to boron. Thus, ammonia forms a coordinate bond with boron trifluoride.

iv. Once formed, a coordinate covalent bond is the same as any other covalent bond.

Question E.
Give reasons for need of Hybridisation.
Answer:
The concept of hybridization was introduced because the valence bond theory failed to explain the following points:
i. Valencies of certain elements:
The maximum number of covalent bonds which an atom can form equals the number of unpaired electrons present in its valence shell. However, valence bond theory failed to explain how beryllium, boron and carbon forms two, three and four covalent bonds respectively.
a. Beryllium: The electronic configuration of beryllium is 1s² 2s². The expected valency is zero (as there is no unpaired electron) but the observed valency is 2 as in BeCl₂.
b. Boron: The electronic configuration of boron is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\). The valency is expected to be 1 but it is 3 as in BF₃.
c. Carbon: The electronic configuration of carbon is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) . The valency is expected to be 2, but observed valency is 4 as in CH₄.

ii. The shapes and geometry of certain molecules:
The valence bond theory cannot explain shapes, geometries and bond angles in certain molecules,
e.g. a. Tetrahedral shape of methane molecule.
b. Bond angles in molecules like NH₃ (107°18′) and H₂O (104°35′).
However, the valency of the above elements and the observe structural properties of the above molecules can be explained by the concept of hybridization. These are the reasons for need of the concept of hybridization.

Question F.
Explain geometry of methane molecule on the basis of Hybridisation.
Answer:
Formation of methane (CH₄) molecule on the basis of sp³ hybridization:
i. Methane molecule (CH₄) has one carbon atom and four hydrogen atoms.
ii. The ground state electronic configuration of C (Z = 6) is 1s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) \(2 \mathrm{p}_{\mathrm{z}}^{1}\);
Electronic configuration of carbon:

iii. In order to form four equivalent bonds with hydrogen, the 2s and 2p orbitals of C-atom undergo sp³ hybridization.
iv. One electron from the 2s orbital of carbon atom is excited to the 2pz orbital. Then the four orbitals 2s, px, py and pz mix and recast to form four new sp³ hybrid orbitals having same shape and equal energy. They are maximum apart and have tetrahedral geometry with H-C-H bond angle of 109°28′. Each hybrid orbital contains one unpaired electron.
v. Each of these sp³ hybrid orbitals with one electron overlap axially with the 1s orbital of hydrogen atom to form one C-H sigma bond. Thus, in CH₄ molecule, there are four C-H bonds formed by the sp³-s overlap.
Diagram:

Question G.
In Ammonia molecule the bond angle is 107°18 and in water molecule it is 104°35′, although in both the central atoms are sp³ hybridized Explain.
Answer:
i. The ammonia molecule has sp³ hybridization. The expected bond angle is 109°28′. But the actual bond angle is 107°28′. It is due to the following reasons.

• One lone pair and three bond pairs are present in ammonia molecule.
• The strength of lone pair-bond pair repulsion is much higher than that of bond pair-bond pair repulsion.
• Due to these repulsions, there is a small decrease in bond angle (~2°) from 109°28′ to 107°18′.

ii. The water molecule has sp³ hybridization. The expected bond angle is 109°28′. But the actual bond angle is 104°35′. It is due to the following reasons.

• Two lone pairs and two bond pairs are present in water molecule.
• The decreasing order of the repulsion is Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair.
• Due to these repulsions, there is a small decrease in bond angle (~5°) from 109°28′ to 104°35′.

Question H.
Give reasons for:
a. Sigma (σ) bond is stronger than Pi (π) bond.
b. HF is a polar molecule
c. Carbon is a tetravalent in nature.
Answer:
a. i. The strength of the bond depends on the extent of overlap of the orbitals. Greater the overlap, stronger is the bond.
ii. A sigma bond is formed by the coaxial overlap of the atomic orbitals which are oriented along the internuclear axis, hence the extent of overlap is maximum.
iii. A pi bond is formed by the lateral overlap of the atomic orbitals which are oriented perpendicular to the internuclear axis, hence the extent of orbital overlapping in side wise manner is less.
Hence, sigma bond is stronger than pi bond.

b. i. When a covalent bond is formed between two atoms of different elements that have different electronegativities, the shared electron pair does not remain at the centre. The electron pair is pulled towards the more electronegative atom resulting in the separation of charges.
ii. In H-F, fluorine is more electronegative than hydrogen. Therefore, the shared electron pair is pulled towards fluorine and fluorine acquires partial -ve charge and simultaneously hydrogen acquires partial +ve charge. This gives rise to dipole and H-F bond becomes polar. Hence, H-F is a polar molecule.

c. The electronic configuration of carbon is:
1s² 2s² 2p_x¹ 2p_y¹
One electron from ‘2s’ orbital is promoted to the empty ‘2p’ orbital.
Thus, in excited state, carbon has four half-filled orbitals.

Hence, carbon can form 4 bonds and is tetravalent in nature.

Question I.
Which type of hybridization is present in ammonia molecule? Write the geometry and bond angle present in ammonia.
Answer:
The type of hybridization present in ammonia (NH₃) molecule is sp³.
Geometry of ammonia molecule is pyramidal or distorted tetrahedral.
Bond angle in ammonia molecule is 107°18′.

Question J.
Identify the type of orbital overlap present in
a. H₂
b. F₂
c. H-F molecule.
Explain diagramatically.
Answer:
i. s-s σ overlap:
a. The overlap between two half-filled s orbitals of two different atoms containing unpaired electrons with opposite spins is called s-s overlap.
e.g. Formation of H₂ molecule by s-s overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹. The 1s¹ orbitais of two hydrogen atoms overlap along the internuclear axis to form a σ bond between the atoms in H₂ molecule.
b. Diagram:

ii. p-p σ overlap:
a. This type of overlap takes place when two p orbitals from different atoms overlap along the internuclear axis.
e.g. Formation of F₂ molecule by p-p overlap:
Fluorine atom (Z = 9) has electronic configuration 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{2}\) \(2 \mathrm{p}_{\mathrm{z}}^{2}\).
During the formation of F₂ molecule, half-filled 2p_z orbital of one F atom overlaps with similar half-filled 2p_z orbital containing electron with opposite spin of another F atom axially and a p-p σ bond is formed.
b. Diagram:

iii. s-p σ overlap:
a. In this type of overlap one half filled s orbital of one atom and one half filled p orbital of another orbital overlap along the internuclear axis.
e.g. Formation of HF molecule by s-p overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹ and fluorine atom (Z = 9) has electronic configuration 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{2}\) \(2 \mathrm{p}_{\mathrm{z}}^{2}\). During the formation of HF molecule, half-filled Is orbital of hydrogen atom overlaps coaxially with half-filled 2p_z orbital of fluorine atom with opposite electron spin and an s-p σ bond is formed.
b. Diagram:

Question K.
F-Be-F is a liner molecule but H-O-H is angular. Explain.
Answer:
i. In the BeF₂ molecule, the central beryllium atom undergoes sp hybridization giving rise to two sp hybridized orbitals placed diagonally opposite with an angle of 180°. Thus, F-Be-F is a linear molecule.

ii. In the H₂O molecule, the central oxygen atom undergoes sp³ hybridization giving rise to four sp³ hybridized orbitals directed towards four comers of a tetrahedron. There are two lone pairs of electrons in two of the sp³ hybrid orbitals of oxygen. The lone pair-lone pair repulsion distorts the structure. Hence, H-O-H is angular or V-shaped.

Question L.
BF₃ molecule is planar but NH₃ pyramidal. Explain.
Answer:
i. In the BF₃ molecule, the central boron atom undergoes sp² hybridization giving rise to three sp² hybridized orbitals directed towards three comers of an equilateral triangle. Thus, the geometry is trigonal planar.

ii. In the NH₃ molecule, the central nitrogen atom undergoes sp³ hybridization giving rise to four sp³ hybridized orbitals directed towards four comers of a tetrahedron. The expected geometry of NH₃ molecule is regular tetrahedral with bond angle 109°28′. There is one lone pair of electrons in one of the sp³ hybrid orbitals of nitrogen. The lone pair-bond pair repulsion distorts the bond angle. Hence, the structure of NH₃ is distorted and it has pyramidal geometry.

Question M.
In case of bond formation in Acetylene molecule :
a. How many covalend bonds are formed ?
b. State number of sigma and pi bonds formed.
c. Name the type of Hybridisation.
Answer:
a. In acetylene molecule, there are five covalent bonds.
b. In acetylene molecule, there are three sigma bonds and two pi bonds.
c. In acetylene molecule, each carbon atom undergoes sp hybridization.

Question N.
Define :
a. Bond Enthalpy
b. Bond Length
Answer:
a. Bond Enthalpy:
Bond enthalpy is defined as the amount of energy required to break one mole of a bond of one type, present between two atoms in a gaseous state.

b. Bond Length:
Bond length is defined as the equilibrium distance between the nuclei of two covalently bonded atoms in a molecule.

Question O.
Predict the shape and bond angles in the following molecules:
a. CF₄
b. NF₃
c. HCN
d. H₂S
Answer:
a. CF₄: There are four bond pairs on the central atom. Hence, shape of CF₄ is tetrahedral and F-C-F bond angle is 109° 28′.
b. NF₃: There are three bond pairs and one lone pair on the central atom. Hence, shape of NF₃ is trigonal pyramidal and F-N-F bond angle is less than 109° 28′.
c. HCN: There are two bond pairs on the central atom. Hence, shape of HCN is linear and H-C-N bond angle is 180°.
d. H₂S: There are two bond pairs and two lone pairs on the central atom. Hence, shape of H₂S is bent or V-shaped and H-S-H bond angle is slightly less than 109° 28′.

4. Using data from the Table, answer the following :

a. What happens to the bond length when unsaturation increases?
b. Which is the most stable compound?
c. Indicate the relation between bond strength and Bond enthalpy.
d. Comment on overall relation between Bond length, Bond Enthalpy and Bond strength and stability.
Answer:
a. When unsaturation increases, the bond length decreases.
b. The stable compound is ethyne (C₂H₂).
c. Bond strength ∝ Bond enthalpy
Larger the bond enthalpy, stronger is the bond.
d. As bond length decreases, bond enthalpy, bond strength and stability increase.

5. Complete the flow chart

Answer:

6. Complete the following Table

Answer:

7. Answer in one sentence:

Question A.
Indicate the factor on which stalility of ionic compound is measured?
Answer:
The stability of an ionic compound is measured by the amount of energy released during lattice formation.

Question B.
Arrange the following compounds on the basis of lattice energies in decreasing (descending) order: BeF₂, AlCl₃, LiCl, CaCl₂, NaCl.
Answer:
AlCl₃ > BeF₂ > CaCl₂ > LiCl > NaCl

Question C.
Give the total number of electrons around sulphur (S) in SF₆ compound.
Answer:
The total number of electrons around sulphur (S) in SF₆ is 12.

Question D.
Covalant bond is directional in nature. Justify.
Answer:
Covalent bond is formed by the overlap of two half-filled atomic orbitals. The atomic orbitals are oriented in specific directions in space (except s-orbital which is spherical). Hence, covalent bond is directional in nature.

Question E.
What are the interacting forces present during formation of a molecule of a compound ?
Answer:
a. Forces of attraction: The nucleus of one atom attracts the electrons of the other atom and vice-versa.
b. Forces of repulsion: The electron of one atom repels the electron of the other atom and vice-versa (as electrons are negatively charged). There is repulsion between the two nuclei (as the nuclei are positively charged).

Question F.
Give the type of overlap by which pi (π) bond is formed.
Answer:
The type of overlap by which pi (π) bond is formed is p-p lateral overlap.

Question G .
Mention the steps involved in Hybridization.
Answer:
The steps involved in hybridization are:

• formation of the excited state and
• mixing and recasting of orbitals.

Question H.
Write the formula to calculate bond order of molecule.
Answer:
Bond order of a molecule = \(\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}\)
where, N_b is the number of electrons present in bonding MOs and N_a is the number of electrons present in antibonding MOs.

Question I.
Why is O₂ molecule paramagnetic?
Answer:
The electronic configuration of O₂ molecule is (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (σ2p_z)² (π2p_x)² (π2p_y)² (π*2p_x)¹ (π*2p_y)¹
Since the oxygen molecule contains two unpaired electrons, it is paramagnetic.

Question J.
What do you mean by formal charge ? Explain its significance with the help of suitable example.
Answer:
Formal charge is the charge assigned to an atom in a molecule, assuming that all electrons are shared equally between atoms, regardless of their relative electronegativities.

Structure (I):

Structure (II):

Structure (III):

While determining the best Lewis structure per molecule, the structure is chosen such that the formal charge is as close to zero as possible. The structure having the lowest formal charge has the lowest energy.

In structure (I), the formal charge on each atom is 0 while in structures (II) and (III) formal charge on carbon is 0 while oxygens have formal charge -1 or +1. Hence, the possible structure with the lowest energy will be structure (I). Thus, formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species.

11th Chemistry Digest Chapter 5 Chemical Bonding Intext Questions and Answers

(Textbook Page No. 55)

Question 1.
Why are atoms held together in chemical compounds?
Answer:
Atoms are held together in chemical compounds due to chemical bonds.

Question 2.
How are chemical bonds formed between two atoms?
Answer:
There are two ways of formation of chemical bonds:

1. by loss and gain of electrons
2. by sharing a pair of electrons between the two atoms.

In either process of formation of chemical bond, each atom attains a stable noble gas electronic configuration.

Question 3.
Which electrons are involved in the formation of chemical bonds?
Answer:
The electrons present in the outermost shell of an atom are involved in the formation of a chemical bond.

Internet my friend (Textbook Page No. 55)

Question 1.
Search more atoms, which complete their octet during chemical combinations.
Answer:
In compounds like KCl, MgCl₂, CaO, NaF, etc, the constituent atoms complete their octet by lose or gain of electrons.
e.g. K → K⁺ + e^–
Cl + e^– → Cl^–
K⁺ + Cl^– → KCl
[Note: Students are expected to search more atoms on their own.]

Use your brainpower. (Textbook Page No. 60)

Question 1.
Which atom in \(\mathrm{NH}_{4}^{+}\) will have formal charge +1?
Answer:
In \(\mathrm{NH}_{4}^{+}\), nitrogen atom (N) will have formal charge of+1.

Use your brainpower. (Textbook Page No. 61)

Question 1.
How many electrons will be around I in the compound IF₇?
Answer:
Lewis structure of IF₇ is:

In IF₇, iodine (I) atom will be surrounded by 14 electrons.

Question 2.
Why is H₂ stable even though it never satisfies the octet rule?
Answer:
The valence shell configuration of hydrogen atom is 1s¹. Two hydrogen atoms approach each other and share their valence electrons. By having two electrons in its valence shell, H atom attains the nearest noble gas configuration of He. H₂ molecule attains stability due to duplet formation. Hence, H₂ is stable even though it never satisfies the octet rule.

(Textbook Page No. 64)

Question 1.
Lowering of energy takes during bond formation. How does this happen?
Answer:
i. When two combining atoms approach each other to form a covalent bond, the following interacting forces come into play.

• Forces of attraction: The nucleus of one atom attracts the electrons of the other atom and vice-versa.
• Forces of repulsion: The electron of one atom repels the electron of the other atom and vice-versa (as electrons are negatively charged). There is repulsion between the two nuclei (as the nuclei are positively charged).

ii. The balance between attractive and repulsive forces decide whether the bond will be formed or not.
iii. When the magnitude of attractive forces is more than the magnitude of repulsive forces, the energy of the system decreases and a covalent bond is formed.
iv. When the magnitude of repulsive forces becomes more than that of attraction, the total energy of the system increases, and a covalent bond is not formed.
Hence, lowering of energy takes during bond formation.

Can you tell? (TextBook Page No. 76)

Question 1.
Which molecules are polar?
H-I, H-O-H, H-Br, Br₂, N₂, I₂, NH₃
Answer:
i. H-I: Polar
ii. H-O-H: Polar
iii. H-Br: Polar
iv. Br₂: Nonpolar
v. N₂: Nonpolar
vi. I₂: Nonpolar
vii. NH₃: Polar

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 4 Structure of Atom Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 4 Structure of Atom

1. Choose the correct option.

Question A.
The energy difference between the shells goes on ……….. when moved away from the nucleus.
a. Increasing
b. decreasing
c. equalizing
d. static
Answer:
b. decreasing

Question B.
The value of Plank’s constant is
a. 6.626× 10^-34 Js
b. 6.023× 10^-24 Js
c. 1.667 × 10^-28 Js
d. 6.626× 10^-28 Js
Answer:
a. 6.626× 10^-34 Js

Question C.
p-orbitals are ……. in shape.
a. spherical
b. dumbbell
c. double dumbbell
d. diagonal
Answer:
b. dumbbell

Question D.
“No two electrons in the same atoms can have an identical set of four quantum numbers”. This statement is known as
a. Pauli’s exclusion principle
b. Hund’s rule
c. Aufbau rule
d. Heisenberg uncertainty principle
Answer:
a. Pauli’s exclusion principle

Question E.
Principal Quantum number describes
a. shape of orbital
b. size of the orbital
c. spin of electron
d. orientation of in the orbital electron cloud
Answer:
b. size of the orbital

2. Make the pairs:

	A		B
a.	Neutrons	i.	six electrons
b.	p-orbital	ii.	-1.6 × 10-19 C
c.	charge on electron	iii.	Ultraviolet region
d.	Lyman series	iv.	Chadwick

Answer:
a – iv,
b – i,
c – ii,
d – iii

3. Complete the following information about the isotopes in the chart given below :

(Hint: Refer to Periodic Table if required)
Answer:

4. Match the following :

Answer:
a – iv,
b – iii,
c – ii,
d – i

5. Answer in one sentence :

Question A.
If an element ‘X’ has mass number 11 and it has 6 neutrons, then write its representation.
Answer:
The representation of the given element is \({ }_{5}^{11} \mathrm{X}\).

Question B.
Name the element that shows simplest emission spectrum.
Answer:
The element that shows simplest emission spectrum is hydrogen.

Question C.
State Heisenberg uncertainty principle.
Answer:
Heisenberg uncertainty principle states that “It is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron”.

Question D.
Give the names of quantum numbers.
Answer:
The four quantum numbers are: principal quantum number (n), azimuthal or subsidiary quantum number (l), magnetic quantum number (m_l) and electron spin quantum number (m_s).

Question E.
Identify from the following the isoelectronic species:
Ne, O^2-, Na⁺ OR Ar, Cl^2-, K⁺
Answer:
Atoms and ions having the same number of electrons are isoelectronic.

Species	No. of electrons
Ne	10
O2-	8 + 2 = 10
Na+	11 – 1 = 10
Ar	18
Cl2-	17 + 2 = 19
K+	19 – 1 = 18

Hence, Ne, O^2-, Na⁺ are isoelectronic species.

6. Answer the following questions.

Question A.
Differentiate between Isotopes and Isobars.
Answer:

No.	Isotopes	Isobars
i.	Isotopes are atoms of same element.	Isobars are atoms of different elements.
ii.	They have same atomic number but different atomic mass number.	They have same atomic mass number but different atomic numbers.
iii.	They have same number of protons but different number of neutrons.	They have different number of protons and neutrons.
iv.	They have same number of electrons.	They have different number of electrons.
V.	They occupy same position in the modem periodic table.	They occupy different positions in the modem periodic table.
vi.	They have similar chemical properties.	They have different chemical properties.
e.g.	\({ }_{6}^{12} \mathrm{C}\) and \({ }_{6}^{14} \mathrm{C}\)	\({ }_{6}^{14} \mathrm{C}\) and \({ }_{7}^{14} \mathrm{~N}\)

Question B.
Define the terms:
i. Isotones
ii. Isoelectronic species
iii. Electronic configuration
Answer:
i. Isotones: Isotones are defined as the atoms of different elements having same number of neutrons in their nuclei. e.g. \({ }_{5}^{11} \mathrm{B}\) and \({ }_{6}^{12} \mathrm{C}\) having 6 neutrons each are isotones.

ii. Isoelectronic species:
soelectronic species are defined as atoms and ions having the same number of electrons.
e. g. Ar, Ca²⁺ and K⁺ containing 18 electrons each.

iii. Electronic configuration:
Electronic configuration of an atom is defined as the distribution of its electrons in orbitals.

Question C.
State and explain Pauli’s exclusion principle.
Answer:
Pauli’s exclusion principle:
i. Statement: “No two electrons in an atom can have the same set of four quantum numbers”. OR “Only two electrons can occupy the same orbital and they must have opposite spins. ”
ii. The capacity of an orbital to accommodate electrons is decided by Pauli’s exclusion principle.
iii. According to this principle, for an electron belonging to the same orbital, the spin quantum number must be different since the other three quantum numbers are the same.
iv. The spin quantum number can have two values: +\(\frac {1}{2}\) and –\(\frac {1}{2}\).
v. Example, consider helium (He) atom with electronic configuration 1 s².
For the two electrons in Is orbital, the four quantum numbers are as follows:
Electron number Quantum number Set of values of quantum numbers

Thus, in an atom, any two electrons can have the same three quantum numbers, but the fourth quantum number must be different.
vi. This leads to the conclusion that an orbital can accommodate maximum of two electrons and if it has two electrons, they must have opposite spin.

Question D.
State Hund’s rule of maximum multiplicity with suitable example.
Answer:
Hund’s rule of maximum multiplicity:
i. Statement: “Pairing of electrons in the orbitals belonging to the same subshell does not occur unless each orbital belonging to that subshell has got one electron each.”
ii. Example, according to Hund’s rule, each of the three-degenerate p-orbitals must get one electron of parallel spin before any one of them receives the second electron of opposite spin. Therefore, the configuration of four electrons occupying p-orbitals is represented as

iii. As a result of Hund’s rule, the atom with fully filled and half-filled set of degenerate orbitals has extra stability.

Question E.
Write the drawbacks of Rutherford’s model of an atom.
Answer:
Drawbacks of Rutherford’s model of an atom:
i. Rutherford’s model of an atom resembles the solar system with the nucleus playing the role of the massive sun and the electrons are lighter planets. Thus, according to this model, electrons having negative charge revolve in various orbits around the nucleus. However, the electrons revolving about the nucleus in fixed orbits pose a problem. Such orbital motion is an accelerated motion accompanied by a continuous change in the velocity of electron as noticed from the continuously changing direction. According to Maxwell’s theory of electromagnetic radiation, accelerated charged particles would emit electromagnetic radiation. Hence, an electron revolving around the nucleus should continuously emit radiation and lose equivalent energy. As a result, the orbit would shrink continuously and the electron would come closer to the nucleus by following a spiral path. It would ultimately fall into the nucleus. Thus, Rutherford’s model has an intrinsic instability of atom. However, real atoms are stable.

ii. Rutherford’s model of an atom does not describe the distribution of electrons around the nucleus and their energies.

Question F.
Write postulates of Bohr’s Theory of hydrogen atom.
Answer:
Postulates of Bohr’s theory of hydrogen atom:
i. The electron in the hydrogen atom can move around the nucleus in one of the many possible circular paths of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states. These orbits are arranged concentrically around the nucleus in an increasing order of energy.

ii. The energy of an electron in the orbit does not change with time. However, the electron will move from a lower stationary state to a higher stationary state if and when the required amount of energy is absorbed by the electron. Energy is emitted when electron moves from a higher stationary state to a lower stationary state. The energy change does not take place in a continuous manner.

iii. The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ in energy by ΔE is given by the following expression:
ν = \(\frac{\Delta E}{h}=\frac{E_{2}-E_{1}}{h}\) ………….(1)
Where E₁ and E₂ are the energies of the lower and higher allowed energy states respectively. This expression is commonly known as Bohr’s frequency rule.

iv. The angular momeñtum of an electron in a given stationary state can be expressed as mvr = n × h/2π
where, n 1,2, 3
Thus, an electron can move only in those orbits for which its angular momentum is integral multiple of h/2π.
Thus, only certain fixed orbits are allowed.

Question G.
Mention demerits of Bohr’s Atomic model.
Answer:
Demerits of Bohr’s atomic model:

• Bohr’s atomic model (theory) failed to account for finer details of the atomic spectrum of hydrogen as observed in sophisticated spectroscopic experiments.
• Bohr’s atomic model (theory) was unable to explain the spectrum of atoms other than hydrogen.
• Bohr’s atomic model (theory) could not explain the splitting of spectral lines in the presence of a magnetic field (Zeeman effect) or electric field (Stark effect).
• Bohr’s atomic model (theory) failed to explain the ability of atoms to form molecules by chemical bonds.

Question H.
State the order of filling atomic orbitals following Aufbau principle.
Answer:
Aufbau principle:
i. Aufbau principle gives the sequence in which various orbitals are filled with electrons.
ii. In the ground state of an atom, the orbitals are filled with electrons based on increasing order of energies of orbitals, Pauli’s exclusion principle and Hund’s rule of maximum multiplicity.
iii. Increasing order of energies of orbitals:

• Orbitals are filled in order of increasing value of (n + l)
• In cases where the two orbitals have same value of (n + l), the orbital with lower value of n is filled first.

iv. The increasing order of energy of different orbitals in a multi-electron atom is:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s and so on.

Question I.
Explain the anomalous behavior of copper and chromium.
Answer:
i. Copper:

• Copper (Cu) has atomic number 29.
• Its expected electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁹.
• The 3d orbital is neither half-filled nor fully filled. Hence, it has less stability.
• Due to interelectronic repulsion forces, one 4s electron enters into 3d orbital. This makes 3d orbital completely filled and 4s orbital half-filled which gives extra stability and the electronic configuration of Cu becomes, 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰.

ii. Chromium:

• Chromium (Cr) has atomic number 24.
• Its expected electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²3d⁴.
• The 3d orbital is less stable as it is not half-filled.
• Due to inter electronic repulsion forces, one 4s electron enters into 3d orbital. This makes 4s and 3d orbitals half-filled which gives extra stability and the electronic configuration of Cr becomes, 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵.

Question J.
Write orbital notations for electrons in orbitals with the following quantum numbers.
a. n = 2, l =1
b. n = 4, l = 2
c. n = 3, l = 2
Answer:
i. 2p
ii. 4d
iii. 3d

Question K.
Write electronic configurations of Fe, Fe²⁺, Fe³⁺
Answer:

Species	Orbital notation
Fe	1s2 2s2 2p63s2 3p6 4s2 3d6 OR [Ar] 4s2 3d6
Fe2+	Is2 2s2 2p6 3s2 3p6 3d6 OR [Ar] 3d6
Fe3+	1s2 2s2 2p6 3s2 3p6 3d5 OR [Ar] 3d5

Question L.
Write condensed orbital notation of electonic configuration of the following elements:
a. Lithium (Z = 3)
b. Carbon (Z=6)
c. Oxygen (Z = 8)
d. Silicon (Z = 14)
e. Chlorine (Z = 17)
f. Calcium (Z = 20)
Answer:

No.	Element	Condensed orbital notation
i.	Lithium (Z = 3)	[He] 2s1
ii.	Carbon (Z = 6)	[He] 2s2 2p2
iii.	Oxygen (Z = 8)	[He] 2s2 2p4
iv.	Silicon (Z = 14)	[Ne] 3s2 3p2
v.	Chlorine (Z = 17)	[Ne] 3s2 3p5
vi.	Calcium (Z = 20)	[Ar] 4s2

Question M.
Draw shapes of 2s and 2p orbitals.
Answer:
2s orbital:

2p orbital:

Question N.
Explain in brief, the significance of azimuthal quantum number.
Answer:
Azimuthal quantum number (l):

• Azimuthal quantum number is also known as subsidiary quantum number and is represented by letter l.
• It represents the subshell to which the electron belongs. It also defines the shape of the orbital that is occupied by the electron.
• Its value depends upon the value of principal quantum number ‘n’. It can have only positive values between 0 and (n – 1).
• Atomic orbitals with the same value of ‘n’ but different values of ‘l’ constitute a subshell belonging to the shell for the given ‘n’ The azimuthal quantum number gives the number of subshells in a principal shell. The subshells have l to be 0, 1, 2,3 … which are represented by symbols s, p, d, f, … respectively.

Question O.
If n = 3, what are the quantum number l and m_l?
Answer:
: For a given n, l = 0 to (n – 1) and for given l, m_l = -l……, 0…….. + l
Therefore, the possible values of l and m_l for n = 3 are:

Question P.
The electronic configuration of oxygen is written as 1s² 2s² 2p_x² 2p_y¹ 2p_z¹ and not as 1s² 2s² 2p_x² 2p_y² 2p_z⁰. Explain.
Answer:

• According to Hund’s rule of maximum multiplicity “Pairing of electrons in the orbitals belonging to the same subshell does not occur unless each orbital belonging to that subshell has got one electron each.”
• Oxygen has 8 electrons. The first two electrons will pair up in the Is orbital, the next two electrons will pair up in the 2s orbital and this leaves 4 electrons, which must be placed in the 2p orbitals.
• Each of the three degenerate p-orbitals must get one electron of parallel spin before any one of them receives the second electron of opposite spin. Therefore, two p orbitals have one electron each and one p-orbital will have two electrons.

Thus, the electronic configuration of oxygen is written as 1s² 2s² 2p_x² 2p_y¹ 2p_z¹ and not as 1s² 2s² 2p_x² 2p_y² 2p_z⁰.

Question Q.
Write note on ‘Principal Quantum number.
Answer:
Principal quantum number (n):
i. Principal quantum number indicates the principal shell or main energy level to which the electron belongs.
ii. It is denoted by ‘n’ and is a positive integer with values 1, 2, 3, 4, 5, 6, ….
iii. A set of atomic orbitals with given value of ‘n’ constitutes a single shell. These shells are also represented by the letters K, L, M, N, etc.
iv. With increase of ‘n’, the number of allowed orbitals in that shell increases and is given by n².
v. The allowed orbitals in the first four shells are given below:

vi. As the value of ‘n’ increases, the distance of the shell from the nucleus increases and the size of the shell increases. Its energy also goes on increasing.

Question R.
Using concept of quantum numbers, calculate the maximum numbers of electrons present in the ‘M’ shell. Give their distribution in shells, subshells and orbitals.
Answer:
i. Each main shell contains a maximum of 2n² electrons.
For ‘M’ shell, n = 3.
Therefore, the maximum numbers of electrons present in the ‘M’ shell = 2 × (3)² = 18.

ii. The distribution of these electrons in shells, subshells and orbitals can be given as follows:

Note: Orbital distribution in the first four shells:

Question S.
Indicate the number of unpaired electrons in :
a. Si (Z = 14)
b. Cr (Z = 24)
Answer:
i. . Si (Z = 14): 1s² 2s² 2p⁶ 3s² 3p²
Orbital diagram:

Number of unpaired electrons = 2

ii. Cr (Z = 24): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵
Orbital diagram:

Number of unpaired electrons = 6

Question T.
An atom of an element contains 29 electrons and 35 neutrons. Deduce-
a. the number of protons
b. the electronic configuration of that element.
Answer:
a. In an atom, number of protons is equal to number of electrons.
The given atom contains 29 electrons.
∴ Number of protons = 29

b. The electronic configuration of an atom of an element containing 29 electrons is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰
[Note: Given element is copper (Cu) with Z = 29]

11th Chemistry Digest Chapter 4 Structure of Atom Intext Questions and Answers

Can you recall? (Textbook Page No. 35)

Question i.
What is the smallest unit of matter?
Answer:
The smallest unit of matter is atom.

Question ii.
What is the difference between molecules of an element and those of a compound?
Answer:
The molecules of an element are made of atoms of same element while the molecules of a compound are made of atoms of different elements.

Question iii.
Does an atom have any internal structure or is it indivisible?
Answer:
Yes, an atom has internal structure. Different subatomic particles such as protons, electrons and neutrons constitute an atom. So, it is divisible.

Question iv.
Which particle was identified by J. J. Thomson in the cathode ray tube experiment?
Answer:
Electron was identified by J.J. Thomson in the cathode ray tube experiment.

Question v.
Which part of an atom was discovered by Ernest Rutherford from the experiment of scattering of α-particles by gold foil?
Answer:
Nucleus of an atom was discovered by Ernest Rutherford from the experiment of scattering of α-particles by gold foil.

Just Think (Textbook Page No. 41)

Question 1.
What does the negative sign of electron energy convey?
Answer:
Negative sign for the energy of an electron in any orbit in a hydrogen atom indicates that the energy of the electron in the atom is lower than the energy of a free electron at rest. A free-electron at rest is an electron that is infinitely far away from the nucleus and is assigned the energy value of zero.

As the electron gets close to the nucleus, value of ‘n’ decreases and E_n becomes large in absolute value and more negative. The negative sign corresponds to attractive forces between electron and nucleus.

Internet my friend (Textbook Page No. 44)

Question 1.
Collect information about the structure of atom.
Answer:
Students can use links given below as references and collect information about structure of atom on their own.
https://www.livescience.com/65427-fundamental-elementary-particles.html http://www.chemistryexplained.com/Ar-Bo/Atomic-Structure.html
https://www.thoughtco.com/basic-model-of-the-atom-603799

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 3 Basic Analytical Techniques Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 3 Basic Analytical Techniques

1. Choose the correct option

Question A.
Which of the following methods can be used to separate two compounds with different solubilities in the same solvent?
a. Fractional crystallization
b. Crystallization
c. Distillation
d. Solvent extraction
Answer:
a. Fractional crystallization

Question B.
Which of the following techniques is used for the separation of glycerol from soap in the soap industry?
a. Distillation under reduced pressure
b. Fractional distillation
c. Filtration
d. Crystallization
Answer:
a. Distillation under reduced pressure

Question C.
Which technique is widely used in industry to separate components of the mixture and also to purify them?
a. Steam distillation
b. Chromatography
c. Solvent extraction
d. Filtration
Answer:
b. Chromatography

Question D.
A mixture of acetone and benzene can be separated by the following method :
a. Simple distillation
b. Fractional distillation
c. Distillation under reduced pressure
d. Sublimation
Answer:
b. Fractional distillation

Question E.
Colourless components on chromatogram can not be observed by the following :
a. Using UV light
b. Using iodine chamber
c. Using the spraying reagent
d. Using infrared light
Answer:
d. Using infrared light

2. Answer the following

Question A.
Which of the following techniques is used for purification of solid organic compounds?
a. Crystallisation
b. Distillation
Answer:
Solid (crude/impure) organic compounds can be purified by crystallization.

Question B.
What do you understand by the terms
a. residue
b. filtrate.
Answer:
a. Residue: In the process of filtration, the insoluble (undissolved) impurities which remain on the filter paper are called residue.

b. Filtrate: In the process of filtration, the liquid which pass through the filter paper and collected in the beaker is called filtrate.

Question C.
Why is a condenser used in distillation process?
Answer:
In the process of distillation, a liquid is converted into its vapour and the vapour is then condensed back to liquid on cooling. The condenser has a jacket with two outlets through which water is circulated. Hence, to provide efficient cooling, a condenser is used.

Question D.
Why is paper moistened before filtration?
Answer:
Before filtration, filter paper is moistened with appropriate solvent to ensure that it sticks to the funnel and does not let the air to pass through the leaks.

Question E.
What is the stationary phase in Paper Chromatography?
Answer:
Paper chromatography is a type of partition chromatography in which a special quality paper, namely Whatman paper 1 is used. The water trapped in the fibres of the paper acts as stationary phase.

Question F.
What will happen if the upper outlet of the condenser is connected to the tap instead of the lower outlet?
Answer:

• If water enters through upper outlet of condenser, the water will quickly flow down under the influence of gravity. This allows only a small section of the condenser to be cooled enough.
• If water enters through lower outlet of condenser, the entire condenser will be filled with water before it leaves out providing maximum cooling to the condenser. This results in maximum recovery of purified liquid.

Hence, water must be allowed to enter through lower outlet of condenser during distillation process.

Question G.
Give names of two materials used as stationary phase in chromatography.
Answer:

1. Alumina
2. Silica gel

Question H.
Which properties of solvents are useful for solvent extraction?
Answer:

• Organic compound must be more soluble in the organic solvent, than in water.
• Solvent should be immiscible with water and be able to form two distinct layers.

Question I.
Why should spotting of mixture be done above the level of mobile phase ?
Answer:

• If spotting of a mixture is done at the level of mobile phase, then solvent will come in contact with the sample spot.
• Sample spot will dissolve in the mobile phase and its components will move all over the plate resulting in no distinct separation.

Hence, spotting of mixture should be done above the level of mobile phase.

Question J.
Define : a. Stationary phase b. Saturated solution
Answer:
a. Stationary phase:
Stationary phase is a solid or a liquid supported on a solid which remains fixed in a place and on which different solutes are adsorbed to a different extent.

b. Saturated solution:
A saturated solution is a solution which cannot dissolve additional quantity of a solute.

Question K.
What is the difference between simple distillation and fractional distillation?
Answer:

No.	Simple distillation	Fractional distillation
i.	If in a mixture the difference in boiling points of two liquids is appreciable/large, they are separated from each other using the simple distillation.	If in a mixture the difference in boiling points of two liquids is not appreciable/large, they are separated from each other using the fractional distillation.
ii.	Simple distillation assembly is used.	fractionating column is fitted in distillation assembly.
e.g.	Mixture of acetone (b.p. 329 K) and water (b.p. 373 K) can be separated by this method.	Mixture of acetone (b.p. 329 K) and methanol (b.p. 337.7 K) can be separated by this method.

Question L.
Define a. Solvent extraction
b. Distillation.
Answer:
a. Solvent extraction:
Solvent extraction is a method used to separate an organic compound present in an aqueous solution, by shaking it with a suitable organic solvent in which the compound is more soluble than water.

b. Distillation:
The process in which liquid is converted into its vapour phase at its boiling point and the vapour is then condensed back to liquid on cooling is known as distillation.

Question M.
List the properties of solvents which make them suitable for crystallization.
Answer:
The solvent to be used for crystallization should have following properties:

• The compound to be crystallized should be least or sparingly soluble in the solvent at room temperature but highly soluble at high temperature.
• Solvent should not react chemically with the compound to be purified.
• Solvent should be volatile so that it can be removed easily.

Question N.
Name the different types of Chromatography and explain the principles underlying them.
Answer:
Depending on the nature of the stationary phase i.e., whether it is a solid or a liquid, chromatography is classified into adsorption chromatography and partition chromatography.
i. Adsorption chromatography: This technique is based on the principle of differential adsorption. Different solutes are adsorbed on an adsorbent to different extent.

Adsorption chromatography is further classified into two types:

1. Column chromatography
2. Thin-layer chromatography

ii. Partition chromatography: This technique is based on continuous differential partitioning of components of a mixture between stationary and mobile phases. For example, paper chromatography

Question O.
Why do we see bands separating in column chromatography?
Answer:

• In column chromatography, the solutes get adsorbed on the stationary phase and depending on the degree to which they are adsorbed, they get separated from each other.
• The component which is readily adsorbed are retained on the column and others move down the column to various distances forming distinct bands.

Hence, we see bands separating in column chromatography.

Question P.
How do you visualize colourless compounds after separation in TLC and Paper Chromatography?
Answer:
i. Thin-layer chromatography (TLC): If components are colourless but have the property of fluorescence then they can be visualized under UV light, or the plate can be kept in a chamber containing a few iodine crystals. The iodine vapours are adsorbed by the components and the spots appear brown. Also, spraying agent like ninhydrin can also be used (for amino acids).

ii. Paper Chromatography: The spots of the separated colourless components may be observed either under ultra-violet light or by the use of an appropriate spraying agent.

Question Q.
Compare TLC and Paper Chromatography techniques.
Answer:

Chromatography technique	TLC	Paper chromatography
Principle	It is based on the principle of differential adsorption. Different solutes are adsorbed on an adsorbent to different extent.	It is based on continuous differential partitioning of components of a mixture between stationary and mobile phases.
Stationary phase	Solid (adsorbent like silica gel or alumina over a glass plate)	Liquid (water trapped in the fibres of a Paper)
Mobile phase	Liquid (single solvent/mixture of solvents)	Liquid (single solvent/mixture of solvents)
Visualization of components of a mixture	Similar to TLC the coloured components are visible as coloured spots and the colourless components are observed under UV light or using a spraying agent.

3. Label the diagram and explain the process in your words.

Answer:
When filtration is carried out using a vacuum pump it is called filtration under suction. It is a faster and more efficient technique than simple filtration. The diagram is as follows:

ii. Procedure:

• The assembly for filtration under suction consists of a thick wall conical flask with a sidearm (Buchner flask).
• The flask is connected to a safety bottle by rubber tube through the side arm.
• Buchner funnel (a special porcelain funnel with a porous circular bottom) is fitted on the conical flask with the help of a rubber cork.
• A circular filter paper of correct size is placed on the circular porous bottom of the Buchner funnel and the funnel is placed on the flask.
• Filter paper is moistened with a few drops of water or solvent.
• Suction is created by starting the pump and filtration is carried out.

iii. Crystals are collected on the filter paper and filtrate in the flask.

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 2 Introduction to Analytical Chemistry Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 2 Introduction to Analytical Chemistry

1. Choose the correct option

Question A.
The branch of chemistry which deals with the study of separation, identification, and quantitative determination of the composition of different substances is called………………..
a. Physical chemistry
b. Inorganic chemistry
c. Organic chemistry
d. Analytical chemistry
Answer:
d. Analytical chemistry

Question B.
Which one of the following property of matter is Not quantitative in nature ?
a. Mass
b. Length
c. Colour
d. Volume
Answer:
c. Colour

Question C.
SI unit of mass is ……..
a. kg
b. mol
c. pound
d. m³
Answer:
a. kg

Question D.
The number of significant figures in 1.50 × 10⁴ g is ………..
a. 2
b. 3
c. 4
d. 6
Answer:
b. 3

Question E.
In Avogadro’s constant 6.022 × 10²³ mol^-1, the number of significant figures is ……….
a. 3
b. 4
c. 5
d. 6
Answer:
b. 4

Question F.
By decomposition of 25 g of CaCO₃, the amount of CaO produced will be ……………….
a. 2.8 g
b. 8.4 g
c. 14.0 g
d. 28.0 g
Answer:
c. 14.0 g

Question G.
How many grams of water will be produced by complete combustion of 12g of methane gas
a. 16
b. 27
c. 36
d. 56
Answer:
b. 27

Question H.
Two elements A (At. mass 75) and B (At. mass 16) combine to give a compound having 75.8 % of A. The formula of the compound is
a. AB
b. A₂B
c. AB₂
d. A₂B₃
Answer:
d. A₂B₃

Question I.
The hydrocarbon contains 79.87 % carbon and 20.13 % of hydrogen. What is its empirical formula ?
a. CH
b. CH₂
c. CH₃
d. C₂H₅
Answer:
c. CH₃

Question J.
How many grams of oxygen will be required to react completely with 27 g of Al? (Atomic mass : Al = 27, O = 16)
a. 8
b. 16
c. 24
d. 32
Answer:
c. 24

Question K.
In CuSO₄.5H₂O the percentage of water is ……
(Cu = 63.5, S = 32, O = 16, H = 1)
a. 10 %
b. 36 %
c. 60 %
d. 72 %
Answer:
b. 36 %

Question L.
When two properties of a system are mathematically related to each other, the relation can be deduced by
a. Working out mean deviation
b. Plotting a graph
c. Calculating relative error
d. all the above three
Answer:
b. Plotting a graph

2. Answer the following questions

Question A.
Define : Least count
Answer:
The smallest quantity that can be measured by the measuring equipment is called least count.

Question B.
What do you mean by significant figures? State the rules for deciding significant figures.
Answer:
i. The significant figures in a measurement or result are the number of digits known with certainty plus one uncertain digit.
ii. Rules for deciding significant figures:
a. All non-zero digits are significant.
e.g. 127.34 g contains five significant figures which are 1, 2, 7, 3 and 4.
b. All zeros between two non-zero digits are significant, e.g. 120.007 m contains six significant figures.
c. Zeros on the left of the first non-zero digit are not significant. Such a zero indicates the position of the decimal point.
e.g. 0.025 has two significant figures, 0.005 has one significant figure.
d. Zeros at the end of a number are significant if they are on the right side of the decimal point,
e. g. 0.400 g has three significant figures and 400 g has one significant figure.
e. In numbers written is scientific notation, all digits are significant.
e.g. 2.035 × 10² has four significant figures and 3.25 × 10^-5 has three significant figures.

Question C.
Distinguish between accuracy and precision.
Answer:
Accuracy:

1. Accuracy refers to nearness of the measured value to the true value.
2. Accuracy represents the correctness of the measurement.
3. Accuracy is expressed in terms of absolute error and relative error.
4. Accuracy takes into account the true or accepted value.
5. Accuracy can be determined by a single measurement.
6. High accuracy implies smaller error.

Precision:

1. Precision refers to closeness of multiple readings of the same quantity.
2. Precision represents the agreement between two or more measured values.
3. Precision is expressed in terms of absolute deviation and relative deviation.
4. Precision does not take into account the true or accepted value.
5. Several measurements are required to determine precision.
6. High precision implies reproducibility of the readings.

Question D.
Explain the terms percentage composition, empirical formula and molecular formula.
Answer:
Percentage Composition:

• The percentage composition of a compound is the percentage by weight of each element present in the compound.
• Quantitative determination of the constituent elements by suitable methods provides the percent elemental composition of a compound.
• If the percent total is not 100, the difference is considered as percent oxygen.
• From the percentage composition, the ratio of the atoms of the constituent elements in the molecule is calculated.

Empirical Formula:
The simplest ratio of atoms of the constituent elements in a molecule is called the empirical formula of that compound.
e.g. The empirical formula of benzene is CH.

Molecular Formula:
1. Molecular formula of a compound is the formula which indicates the actual number of atoms of the constituent elements in a molecule.
e.g. The molecular formula of benzene is C6H6.
2. It can be obtained from the experimentally determined values of percent elemental composition and molar mass of that compound.
3. Molecular formula can be obtained from the empirical formula if the molar mass is known.
Molecular formula = r × Empirical formula

Question E.
What is a limiting reagent ? Explain.
Answer:
Limiting reagent:

• The reactant which gets consumed and limits the amount of product formed is called the limiting reagent.
• When a chemist carries out a reaction, the reactants are not usually present in exact stoichiometric amounts, that is, in the proportions indicated by the balanced equation.
• This is because the goal of a reaction is to produce the maximum quantity of a useful compound from the starting materials. Frequently, a large excess of one reactant is supplied to ensure that the more expensive reactant is completely converted into the desired product.
• The reactant which is present in lesser amount gets consumed after some time and subsequently, no further reaction takes place, whatever be the amount left of the other reactant present.

Hence, limiting reagent is the reactant that gets consumed entirely and limits the reaction.

Question F.
What do you mean by SI units ? What is the SI unit of mass ?
Answer:
i. In 1960, the general conference of weights and measures proposed revised metric system, called International system of Units i.e. SI units, abbreviated from its French name.
ii. The SI unit of mass is kilogram (kg).

Question G.
Explain the following terms
(a) Mole fraction
(b) Molarity
(c) Molality
Answer:
(a) Mole fraction: Mole fraction is the ratio of number of moles of a particular component of a solution to the total number of moles of the solution.

If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are n_A and n_B, respectively, then the mole fraction of A and B are given as:

(b) Molarity: Molarity is defined as the number of moles of the solute present in 1 litre of the solution. It is the most widely used unit and is denoted by M.
Molarity is expressed as follows:
Molarity (M) = \(\frac{\text { Number of moles of solute }}{\text { Volume of solution in litres }}\)

Molality: Molality is the number of moles of solute present in 1 kg of solvent. It is denoted by m. Molality is expressed as follows:
Molality (m) = \(\frac{\text { Number of moles of solute }}{\text { Mass of solvent in kilograms }}\)

Question H.
Define : Stoichiometry
Answer:
The study of quantitative relations between the amount of reactants and/or products is called stoichiometry.

Question I.
Why there is a need of rounding off figures during calculation ?
Answer:

• When performing calculations with measured quantities, the rule is that the accuracy of the final result is limited to the accuracy of the least accurate measurement.
• In other words, the final result cannot be more accurate than the least accurate number involved in the calculation.
• Sometimes, the final result of a calculation often contains figures that are not significant.
• When this occurs, the final result is rounded off.

Question J.
Why does molarity of a solution depend upon temperature ?
Answer:

• Molarity is the number of moles of the solute present in 1 litre of the solution. Therefore, molarity depends on the volume of the solution.
• Volume of the solution varies with the change in temperature.

Hence, molarity of a solution depends upon temperature.

Question M.
Define Analytical chemistry. Why is accurate measurement crucial in science?
Answer:
The branch of chemistry which deals with the study of separation, identification, qualitative and quantitative determination of the compositions of different substances, is called analytical chemistry.

1. The accuracy of measurement is of great concern in analytical chemistry. This is because faulty equipment, poor data processing, or human error can lead to inaccurate measurements. Also, there can be intrinsic errors in analytical measurement.
2. When measurements are not accurate, this provides incorrect data that can lead to wrong conclusions. For example, if a laboratory experiment requires a specific amount of a chemical, then measuring the wrong amount may result in an unsafe or unexpected outcome.
3. Hence, the numerical data obtained experimentally are treated mathematically to reach some quantitative conclusion.
4. Also, an analytical chemist has to know how to report the quantitative analytical data, indicating the extent of the accuracy of measurement, perform the mathematical operation, and properly express the quantitative error in the result.

3. Solve the following questions

Question A.
How many significant figures are in each of the following quantities ?
a. 45.26 ft
b. 0.109 in
c. 0.00025 kg
d. 2.3659 × 10^-8 cm
e. 52.0 cm³
f. 0.00020 kg
g. 8.50 × 10⁴ mm
h. 300.0 cg
Answer:
a. 4
b. 3
c. 2
d. 5
e. 3
f. 2
g. 3
h. 4

Question B.
Round off each of the following quantities to two significant figures :
a. 25.55 mL
b. 0.00254 m
c. 1.491 × 10⁵ mg
d. 199 g
Answer:
a. 26 mL
b. 0.0025 m
c. 1.5 × 10⁵ mg
d. 2.0 × 10² g

Question C.
Round off each of the following quantities to three significant figures :
a. 1.43 cm³
b. 458 × 10² cm
c. 643 cm²
d. 0.039 m
e. 6.398 × 10^-3 km
f. 0.0179 g
g. 79,000 m
h. 42,150
i. 649.85
j. 23,642,000 mm
k. 0.0041962 kg
Answer:
a. 43 cm³
b. 4.58 × 10⁴ cm
c. 643 cm² (or 6.43 × 10² cm²)
d. 0.0390 m (or 3.90 × 10^-2 m)
e. 6.40 × 10^-3 km
f. 0.0179 g (or 1.79 × 10^-2 m)
g. 7.90 × 10⁴ m
h. 4.22 × 10⁴ (or 42,200)
i. 6.50 × 10²
j. 2.36 × 10⁷ mm
k. 0.00420 kg (or 4.20 × 10^-3 kg)

Question D.
Express the following sum to appropriate number of significant figures :
a. 2.3 × 10³ mL + 4.22 × 10⁴ mL + 9.04 × 10³ mL + 8.71 × 10⁵ mL;
b. 319.5 g – 20460 g – 0.0639 g – 45.642 g – 4.173 g
Answer:
To perform addition/subtraction operation, first the numbers are written in such a way that they have the same exponent. The coefficients are then added/subtracted.
a. (0.23 × 10⁴ mL) + (4.22 × 10⁴ mL) +(0.904 × 10⁴ mL) + (87.1 × 10⁴ mL)
= (0.23 + 4.22 + 0.904 + 87.1) × 10⁴ mL
= 92.454 × 10⁴ mL
= 9.2454 × 10⁵
= 9.2 × 10⁵ mL
b. 319.5 g – 20460 g – 0.0639 g – 45.642 g – 4.173 g
= – 20190.3789 g
= – 20190 g
Ans: Sum to appropriate number of significant figures = 9.2 × 10⁵ mL
ii. Sum to appropriate number of significant figures = – 20190 g
[Note: In addition and subtraction, the final answer is rounded to the minimum number of decimal point of the number taking part in calculation. If there is no decimal point, then the final answer will have no decimal point.]

4. Solve the following problems

Question A.
Express the following quantities in exponential terms.
a. 0.0003498
b. 235.4678
c. 70000.0
d. 1569.00
Answer:
a. 0.0003498 = 3.498 × 10^-4
b. 235.4678 = 2.354678 × 10²
c. 70000.0 = 7.00000 × 10⁴
d. 1569.00 = 1.56900 × 10³

Question B.
Give the number of significant figures in each of the following
a. 1.230 × 10⁴
b. 0.002030
c. 1.23 × 10⁴
d. 1.89 × 10^-4
Answer:
a. 4
b. 4
c. 3
d. 3

Question C.
Express the quantities in above (B) with or without exponents as the case may be.
Answer:
a. 12300
b. 2.030 × 10^-3
c. 12300
d. 0.000189

Question D.
Find out the molar masses of the following compounds :
a. Copper sulphate crystal (CuSO₄.5H₂O)
b. Sodium carbonate, decahydrate (Na₂CO₃.10H₂O)
c. Mohr’s salt [FeSO₄(NH₄)₂SO₄.6H₂O]
(At. mass : Cu = 63.5; S = 32; O = 16; H = 1; Na = 23; C = 12; Fe = 56; N = 14)
Answer:
a. Molar mass of CuSO₄.5H₂O
= (1 × At. mass Cu) + (1 × At. mass S) + (9 × At. mass O) + (10 × At. mass H)
= (1 × 63.5) + (1 × 32) + (9 × 16) + (10 × 1)
= 63.5 + 32 + 144 + 10
= 249.5 g mol^-1
Molar mass of CuSO₄.5H₂O = 249.5 g mol^-1

b. Molar mass of Na₂CO₃.10H₂O
= (2 × At. mass Na) + (1 × At. mass C) + (13 × At. mass O) + (20 × At. mass H)
= (2 × 23) + (1 × 12) + (13 × 16) + (20 × 1)
= 46 + 12 + 208 + 20
= 286 g mol^-1
Molar mass of Na₂CO₃.10H₂O = 286 g mol^-1

c. Molar mass of [FeSO₄(NH₄)₂SO₄.6H₂O]
= (1 × At. mass Fe) + (2 × At. mass S) + (2 × At. mass N) + (14 × At. mass O) + (20 × At. mass H)
= (1 × 56) + (2 × 32) + (2 × 14) + (14 × 16) + (20 × 1)
= 56 + 64 + 28 + 224 + 20
= 392 g mol^-1
Molar mass of [FeSO₄(NH₄)₂SO₄.6H₂O] = 392 g mol^-1

Question E.
Work out the percentage composition of constituents elements in the following compounds :
a. Lead phosphate [Pb₃(PO₄)₂],
b. Potassium dichromate (K₂Cr₂O₇),
c. Macrocosmic salt – Sodium ammonium hydrogen phosphate, tetrahydrate (NaNH₄HPO₄.4H₂O)
(At. mass : Pb = 207; P = 31; O = 16; K = 39; Cr = 52; Na = 23; N = 14)
Answer:
Given: Atomic mass: Pb = 207; P = 31; O = 16; K = 39; Cr = 52; Na = 23; N = 14
To find: The percentage composition of constituent elements
Formula:

Calculation:
i. Lead phosphate [Pb₃(PO₄)₂]
Molar mass of Pb₃(PO₄)₂ = 3 × (207) + 2 × (31) + 8 × (16) = 621 + 62 + 128 = 811 g mol^-1
Percentage of Pb = \(\frac {621}{811}\) × 100 = 76.57%
Percentage of P = \(\frac {621}{811}\) × 100 = 7.64%
Percentage of O = \(\frac {128}{811}\) × 100 = 15.78%

ii. Potassium dichromate (K₂Cr₂O₇)
Molar mass of K₂Cr₂O₇ = 2 × (39) + 2 × (52) + 7 × (16) = 78 + 104 + 112 = 294 g mol^-1
Percentage of K = \(\frac {78}{294}\) × 100 = 26.53%
Percentage of Cr = \(\frac {104}{294}\) × 100 = 35.37%
Percentage of O = \(\frac {112}{294}\) × 100 = 38.10%

iii. Microcosmic salt – Sodium ammonium hydrogen phosphate, tetrahydrate (NaNH₄HPO₄.4H₂O)
Molar mass of NaNH₄HPO₄.4H₂O = 1 × (23) + 1 × (14) + 1 × (31) + 13 × (1) + 8 × (16)
= 23 + 14 + 31 + 13 + 128 = 209 g mol^-1
Percentage of Na = \(\frac {23}{209}\) × 100 = 11.00%
Percentage of N = \(\frac {14}{209}\) × 100 = 6.70%
Percentage of P = \(\frac {31}{209}\) × 100 = 14.83%
Percentage of H = \(\frac {13}{209}\) × 100 = 6.22%
Percentage of O = \(\frac {128}{209}\) × 100 = 61.24%
Ans: i. Mass percentage of Pb, P and O in lead phosphate [Pb₃(PO₄)₂] are 76.57%, 7.64% and 15.78% respectively.
ii. Mass percentage of K, Cr and O in potassium dichromate (K₂Cr₂O₇) are 26.53%, 35.37% and 38.10% respectively.
iii. Mass percentage of Na, N, P, H and O in NaNH₄HPO₄.4H₂O are 11.00%, 6.70%, 14.83%, 6.22% and 61.24% respectively.

Question F.
Find the percentage composition of constituent green vitriol crystals (FeSO₄.7H₂O). Also find out the mass of iron and the water of crystallisation in 4.54 kg of the crystals. (At. mass : Fe = 56; S = 32; O = 16)
Answer:
Given: i. Atomic mass: Fe = 56; S = 32; O = 16
ii. Mass of crystal = 4.54 kg
To find: i. Mass percentage of Fe, S, H and O
ii. Mass of iron and water of crystallisation in 4.54 kg of crystal
Formula:

i. Molar mass of FeSO₄.7H₂O = 1 × (56) + 1 × (32) + 14 × (1) + 11 × (16)
= 56 + 32 + 14+ 176
= 278 g mol^-1
Percentage of Fe = \(\frac {56}{278}\) × 100 = 20.14%
Percentage of S = \(\frac {32}{278}\) × 100 = 11.51%
Percentage of H = \(\frac {14}{278}\) × 100 = 5.04%
Percentage of O = \(\frac {176}{278}\) × 100 = 63.31%

ii. 278 kg green vitriol = 56 kg iron
∴ 4.54 kg green vitriol = x
∴ x = \(\frac{56 \times 4.54}{278}\)
Mass of 7H₂O in 278 kg green vitriol = 7 × 18 = 126 kg
∴ 4.54 kg green vitriol = y
∴ y = \(\frac{126 \times 4.54}{278}\)
Ans: i. Mass percentage of Fe, S, H and O in FeSO₄.7H₂O are 20.14%, 11.51%, 5.04% and 63.31% respectively.
ii. Mass of iron in 4.54 kg green vitriol = 0.915 kg
Mass of water of crystallisation in 4.54 kg green vitriol = 2.058 kg

Question G.
The red colour of blood is due to a compound called “haemoglobin”. It contains 0.335 % of iron. Four atoms of iron are present in one molecule of haemoglobin. What is its molecular weight ? (At. mass : Fe = 55.84)
Answer:
Given: Iron percentage in haemoglobin = 0.335%
To find: Molecular weight of haemoglobin
Calculation: There are four atoms of iron in a molecule of haemoglobin. Four atoms of iron contribute 0.335% mass to a molecule of haemoglobin.
Mass of one Fe atom = 55.84 u
∴ Mass of 4 Fe atoms = 55.84 × 4 = 223.36 u = 0.335%
Let molecular weight of haemoglobin be x.
Hence,
\(\frac{223.36}{x}\) × 100 = 0.335%
∴ x = \(\frac{223.36}{0.335}\) × 100 = 66674.6 g mol^-1
Ans: Molecular weight of haemoglobin = 66674.6 g mol^-1

Question H.
A substance, on analysis, gave the following percent composition:
Na = 43.4 %, C = 11.3 % and O = 45.3 %. Calculate the empirical formula. (At. mass Na = 23 u, C = 12 u, O = 16 u).
Answer:
Given: Atomic mass of Na = 23 u, C = 12 u, and O = 16 u
Percentage of Na, C and O = 43.4%, 11.3% and 45.3% respectively.
To find: The empirical formula of the compound
Calculation:

Hence, empirical formula is Na₂CO₃.
Ans: Empirical formula of the compound = Na₂CO₃

Question I.
Assuming the atomic weight of a metal M to be 56, find the empirical formula of its oxide containing 70.0% of M.
Answer:
Given: Atomic mass of M = 56
Percentage of M = 70.0%
To find: The empirical formula of the compound
Calculation: % M = 70.0%
Hence, % O = 30.0%, Atomic mass of O = 16 u

Convert the ratio into whole number by multiplying by the suitable coefficient, i.e., 2.
Therefore, the ratio of number of moles of M : O is 2 : 3.
Hence, the empirical formula is M₂O₃.
Ans: Empirical formula of the compound = M₂O₃

Question J.
1.00 g of a hydrated salt contains 0.2014 g of iron, 0.1153 g of sulfur, 0.2301 g of oxygen and 0.4532 g of water of crystallisation. Find the empirical formula. (At. wt. : Fe = 56; S = 32; O = 16)
Answer:
Given: Atomic mass of Fe = 56, S = 32, and O = 16
Mass of iron, sulphur, oxygen and water = 0.2014 g, 0.1153 g, 0.2301 g and 0.4532 respectively.
To find: The empirical formula of the compound
Calculation: Since the mass of crystal is 1 g, the % iron, sulphur, oxygen and water = 20.14%, 11.53%, 23.01% and 4.32 % respectively.

Hence, empirical formula is FeSO₄.7H₂O.
Ans: Empirical formula of the compound = FeSO₄.7H₂O.

Question K.
An organic compound containing oxygen, carbon, hydrogen and nitrogen contains 20 % carbon, 6.7 % hydrogen and 46.67 % nitrogen. Its molecular mass was found to be 60. Find the molecular formula of the compound.
Answer:
Given: Percentage of carbon, hydrogen, nitrogen = 20%, 6.7%, 46.67% respectively.
Molar mass of the compound = 60 g mol^-1
To find: The molecular formula of the compound
Calculation: % carbon + % hydrogen + % nitrogen = 20 + 6.7 + 46.67 = 73.37%
This is less than 100%. Hence, compound contains adequate oxygen so that the total percentage of elements is 100%.
Hence, % of oxygen = 100 – 73.37 = 26.63%

Hence, empirical formula is CH₄N₂O.
Empirical formula mass = 12 + 4 + 28 + 16 = 60 g mol^-1
Hence,
Molar mass = Empirical formula mass
∴ Molecular formula = Empirical formula = CH₄N₂O
Ans: Molecular formula of the compound = CH₄N₂O

Question L.
A compound on analysis gave the following percentage composition by mass : H = 9.09; O = 36.36; C = 54.55. Mol mass of compound is 88. Find its molecular formula.
Answer:
Given: Percentage of H, O, C = 9.09%, 36.36%, 54.55% respectively.
Molar mass of the compound = 88 g mol^-1
To find: The molecular formula of the compound
Calculation:

Hence, empirical formula is C₂H₄O.
Empirical formula mass = 24 + 4 + 16 = 44 g mol^-1
Hence,

Molecular formula = r × empirical formula
Molecular formula = 2 × C₂H₂O = C₄H₈O₂
Ans: Molecular formula of the compound = C₄H₈O₂

Question M.
Carbohydrates are compounds containing only carbon, hydrogen and oxygen. When heated in the absence of air, these compounds decompose to form carbon and water. If 310 g of a carbohydrate leave a residue of 124 g of carbon on heating in absence of air, what is the empirical formula of the carbohydrate ?
Answer:
Given: Mass of carbon residue = 124 g, mass of carbohydrate = 310 g
To find: Empirical formula of the carbohydrate
Calculation: Since the 310 g of compound decomposes to carbon and water and the mass of carbon produced is 124 g, the remaining mass would be of water.
∴ Molar mass of water = 310 – 124 = 186 g

The ratio of number of moles of C : water = C : H₂O = 1 : 1
Hence, empirical formula = CH₂O
Ans: Empirical formula of the carbohydrate = CH₂O

Question N.
Write each of the following in exponential notation :
a. 3,672,199
b. 0.000098
c. 0.00461
d. 198.75
Answer:
a. 3,672,199 = 3.672199 × 10⁶
b. 0.000098 = 9.8 × 10^-5
c. 0.00461 = 4.61 × 10^-3
d. 198.75 = 1.9875 × 10²

Question O.
Write each of the following numbers in ordinary decimal form :
a. 3.49 × 10^-11
b. 3.75 × 10^-1
c. 5.16 × 10⁴
d. 43.71 × 10^-4
e. 0.011 × 10^-3
f. 14.3 × 10^-2
g. 0.00477 × 10⁵
h. 5.00858585
Answer:
a. 3.49 × 10^-11 = 0.0000000000349
b. 3.75 × 10^-1 = 0.375
c. 5.16 × 10⁴ = 51,600
d. 43.71 × 10^-4 = 0.004371
e. 0.011 × 10^-3 = 0.000011
f. 14.3 × 10^-2 = 0.143
g. 0.00477 × 10⁵ = 477
h. 5.00858585 = 5.00858585

Question P.
Perform each of the following calculations. Round off your answers to two digits.

Answer:

Question Q.
Perform each of the following calculations. Round off your answers to three digits.
a. (3.26 × 10⁴) (1.54 × 10⁶)
b. (8.39 × 10⁷) (4.53 × 10⁹)
c. \(\frac{8.94 \times 10^{6}}{4.35 \times 10^{4}}\)
d. \(\frac{\left(9.28 \times 10^{9}\right) \times\left(9.9 \times 10^{-7}\right)}{(511) \times\left(2.98 \times 10^{-6}\right)}\)
Answer:
i. (3.26 × 10⁴) (1.54 × 10⁶) = 5.0204 × 10⁴⁺⁶ = 5.02 × 10¹⁰
ii. (8.39 × 10⁷) (4.53 × 10⁹) = 38.0067 × 10⁷⁺⁹ = 38.0067 × 10¹⁶ = 3.80 x 10¹⁷

Question R.
Perform the following operations :
a. 3.971 × 10⁷ + 1.98 × 10⁴;
b. 1.05 × 10^-4 – 9.7 × 10^-5;
c. 4.11 × 10^-3 + 8.1 × 10^-4;
d. 2.12 × 10⁶ – 3.5 × 10⁵.
Answer:
Solution:
To perform addition/subtraction operation, first the numbers are written in such a way that they have the same exponent. The coefficients are then added/subtracted.
a. 3.971 × 10⁷ + 1.98 × 10⁴ = 3.971 × 10⁷ + 0.00198 × 10⁷ = (3.971 + 0.00198) × 10⁷
= 3.97298 × 10⁷
b. 1.05 × 10^-4 – 9.7 × 10^-5 = 10.5 × 10^-5 – 9.7 × 10^-5 = (10.5 – 9.7) × 10^-5 = 0.80 × 10^-5
= 8.0× 10^-6
c. 4.11 × 10^-3 + 8.1 × 10^-4 = 41.1 × 10^-4 + 8.1 × 10^-4 = (41.1 + 8.1) × 10^-4 = 49.2 × 10^-4
= 4.92 × 10^-3
d. 2.12 × 10⁶ – 3.5 × 10⁵ = 21.2 × 10⁵ – 3.5 × 10⁵ = (21.2 – 3.5) × 10⁵ = 17.7 × 10⁵
= 1.77 × 10⁶

Question S.
A 1.000 mL sample of acetone, a common solvent used as a paint remover, was placed in a small bottle whose mass was known to be 38.0015 g. The following values were obtained when the acetone – filled bottle was weighed : 38.7798 g, 38.7795 g and 38.7801 g. How would you characterise the precision and accuracy of these measurements if the actual mass of the acetone was 0.7791 g ?
Answer:
Precision:

Measurement	Mass of acetone observed (g)
1	38.7798 – 38.0015 = 0.7783
2	38.7795 – 38.0015 = 0.7780
3	38.7801 – 38.0015 = 0.7786

Mean = \(\frac{0.7783+0.7780+0.7786}{3}\) = 0.7783 g

Measurement	Mass of acetone observed (g)	Absolute deviation (g) = | Observed value – Mean |
1	0.7783	0
2	0.7780	0.0003
3	0.7786	0.0003

Mean absolute deviation = \(\frac{0+0.0003+0.0003}{3}\) = 0.0002
∴ Mean absolute deviation = ±0.0002 g

ii. Accuracy:
Actual mass of acetone = 0.7791 g
Observed value (average) = 0.7783 g
a. Absolute error = Observed value – True value
= 0.7783 – 0.7791
= – 0.0008 g

Ans: These observed values are close to each other and are also close to the actual mass. Therefore, the results are precise and as well accurate.
i. Relative deviation = 0.0257%
ii. Relative error = 0.1027%
[Note: i. As per the method given in textbook, the calculated value of relative deviation is 0.0257%.
ii. The negative sign in -0.1027% indicates that the experimental result is lower than the true value.]

Question T.
Your laboratory partner was given the task of measuring the length of a box (approx 5 in) as accurately as possible, using a metre stick graduated in milimeters. He supplied you with the following measurements: 12.65 cm, 12.6 cm, 12.65 cm, 12.655 cm, 126.55 mm, 12 cm.
a. State which of the measurements you would accept, giving the reason.
b. Give your reason for rejecting each of the others.
Answer:
a. The metre stick is graduated in millimetres i.e. 1 mm to 1000 mm, and 1 mm = 0.1 cm. Therefore, if the length is measured in centimetres, the least count of metre stick is 0.1 cm. The results 12.6 cm has the least count of 0.1 cm and is an acceptable result.

b. Since, the least count of metre stick is 0.1 cm or 1mm, the results such as 12.65 cm, 12.655 cm, 126.55 mm cannot be measured using this stick, and hence, these results are rejected. The result, 12 cm doesn’t include the least count and is rejected.

Question U.
What weight of calcium oxide will be formed on heating 19.3 g of calcium carbonate?
(At. wt. : Ca = 40; C = 12; O = 16)
Answer:
Given: Mass of CaCO₃ consumed in reaction = 19.3 g
To find: Mass of CaO formed
Calculation: Calcium carbonate decomposes according to the balanced equation,

So, 100 g of CaCO3 produce 56 g of CaO.

Ans: Mass of CaO formed = 10.81 g

[Calculation using log table:
56 × 0.193
= Antilog₁₀ [log₁₀ (56) + log₁₀ (0.193)]
= Antilog₁₀ [1.7482 + \(\overline{1} .2856\)]
= Antilog₁₀ [1.0338] = 10.81]

Question V.
The hourly energy requirements of an astronaut can be satisfied by the energy released when 34 grams of sucrose are “burnt” in his body. How many grams of oxygen would be needed to be carried in a space capsule to meet his requirement for one day?
Answer:
34 g of sucrose provides energy for an hour.
Hence, for a day, the mass of sucrose needed = 34 × 24 = 816g
The balanced equation is,

Thus, 342 g of sucrose requires 384 g of oxygen.
∴ 816 g of sucrose will require = \(\frac{816}{342}\) × 384 = 916 g of O₂
Ans: Astronaut needs to carry 916 g of O₂.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Trigonometry – II Miscellaneous Exercise 3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Miscellaneous Exercise 3

I. Select the correct option from the given alternatives.

Question 1.
The value of sin(n + 1) A sin(n + 2) A + cos(n + 1) A cos(n + 2) A is equal to
(a) sin A
(b) cos A
(c) -cos A
(d) sin 2A
Answer:
(b) cos A
Hint:
L.H.S. = sin [(n + 1)A] . sin [(n + 2)A] + cos [(n + 1)A] . cos [(n + 2)A]
= cos [(n + 2)A] . cos [(n + 1)A] + sin [(n + 2)A] . sin [(n + 1)A]
Let (n + 2)A = a and (n + 1)A = b … (i)
∴ L.H.S. = cos a . cos b + sin a . sin b
= cos (a – b)
= cos [(n + 2)A – (n + 1)A] ……..[From (i)]
= cos [(n + 2 – n – 1)A]
= cos A
= R.H.S.

Question 2.
If tan A – tan B = x and cot B – cot A = y, then cot (A – B) = ________
(a) \(\frac{1}{y}-\frac{1}{x}\)
(b) \(\frac{1}{x}-\frac{1}{y}\)
(c) \(\frac{1}{x}+\frac{1}{y}\)
(d) \(\frac{x y}{x-y}\)
Answer:
(c) \(\frac{1}{x}+\frac{1}{y}\)
Hint:

Question 3.
If sin θ = n sin(θ + 2α), then tan(θ + α) is equal to
(a) \(\frac{1+n}{2-n}\) tan α
(b) \(\frac{1-n}{1+n}\) tan α
(c) tan α
(d) \(\frac{1+n}{1-n}\) tan α
Answer:
(d) \(\frac{1+n}{1-n}\) tan α
Hint:

Question 4.
The value of \(\frac{\cos \theta}{1+\sin \theta}\) is equal to ________
(a) \(\tan \left(\frac{\theta}{2}-\frac{\pi}{4}\right)\)
(b) \(\tan \left(-\frac{\pi}{4}-\frac{\theta}{2}\right)\)
(c) \(\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\)
(d) \(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\)
Answer:
(c) \(\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\)
Hint:

Question 5.
The value of cos A cos (60° – A) cos (60° + A) is equal to ________
(a) \(\frac{1}{2}\) cos 3A
(b) cos 3A
(c) \(\frac{1}{4}\) cos 3A
(d) 4cos 3A
Answer:
(c) \(\frac{1}{4}\) cos 3A
Hint:

Question 6.
The value of \(\sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \sin \frac{7 \pi}{14} \sin \frac{9 \pi}{14} \sin \frac{11 \pi}{14} \sin \frac{13 \pi}{14}\) is ________
(a) \(\frac{1}{16}\)
(b) \(\frac{1}{64}\)
(c) \(\frac{1}{128}\)
(d) \(\frac{1}{256}\)
Answer:
(b) \(\frac{1}{64}\)
Hint:

Question 7.
If α + β + γ = π, then the value of sin² α + sin² β – sin² γ is equal to ________
(a) 2 sin α
(b) 2 sin α cos β sin γ
(c) 2 sin α sin β cos γ
(d) 2 sin α sin β sin γ
Answer:
(c) 2 sin α sin β cos γ
Hint:
sin² α + sin² β – sin² γ
= \(\frac{1-\cos 2 \alpha}{2}+\frac{1-\cos 2 \beta}{2}-\sin ^{2} \gamma\)
= 1 – \(\frac{1}{2}\) (cos 2α + cos 2β) – 1 + cos² γ
= \(\frac{-1}{2}\) × 2 cos(α + β) cos(α – β) + cos² γ
= cos γ cos (α – β) + cos² γ …..[∵ α + β + γ = π]
= cos γ [cos (α – β) + cos γ]
= cos γ [cos (α – β) – cos (α + β)]
= 2 sin α sin β cos γ

Question 8.
Let 0 < A, B < \(\frac{\pi}{2}\) satisfying the equation 3sin² A + 2sin² B = 1 and 3sin 2A – 2sin 2B = 0, then A + 2B is equal to ________
(a) π
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{4}\)
(d) 2π
Answer:
(b) \(\frac{\pi}{2}\)
Hint:
3 sin 2A – 2sin 2B = 0
sin 2B = \(\frac{3}{2}\) sin 2A …….(i)
3 sin² A + 2 sin² B = 1
3 sin² A = 1 – 2 sin² B
3 sin² A = cos 2B ……(ii)
cos(A + 2B) = cos A cos 2B – sin A sin 2B
= cos A (3 sin² A) – sin A (\(\frac{3}{2}\) sin 2A) …..[From (i) and (ii)]
= 3 cos A sin² A – \(\frac{3}{2}\) (sin A) (2 sin A cos A)
= 3 cos A sin² A – 3 sin² A cos A
= 0
= cos \(\frac{\pi}{2}\)
∴ A + 2B = \(\frac{\pi}{2}\) ……..[∵ 0 < A + 2B < \(\frac{3 \pi}{2}\)]

Question 9.
In ∆ABC if cot A cot B cot C > 0, then the triangle is ________
(a) acute-angled
(b) right-angled
(c) obtuse-angled
(d) isosceles right-angled
Answer:
(a) acute angled
Hint:
cot A cot B cot C > 0
Case I:
cot A, cot B, cot C > 0
∴ cot A > 0, cot B > 0, cot C > 0
∴ 0 < A < \(\frac{\pi}{2}\), 0 < B < \(\frac{\pi}{2}\), 0 < C < \(\frac{\pi}{2}\)
∴ ∆ABC is an acute angled triangle.
Case II:
Two of cot A, cot B, cot C < 0
0 < A, B, C < π and two of cot A, cot B, cot C < 0
∴ Two angles A, B, C are in the 2nd quadrant which is not possible.

Question 10.
The numerical value of tan 20° tan 80° cot 50° is equal to ________
(a) √3
(b) \(\frac{1}{\sqrt{3}}\)
(c) 2√3
(d) \(\frac{1}{2 \sqrt{3}}\)
Answer:
(a) √3
Hint:
L.H.S. = tan 20° tan 80° cot 50°
= tan 20° tan 80° cot (90° – 40°)
= tan 20° tan 80° tan 40°
= tan 20° tan (60° + 20°) tan (60° – 20°)

= tan 3(20°)
= tan 60°
= √3
= R.H.S.

II. Prove the following.

Question 1.
tan 20° tan 80° cot 50° = √3
Solution:
L.H.S. = tan 20° tan 80° cot 50°
= tan 20° tan 80° cot (90° – 40°)
= tan 20° tan 80° tan 40°
= tan 20° tan (60° + 20°) tan (60° – 20°)

= tan 3(20°)
= tan 60°
= √3
= R.H.S.

Question 2.
If sin α sin β – cos α cos β + 1 = 0, then prove that cot α tan β = -1.
Solution:
sin α sin β – cos α cos β + 1 = 0
∴ cos α cos β – sin α sin β = 1
∴ cos (α + β) = 1
∴ α + β = 0 ……[∵ cos 0 = 1]
∴ β = -α
L.H.S. = cot α tan β
= cot α tan(-α)
= -cot α tan α
= -1
= R.H.S.

Question 3.
\(\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}=\frac{1}{16}\)
Solution:

Question 4.
\(\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)=\frac{1}{8}\)
Solution:

Question 5.
cos 12° + cos 84° + cos 156° + cos 132° = \(-\frac{1}{2}\)
Solution:

Question 6.
\(\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)=\sqrt{2} \cos x\)
Solution:

Question 7.
\(\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x\)
Solution:

Question 8.
sin² 6x – sin² 4x = sin 2x sin 10x
Solution:

Question 9.
cos² 2x – cos² 6x = sin 4x sin 8x
Solution:

Question 10.
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Solution:

Question 11.
\(\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}\)
Solution:

Question 12.
If sin 2A = λ sin 2B, then prove that \(\frac{\tan (A+B)}{\tan (A-B)}=\frac{\lambda+1}{\lambda-1}\)
Solution:

Question 13.
\(\frac{2 \cos 2 A+1}{2 \cos 2 A-1}\) = tan (60° + A) tan (60° – A)
Solution:

Question 14.
tan A + tan (60° + A) + tan (120° + A) = 3 tan 3A
Solution:

Question 15.
3 tan⁶ 10° – 27 tan⁴ 10° + 33 tan² 10° = 1
Solution:

Question 16.
cosec 48° + cosec 96° + cosec 192° + cosec 384° = 0
Solution:
L.H.S. = cosec 48° + cosec 96° + cosec 192° + cosec 384°
= cosec 48° + cosec (180° – 84°) + cosec (180° + 12°) + cosec (360° + 24°)
= cosec 48° + cosec 84° + cosec (-12°) + cosec 24°

Question 17.
3(sin x – cos x)⁴ + 6(sin x + cos x)² + 4(sin⁶ x + cos⁶ x) = 13
Solution:
(sin x – cos x)⁴
= [(sin x – cos x)²]²
= (sin² x + cos² x – 2 sin x cos x)²
= (1 – 2 sin x cosx)²
= 1 – 4 sin x cos x + 4 sin² x cos² x
(sin x + cos x)² = sin² x + cos² x + 2 sin x cos x = 1 + 2 sin x cos x
sin⁶ x + cos⁶ x
= (sin² x)³ + (cos² x)³
= (sin² x + cos² x)³ – 3 sin² x cos² x (sin² x + cos² x) …..[∵ a³ + b³ = (a + b)³ – 3ab(a + b)]
= 1³ – 3 sin² x cos² x (1)
= 1 – 3 sin² x cos² x
L.H.S. = 3(sin x – cos x)⁴ + 6(sin x + cos x)² + 4(sin⁶ x + cos⁶ x)
= 3(1 – 4 sin x cos x + 4 sin² x cos² x) + 6(1 + 2 sin x cos x) + 4(1 – 3 sin² x cos² x)
= 3 – 12 sin x cos x + 12 sin² x cos² x + 6 + 12 sin x cos x + 4 – 12 sin² x cos² x
= 13
= R.H.S.

Question 18.
tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A
Solution:
We have to prove that,
tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A
i.e., to prove,
cot A – tan A – 2 tan 2A – 4 tan 4A – 8 cot 8A = 0

∴ cot θ – tan θ = 2 cot 2θ …..(i)
L.H.S. = cot A – tan A – 2 tan 2A – 4 tan 4A – 8 cot 8A
= 2 cot 2A – 2 tan 2A – 4 tan 4A – 8 cot 8A …..[From (i)]
= 2(cot 2A – tan 2A) – 4 tan 4A – 8 cot 8A
= 2 × 2 cot 2(2A) – 4 tan 4A – 8 cot 8A ……[From (i)]
= 4(cot 4A – tan 4A) – 8 cot 8A
= 4 × 2 cot 2(4A) – 8 cot 8A ……[From (i)]
= 8 cot 8A – 8 cot 8A = 0
= R.H.S.
Alternate Method:

Question 19.
If A + B + C = \(\frac{3 \pi}{2}\), then cos 2A + cos 2B + cos 2C = 1 – 4 sin A sin B sin C
Solution:

Question 20.
In any triangle ABC, sin A – cos B = cos C. Show that ∠B = \(\frac{\pi}{2}\).
Solution:
sin A – cos B = cos C
∴ sin A = cos B + cos C

A = B – C ………(i)
In ∆ABC,
A + B + C = π
∴ B – C + B + C = π
∴ 2B = π
∴ B = \(\frac{\pi}{2}\)

Question 21.
\(\frac{\tan ^{3} x}{1+\tan ^{2} x}+\frac{\cot ^{3} x}{1+\cot ^{2} x}\) = sec x cosec x – 2 sin x cos x
Solution:

Question 22.
sin 20° sin 40° sin 80° = \(\frac{\sqrt{3}}{8}\)
Solution:
L.H.S. = sin 20°. sin 40°. sin 80°
= sin 20°. sin 40°. sin 80°
= \(\frac{1}{2}\) (2 . sin 40°. sin 20°) . sin 80°
= \(\frac{1}{2}\) [cos(40° – 20°) – cos (40° + 20°)] . sin 80°
= \(\frac{1}{2}\) (cos 20° – cos 60°) sin 80°
= \(\frac{1}{2}\) . cos 20° . sin 80° – \(\frac{1}{2}\) . cos 60° . sin 80°
= \(\frac{1}{2 \times 2}\) (2 sin 80° . cos 20°) – \(\frac{1}{2 \times 2}\) . sin 80°
= \(\frac{1}{4}\) [sin(80° + 20°) + sin (80° – 20°)] – \(\frac{1}{2}\) . sin 80°

Question 23.
sin 18° = \(\frac{\sqrt{5}-1}{4}\)
Solution:
Let θ = 18°
∴ 5θ = 90°
∴ 2θ + 3θ = 90°
∴ 2θ = 90° – 3θ
∴ sin 2θ = sin (90° – 3θ)
∴ sin 2θ = cos 3θ
∴ 2 sin θ cos θ = 4 cos³ θ – 3 cos θ
∴ 2 sin θ = 4 cos² θ – 3 …..[∵ cos θ ≠ 0]
∴ 2 sin θ = 4 (1 – sin² θ) – 3
∴ 2 sin θ = 1 – 4 sin² θ
∴ 4 sin² θ + 2 sin θ – 1 = 0
∴ sin θ = \(\frac{-2 \pm \sqrt{4+16}}{8}\)
= \(\frac{-2 \pm 2 \sqrt{5}}{8}\)
= \(\frac{-1 \pm \sqrt{5}}{4}\)
Since, sin 18° > 0
∴ sin 18°= \(\frac{\sqrt{5}-1}{4}\)

Question 24.
cos 36° = \(\frac{\sqrt{5}+1}{4}\)
Solution:
We know that,
cos 2θ = 1 – 2 sin² θ
cos 36° = cos 2(18°)
= 1 – 2 sin² 18°

∴ cos 36° = \(\frac{\sqrt{5}+1}{4}\)

Question 25.
sin 36° = \(\frac{\sqrt{10-2 \sqrt{5}}}{4}\)
Solution:
We know that, sin² θ = 1 – cos² θ
sin² 36° = 1 – cos² 36°
= 1 – \(\left(\frac{\sqrt{5}+1}{4}\right)^{2}\)
= \(\frac{16-(5+1+2 \sqrt{5})}{16}\)
= \(\frac{10-2 \sqrt{5}}{16}\)
∴ sin 36° = \(\frac{\sqrt{10-2 \sqrt{5}}}{4}\) ……[∵ sin 36° is positive]

Question 26.
\(\sin \frac{\pi^{c}}{8}=\frac{1}{2} \sqrt{2-\sqrt{2}}\)
Solution:

Question 27.
tan \(\frac{\pi}{8}\) = √2 – 1
Solution:

Question 28.
tan 6° tan 42° tan 66° tan 78° = 1
Solution:

Question 29.
sin 47° + sin 61° – sin 11° – sin 25° = cos 7°
Solution:

Question 30.
√3 cosec 20° – sec 20° = 4
Solution:

Question 31.
In ∆ABC, ∠C = \(\frac{2 \pi}{3}\), then prove that cos² A + cos² B – cos A cos B = \(\frac{3}{4}\).
Solution:

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 1 Some Basic Concepts of Chemistry Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

1. Choose the most correct option.

Question A.
A sample of pure water, whatever the source always contains …………. by mass of oxygen and 11.1 % by mass of hydrogen.
a. 88.9
b. 18
c. 80
d. 16
Answer:
a. 88.9

Question B.
Which of the following compounds can NOT demonstrate the law of multiple proportions?
a. NO, NO₂
b. CO, CO₂
c. H₂O, H₂O₂
d. Na₂S, NaF
Answer:
d. Na₂S, NaF

Question C.
Which of the following temperature will read the same value on celsius and Fahrenheit scales.
a. – 40°
b. + 40°
c. – 80°
d. – 20°
Answer:
a. – 40°

Question D.
SI unit of the quantity electric current is
a. Volt
b. Ampere
c. Candela
d. Newton
Answer:
b. Ampere

Question E.
In the reaction N₂ + 3H₂ → 2NH₃, the ratio by volume of N₂, H₂ and NH₃ is 1 : 3 : 2 This illustrates the law of
a. definite proportion
b. reciprocal proportion
c. multiple proportion
d. gaseous volumes
Answer:
d. gaseous volumes

Question F.
Which of the following has maximum number of molecules ?
a. 7 g N₂
b. 2 g H₂
c. 8 g O₂
d. 20 g NO₂
Answer:
b. 2 g H₂

Question G.
How many g of H₂O are present in 0.25 mol of it ?
a. 4.5
b. 18
c. 0.25
d. 5.4
Answer:
a. 4.5

Question H.
The number of molecules in 22.4 cm³ of nitrogen gas at STP is
a. 6.022 × 10²⁰
b. 6.022 × 10²³
c. 22.4 × 10²⁰
d. 22.4 × 10²³
Answer:
a. 6.022 × 10²⁰

Question I.
Which of the following has the largest number of atoms ?
a. 1g Au_(s)
b. 1g Na_(s)
c. 1g Li_(s)
d. 1g Cl_2(g)
Answer:
c. 1g Li_(s)

2. Answer the following questions.

Question A.
State and explain Avogadro’s law.
Answer:
i. In the year 1811, Avogadro made a distinction between atoms and molecules and thereby proposed Avogadro’s law.

ii. Avogadro proposed that, “Equal volumes of all gases at the same temperature and pressure contain equal number of molecules”.
e.g. Hydrogen gas combines with oxygen gas to produce water vapour as follows:

According to Avogadro’s law, if 1 volume contains n molecules, then 2n molecules of hydrogen combine with n molecules of oxygen to give 2n molecules of water, i.e., 2 molecules of hydrogen gas combine with 1 molecule of oxygen to give 2 molecules of water vapour as represented below:

Question B.
Point out the difference between 12 g of carbon and 12 u of carbon.
Answer:
12 g of carbon is the molar mass of carbon while 12 u of carbon is the mass of one carbon atom.

Question C.
How many grams does an atom of hydrogen weigh ?
Answer:
The mass of a hydrogen atom is 1.6736 × 10^-24 g.

Question D.
Calculate the molecular mass of the following in u.
a. NH₃
b. CH₃COOH
c. C₂H₅OH
Answer:
i. Molecular mass of NH₃ = (1 × Average atomic mass of N) + (3 × Average atomic mass of H)
= (1 × 14.0 u) +(3 × 1.0 u)
= 17 u

ii. Molecular mass of CH₃COOH = (2 × Average atomic mass of C) + (4 × Average atomic mass of H) + (2 × Average atomic mass of O)
= (2 × 12.0 u) + (4 × 1.0 u) + (2 × 16.0 u)
= 60 u

iii. Molecular mass of C₂H₅OH = (2 × Average atomic mass of C) + (6 × Average atomic mass of H) + (1 × Average atomic mass of O)
= (2 × 12.0 u) + (6 × 1.0 u) + (1 × 16.0 u)
= 46 u
Ans: i. The molecular mass of NH₃ = 17 u
ii. The molecular mass of CH₃COOH = 60 u
iii. The molecular mass of C₂H₅OH = 46 u

Question E.
How many particles are present in 1 mole of a substance ?
Answer:
The number of particles in one mole is 6.0221367 × 10²³.

Question F.
What is the SI unit of amount of a substance ?
Answer:
The SI unit for the amount of a substance is mole (mol).

Question G.
What is meant by molar volume of a gas ?
Answer:
The volume occupied by one mole of a gas at standard temperature (0 °C) and pressure (1 atm) (STP) is called as molar volume of a gas. The molar volume of a gas at STP is 22.4 dm³.

Question H.
State and explain the law of conservation of mass.
Answer:
Law of conservation of mass:

• The law of conservation of mass states that, “Mass can neither be created nor destroyed” during chemical combination of matter.
• Antoine Lavoisier who is often referred to as the father of modem chemistry performed careful experimental studies for various combustion reactions, namely burning of phosphorus and mercury in the presence of air.
• Both his experiments resulted in increased weight of products.
• After several experiments, in burning of phosphorus, he found that the weight gained by the phosphoms was exactly the same as the weight lost by the air. Hence, total mass of reactants = total mass of products.
• When hydrogen gas bums and combines with oxygen to form water, the mass of the water formed is equal to the mass of the hydrogen and oxygen consumed. Thus, this is in accordance with the law of conservation of mass.

Question I.
State the law of multiple proportions.
Answer:
The law states that, “When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers”.

3. Give one example of each

Question A.
Homogeneous mixture
Answer:
Homogeneous mixture: Solution (An aqueous solution of sugar)

Question B.
Heterogeneous mixture
Answer:
Heterogeneous mixture: Suspension (of sand in water)

Question C.
Element
Answer:
Element: Gold

Question D.
Compound
Answer:
Compound: Distilled water.

4. Solve problems :

Question A.
What is the ratio of molecules in 1 mole of NH₃ and 1 mole of HNO₃.
Answer:
One mole of any substance contains particles equal to 6.022 × 10²³.
1 mole of NH₃ = 6.022 × 10²³ molecules of NH₃
I mole of HNO₃ = 6.022 × 10²³ molecules of HNO₃
∴ Ratio = \(\frac{6.022 \times 10^{23}}{6.022 \times 10^{23}}\) = 1 : 1
Ans: The ratio of molecules is = 1 : 1.

Question B.
Calculate number of moles of hydrogen in 0.448 litre of hydrogen gas at STP.
Answer:
Given: Volume of hydrogen at STP = 0.448 L
To find: Number of moles of hydrogen

Molar volume of a gas = 22.4 dm³ mol^-1 = 22.4 L at STP

Ans: Number of moles of hydrogen = 0.02 mol

Question C.
The mass of an atom of hydrogen is 1.008 u. What is the mass of 18 atoms of hydrogen.
Answer:
Mass of 1 atom of hydrogen = 1.008 u
∴ Mass of 18 atoms of hydrogen = 18 × 1.008 u = 18.144 u
Ans: The mass of 18 atoms of hydrogen = 18.144 u

Question D.
Calculate the number of atom in each of the following (Given : Atomic mass of I = 127 u).
a. 254 u of iodine (I)
b. 254 g of iodine (I)
Answer:
a. 254 u of iodine (I) = x atoms
Atomic mass of iodine (I) = 127 u
∴ Mass of one iodine atom = 127 u
∴ x = \(\frac{254 \mathrm{u}}{127 \mathrm{u}}\) = 2 atoms

b. 254 g of iodine (I)
Atomic mass of iodine = 127 u
∴ Molar mass of iodine = 127 g mol^-1
Now,

Now,
Number of atoms = Number of moles × Avogadro’s constant
= 2 mol × 6.022 × 10²³ atoms/mol
= 12.044 × 10²³ atoms
= 1.2044 × 10²⁴ atoms
Ans. i.Number of iodine atoms in 254 u = 2 atoms
ii. Number of iodine atoms in 254 g = 1.2044 × 10²⁴ atoms

Question E.
A student used a carbon pencil to write his homework. The mass of this was found to be 5 mg. With the help of this calculate.
a. The number of moles of carbon in his homework writing.
b. The number of carbon atoms in 12 mg of his homework writting.
Answer:
a. 5 mg carbon = 5 × 10^-3 g carbon
Atomic mass of carbon = 12 u
∴ Molar mass of carbon 12 g mol^-1

b. 12 mg carbon = 12 × 10^-3 g carbon

Number of atoms = Number of moles × Avogadro’s constant
Number of atoms of carbon = 1 × 10^-3 mol × 6.022 × 10²³ atoms/mol
= 6.022 × 10²⁰ atoms
Ans: Number of moles of carbon in his homework writing = 4.167 × 10^-4 mol
Number of atoms of carbon in 12 mg homework writing = 6.022 × 10²⁰ atoms

Question F.
Arjun purchased 250 g of glucose (C₆H₁₂O₆) for Rs 40. Find the cost of glucose per mole.
Answer:
Given: Mass of urea = 250 g, cost for 250 g glucose = Rs 40, molecular formula of glucose = C₆H₁₂O₆
To find: Cost per mole of glucose
Calculation: Molecular formula of glucose is (C₆H₁₂O₆).
Molecular mass of glucose
= (6 × Average atomic mass of C) + (12 × Average atomic mass of H) + (6 × Average atomic mass of O)
= (6 × 12 u) + (12 × 1 u) + (6 × 16 u)
=180 u
∴ Molar mass of glucose = 180 g mol^-1

Now,
\(\frac {250}{180}\) mol of glucose cost = Rs 40
1 mol glucose cost = x
∴ x = \(\frac{40 \times 180}{250}\) = Rs 28.8/mol of glucose
Ans. The cost of glucose per mole is Rs 28.8.

[ Calculation using log table:
\(\frac{40 \times 180}{250}\)
= Antilog₁₀ [log₁₀(40) + log₁₀(180) + log₁₀(250)]
= Antilog₁₀ [1.6021 + 2.2553 – 2.3979]
= Antilog₁₀ [1.4595] = 28.80 ]

Question G.
The natural isotopic abundance of ¹⁰B is 19.60% and ¹¹B is 80.40 %. The exact isotopic masses are 10.13 and 11.009 respectively. Calculate the average atomic mass of boron.
Answer:
Average atomic mass of Boron(B)

Ans. Average atomic mass of boron = 10.84 u

Question H.
Convert the following degree Celsius temperature to degree Fahrenheit.
a. 40 °C
b. 30 °C
Answer:
a. Given: Temperature in degree Celsius =40°C
To find: Temperature in degree Fahrenheit
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 40 °C in the formula,
°F = \(\frac {9}{5}\) (°C)+32
= \(\frac {9}{5}\) (40) + 32
= 72 + 32
= 104 °F

b. Given: Temperature in degree Celsius = 30 °C
To find: Temperature in degree Fahrenheit
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 30 °C in the formula,
°F = \(\frac {9}{5}\)(°C) + 32
= \(\frac {9}{5}\)(30) + 32
= 54 + 32
= 86 °F
Ans: i. The temperature 40 °C corresponds to 104 °F.
ii. The temperature 30 °C corresponds to 86 °F.

Question I.
Calculate the number of moles and molecules of acetic acid present in 22 g of it.
Answer:
Given: Mass of acetic acid = 22 g
To find: The number of moles and molecules of acetic acid
Formulae: Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
ii. Number of molecules = Number of moles × Avogadro’s constant
Calculator: Mass of acetic acid = 22 g
Molecular mass of acetic acid, CH₃COOH
= (2 × Average atomic mass of C) + (4 × Average atomic mass of H) + (2 × Average atomic mass of O)
= (2 × 12 u) + (4 × 1 u) + (2 × 16 u) = 60 u
∴ Molar mass of acetic acid = 60 g mol^-1

Now,
Number of molecules of acetic acid = Number of moles × Avogadro’s constant
= 0.367 mol × 6.022 × 10²³ molecules/mol
= 2.210 × 10²³ molecules
Ans: Number of moles = 0.367 mol
Number of molecules of acetic acid = 2.210 × 10²³ molecules

Question J.
24 g of carbon reacts with some oxygen to make 88 grams of carbon dioxide. Find out how much oxygen must have been used.
Answer:
Given: Mass of carbon (reactant) = 24 g, mass of carbon dioxide (product) = 88 g
To find: Mass of oxygen (reactant)
Calculation: 12 g of carbon combine with 32 g oxygen to form 44 g of carbon dioxide as follows:

Hence, (2 × 12 = 24 g) of carbon will combine with (2 × 32 = 64 g) of oxygen to give (2 × 44 = 88 g) carbon dioxide.
Ans: Mass of oxygen used = 64 g

Question K.
Calculate number of atoms is each of the following. (Average atomic mass : N = 14 u, S = 32 u)
a. 0.4 mole of nitrogen
b. 1.6 g of sulfur
Answer:
a. 0.4 mole of nitrogen (N)
Number of atoms of N = Number of moles × Avogadro’s constant
= 0.4 mol × 6.022 × 10²³ atoms/mol
= 2.4088 × 10²³ atoms of N

b. 1.6 g of Sulphur (S)
Molar mass of sulphur = 32 g mol^-1

Number of atoms of S = Number of moles × Avogadro’s constant
= 0.05 mol × 6.022 × 10²³ atoms/mol
= 0.3011 × 10²³ atoms
= 3.011 × 10²² atoms of S
Ans: a. Number of nitrogen atoms in 0.4 mole = 2.4088 × 10²³ atoms of N
b. Number of sulphur atoms in 1.6 g = 3.011 × 10²² atoms of S

Question L.
2.0 g of a metal burnt in oxygen gave 3.2 g of its oxide. 1.42 g of the same metal heated in steam gave 2.27 of its oxide. Which law is verified by these data ?
Answer:
Here, metal oxide is obtained by two different methods; reactions of metal with oxygen and reaction of metal with water vapour (steam).
In first reaction (reaction with oxygen),
The mass of oxygen in metal oxide = 3.2 – 2.0 = 1.2 g
% of oxygen = \(\frac{1.2}{3.2}\) × 100 = 37.5%
% of metal = \(\frac{2.0}{3.2}\) × 100 = 62.5%
In second reaction (reaction with steam),
The mass of oxygen in metal oxide = 2.27 – 1.42 = 0.85 g
% of oxygen = \(\frac{0.85}{2.27}\) × 100 = 37.44 ≈ 37.5%
% of metal = \(\frac{1.42}{2.27}\) × 100 = 62.56 ≈ 62.5%
Therefore, irrespective of the source, the given compound contains same elements in the same proportion. The law of definite proportions states that “A given compound always contains exactly the same proportion of elements by weight”. Hence, the law of definite proportions is verified by these data.
Ans: The law of definite proportions is verified by given data.

Question M.
In two moles of acetaldehyde (CH₃CHO) calculate the following
a. Number of moles of carbon
b. Number of moles of hydrogen
c. Number of moles of oxygen
d. Number of molecules of acetaldehyde
Answer:
Molecular formula of acetaldehyde: C₂H₄O
Moles of acetaldehyde = 2 mol
a. Number of moles of carbon atoms = Moles of acetaldehyde × Number of carbon atoms
= 2 × 2
= 4 moles of carbon atoms

b. Number of moles of hydrogen atoms = Moles of acetaldehyde × Number of hydrogen atoms
= 2 × 4
= 8 moles of hydrogen atoms

c. Number of moles of oxygen atoms = Moles of acetaldehyde × Number of oxygen atoms
= 2 × 1
= 2 moles of oxygen atoms

d. Number of molecules of acetaldehyde = Moles of acetaldehyde × Avogadro number
= 2 mol × 6.022 × 10²³ molecules/mol
= 12.044 × 10²³ molecules of acetaldehyde
Ans: i. Number of moles of carbon, hydrogen and oxygen are 4, 8, 2 respectively,
ii. Number of molecules of acetaldehyde = 12.044 × 10²³

Question N.
Calculate the number of moles of magnesium oxide, MgO in
i. 80 g and
ii. 10 g of the compound.
(Average atomic masses of Mg = 24 and O = 16)
Answer:
Given: i. Mass of MgO = 80 g
ii. Mass of MgO = 10 g
To find: Number of moles of MgO
Formulae: Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
Calculation: i. Molecular mass of MgO = (1 × Average atomic mass of Mg) + (1 × Average atomic mass of O)
= (1 × 24u) + (1 × 16 u)
= 40 u
∴ Molar mass of MgO = 40 g mol^-1
Mass of MgO = 80 g
Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{80 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 2 mol

ii. Mass of MgO = 10 g, Molar mass of MgO = 40 g mol^-1
Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{10 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 0.25 mol
Ans: i. The number of moles in 80 g of magnesium oxide, MgO = 2 mol
ii. The number of moles in 10 g of magnesium oxide, MgO = 0.25 mol

Question O.
What is volume of carbon dioxide, CO₂ occupying by i. 5 moles and ii. 0.5 mole of CO₂ gas measured at STP.
Answer:
Given: i. Number of moles of CO₂ = 5 mol
ii. Number of moles of CO₂ = 0.5 mol
To find: Volume at STP
Formula: Number of moies of a gas (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
Calculation: Molar volume of a gas 22.4 dm³ mol^-1 at STP.
Number of moles of a gas (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
∴ i. Volume of the gas at STP = Number of moles of a gas (n) × Molar volume of a gas
= 5mol × 22.4 dm³ mol^-1 = 112 dm³
ii. Volume of the gas at STP Number of moles of a gas (n) × Molar volume of a gas
= 0.5 mol × 22.4 dm³ mol^-1 = 11.2 dm³
Ans: i. Volume of 5 mol of CO₂ = 112 dm³
ii. Volume of 0.5 mol of CO₂ = 11.2 dm³

Question P.
Calculate the mass of potassium chlorate required to liberate 6.72 dm³ of oxygen at STP. Molar mass of KClO₃ is 122.5 g mol^-1.
Answer:
The molecular formula of potassium chlorate is KClO₃.
Required chemical equation:

2 moles of KClO₃ = 2 × 122.5 = 245 g
3 moles of O₂ at STP occupy = (3 × 22.4 dm³) = 67.2 dm³
Thus, 245 g of potassium chlorate will liberate 67.2 dm³ of oxygen gas.
Let ‘x’ gram of KClO₃ liberate 6.72 dm³ of oxygen gas at S.T.P.
∴ x = \(\frac{245 \times 6.72}{67.2}\) = 24.5 g
Ans: Mass of potassium chlorate required = 24.5 g

Question Q.
Calculate the number of atoms of hydrogen present in 5.6 g of urea, (NH₂)₂CO. Also calculate the number of atoms of N, C and O.
Answer:
Given: Mass of urea = 5.6 g
To find: The number of atoms of hydrogen, nitrogen, carbon and oxygen
Calculation: Molecular formula of urea: (NH₂)₂CO
Molar mass of urea = 60 g mol^-1

∴ Moles of urea = 0.0933 mol
Number of atoms = Number of moles × Avogadro’s constant
Now, 1 molecule of urea has total 8 atoms, out of which 4 atoms are of H, 2 atoms are of N, 1 of C and 1 of O.
∴ Number of H atoms in 5.6 g of urea = (4 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 2.247 × 10²³ atoms of hydrogen
∴ Number of N atoms in 5.6 g of urea = (2 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 1.124 × 10²³ atoms of nitrogen
∴ Number of C atoms in 5.6 g of urea = (1 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 0.562 × 10²³ atoms of carbon
∴ Number of O atoms in 5.6 g of urea = (1 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 0.562 × 10²³ atoms of oxygen
Ans: 5.6 g of urea contain 2.247 × 10²³ atoms of H, 1.124 × 10²³ atoms of N, 0.562 × 10²³ atoms of C and 0.562 × 10²³ atoms of O.

Question R.
Calculate the mass of sulfur dioxide produced by burning 16 g of sulfur in excess of oxygen in contact process. (Average atomic mass : S = 32 u, O = 16 u)
Answer:
Given: Mass of sulphur (reactant) = 16 g
To find: Mass of sulphur dioxide (product)
Calculation: 32 g of sulphur combine with 32 g oxygen to form 64 g of sulphur dioxide as follows:

Hence, (0.5 × 32 = 16 g) of sulphur will combine with (0.5 × 32 = 16 g) of oxygen to give (0.5 × 64 = 32 g) sulphur dioxide.
Ans: Mass of sulphur dioxide produced = 32 g

5. Explain

Question A.
The need of the term average atomic mass.
Answer:

• Several naturally occurring elements exist as a mixture of two or more isotopes.
• Isotopes have different atomic masses.
• The atomic mass of such an element is the average of atomic masses of its isotopes.
• For this purpose, the atomic masses of isotopes and their relative percentage abundances are considered.

Hence, the term average atomic mass is needed to express atomic mass of elements containing mixture of two or more isotopes.

Question B.
Molar mass.
Answer:
i. The mass of one mole of a substance (element/compound) in grams is called its molar mass.
ii. The molar mass of any element in grams is numerically equal to atomic mass of that element in u.
e.g.

Element	Atomic mass (u)	Molar mass (g mol-1)
H	1.0	1 0
C	12.0	12.0
O	16.0	16.0

iii. Similarly, molar mass of polyatomic molecule, in grams is numerically equal to its molecular mass or formula mass in u.
e.g.

Polyatomic substance	Molecular/formula mass (u)	Molar mass (g mol-1)
O2	32.0	32.0
H2O	18.0	18.0
NaCl	58.5	58.5

Question C.
Mole concept.
Answer:

• Even a small amount of any substance contains very large number of atoms or molecules. Therefore, a quantitative adjective ‘mole’ is used to express the large number of sub-microscopic entities like atoms, molecules, ions, electrons, etc. present in a substance.
• Thus, one mole is the amount of a substance that contains as many entities or particles as there are atoms in exactly 12 g (or 0.012 kg) of the carbon -12 isotope.
• One mole is the amount of substance which contains 6.0221367 × 10²³ particles/entities.

Question D.
Formula mass with an example.
Answer:

• The formula mass of a substance is the sum of atomic masses of the atoms present in the formula.
• In substances such as sodium chloride, positive (sodium), and negative (chloride) entities are arranged in a three-dimensional structure in a way that one sodium (Na⁺) ion is surrounded by six chlorides (Cl^–) ions, all at the same distance from it and vice versa. Thus, sodium chloride does not contain discrete molecules as the constituent units.
• Therefore, NaCl is just the formula that is used to represent sodium chloride though it is not a molecule.
• In such compounds, the formula (i.e., NaCl) is used to calculate the formula mass instead of molecular mass.

e.g. Formula mass of sodium chloride = atomic mass of sodium + atomic mass of chlorine
= 23.0 u + 35.5 u = 58.5 u

Question E.
Molar volume of gas.
Answer:
i. It is more convenient to measure the volume rather than mass of the gas.
ii. It is found from Avogadro law that one mole of any gas occupies a volume of 22.4 dm³ at standard temperature (0 °C) and pressure (1 atm) (STP).
iii. The volume of 22.4 dm³ at STP is known as molar volume of a gas.
iv. The relationship between number of moles and molar volume can be expressed as follows:

[Note: IUPAC has recently changed the standard pressure to 1 bar. Under these new STP conditions the molar volume of a gas is 22.71 L mol^-1]

Question F.
Types of matter (on the basis of chemical composition).
Answer:
Matter on the basis of chemical composition can be classified as follows:
i. Pure substances: They always have a definite chemical composition. They always have the same properties regardless of their origin.
e.g. Pure metal, distilled water, etc.

They are of two types:
a. Elements: They are pure substances, which cannot be broken down into simpler substances by ordinary chemical changes.
Elements are further classified into three types:
1. Metals:

• They have a lustre (a shiny appearance).
• They conduct heat and electricity.
• They can be drawn into wire (ductile).
• They can be hammered into thin sheets (malleable).
• e.g. Gold, silver, copper, iron. Mercury is a liquid metal at room temperature.

2. Nonmetals:

• They have no lustre, (except diamond, iodine)
• They are poor conductors of heat and electricity, (except graphite)
• They cannot be hammered into sheets or drawn into wire, because they are brittle. e.g. Iodine

3. Metalloids: Some elements have properties that are intermediate between metals and nonmetals and are called metalloids or semimetals.
e.g. Arsenic, silicon and germanium.
b. Compounds: They are the pure substances which are made up of two or more elements in fixed proportion.
e.g. Water, ammonia, methane, etc.

ii. Mixtures: They have no definite chemical composition and hence no definite properties. They can be separated by physical methods.
e.g. Paint (mixture of oils, pigment, additive), concrete (a mixture of sand, cement, water), etc.

Mixtures are of two types:

• Homogeneous mixture: In homogeneous mixture, constituents remain uniformly mixed throughout its bulk.
  e.g. Solution, in which solute and solvent molecules are uniformly mixed throughout its bulk.
• Heterogeneous mixture: In heterogeneous mixture, constituents are not uniformly mixed throughout its bulk.
  e.g. Suspension, which contains insoluble solid in a liquid.

11th Chemistry Digest Chapter 1 Some Basic Concepts of Chemistry Intext Questions and Answers

Can you tell? (Textbook Page No. 1)

Question 1.
Which are mixtures and pure substances from the following?
i. Sea water
ii. Gasoline
iii. Skin
iv. A rusty nail
v. A page of textbook
vi. Diamond
Answer:

No.	Material	Pure substance or mixture
i.	Seawater	Mixture
ii.	Gasoline	Mixture
iii.	Skin	Mixture
iv.	A rusty nail	Mixture
V.	A page of textbook	Mixture
vi.	Diamond	Pure substance

Can you tell? (Textbook Page No. 2)

Question 1.
Classify the following as element and compound.
i. Mercuric oxide
ii. Helium gas
iii. Water
iv. Table salt
v. Iodine
vi. Mercury
vii. Oxygen
viii. Nitrogen
Answer:

No.	Material	Element or compound
i.	Mercuric oxide	Compound
ii.	Helium gas	Element
iii.	Water	Compound
iv.	Table salt	Compound
V.	Iodine	Element
vi.	Mercury	Element
vii.	Oxygen	Element
viii.	Nitrogen	Element

Can you tell? (Textbook Page No. 6)

Question 1.
If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, how many volumes of water vapour would be produced?
Answer:

If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, then 10 volumes of water vapour would be produced.

Can you recall? (Textbook Page No. 6)

Question 1.
What is an atom and molecule? What is the order of magnitude of mass of one atom? What are isotopes?
Answer:

• The smallest indivisible particle of an element is called an atom.
• A molecule is an aggregate of two or more atoms of definite composition which are held together by chemical bonds.
• Every atom of an element has definite mass. The order of magnitude of mass of one atom is 10^-27 kg.
• Isotopes are the atoms of the same element having same atomic number but different mass number.

Try this (Textbook Page No. 8)

Question 1.
Find the formula mass of CaSO₄, if atomic mass of Ca = 40.1 u, S =32.1 u and O = 16.0 u.
Solution:
Formula mass of CaSO₄
= Average atomic mass of Ca + Average atomic mass of S + Average atomic mass of four O
= (40.1) + 32.1 + (4 × 16.0) = 136.2 u
Ans: Formula mass of CaSO₄ = 136.2 u

Can you recall? (Textbook Page No. 8)

Question 1.
i. One dozen means how many items?
ii. One gross means how many items?
Answer:
i. One dozen means 12 items.
ii. One gross means 144 items.

Try this (Textbook Page No. 10)

Question 1.
Calculate the volume in dm³ occupied by 60.0 g of ethane at STP.
Solution:
Given: Mass of ethane at STP = 60.0 g
To find: Volume of ethane
Formulae:

Calculation: Molar volume of a gas = 22.4 dm³ mol^-1 at STP
Molecular mass of ethane = 30 g mol^-1

∴ Volume of the gas at STP = Number of moles of a gas (n) × Molar volume of a gas
= 2 mol × 22.4 dm³ mol^-1 = 44.8 dm³
Ans: Volume of ethane = 44.8 dm³

Activity :

Activity 1.
Collect information from various scientists and prepare charts of their contributions to chemistry.
Answer:

Scientists		Contributions
Joseph Louis Gay-Lussac (1778 – 1850) (French chemist and physicist)	i.	Formulated the gas law.
	ii.	Collected samples of air at different heights and recorded temperatures and moisture contents.
	iii.	Discovered that the composition of the atmosphere does not change with increasing altitude.
Amedeo Avogadro (1776 – 1856) (Italian scholar)	i.	Published article in French journal on determining the relative masses of elementary particles of bodies and proportions by which they enter combinations.
	ii.	Published a research paper titled “New considerations on the theory of proportions and on the determination of the masses of atoms.”

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 14 Semiconductors Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 14 Semiconductors

1. Choose the correct option.

Question 1.
Electric conduction through a semiconductor is due to:
(A) Electrons
(B) holes
(C) none of these
(D) both electrons and holes
Answer:
(D) both electrons and holes

Question 2.
The energy levels of holes are:
(A) in the valence band
(B) in the conduction band
(C) in the band gap but close to valence band
(D) in the band gap but close to conduction band
Answer:
(C) in the band gap but close to valence band

Question 3.
Current through a reverse biased p-n junction, increases abruptly at:
(A) Breakdown voltage
(B) 0.0 V
(C) 0.3V
(D) 0.7V
Answer:
(A) Breakdown voltage

Question 4.
A reverse biased diode, is equivalent to:
(A) an off switch
(B) an on switch
(C) a low resistance
(D) none of the above
Answer:
(A) an off switch

Question 5.
The potential barrier in p-n diode is due to:
(A) depletion of positive charges near the junction
(B) accumulation of positive charges near the junction
(C) depletion of negative charges near the junction,
(D) accumulation of positive and negative charges near the junction
Answer:
(D) accumulation of positive and negative charges near the junction

2. Answer the following questions.

Question 1.
What is the importance of energy gap in a semiconductor?
Answer:

1. The gap between the bottom of the conduction band and the top of the valence band is called the energy gap or the band gap.
2. This band gap is present only in semiconductors and insulators.
3. Magnitude of the band gap plays a very important role in the electronic properties of a solid.
4. Band gap in semiconductors is of the order of 1 eV.
5. If electrons in valence band of a semiconductor are provided with energy more than band gap energy (in the form of thermal energy or electrical energy), then the electrons get excited and occupy energy levels in conduction band. These electrons can easily take part in conduction.

Question 2.
Which element would you use as an impurity to make germanium an n-type semiconductor?
Answer:
Germanium can be made an n-type semiconductor by doping it with pentavalent impurity, like phosphorus (P), arsenic (As) or antimony (Sb).

Question 3.
What causes a larger current through a p-n junction diode when forward biased?
Answer:
In case of forward bias the width of the depletion region decreases and the p-n junction offers a low resistance path allowing a high current to flow across the junction.

Question 4.
On which factors does the electrical conductivity of a pure semiconductor depend at a given temperature?
Answer:
For pure semiconductor, the number density of free electrons and number density of holes is equal. Thus, at a given temperature, the conductivity of pure semiconductor depends on the number density of charge carriers in the semiconductor.

Question 5.
Why is the conductivity of a n-type semiconductor greater than that of p-type semiconductor even when both of these have same level of doping?
Answer:

1. In a p-type semiconductor, holes are majority charge carriers.
2. When a p-type semiconductor is connected to terminals of a battery, holes, which are not actual charges, behave like a positive charge and get attracted towards the negative terminal of the battery.
3. During transportation of hole, there is an indirect movement of electrons.
4. The drift speed of these electrons is less than that in the n-type semiconductors. Mobility of the holes is also less than that of the electrons.
5. As, electrical conductivity depends on the mobility of charge carriers, the conductivity of a n-type semiconductor is greater than that of p-type semiconductor even when both of these have same level of doping.

3. Answer in detail.

Question 1.
Explain how solids are classified on the basis of band theory of solids.
Answer:
i. The solids can be classified into conductors, insulators and semiconductors depending on the distribution of electron energies in each atom.

ii. As an outcome of the small distances between atoms, the resulting interaction amongst electrons and the Pauli’s exclusion principle, energy bands are formed in the solids.

iii. In metals, conduction band and valence band overlap. However, in a semiconductor or an insulator, there is gap between the bottom of the conduction band and the top of the valence band. This is called the energy gap or the band gap.

iv. For metals, the valence band and the conduction band overlap and there is no band gap as shown in figure (b). Therefore, electrons can easily gain electrical energy when an external electric field is applied and are easily available for conduction.

v. In case of semiconductors, the band gap is fairly small, of the order of 1 eV or less as shown in figure (c). Hence, with application of external electric field, electrons get excited and occupy energy levels in conduction band. These can take part in conduction easily.

vi. Insulators, on the contrary, have a wide gap between valence band and conduction band of the order of 5 eV (for diamond) as shown in figure (d). Therefore, electrons find it very difficult to gain sufficient energy to occupy energy levels in conduction band.

vii. Thus, an energy band gap plays an important role in classifying solids into conductors, insulators and semiconductors based on band theory of solids.

Question 2.
Distinguish between intrinsic semiconductors and extrinsic semiconductors
Answer:

Intrinsic semiconductors	Extrinsic semiconductors
1. A pure semiconductor is known as intrinsic semiconductors.	The semiconductor, resulting
2. Their conductivity is low	Their conductivity is high even at room temperature.
3. Its electrical conductivity is a function of temperature alone.	Its electrical conductivity depends upon the temperature as well as on the quantity of impurity atoms doped in the structure.
4. The number density of holes (nh) is same as the number density of free electron (ne) (nh = ne).	The number density of free electrons and number density of holes are unequal.

Question 3.
Explain the importance of the depletion region in a p-n junction diode.
Answer:
i. The region across the p-n junction where there are no charges is called the depletion layer or the depletion region.

ii. During diffusion of charge carriers across the junction, electrons migrate from the n-side to the p-side of the junction. At the same time, holes are transported from p-side to n-side of the junction.

iii. As a result, in the p-type region near the junction there are negatively charged acceptor ions, and in the n-type region near the junction there are positively charged donor ions.

iv. The potential barrier thus developed, prevents continuous flow of charges across the junction. A state of electrostatic equilibrium is thus reached across the junction.

v. Free charge carriers cannot be present in a region where there is a potential barrier. This creates the depletion region.

vi. In absence of depletion region, all the majority charge carriers from n-region (i.e., electron) will get transferred to the p-region and will get combined with the holes present in that region. This will result in the decreased efficiency of p-n junction.

vii. Hence, formation of depletion layer across the junction is important to limit the number of majority carriers crossing the junction.

Question 4.
Explain the I-V characteristic of a forward biased junction diode.
Answer:

1. Figure given below shows the I-V characteristic of a forward biased diode.
2. When connected in forward bias mode, initially, the current through diode is very low and then there is a sudden rise in the current.
3. The point at which current rises sharply is shown as the ‘knee’ point on the I-V characteristic curve.
4. The corresponding voltage is called the knee voltage. It is about 0.7 V for silicon and 0.3 V for germanium.
5. A diode effectively becomes a short circuit above this knee point and can conduct a very large current.
6. To limit current flowing through the diode, resistors are used in series with the diode.
7. If the current through a diode exceeds the specified value, the diode can heat up due to the Joule’s heating and this may result in its physical damage.

Question 5.
Discuss the effect of external voltage on the width of depletion region of a p-n junction.
Answer:

1. A p-n junction can be connected to an external voltage supply in two possible ways.
2. A p-n junction is said to be connected in a forward bias when the p-region connected to the positive terminal and the n-region is connected to the negative terminal of an external voltage source.
3. In forward bias connection, the external voltage effectively opposes the built-in potential of the junction. The width of depletion region is thus reduced.
4. The second possibility of connecting p-n junction is in reverse biased electric circuit.
5. In reverse bias connection, the p-region is connected to the negative terminal and the n-region is connected to the positive terminal of the external voltage source. This external voltage effectively adds to the built-in potential of the junction. The width of potential barrier is thus increased

11th Physics Digest Chapter 14 Semiconductors Intext Questions and Answers

Internet my friend (Textbookpage no. 256)

i. https://www.electronics-tutorials.ws/diode
ii. https://www.hitachi-hightech.com
iii. https://nptel.ac.in/courses
iv. https://physics.info/semiconductors
v. http://hyperphysics.phy- astr.gsu.edu/hbase/Solids/semcn.html

[Students are expected to visit above mentioned links and collect more information regarding semiconductors.]

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 13 Electromagnetic Waves and Communication System Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

1. Choose the correct option.

Question 1.
The EM wave emitted by the Sun and responsible for heating the Earth’s atmosphere due to the greenhouse effect is
(A) Infra-red radiation
(B) X-ray
(C) Microwave
(D) Visible light
Answer:
(A) Infra-red radiation

Question 2.
Earth’s atmosphere is richest in
(A) UV
(B) IR
(C) X-ray
(D) Microwaves
Answer:
(B) IR

Question 3.
How does the frequency of a beam of ultraviolet light change when it travels from air into glass?
(A) depends on the values of p and e
(B) increases
(C) decreases
(D) remains same
Answer:
(D) remains same

Question 4.
The direction of EM wave is given by
(A) \(\bar{E}\) × \(\bar{B}\)
(B) \(\bar{E}\).\(\bar{B}\)
(C) along \(\bar{E}\)
(D) along \(\bar{B}\)
Answer:
(A) \(\bar{E}\) × \(\bar{B}\)

Question 5.
The maximum distance upto which TV transmission from a TV tower of height h can be received is proportional to
(A) h½
(B) h
(C) h^3/2
(D) h²
Answer:
(A) h½

Question 6.
The waves used by artificial satellites for communication purposes are
(A) Microwave
(B) AM radio waves
(C) FM radio waves
(D) X-rays
Answer:
(A) Microwave

Question 7.
If a TV telecast is to cover a radius of 640 km, what should be the height of transmitting antenna?
(A) 32000 m
(B) 53000 m
(C) 42000 m
(D) 55000 m
Answer:
(A) 32000 m

2. Answer briefly.

Question 1.
State two characteristics of an EM wave.
Answer:
i. The electric and magnetic fields, \(\vec{E}\) and \(\vec{B}\) are always perpendicular to each other and also to the direction of propagation of the EM wave. Thus, the EM waves are transverse waves.

ii. The cross product (\(\vec{E}\) × \(\vec{B}\)) gives the direction in which the EM wave travels. (\(\vec{E}\) × \(\vec{B}\)) also gives the energy carried by EM wave.

Question 2.
Why are microwaves used in radar?
Answer:
Microwaves are used in radar systems for identifying the location of distant objects like ships, aeroplanes etc.

Question 3.
What are EM waves?
Answer:
Waves that are caused by the acceleration of charged particles and consist of electric and magnetic fields vibrating sinusoidally at right angles to each other and to the direction of propagation are called EM waves or EM radiation.

Question 4.
How are EM waves produced?
Answer:

1. According to quantum theory, an electron, while orbiting around the nucleus in a stable orbit does not emit EM radiation even though it undergoes acceleration.
2. It will emit an EM radiation only when it falls from an orbit of higher energy to one of lower energy.
3. EM waves (such as X-rays) are produced when fast moving electrons hit a target of high atomic number (such as molybdenum, copper, etc.).
4. An electric charge at rest has an electric field in the region around it but has no magnetic field.
5. When the charge moves, it produces both electric and magnetic fields.
6. If the charge moves with a constant velocity, the magnetic field will not change with time and hence, it cannot produce an EM wave.
7. But if the charge is accelerated, both the magnetic and electric fields change with space and time and an EM wave is produced.
8. Thus, an oscillating charge emits an EM wave which has the same frequency as that of the oscillation of the charge.

Question 5.
Can we produce a pure electric or magnetic wave in space? Why?
Answer:
No.
In vacuum, an electric field cannot directly induce another electric field so a “pure” electric field wave cannot exist and same can be said for a “pure” magnetic wave.

Question 6.
Does an ordinary electric lamp emit EM waves?
Answer:
Yes, ordinary electric lamp emits EM waves.

Question 7.
Why light waves travel in vacuum whereas sound wave cannot?
Answer:
Light waves are electromagnetic waves which can travel in vacuum whereas sound waves travel due to the vibration of particles of medium. Without any particles present (like in a vacuum) no vibrations can be produced. Hence, the sound wave cannot travel through the vacuum.

Question 8.
What are ultraviolet rays? Give two uses.
Answer:
Production:

1. Ultraviolet rays can be produced by the mercury vapour lamp, electric spark and carbon arc lamp.
2. They can also be obtained by striking electrical discharge in hydrogen and xenon gas tubes.
3. The Sun is the most important natural source of ultraviolet rays, most of which are absorbed by the ozone layer in the Earth’s atmosphere.

Uses:

1. Ultraviolet rays destroy germs and bacteria and hence they are used for sterilizing surgical instruments and for purification of water.
2. Used in burglar alarms and security systems.
3. Used to distinguish real and fake gems.

Question 9.
What are radio waves? Give its two uses.
Answer:

1. Radio waves are produced by accelerated motion of charges in a conducting wire. The frequency of waves produced by the circuit depends upon the magnitudes of the inductance and the capacitance.
2. Thus, by choosing suitable values of the inductance and the capacitance, radio waves of desired frequency can be produced.

Uses:

1. Radio waves are used for wireless communication purpose.
2. They are used for radio broadcasting and transmission of TV signals.
3. Cellular phones use radio waves to transmit voice communication in the ultra high frequency (UHF) band.

Question 10.
Name the most harmful radiation entering the Earth’s atmosphere from the outer space.
Answer:
Ultraviolet radiation.

Question 11.
Give reasons for the following:
i. Long distance radio broadcast uses short wave bands.
ii. Satellites are used for long distance TV transmission.
Answer:
i. Long distance radio broadcast uses short wave bands because electromagnetic waves only in the frequency range of short wave bands only are reflected by the ionosphere.

ii. a. It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere.
b. Hence, satellites are helpful in long distance TV transmission.

Question 12.
Name the three basic units of any communication system.
Answer:
Three basic (essential) elements of every communication system are transmitter, communication channel and receiver.

Question 13.
What is a carrier wave?
Answer:
The high frequency waves on which the signals to be transmitted are superimposed are called carrier waves.

Question 14.
Why high frequency carrier waves are used for transmission of audio signals?
Answer:
An audio signal has low frequency (<20 kHz) and low frequency signals cannot be transmitted over large distances. Because of this, a high frequency carrier waves are used for transmission.

Question 15.
What is modulation?
Answer:
The signals in communication system (e.g. music, speech etc.) are low frequency signals and cannot be transmitted over large distances. In order to transmit the signal to large distances, it is superimposed on a high frequency wave (called carrier wave). This process is called modulation.

Question 16.
What is meant by amplitude modulation?
Answer:
When the amplitude of carrier wave is varied in accordance with the modulating signal, the process is called amplitude modulation.

Question 17.
What is meant by noise?
Answer:

1. A random unwanted signal is called noise.
2. The source generating the noise may be located inside or outside the system.
3. Efforts should be made to minimize the noise level in a communication system.

Question 18.
What is meant by bandwidth?
Answer:
The bandwidth of an electronic circuit is the range of frequencies over which it operates efficiently.

Question 19.
What is demodulation?
Answer:
The process of regaining signal from a modulated wave is called demodulation. This is the reverse process of modulation.

Question 20.
What type of modulation is required for television broadcast?
Answer:
Amplitude modulation is required for television broadcast.

Question 21.
How does the effective power radiated by an antenna vary with wavelength?
Answer:

1. To transmit a signal, an antenna or an aerial is needed.
2. Power radiated from a linear antenna of length l is, P ∝ (\(\frac {l}{λ}\))²
   where, λ is the wavelength of the signal.

Question 22.
Why should broadcasting programs use different frequencies?
Answer:
If broadcasting programs run on same frequency, then the information carried by these waves will get mixed up with each other. Hence, different broadcasting programs should run on different frequencies.

Question 23.
Explain the necessity of a carrier wave in communication.
Answer:

1. Without a carrier wave, the input signals could be carried by very low frequency electromagnetic waves but it will need quite a bit of amplification in order to transmit those very low frequencies.
2. The input signals themselves do not have much power and need a fairly large antenna in order to transmit the information.
3. Hence, it is necessary to impose the input signal on carrier wave as it requires less power in order to transmit the information.

Question 24.
Why does amplitude modulation give noisy reception?
Answer:
i. In amplitude modulation, carrier is varied in accordance with the message signal.

ii. The higher the amplitude, the greater is magnitude of the signal. So even if due to any reason, the magnitude of the signal changes, it will lead to variation in the amplitude of the signal. So its easy for noise to disturb the amplitude modulated signal.

Question 25.
Explain why is modulation needed.
Answer:
Modulation helps in avoiding mixing up of signals from different transmitters as different carrier wave frequencies can be allotted to different transmitters. Without the use of these waves, the audio signals, if transmitted directly by different transmitters, would get mixed up.

3. Solve the numerical problem.

Question 1.
Calculate the frequency in MHz of a radio wave of wavelength 250 m. Remember that the speed of all EM waves in vacuum is 3.0 × 10⁸ m/s.
Answer:
Given: λ = 250 m, c = 3 × 10⁸ m/s
To find: Frequency (v)
Formula: c = v⁸
Calculation: From formula,
v = \(\frac {c}{λ}\) = \(\frac {3×10^8}{250}\) = 1.2 × 10⁶ Hz
= 1.2 MHz

Question 2.
Calculate the wavelength in nm of an X-ray wave of frequency 2.0 × 10¹⁸ Hz.
Solution:
Given: c = 3 × 10⁸, v = 2 × 10¹⁸ Hz
To find: Wavelength (λ)
Formula: c = vλ
Calculation. From formula,
λ = \(\frac {c}{v}\) = \(\frac {3×10^8}{2×10^{18}}\) = 1.5 × 10^-10
= 0.15 nm

Question 3.
The speed of light is 3 × 10⁸ m/s. Calculate the frequency of red light of wavelength of 6.5 × 10^-7 m.
Answer:
Given: c = 3 × 10⁸ m/s, λ = 6.5 × 10^-7 m
To find: Frequency (v)
Formula: c = vλ
Calculation: From formula,
v = \(\frac {c}{λ}\) = \(\frac {3×10^8}{6.5×10^{-7}}\) = 4.6 × 10¹⁴ Hz

Question 4.
Calculate the wavelength of a microwave of frequency 8.0 GHz.
Answer:
Given: v = 8 GHz = 8 × 10⁹ Hz,
c = 3 × 10⁸ m/s
To find: Wavelength (λ)
Formula: c = vλ
Calculation: From formula,
λ = \(\frac {c}{λ}\) = \(\frac {3×10^8}{8×10^9}\) = 3.75 × 10^-2
= 3.75 cm

Question 5.
In a EM wave the electric field oscillates sinusoidally at a frequency of 2 × 10¹⁰ What is the wavelength of the wave?
Answer:
Given: v = 2 × 10¹⁰ Hz, c = 3 × 10⁸ m
To find: Wavelength (λ)
Formula: c = vλ
Calculation: From formula,
λ = \(\frac {c}{λ}\) = \(\frac {3×10^8}{2×10^{10}}\) = 1.5 × 10^-2

Question 6.
The amplitude of the magnetic field part of a harmonic EM wave in vacuum is B0 = 5 X 10-7 T. What is the amplitude of the electric field part of the wave?
Answer:
Given: B₀ = 5 × 10^-7 T, c = 3 × 10⁸
To find: Amplitude of electric field (E₀)
Formula: c = \(\frac {E_0}{B_0}\)
Calculation /From formula,
E₀ = c × B₀
= 3 × 10⁸ × 5 × 10^-7
= 150 V/m

Question 7.
A TV tower has a height of 200 m. How much population is covered by TV transmission if the average population density around the tower is 1000/km²? (Radius of the Earth = 6.4 × 10⁶ m)
Answer:
Given: h = 200 m,
Population density (n)
= 1000/km² = 1000 × 10^-6/m² = 10^-3/m²
R = 6.4 ×10⁶ m
To find: Population covered
Formulae: i. A = πd² = π(\(\sqrt{2Rh}\))² = 2πRh
ii. Population covered = nA
Calculation /From formula (i),
A = 2πRh
= 2 × 3.142 × 6.4 × 10⁶ × 200
≈ 8 × 10⁹ m²
From formula (ii),
Population covered = nA
= 10^-3 × 8 × 10⁹
= 8 × 10⁶

Question 8.
Height of a TV tower is 600 m at a given place. Calculate its coverage range if the radius of the Earth is 6400 km. What should be the height to get the double coverage area?
Answer:
Given: h = 600 m, R = 6.4 × 10⁶ m
To find: Range (d)
Height to get the double coverage (h’)
Formula: d = \(\sqrt{2hR}\)
Calculation: From formula,
d = \(\sqrt{2×600×6.4×10^6}\) = 87.6 × 10³ = 87.6 km
Now, for A’ = 2A
π(d’)² = 2 (πd²)
∴ (d’)² = 2d²
From formula,
h’ = \(\frac{(d’)^2}{2R}\)
= \(\frac{2d^2}{2R}\)
= 2 × h ……….. (∵ h = \(\frac{d^2}{2R}\))
= 2 × 600
=1200 m

Question 9.
A transmitting antenna at the top of a tower has a height 32 m and that of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in line of sight mode? Given radius of Earth is 6.4 × 10⁶ m.
Answer:
Given: h_t = 32 m, h_r = 50 m, R = 6.4 × 10⁶ m
To find: Maximum distance or range (d)
Formula: d = \(\sqrt{2Rh}\)
Calculation: From formula,
d_t = \(\sqrt{2Rh_t}\) = \(\sqrt{2×6.4×10^6×32}\)
= 20.238 × 10³ m
= 20.238 km
d_r = \(\sqrt{2Rh_t}\)
= \(\sqrt{2×6.4×10^6×50}\)
= 25.298 × 10³ m
= 25.298 km
Now, d = d_t + d_r
= 20.238 + 25.298
= 45.536 km

11th Physics Digest Chapter 13 Electromagnetic Waves and Communication System Intext Questions and Answers

Can you recall? (Textbookpage no. 229)

Question 1.
i. What is a wave?
Answer:
Wave is an oscillatory disturbance which travels through a medium without change in its form.

ii. What is the difference between longitudinal and transverse waves?
Answer:
a. Transverse wave: A wave in which particles of the medium vibrate in a direction perpendicular to the direction of propagation of wave is called transverse wave.
b. Longitudinal wave: A wave in which particles of the medium vibrate in a direction parallel to the direction of propagation of wave is called longitudinal wave.

iii. What are electric and magnetic fields and what are their sources?
Answer:
a. Electric field is the force experienced by a test charge in presence of the given charge at the given distance from it.
b. A magnetic field is produced around a magnet or around a current carrying conductor.

iv. By which mechanism heat is lost by hot bodies?
Answer:
Hot bodies lose the heat in the form of radiation.

Question 2.
What are Lenz’s law, Ampere’s law and Faraday’s law?
Answer:
Lenz’s law:
Whereas, Lenz’s law states that, the direction of the induced emf is such that the change is opposed.

Ampere’s law:
Ampere’s law describes the relation between the induced magnetic field associated with a loop and the current flowing through the loop.

Faraday’s law:
Faraday’s law states that, time varying magnetic field induces an electromotive force (emf) and an electric field.

Internet my friend. (Tpxtboakpage no. 240)

https//www.iiap.res.in/centers/iao
[Students are expected to visit the above mentioned website and collect more information about different EM wave propagations used by astronomical observatories.]

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 14 Basic Principles of Organic Chemistry Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 14 Basic Principles of Organic Chemistry

1. Answer the following :

Question A.
Write condensed formulae and bond line formulae for the following structures.

Answer:

Question B.
Write dash formulae for the following bond line formulae.

Answer:

Question C.
Write bond-line formulae and condensed formulae for the following compounds
a. 3-methyloctane
b. hept-2-ene
c. 2, 2, 4, 4- tetramethylpentane
d. octa-1,4-diene
e. methoxy ethane
Answer:

Question D.
Write the structural formulae for the following names and also write correct IUPAC names for them.
a. 5-ethyl-3-methylheptane
b. 2,4,5-trimethylthexane
c. 2,2,3-trimethylpentan-4-01
Answer:

Question E.
Identify more favourable resonance structure from the following. Justify.

Answer:
a.

Structure (I) will be more favourable resonance structure as structure (II) involves separation of opposite charges and the electronegative oxygen atom has a positive charge.

Both structures (I) and (II) involves separation of opposite charges, but structure (I) has a positive charge on the more electropositive ‘C’ and a negative charge on more electronegative ‘O’. Thus, structure (I) will be more favourable resonance structure.

Question F.
Find out all the functional groups present in the following polyfunctional compounds.
a. Dopamine a neurotransmitter that is deficient in Parkinson’s disease.

b. Thyroxine the principal thyroid hormone.

c. Penicillin G, a naturally occurring antibiotic

Answer:
i. Functional groups: Phenolic -OH group (Ar-OH) and primary amine (-NH₂) group are present in dopamine.
ii. Functional groups: Phenolic -OH group (Ar-OH), halide (-I), ether (Ar-O-Ar), primary amine (-NH₂) carboxylic acid (-COOH) groups are present in thyroxine.
iii. Functional groups: Secondary amide
,
carboxylic acid (-COOH), tertiary amide
,
thioether (R-S-R) groups are present in penicillin G.

Question G.
Find out the most stable species from the following. Justify.

Answer:
a. The most stable species from the given species is \(\left(\mathrm{H}_{3} \mathrm{C}\right)_{3} \dot{\mathrm{C}}\) i.e., tert-butyl radical.
This is because it has greater number of alkyl groups attached to the C-atom having unpaired electron. More the number of the alkyl groups, the greater will be +1 inductive (electron releasing) effect, and thereby greater will be the stability of the free radical.

b. The most stable species from the given species is \(\mathrm{CBr}_{3}^{-}\).
This is because it contains 3 -Br atoms, which exhibits electron withdrawing inductive effect. Carbanions are stabilized by -I inductive (electron withdrawing) effect. Larger the number of -I groups attached to the negatively charged carbon atom, lower will be the electron density on the carbon atom and higher will be its stability.

c. The most stable species from the given species is \(\stackrel{+}{\mathbf{C}} \mathbf{H}_{3}\).
This because it does not contain Cl atom, which exhibits electron withdrawing inductive effect. Carbocations are destabilized by -I inductive (electron withdrawing) effect. When more number of-I groups are attached to the positively charged carbon atom, the positive charge on the carbon atom increases further, thus destabilizing the species. Hence, the species with no -I groups will be most stable.

Question H.
Identify the α-carbons in the following species and give the total number of α-hydrogen in each.

Answer:
a.

In structure (i), C-2 and C-4 are α-carbon atoms.
Hydrogen atoms(s) attached to α-C atoms is a α-H atom. Thus, structure (i) contains 4 α-H atoms.
b.

In structure (ii), carbon atoms adjacent to C-2 are α-carbon atoms (as shown in the structure).
Thus, structure (ii) contains 6 α-H atoms.

c.

C-3 carbon atom, that is, C-atom next to (H₂C=CH-) is a α-C atom.
Thus, structure (iii) contains 2 α-H atoms.

Question I.
Identify primary, secondary, tertiary and quaternary carbon in the following compounds.

Answer:

2. Match the pairs

	Column ‘A’		Column ‘B’
i.	Inductive effect	a.	Delocalization of π  electrons
ii.	Hyperconjugation	b.	Displacement of π electrons
iii.	Resonance effect	c.	Delocalization of σ electrons
		d.	Displacement of σ electrons

Answer:
i – d,
ii – c,
iii – a

3. What is meant by homologous series ? Write the first four members of homologous series that begins with
A. CH₃CHO
B. H-C≡C-H
Also write down their general molecular formula.
Answer:
Homologous series: A series of compounds of the same family in which each member has the same type of carbon skeleton and functional group, and differs from the next member by a constant difference of one methylene group (-CH₂-) in its molecular and structural formula is called as homologous series.
A. CH₃CHO :

Comparing these molecular formulae and assigning the number of carbon atoms as ‘n’, the following general formula is deduced: C_nH_2nO/C_nH_2n-1CHO (where n = 1, 2, 3, …).

B. H-C≡C-H :

Comparing these molecular formulae and assigning the number of carbon atoms as ‘n’, the following general formula is deduced: C_nH_2n-2 (where n = 2, 3,4,….).

4. Write IUPAC names of the following

Answer:

5. Find out the type of isomerism exhibited by the following pairs.


Answer:
A. Metamerism
B. Functional group isomerism
C. Tautomerism
D. Tautomerism

6. Draw resonance srtuctures of the following :

A. Phenol
B. Benzaldehyde
C. Buta-1,3-diene
D. Acetate ion
Answer:
A. Resonance structures for phenol:

B. Resonance structures of benzaldehyde:

C. Resonance structures of Buta-1,3-diene:

D. Resonance structures of acetate ion:

7. Distinguish :

Question A.
Inductive effect and resonance effect
Answer:
Inductive effect:

1. Presence of polar covalent bond is required.
2. The polarity is induced in adjacent carbon- carbon single (covalent) bond due to a presence of influencing group (more electronegative atom than carbon).
3. Depending on the nature of influencing group it is differentiated as +I effect and -I effect.
4. The direction of the arrow head denotes the direction of the permanent electron displacement.

Resonance effect:

1. Presence of conjugated n electron system or species having an atom carrying p orbital attached to a multiple bond is required.
2. The polarity is produced in the molecule by the interaction of conjugated π bonds (or that between π bond and p orbital on the adjacent atom).
3. Depending on the nature of influencing group it is differentiated as +R and -R effect.
4. The delocalisation of n electrons is denoted by using curved arrows.

Question B.
Electrophile and nucleophile
Answer:
Electrophile:

1. Electrophile is an electron deficient species.
2. It is attracted towards negative charge (electron seeking).
3. It attacks a nucleophilic centre in the substrate and brings about an electrophilic reaction
4. It is an electron pair acceptor. (Lewis acid)
5. It can be a positively charged ion or a neutral species having a vacant orbital.
   e.g. H⁺, Br , \(\mathrm{NO}_{2}^{+}\), BF₃, AlCl₃, etc.

Nucleophile:

• Nucleophile is an electron rich species.
• It is attracted towards positive charge (nucleus seeking).
• It attacks the electrophilic centre in the substrate and brings about a nucleophilic reaction.
• It is an electron pair donor. (Lewis base)
• It can be negatively charged ion or neutral species having at least one lone pair of electrons.
  

C. Carbocation and carbanion
Answer:
Carbocation:

• It is a species in which carbon carries a positive charge.
• Positively charged carbon is sp² hybridized.
• It is electron-deficient.
• e.g. tert-Butyl carbocation, (CH₃)₃C⁺

Carbanion:

• It is a species in which carbon carries a negative charge.
• Negatively charged carbon is sp³/sp² hybridized.
• It is electron-rich.
• e.g.Methyl carbanion,
  

D. Homolysis and heterolysis
Answer:

8. Write true or false. Correct the false stament
A. Homolytic fission involves unsymmetrical breaking of a covalent bond.
B. Heterolytic fission results in the formation of free radicals.
C. Free radicals are negatively charged species
D. Aniline is heterocyclic compound.
Answer:
A. False
Homolytic fission involves symmetrical breaking of a covalent bond.
B. False
Heterolytic fission results in the formation of charged ions like cation and anion.
C. False
Free radicals are electrically neutral/uncharged species.
D. False
Aniline is a homocyclic aromatic compound.

9. Phytane is naturally occuring alkane produced by the alga spirogyra and is a constituent of petroleum. The IUPAC name for phytane is 2, 6, 10, 14-tetramethyl hexadecane. Write zig-zag formula for phytane. How many primary, secondary, tertiary and quaternary carbons are present in this molecule.
Answer:
Zig-zag formula of phytane (2,6,10,14-tetramethyl hexadecane) is as follows:

Dash formula to represent types of C-atom:

In phytane, six 1° C-atoms, ten 2° C-atoms, four 3° C-atoms are present. Phytane does not contain any quaternary carbon atom in its structure.

10. Observe the following structures and answer the questions given below.

a. What is the relation between (i) and (ii) ?
b. Write IUPAC name of (ii).
c. Draw the functional group isomer of (i).
Answer:
a. (a) and (b) are chain isomers of each other.
b. IUPAC name of structure (b) is 2-methylpropanal.
c. Functional group isomer of (a) is butanone.

11. Observe the following and answer the questions given below

a. Name the reactive intermediae produced
b. Indicate the movement of electrons by suitable arrow to produce this intermediate
c. Comment on stability of this intermediate produced.
Answer:
i. The reactive intermediates produced are methyl free radicals:

ii. Stability order of alkyl free radicals is: \(\dot{\mathrm{C}} \mathrm{H}_{3}\) < 1° <2° <3°
Hence, \(\dot{\mathrm{C}} \mathrm{H}_{3}\) produced in the above reaction is least stable and highly reactive.

12. An electronic displacement in a covalent bond is represented by following notation.

A. Identify the effect
B. Is the displacement of electrons in a covalent bond temporary or permanent.
Answer:
A. The electronic displacement represented above is inductive effect (-I effect).
B. Inductive effect is a permanent electronic effect as it depends on the electronegativity of the atoms. In the given example, the displacement of electrons is permanent as Cl is more electronegative than C.

13. Draw all the no-bond resonance structures of isopropyl carbocation.
Answer:

14. A covalent bond in tert-butyl bromide breaks in a suitable polat solvent to give ions.
A. Name the anion produced by this breaking of a covalent bond.
B. Indicate the type of bond breaking in this case.
C. Comment on geometry of the cation formed by such bond cleavage.
Answer:
A. The anion produced by breaking of the covalent C – Br bond is bromide

B. Heterolytic cleavage/fission takes place as charged ions are produced.
C. tert-Butyl carbocation formed in the given cleavage has trigonal planar geometry.

15. Choose correct options

A. Which of the following statements are true with respect to electronic displacement in covalent bond ?
a. Inductive effect operates through π bond
b. Resonance effect operates through σ bond
c. Inductive effect operates through σ bond
d. Resonance effect operates through π bond
i. a. and b
ii. a and c
iii. c and d
iv. b and c
Answer:
iii. c and d

B. Hyperconjugation involves overlap of …………. orbitals
a. σ – σ
b. σ – p
c. p – p
d. π – π
Answer:
b. σ – p

C. Which type of isomerism is possible in CH₃CHCHCH₃?
a. Position
b. Chain
c. Geometrical
d. Tautomerism
Answer:
a. Position

D. The correct IUPAC name of the compound

is ……………
a. hept-3-ene
b. 2-ethylpent-2-ene
c. hex-3-ene
d. 3-methylhex-3-ene
Answer:
d. 3-methylhex-3-ene

E. The geometry of a carbocation is …………
a. linear
b. planar
c. tetrahedral
d. octahedral
Answer:
b. planar

F. The homologous series of alcohols has general molecular formula ………..
a. C_nH_2n+1OH
b. C_nH_2n+2OH
c. C_nH_2n-2OH
d. C_nH_2nOH
Answer:
a. C_nH_2n+1OH

G. The delocaalization of electrons due to overlap between p-orbital and sigma bond is called …………….
a. Inductive effect
b. Electronic effect
c. Hyperconjugation
d. Resonance
Answer:
c. Hyperconjugation

11th Chemistry Digest Chapter 14 Basic Principles of Organic Chemistry Intext Questions and Answers

Can you recall? (Textbook Page No. 204)

Question i.
Which is the essential element in all organic compounds?
Answer:
Carbon is the essential element in all organic compounds.

Question ii.
What is the unique property of carbon that makes organic chemistry a separate branch of chemistry?
Answer:

• All organic compounds contain carbon.
• Carbon atoms show catenation property in which carbon atoms combine with other carbon atoms to form long chains and rings.
• Carbon atom can also form multiple bonds with other carbon atoms and with atoms of other elements.
• Due to this property of self-linking of carbon, a large number of organic compounds like proteins, DNA, sugar, oils, etc., are formed.

Thus, the unique property of catenation of carbon makes organic chemistry a separate branch of chemistry.

Question iii.
Which classes of organic compounds are often used in our daily diet?
Answer:
Carbohydrates (sugars), proteins (pulses), fats (edible plant and animal oil) and vitamins are the major classes of organic compounds often used in our daily diet.

Try this. (Textbook Page No. 204)

Question 1.
Find out the structures of glucose, vanillin, camphor and paracetamol using internet. Mark the carbon atoms present in them. Assign the hybridization state to each of the carbon and oxygen atom. Identify sigma (σ) and pi (π) bonds in these molecules.
Answer:
i. Structure of glucose:

a. Hybridization of carbon: In glucose, only carbon at position C-1 is sp² hybridized. On the other hand, carbons at C-2, C-3, C-4, C-5 and C-6 positions are sp³ hybridized.
b. Hybridization of oxygen: Oxygen atom attached to C-1 is sp² hybridized, rest oxygen atoms attached to carbon at C-2, C-3, C-4, C-5 and C-6 are sp³ hybridized.
[Note: Here, the open chain structure of glucose is used to answer the given questions.]

ii. Structure of vanillin:

a. Hybridization of carbon: In vanillin, carbon atoms C-1 to C-7 are sp² hybridized. Only C-8 carbon is sp³ hybridized.
b. Hybridization of oxygen: Oxygen atom bonded to C-7 sp² hybridized whereas oxygen atom bonded to C-4 and C-8 are sp³ hybridized.

iii. Structure of camphor:

a. Hybridization of carbon: In camphor, all the carbons are sp³ hybridized except the carbonyl carbon which is sp² hybridized.
b. Hybridization of oxygen: The carbonyl oxygen is sp² hybridized.

iv. Structure of paracetamol:

a. Hybridization of carbon: In paracetamol, carbons present in the ring and carbon at C-7 position are sp² hybridized. Only C-8 carbon is sp³ hybridized.
b. Hybridization of oxygen: Oxygen atom attached to carbon at ,C-1 position is sp³ hybridized. Oxygen atom attached to carbon at C-7 position is sp² hybridized.

Question 2.
i. Draw the structural formula of ethane.
ii. Draw electron-dot structure of propane.
Ans:
i. Structural formula of ethane (C₂H₆) can be drawn as follows:

ii. Electron-dot structure of propane is given as,

Where ‘•’ represents valence electrons of carbon and hydrogen.

Try this (Textbook Page No. 205)

Complete the table:

Answer:

Try this. (Textbook Page No. 206)

Question 1.
Draw two Newman projection formulae and two Sawhorse formulae for the propane molecule.
Answer:
Structural formula of propane is:

Structural formula of propane:
i. Newman projection formulae for propane molecule can be given as:

ii. Sawhorse formula for propane molecule can be given as:

Can you tell? (Textbook Page No. 208)

Question 1.
Consider the following reaction:
2CH₃ – CH₂ – CH₂ – OH + 2Na → 2CH₃ – CH₂ – CH₂ – ONa + H₂
Compare the structure of the substrate propanol with that of the product sodium propoxide. Which part of the substrate, the carbon skeleton or the OH group has undergone a change during the reaction?
Answer:
In above reaction, the -OH group of the substrate molecule has undergone a change. The H-atom of hydroxyl group (-OH) is replaced by sodium forming the product.

Activity: (Textbook Page No. 219)

Observe the structural formulae (a) and (b).
i. Find out their molecular formulae.
ii. What is the difference between them?
iii. What is the relation between the two compounds represented by these structural formulae?
Answer:
i. Molecular formula of both (a) and (b) are same i.e., C₃H₆O.
ii. Compound (a), has a ketone (-CO-) functional group (i.e., acetone) and compound (b) has an aldehyde (-CHO) functional group (i.e., propionaldehyde). Both the compounds have different functional groups.
iii. Compound (a) and (b) are isomers of each other.
[Note: Aldehydes and ketones are the functional group isomers of each other.]

Can you tell? (Textbook Page No. 222)

Question 1.
Some bond fissions are described in the following table. For each of them, show the movement of electron/s using curved arrow notation. Classify them as homolysis or heterolysis and identify the intermediate species produced as carbocation, carbanion or free radical.

Answer:

Can you recall? (Textbook Page No. 223)

i. What is meant by ‘reagent’?
ii. Identify the ‘reagent’, ‘substrate’, ‘product’ and ‘byproduct’ in the following reaction.
CH₃COCl + NH₃ → CH₃CONH₂ + HCl
Answer:
i. The reactant which reacts with a substrate to form corresponding products is known reagent.
ii.

Can you recall? (Textbook Page No. 224)

i. How is covalent bond formed between two atoms?
ii. Consider two covalently bonded atoms Q and R where R is more electronegative than Q. Will these atoms share the electron pair equally between them?
iii. Represent the above polar covalent bond between Q and R using fractional charges δ⁺ and δ^–.
Answer:
i. A covalent bond is formed between two atoms by mutual sharing of electrons so as to complete their octets or duplets (in case of elements having only one shell).

ii. A covalent bond is formed between Q and R having different electronegativities, that is, R is more electronegative than Q. In such a case, the atom R with a higher value of electronegativity pulls the shared pair of electrons to a greater extent towards itself as compared to the atom Q with lower value of electronegativity. As a result of this, the shared pair of electrons will get shifted towards atom R. Thus, both the atoms Q and R will not share the electron pair equally between them.

iii. Polar covalent bond between Q and R can be represented as:
\(\mathrm{Q}^{\delta+}-\mathrm{R}^{\delta-}\)

Try this (Textbook Page No. 225)

i. Draw a bond line structure of benzene (C₆H₆).
ii. How many C – C and C = C bonds are there in this structure?
iii. Write down the expected values of the bond lengths of the carbon-carbon bonds in benzene (Refer chapter 5).
Answer:
i. Bond line structure of benzene:

ii. In benzene, there are three alternating C – C single bonds and C = C double bonds.
[Note: In benzene, there are six C – C sigma bonds and three C – C pi bonds.]
iii. The expected values of carbon-carbon bond lengths in benzene are:

Bond	Bond length
C – C	154 pm
C = C	133 pm

Can you recall? (Textbook Page No. 225)

i. Write down two Lewis structures for ozone. (Refer chapter 5)
ii. How are these two Lewis structures related to each other?
iii. What are these two Lewis structures called?
Answer:
i. Lewis structures of ozone can be shown as follows:

ii. In these two Lewis structures, the position of the atoms is same but the position of pair of electrons (or formal charge) is different. These two Lewis structures are considered equivalent to each other.
iii. These two Lewis structures are called as resonating or contributing or canonical structures.

Internet my friend (Textbook Page No. 229)

i. Basic principles of organic chemistry:
https://authors.library.caltech.edu/25034
ii. Collect information about isomerism.
Answer:
i. Students are expected to refer to the book provided in the above link to collect additional information on the basic principles of organic chemistry.

ii. https://www.compoundchem.com/2014/05/22/typesofisomerism/
chemdictionary.org/structural-isomers/
https://en.wikipedia.org/wiki/Structural_isomer
[Note: Students can use the above links as a reference and collect additional information about isomerism on their own.]