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

Maharashtra State Board 12th Chemistry Important Questions Chapter 8 Transition and Inner Transition Elements

Question 1.
What are d-block elements? Give their general electronic configuration.
Answer:
Definition: d-block elements are defined as the elements in which the differentiating electron enters d-orbital of the penultimate shell i.e. (n – 1) d-orbital where ‘n is the last shell.

The general electronic configuration can be represented as, (n – n) d^{n – 10}, ns^{n – 2}

Question 2.
What is the position of the transition elements in the periodic table?
Answer:
The transition elements are placed in periods 4 to 7 and groups 3 to 12 of the periodic table.

Question 3.
In which block of the modern periodic table are the transition elements placed?
Answer:
Transition elements are placed in d-block of the modern periodic table.

Question 4.
Why are most of the d-block elements called transition elements?
Answer:

• d-block elements have electronic configuration,(n – n) d^{n – 10}, ns^{l – 2}. They are all metals.
• In the periodic table, they are placed between the ,s-block and p-block elements, i.e., in the groups between 2 and 13.
• They show characteristic properties which are intermediate between those of the elements of s-block and p-block.
• Hence, they show a transition in the properties from those of the most electropositive .v-block elements and less
• electropositive (or electronegative) p-block elements.
• Therefore, most of the d-block elements are called transition elements.

Question 5.
How many series of d-block elements are present in the long-form periodic table? Give their general electronic configuration.
Answer:
There are four series of d-block elements which are placed between 5 and p-block elements in the long-form periodic table as follows :

Modern periodic table :

Question 6.
Represent the elements in the four series of transition elements.
Answer:

Question 7.
In which period of the periodic table, will an element, be found whose differentiating electron is a 4d-electron?
Answer:
An element whose differentiating electron is a 4d-electron will be present in fifth period of the periodic table.

Question 8.
Write the condensed electronic configuration of each series of transition elements.
Answer:
Condensed Electronic Configuration of Transition Elements

Question 9.
Write expected and observed electronic configuration of 3d-series block elements.
Answer:
Electronic configuration of 3d-series of d-block elements

Question 10.
Explain why transition elements with electronic configuration 3d44s2 and 3d94s2 do not exist.
Answer:
(1) d-orbitals are degenerate orbitals and they acquire extra stability when half-filled (3d⁵) or completely filled (3d¹⁰). Hence 3d⁴ and 3d⁹ electronic configurations are less stable.
(2) The energy difference between 3d and 4.s’ subshells is very low, hence there arises a transfer of one electron from 45 orbital to 3d orbital.
The electronic configuration changes as,
3d⁴, 4s² → 3d⁵ 4s¹
3d⁹, 4s² → 3d¹⁰ 45¹
Therefore transition elements, with electronic configurations 3d⁴, 4s² and 3d⁹, 4s² do not exist.

Question 11.
Write observed electronic configuration of elements from first transition series having half-filled d-orbitals.
Answer:
There are two elements namely Cr and Mn which have half-filled d-orbitals.
₂₄Crls²2s²2p⁶3s²3p⁶3d⁵4s¹
₂₅Mnls²2s²2p⁶3s²3p⁶3d⁵4s²

Question 12.
Explain the variable oxidation states of metals of first transition series.
Answer:

• The transition metals (or, elements) exhibit variable oxidation states due to their electronic configuration, (n – 1) d^{1 – 10} ns^{1 – 2} for the first row.
• They show only positive oxidation states due to loss of electrons from outer 45-orbital and the penultimate 3rf-orbital.
• Loss of one 45 electron forms M+ ion. Loss of two 45 electrons form M²⁺ ion.
• +2 is the common oxidation state of these elements.
• Higher oxidation states are due to loss of 3 d-electrons along with 45 electrons.
• As the number of unpaired electrons increases, the number of oxidation states shown by the element also increases.
• Sc has only one unpaired electron and it shows two oxidation states ( + 2 and + 3)
• Mn with 5 unpaired d electrons show six different oxidation states. They are +2, +3, +4, +5, +6 and + 7. Thus Mn has the highest oxidation state.
• From Fe onwards variable oxidation states decreases as the number of unpaired electron decreases.
• The last element in the series, Zn shows only one oxidation state ( + 2).

Question 13.
Show different oxidation states of 3d-series of transition elements.
Answer:
The following table shows, different oxidation states of 3d-series of transition elements.

Question 14.
Write oxidation states and outer electronic configuration of first transition series elements.
Answer:
Oxidation states of first transition series elements

Question 15.
Zinc shows only one oxidation slate. Explain.
Answer:

• The electronic conliguration of zinc is, ₃₀Zn Is² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² or [Ar] 3d¹⁰ 4s².
• Due lo loss of two electrons from 4s suhshell Zn shows oxidation state +2. with elcctronic configuration. [Ar] ¹⁸3d¹⁰.
• Since Zn⁺² acquires an extra stability of completely fIlled 3d¹⁰ orbital. it shows only one oxidation state + 2.

Question 16.
Why is manganese more stable in the + 2 state than the + 3 state and the reverse is true for iron?
Answer:

• The electronic configuration of Mn is ₂₅Mn [Ar] 3d⁵ 4s²
• In + 2 and + 3 oxidation states, the electronic configuration of Mn is, Mn²⁺ [Ar] 3d⁵ and Mn³⁺ [Ar] 3d⁴
• Since half-filled d-orbital (3d⁵) has more stability and lower energy than 3d⁴, Mn²⁺ is more stable than Mn³⁺.
• The electronic configuration of Fe is ₂₆Fe [Ar] 3d⁶ 4s² In + 2 and + 3 oxidation states of Fe, the electronic configuration is, Fe²⁺ [Ar] 3d⁶ and Fe³⁺ [Ar] 3d⁵ Since half-filled orbital is more stable, + 3 state of Fe is more stable than + 2 state.

Question 17.
What are the electronic configurations of various ions of 3d-elements?
Answer:
Electronic configuration of various ions of 3d elements

Question 18.
Scandium shows only two oxidation states. Explain.
Answer:
Scandium has electronic configuration, ₂₁Sc : Is², 2s², 2p⁶, 3s², 3p⁶, 3d¹, 4s² Sc shows only two oxidation states namely + 2 and + 3.

• Due to the loss of two electrons from the 4s-orbital, Sc acquires + 2 oxidation state Sc^{2 +} : Is² 2s² 2p⁶ 3s² 3p⁶ 3d¹.
• Due to the loss of one more electron from the 3d-orbital, it acquires + 3 oxidation state with the extra stability of an inert element ₁₈Ar. Sc⁺³ : Is² 2s² 2p⁶ 3s² 3p⁶.
• Since Sc³⁺ acquires extra stability of inert element [Ar]¹⁸, it does not form higher oxidation state.

Question 19.
Write different oxidation states of iron.
OR
Write the electronic configurations of
(i) Fe
(ii) Fe²⁺ and
(iii) Fe³⁺.
Answer:
Oxidation states of iron are +2, +3, +4, +5, +6.
(i) ₂₆Fe : ls²2s²2p⁶3s²3p⁶3d⁶4s²
(ii) Fe²⁺ : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
(iii) Fe³⁺ : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁵.

Question 20.
Explain different oxidation states of chromium.
Answer:

• The observed electronic configuration of chromium is, ₂₄Cr [Ar] 3d⁵ 4s¹.
• Different possible oxidation states of Cr are 4-1 (3d⁵), + 2 (3d⁴), + 3 (3d³), + 4 (3d²), + 5 (3d¹) and + 6 (3d°).
• Although in + 1 state, Cr gets extra stability of half-filled 3d⁵-orbital, it does not exhibit + 1 state in common except with pyridine.
• Cr⁺² has few stable salts like CrCl₂, CrSO₄ while Cr⁺³ forms very stable salts like CrCl₃.
• Cr⁺⁴ and Cr ^{+ 5} are unstable oxidation states.
• Cr⁺⁶ is the most stable state due to inert gas [Ar] electronic configuration and form the salts like K₂Cr₂O₇.

Question 21.
Manganese shows variable oxidation states. Give reasons.
Answer:

• Manganese (₂₅Mn) has electronic configuration. ₂₅Mn [Ar]¹⁸ 3d⁵ 4s².
• Mn has stable half-filled d-subshell.
• Due to a small difference in energy between 3d and 4s-orbitals, Mn can lose or share electrons from both the orbitals, hence shows variable oxidation states.
• Mn shows oxidation states ranging from + 2 to + 7.

Question 22.
Write the different oxidation states of manganese. Why is + 2 oxidation state of manganese more stable than Mn³⁺?
Answer:

• The different oxidation states of Mn are +2, +3, +4, +5, + 6 and +7.
• The electronic configuration of Mn is Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²
• + 2 oxidation state is very stable due to higher stability of half-filled 3d orbital.
• Mn³⁺ has electronic configuration, ls²2s² 2p⁶3s²3p⁶3d^A which is less stable.

Question 23.
Write the physical properties of first transition series.
Answer:
Physical properties of first transition series :

• All transition elements of the first series are metals.
• Except Zn, they are very hard and have low volatility.
• They show characteristic properties of metals. They are lustrous, malleable and ductile.
• They are good conductors of heat and electricity.
• They have high melting points and boiling points.
• Except Zn and Mn, they have one or more typical metallic structures at normal temperatures.

Question 24.
Which elements in the transition elements, 3d-series has
(i) the lowest density
(ii) the highest density?
Answer:
In 3d transition elements,
(i) Scandium (Sc) has lowest density and
(ii) Zinc (Zn) has the highest density.

Question 25.
Explain the variation in density of d-block elements.
Answer:
The densities of d-block elements are higher than 5-block elements due to higher nuclear charge which results in reduction in atomic size.

Question 26.
Explain the variation in melting points of the transition elements.
Answer:

1. All transition elements are metals and the strength of metallic bonding increases as the number of unpaired electrons increases.
2. In transition elements as atomic number increases, the number of unpaired electrons increases from (n – 1)d¹ to (n – 1 )d⁵.
   For example in 3d-series, melting points increase from ₂₁Sc to ₂₄Cr in 4d-series from ₃₉Y to ₄₂Mo, and in 5d-series from ₇₂Hf to ₇₄W.
3. After (n – l)d⁵ electronic configuration, the electrons start pairing, hence the number of unpaired electrons decrease, hence metallic character, melting points decrease from (n-1 )d6 to (n – 1)d¹⁰.
4. In all transition series the melting point increases steadily up to d⁵ configuration and after this melting point decreases regularly.
   

Question 27.
The first ionisation enthalpies of third transition series elements are much higher than those of the elements of first and second transition series. Explain.
Answer:

1. Third transition series elements have electronic configuration, 4f¹⁴ 5d^{1 – 10} 6s².
2. Thus, atoms of third series elements possess filled 4f-orbitals.
3. 4f-orbitals due to their diffused shape, exhibit poor shielding effect and give rise to lanthanide contraction. Hence the valence electrons experience greater nuclear attraction and greater amount of energy is required to ionise the elements of third transition series namely (H_f to Au).
4. Therefore the ionisation enthalpies of third transition series elements are much higher than those of the first and second transition series.

Question 28.
Explain the metalic character of transition metals.
Answer:

• All the transition elements are metals.
• They are hard, lustrous, malleable, ductile and they have high tensile strength.
• They have high melting points and boiling points.
• Their metallic character is due to vacant or partially filled (n – 1) d-orbitals, and they involve both metallic and covalent bonding.
• Since the strength of metallic bonds depends upon the number of unpaired electrons, it increases up to middle i.e., up to (n – 1 )d⁵, hence accordingly melting points and boiling points also increase.
• After (n – l)d⁵ configuration, the electrons start pairing, hence the metallic strength, melting points and boiling points decrease with the increase in atomic number.

Question 29.
How does metallic character vary in 3d transition elements?
Answer:

1. In 3d-series elements as atomic number increases from scandium (Sc [Ar]¹⁸ 3d¹ 4s²) the number of unpaired electrons increases up to 3d⁵ in chromium.
2. As the number of unpaired electrons increases, the metallic character increases, hence the melting points and boiling points increase from ₂₁Sc(3d¹) to ₂₄Cr (3d⁵).
3. After chromium the number of unpaired electrons goes on decreasing due to the pairing of electrons, hence metallic character, melting points and boiling points decrease from ₂₅Mn to ₂₉Cu.
4. Zinc has all electrons paired, hence it is soft, has a low melting and boiling points.

Question 30.
Which are the common arrangement of the atoms in the structure of transition metals?
Answer:
Most of the transition metals have simple hexagonal closed packed (hep), cubic closed packed (ccp) or body centred cubic (bcc) lattices.

Question 31.
Why do the compounds of transition metals exhibit magnetic properties?
Answer:
The compounds of transition metals exhibit magnetic properties due to the presence of unpaired electrons in their atoms or ions.

Question 32.
What is the cause of paramagnetism and ferromagnetism?
Answer:
Paramagnetism and ferromagnetism is due to the presence of unpaired electrons in species.

Question 33.
When does species become diamagnetic?
Answer:
When there is no unpaired electron, i.e. all electron spins are paired, the species become diamagnetic.

Question 34.
How do metals Fe, Co, Ni acquire permanent magnetic moment?
Answer:
The transition metals Fe, Co and Ni are ferromagnetic. When the magnetic field is applied, all the unpaired electrons in these metals (and their compounds) align in the direction of the applied magnetic field. Due to this the magnetic susceptibility is enhanced and these metals can be magnetised, that is, they acquire permanent magnetic properties.

Question 35.
In which oxidation state, is vanadium diamagnetic?
Answer:

• The electronic configuration of vanadium is, ₂₃V [Ar] 3d³ 4s².
• In +5 oxidation state, the electronic configuration is, V⁵⁺ [Ar].
• Since in V⁵⁺ state, vanadium has all electrons paired, it is diamagnetic.

Question 36.
How is a magnetic moment expressed?
Answer:
The magnetic moment is expressed in Bohr magneton (B.M.). It is denoted by μ.

Question 37.
What is Bohr magneton (B.M.)?
Answer:
Bohr magneton (B.M.) is a unit of magnetic moment :
\(1 \mathrm{~B} . \mathrm{M} .=\frac{e h}{4 \pi m_{\mathrm{e}} c}\)
where, h : Planck’s constant (h = 6.626 x 10^-34 Js)
e : electronic charge (1.60218 x 10^-19 C)
me : mass of an electron (9.109 x 10^-31 kg)
c : velocity of light. (2.998 x 10⁸ ms^-1)

Question 38.
Explain the magnetic properties of transition (or d-block) elements.
Answer:

• Most of the transition metal ions and their compounds are paramagnetic in nature due to the presence of one or more unpaired electrons in their (n – 1)d-orbitals. Hence they are attracted in the magnetic field.
• As the number of unpaired electrons increases from 1 to 5 in J-orbitals, the paramagnetic character and magnetic moment increase.
• The transition elements or their ions having all electrons paired are diamagnetic and they are repelled in the magnetic field.
• Metals like Fe, Co and Ni possess very high paramagnetism and acquire permanent magnetic moment hence they are ferromagnetic.

Question 39.
Explain the effective magnetic moment of the species.
Answer:

• The magnetic moment in the species arises due to the presence of unpaired electrons.
• The magnetic moment depends upon the sum of orbitals and spin contribution for each unpaired electron present in the species.
• In transition metal ions, the contribution of orbital magnetic moment is suppressed by the electrostatic field of other atoms, molecules or ions surrounding the metal ion in the compound.
• Hence the net or effective magnetic moment arises mainly due to spin of electrons. The effective magnetic moment μ_eff, of a paramagnetic substance is given by 1 spin only’ formula represented as, \(\mu=\sqrt{n(n+2)}\) B.M. where n is the number of unpaired electrons.

Question 40.
What is the importance of magnetic moment (μ)?
Answer:

• From the measurements of the magnetic moment (μ) of the species or metal complexes of the first row of transition elements, the number of unpaired electrons can be calculated with the spin-only formula.
• As magnetic moment is directly related to the number of unpaired electrons, value of μ will vary directly with the number of unpaired electrons.
• In 2nd and 3rd transition series, orbital angular moment is significant. Hence spin-only formula for the complexes of 2nd and 3rd transition series is not useful.

Question 41.
Calculate the magnetic moment of the following species :
(1) Cr³⁺
(2) Co
(3) Co³⁺
(4) Cu^{2 +}.
Answer:

Question 42.
Explain : A slight difference in the calculated and observed values of magnetic moments.
Answer:
Magnetic moments are determined experimentally in solution or in solid state where the central atom or ion is hydrated or bound to ligands. Hence a slight difference is observed in calculated and experimentally obtained values of magnetic moment (μ).

Question 43.
Calculate the magnetic moment of a divalent ion in an aqueous solution, if its atomic number is 24.
Answer:
(1) The electronic configuration of divalent inri M²⁺ having atomic number 24 is.

The ion has number of unpaired electrons. n = 4.
By spin only’ formula, the magnetic μ is given by, \(\mu=\sqrt{n(n+2)}=\sqrt{4(4+2)}=4.90 \mathrm{~B} . \mathrm{M}\)
(This M²⁺ ion is Cr²⁺ ion)

Question 44.
When does a substance appear coloured?
Answer:
A substance appears coloured when it absorbs a portion of visible light. The colour depends upon the wavelength of absorption in the visible region of electromagnetic radiation.

Question 45.
Why do the d-block elements form coloured compounds?
Answer:

• Compounds (or ions) of many d-block elements or transition metals are coloured.
• This is due to the presence of one or more unpaired electrons in (n – 1) d-orbital. The transition metals have incompletely filled (n – 1) cf-orbitals.
• The energy required to promote one or more electrons within the d-orbitals involving d-d transitions is very low.
• The energy changes for d-d transitions lie in visible region of electromagnetic radiation.
• Therefore transition metal ions absorb the radiation in the visible region and appear coloured.
• Colour of ions of d-block elements depends on the number of unpaired electrons in (n – 1) d-orbital. The ions having equal number of unpaired electrons have similar colour.
• The colour of metal ions is complementary to the colour of the radiation absorbed.

Question 46.
How is complementary colour of a compound identified?
Answer:

1. The transition metal ions absorb the radiation in the visible region and appeared coloured.
2. Metal ion absorbs radiation of certain wavelength from the visible region. Remaining light is transmitted and the observed colour corresponds to the complementary colour of the light observed.
3. The complementary colour can be identified (with the diagram given).

For example if red colour is absorbed then transmitted complementary colour is green.

Question 47.
Write outer electronic configuration (d-orbital) and colour of 3d-series of transition metal ions.
Answer:
Colour of 3d-transition metal ions

Question 48.
Mention the factors on which the colour of a transition metal ion depends.
Answer:
The factors on which the colour of transition metal ion depends are as follows :

• The presence of incompletely filled d-orbitals in metal ions. (The compounds with the configuration d° and d’0 are colourless.)
• The presence of unpaired electrons in d-orbitals.
• d → d transitions of electrons due to absorption of radiation in the visible region.
• Nature of groups (anions) (or ligands) linked to the metal ion in the compound or a complex.
• Type of hybridisation in metal ion in the complex.
• Geometry of the complex of the metal ion.

Question 49.
Give reasons : Zinc salts are colourless.
Answer:

• Colour of the ions of d-block elements depends on the number of unpaired electrons in (n – 1) d-orbitals.
• Zinc forms salts of Zn²⁺ ions.
• The electronic configuration of Zn⁺² is [Ar] 3d¹⁰.
• Since Zn⁺² does not have unpaired electrons in 3d-orbital, d→d transition cannot take place, hence, Zn⁺² ions form colourless salts.

Question 50.
Explain : The compounds of Cu(II) are coloured.
Answer:

• The electronic configuration of ₂₉Cu [Ar] 3d¹⁰ 4s¹ and Cu²⁺ [Ar] 3d⁹.
• In copper compounds Cu²⁺ ions have incompletely filled 3d-orbital (3d⁹).
• Due to the presence of one unpaired electron in 3 d-orbital, Cu²⁺ ions absorb red light from visible spectrum and emit blue radiation due to d → d transition. Therefore, copper compounds are coloured.

Question 51.
Explain why the solution of Ti³⁺ salt is purple in colour.
OR
Why is Ti³⁺ coloured? (atomic number Ti = 22)
Answer:

• Ti²⁺ ions in the aqueous solution exist in the hydrated complex form as [Ti(H₂O)₆]²⁺.
• The electronic configuration of Ti is, ₂₂Ti [Ar]¹⁸ 3d² 4s² and Ti³⁺ [Ar]¹⁸ 3d¹. Hence in complex, Ti³⁺ has one unpaired electron in 3d subshell.
• Initially, the 3d electron occupies lower energy d-orbital (in t_2g-orbitals).
• On the absorption of radiations of about 500 nm in yellow green region by a complex, 3d¹ electron is excited to the higher energy d-orbital (e_g-orbitals).
• When the electron returns back to the lower energy d-orbital (t_2g), it transmits radiation of complementary colour i.e. red blue or purple colour. Hence, the solution of hydrated Ti³⁺ is purple.

Question 52.
What will be the colour of Cd²⁺ salts? Explain.
Answer:

• The electronic configuration of, ₄₈Cd [Kr]³⁶ 3d¹⁰ 5s² and Cd²⁺ [Kr]³⁶ 3d¹⁰.
• Cd²⁺ ions have completely filled 3d subshell and there are no unpaired electrons in 3d-orbital.
• Hence d → d transition is not possible.
• Therefore, Cd²⁺ ions do not absorb radiations in the visible region and the salts of Cd²⁺ ions are colourless (or white).

Question 53.
Indicate which of the ions may be coloured- V³⁺, Sc³⁺, Cr³¹, Cu²⁺, Ti³⁺, Cu⁺
Answer:

• V³⁺ [Ar]¹⁸ 3d²-((green)
  Since there are two unpaired electrons available, for d → d transition, it will show a Green colour.
• Sc³⁺ [Ar]¹⁸ 3d° (colourless/white).
  Since there are no unpaired electrons in the 3d subshell, it will not show colour.
• Cr³⁺ [Ar]¹⁸ 3d³ – (violet)
  There are three unpaired electrons in the 3d subshell, hence due to d → d transition, it will show violet colour.
• Cu²⁺ [Ar]¹⁸ 3d⁹ (blue)
  It has one unpaired electron that can undergo a d → d transition, hence it will show the colour blue.
• Ti³⁺ [Ar]¹⁸ 3d¹ (purple)
  It has one unpaired electron that can undergo a d → d transition, hence it will show the colour purple.
• Cu¹⁺ [Ar]¹⁸ 3d¹⁰ (colourless)
  There are no unpaired electrons in the 3d subshell, hence it will not show colour.

Question 54.
Explain why is cobalt chloride pink in colour when dissolved in water but turns deep blue when treated with concentrated hydrochloric acid.
Answer:

• The electronic configuration of ₂₇Co : [Ar] 3d¹4s² and Co²⁺ [Ar] 3d¹.
• When dissolved in water cobalt chloride, Co²⁺ forms pink complex, [Co(H₂O)₆]²⁺.
• The complex has octahedral geometry.
• Due to absorption of radiation in the visible region and d – d transition, it forms pink coloured solution.
• When CoCl₂ solution is treated with concentrated HCl solution it turns deep blue.
• This change is due to the formation of another complex, [CoC1₄]²⁺ which has a tetrahedral geometry.
• Thus due to a change in geometry of the complex formed the colour of the solution changes from pink to deep blue.

Question 55.
Explain the catalytic properties of the rf-block or transition metals.
Answer:

• d-block elements or transition metals and their compounds or complexes influence the rate of a chemical reaction and hence act as catalysts.
• In homogeneous catalysis a catalyst forms an unstable intermediate compound which decomposes into products and regenerates the catalyst. But transition metals involve heterogeneous catalysis.
• The transition metals have incompletely filled d-subshells which adsorb reactants on the surface and provide a large surface area for the reactants to react.
• Since transition metals have variable oxidation states they are very good catalysts.
• Hence, compounds of Fe, Co, Ni, Pt, Pd, Cr etc are used as catalysts in many reactions.

Question 56.
Explain the use of different transition metals as catalysts.
Answer:
The transition metals are very good catalysts.

• MnO₂ is used as a catalyst in the decomposition of KClO₃.
  
• In the manufacture of ammonia by Haber’s process, Mo/Fe is used as a catalyst.
  
• In the synthesis of gasoline by Fischer Tropsch process, Co-Th alloy is used as a catalyst.
• Finely divided Ni (formed by reduction of heated oxide in hydrogen) is very efficient catalyst in hydrogenation of ethene to ethane at 140 °C.
  
• Commercially, hydrogenation with Ninkel as catalyst is used to convert inedible oils into solid fat for the production of margarine.
• In the contact process of industrial production of sulphuric acid, sulphur dioxide and oxygen (from air) react reversibly over a solid catalyst of platinised asbestos.
  
• Carbon dioxide and hydrogen are formed by the reaction of carbon monoxide and steam at 500 °C with Fe-Cr catalyst.
  

Question 57.
What are interstitial compounds of transition metals?
Answer:

• The interstitial compounds of the transition metals are those which are formed when small atoms like H, C or N are trapped inside the interstitial vacant spaces in the crystal lattices of the metals.
• Sometimes, sulphides and oxides are also trapped in the crystal lattices of transition elements.
• Presence of these elements in the crystal lattices of metals provide new properties to the metals.

Question 53.
Give one example of an interstitial compound.
Answer:
Steel and cast iron are examples of interstitial compounds of carbon and iron.

Question 54.
Give examples of interstitial compounds where the property of the transition metal is changed.
Answer:
Steel and cast iron are interstitial compounds of carbon and iron (carbides of iron). Due to the presence of carbon, the malleability and ductility of iron is reduced while its tenacity increases.

Question 55.
What are the properties of the interstitial compounds of transition metals?
Answer:

• The chemical properties of the interstitial compounds are the same as that of parent transition metals.
• They are hard and show the metallic properties like electrical and thermal conductivity, lustre, etc.
• Since metal-non-metal bonds in the interstitial compounds are stronger than metal-metal bonds in pure metals, the compounds have very high melting points, higher than the pure metals.
• They have lower densities than the parent metal.
• The interstitial compounds containing hydrogen (i.e., hydrides of metals) are powerful reducing agents.
• The compounds containing carbon, hence behaving as carbides, are chemically inert and extremely hard like diamond.
• In these compounds, malleability and ductility are changed. For example steel and cast iron.

Question 56.
What are interstitial compounds? Why do these compounds have higher melting points than corresponding pure metals?
Answer:

1. The interstitial compounds of the transition metals are those which are formed when small atoms like H, C or N are trapped inside the interstitial vacant spaces in the crystal lattice of the metals.
2. Since metal-nonmetal bonds in the interstitial compounds are stronger than metal-metal bonds in pure metals, the compounds have very high melting points, higher than the pure metals.

Question 57.
Explain the formation of alloys of transition metals.
Answer:

• The transition metals form a large number of alloys among themselves, which are hard with high melting points.
• During alloy formation atoms of one metal are distributed randomly in the lattice of another metal.
• The metals with similar atomic radii and similar properties readily form alloys.
• These alloys have industrial importance.
• The alloys can be ferrous alloys or nonferrous alloys.

Question 58.
How are the transition metal alloys classIfied?
Answer:
The transition metal alloys are classified into

• Ferrous alloys
• Nonferrous alloys.

Question 59.
Explain what are
(1) ferrous alloys and
(2) nonferrous alloys.
Answer:

1. Ferrous alloys: In ferrous alloys, atoms of other elemems are distributed randomly in atoms of iron in the mixture. As the percentage of iron is more in these alloys, they are termed as ferrous alloys. For expamle : nickel steel, chromium steel, stainless steel, (All steels have abot 2% carbon)
2. onferrous alloys : These are formed by mixing atoms of transition metal other than iron with a non transition elemeni. For example, brass is an alloy of Cu and Zn. Bronze is an alloy of Cu and Sn.

Question 60.
What are the uses of alloys?
Answer:

Question 61.
Write the preparation of potassium permanganate.
Answer:
Potassium permanganate (KMnO₄) is prepared in the following steps,

(1) Chemical Oxidation : When finely divided manganese dioxide (Mn0₂) is heated strongly with fused caustic potash (KOH) and an oxidising agent potassium chlorate (KCIO₃), dark green potassium manganate (K₂MnO₄) is obtained. (In neutral or acidic medium K₂MnO₄ disproportionates.)

The liquid is filtered through glass wool or sintered glass and evaporated. Potassium manganate crystallises as small, blackish crystals.

(2) Oxidation of K₂MnO₄ by
(i) Electrolytic oxidation : An alkaline solution of manganate ion is electrolysed between iron electrodes separated by a diaphragm. Manganate ion \(\left(\mathrm{MnO}_{4}^{2-}\right)\) undergoes oxidation at anode forming permanganate ion \(\left(\mathrm{MnO}_{4}^{-}\right)\). Oxygen evolved at anode converts \(\left(\mathrm{MnO}_{4}^{2-}\right)\) to \(\left(\mathrm{MnO}_{4}^{-}\right)\).

The overall reaction is as follows :
2K₂MnO₄ + H₂O + [O] → 2KMnO₄ + 2KOH

The electrolytic solution is filtered and evaporated to obtain deep purple black crystals of KMn0₄.

(ii) By passing CO₂ through the solution of K₂MnO₄ :
3K₂MnO₄ + 4CO₂ + 2H₂O → 2KMnO₄ + MnO₂ + 4 KHCO₃

Question 62.
What is meant by the disproportionation of an oxidation state? Explain giving example of manganese.
Answer:

1. Disproportionation reaction is a chemical reaction in which atom or an ion of an element forms two or more species having different oxidation states, one lower and one higher.
2. Manganese (Mn) shows different oxidation states + 2 to +7.
3. When one oxidation state, lower or higher oxidation state becomes unstable as compared to another oxidation state, it undergoes disproportionation reaction.
4. For example, + 6 oxidation state of Mn is less stable than + 7 and + 4.
   • Hence, in acidic medium \(\mathrm{Mn}^{6+} \text { in } \mathrm{MnO}_{4}^{2-}\) undergoes disproportionation reaction.
     
   • In neutral medium green K₂MnO₄ disproportionates to KMn04 and MnO₂.
     

Question 63.
Give examples of oxidising reactions of KMnO₄.
Answer:
(1) KMnO₄ in acidic medium :

(2) KMnO₄ in neutral or alkaline medium in neutral or weakly alkaline medium :
(i) Iodide is oxidised to iodate ion.

(ii) Thiosulphate ion is oxidised to sulphate ion.

(iii) Manganous salt is oxidised to MnO₂.

Question 64.
Balance the following equations :
KI + KMnO₄ + H₂SO₄ → K₂SO₄ + MnSO₄ + 8H₂O + I₂
H₂S + KMnO₄ + H₂SO₄ → K₂SO₄ + MnSO₄ + H₂O + S.
Answer:
10 KI + 2KMnO₄ + 8H₂SO_{4 →} 6K₂SO₄ + 2MnSO₄ + 8H₂O + 5I₂
5H₂S + 2KMnO₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 8H₂O + 5S.

Question 65.
Give the uses of potassium permanganate.
Answer:
Uses of potassium permanganate :

• as an antiseptic.
• as a powerful oxidising agent in laboratory and industry.
• in the detection of unsaturation in organic compounds in the laboratory. (Baeyer’s reagent, alkaline KMnO₄).
• for detecting halides in qualitative analysis.
• in volumetric analysis for the estimation of H₂O₂, FeSO₄ etc.)

Question 66.
Write the formula of chromite ore.
Answer:
FeOCr₂O₃.

Question 67.
How is potassium dichromate manufactured from chromite ore (FeOCr₂O₃)?
Answer:
Manufacture of potassium dichromate (K₂Cr₂O₂) from chrome iron ore (FeOCr₂O₃) involves following steps :
(1) Concentration of ore : The chromite ore (FeOCr₂O₃) is powdered and washed with current of water.
(2) Conversion of chromite ore into sodium chromate : The concentrated ore is mixed with anhydrous sodium carbonate (Na₂CO₃) and a flux of lime in excess air and heated in a reverberatory furnace.

Sodium chromate (Na₂CrO₄) formed in the reaction is then extracted with water so that Na₂CrO₄ dissolves into solution and insoluble substances separate out.
(3) Conversion of Na2CrO₄ into sodium dichromate (Na₂Cr₄O₇) : Na₂CrO₄ solution is acidified with concentrated H₂SO₂, so that sodium chromate is converted into sodium dichromate.

Less soluble sodium sulphate crystallises out as Na₂SO₄.10H₂O. which is filtered off.
(4) Conversion of Na₂Cr₂O₇ into K₂Cr₂O₇ : Concentrated solution of Na₂Cr₂O₇ is treated with KCl on by double decomposition, K₂Cr₂O₇ is obtained.
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl
On concentrating and cooling the solution, less soluble orange coloured K₂Cr₂O₇ crystallises out which is filtered and purified by recrystallisation.

Question 68.
What happens when hydrogen sulphide gas (H₂S) is passed through acidified K₂Cr₂O₇ solution?
Answer:
When hydrogen sulphide (H₂S) gas is passed into solution of K₂Cr₂O₇, H₂S is oxidised to a pale yellow solid (precipitate) of sulphur. Orange coloured solution becomes green due to formation of chromic sulphate (green coloured).

In the reaction, H₂S is oxidised to S and K₂Cr₂O₇ is reduced to Cr₂(SO₄)₃.

Question 69.
What are the common physical properties of d-block elements?
Answer:
The common physical properties of d-block elements are :

• All d-block elements are lustrous and shining.
• They are hard and have high density
• They have high melting and boiling points.
• They are good electrical and thermal conductors.
• They have high tensile strength and malleability.
• They can form alloys with transition and nontransition elements
• Many metals and their compounds are paramagnetic.
• Most of the metals are efficient catalysts.

Question 70.
What are the common chemical properties of d-block elements?
Answer:
The common chemical properties of the d-block elements are :

• All d-block elements are electropositive metals.
• They exhibit variable oxidation states and form coloured salts and complexes.
• They are good reducing agents.
• They form insoluble oxides and hydroxides.
• Iron, cobalt, copper, molybdenum and zinc are biologically important metals.
• They catalyse biological reactions.

Question 71.
Give examples to show that elements of first row of d-block elements differ from second and third row with respect to the stabilisation of higher oxidation states.
Answer:

• Highest oxidation state for the first row element is + 7 as in Mn.
  For the second row, the highest oxidation state is + 8 as in Ru (RuO₄).
  For the third row, the highest oxidation state is + 8 as in Os (OsO₄).
• Compounds of Mo(V) of 2nd row and W(VI) of 3rd row of transitional elements are more stable than Cr(VI) and Mn (VIII) of first row elements.

Question 72.
How do metals occur in nature?
Answer:
In nature, few metals occur in earth’s crust in free state or native state while other metals occur in the combined form.
(1) Elements in free state or native state : The metals which are non-reactive with air, water, CO₂ and non-metals occur in free state or native state. For example, gold, platinum, palladium occur in free state. Metals like Cu, Ag and Hg occur partly in the free state.

(2) Combined form : The metals which are reactive occur in the combined state with other elements forming compounds like oxides, sulphides, sulphates, carbonates, silicates, etc.

Question 73.
What are minerals?
Answer:
Minerals : They are naturally occurring chemical substances in the earth’s crust containing metal in free state or in combined form and obtainable from mining are called minerals. For example, haematite Fe203, galena PbS, etc.

Question 74.
What are ores?
Answer:
Ores : The minerals containing a high percentage of metals from which metals can be profitably extracted are called ores.
[Note : Every ore is a mineral but every mineral is not an ore.]

Question 75.
Write names of minerals and ores of Iron, Copper and Zinc.
Answer:

Question 76.
What is metallurgy?
Answer:
Metallurgy : The process of extraction of metal in a pure state from its ore is called metallurgy.

Question 77.
Define the following:
(1) Pyrometallurgy
(2) Hydrometallurgy
(3) Electrometallurgy.
Answer:

1. Pyrometallurgy : It is a process of extraction of metal from metal oxide from concentrated ore by reduction with a suitable reducing agent like carbon, hydrogen, aluminium, etc. at high temperature.
2. Hydrometallurgy : It is a process of extraction of metals by converting their ores into aqueous solutions of metal compounds and reducing them by suitable reducing agents.
3. Electrometallurgy : It is a process of extraction of highly electropositive metals like Na, K, Al, etc. by electrolysis of fused compounds of the metals where metal ions are reduced at cathode forming metals.

Question 78.
What is gangue?
Answer:
Gangue : The earthly and undesired impurities of various substances like sand (SiO₂), metal oxides, etc. present in the ore are called gangue or matrix.

Question 79.
Define concentration of an ore.
Answer:
Concentration : A process of removal of gangue or unwanted impurities from the ore is called concentration of an ore. It is also called benefaction or dressing of an ore.

Question 80.
What are common methods of concentration of an ore?
Answer:
The concentration of an ore involves different methods depending upon the differences in physical properties of compounds or the metal present and the nature of the gangue.

The common methods of concentration of ore are as follows :

1. Gravity separation or hydraulic washing :
   This can be carried out by two processes as follows :
   • Hydraulic washing by using Wilfley’s table method
   • Hydraulic classifier methods.
2. Magnetic separation
3. Froth floatation process.
4. Leaching.

The method depends upon the nature of ore.

Question 81.
What is leaching?
Answer:
Leaching : ft is a (chemical) process used in the concentration of an ore by extracting soluble material from an insoluble solid by dissolving in a suitable solvent. This method is used in the concentration process of ores of Al, Ag, Au, etc.

Question 82.
What is roasting of an ore?
Answer:
Roasting : It is a process of strongly heating a concentrated ore in the excess of air below melting point of metal, to convert it into oxide form. It is used for a sulphide ore. For example, ZnS ore on roasting forms ZnO.

Question 83.
Write an equation to show how zinc blende (ZnS) is converted to ZnO.
Answer:
When zinc blende is roasted, it is converted to ZnO.
\(\mathrm{ZnS}+\mathrm{O}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{ZnO}+\mathrm{SO}_{2}\)

Question 84.
Explain the term : Smelting
Answer:
Smelting : The process of extraction of a metal from its ore by heating and melting at high temperature is called smelting. Reduction of ore is carried out during smelting.

Question 85.
What is calcination?
Answer:
Calcination is a process in which the ore is heated to a high temperature below the melting point of the metal in the absence of air or limited supply of air in a reverberatory furnace.

It is generally used for carbonate and hydrated oxides to convert them into anhydrous oxides.

Question 86.
Define the terms :
(1) Flux
(2) Slag
Answer:
(1) Flux : A flux is a chemical substance which is added to the concentrated ore during smelting in order to remove the gangue or impurities by chemical reaction forming a fusible mass called slag.
(2) Slag : It is a waste product formed by combination of a flux and gangue (or impurities) during the extraction of metals by smelting process.

Iron is the fourth most abundant element in the earth’s crust.

Question 87.
What is the composition of haematite ore?
Answer:
Composition of Haematite ore is Fe2O₃ + SiO₂ + Al₂O₃ + phosphates

Question 88.
Which impurities (gangue) are present in haematite ore?
Answer:
SiO₂ and Al₂O₃ are the impurities present in the haematite ore.

Question 89.
Which reducing agents are used to reduce haematite ore into metallic iron?
Answer:
Haematite ore is reduced using coke and CO. Carbon in the coke is converted to carbon monoxide. Carbon and carbon monoxide together reduce Fe203 to metallic iron.

Fe₂O₃ + 3C → 2Fe + 3CO.
Fe₂O₃ + 3CO → 2Fe + 3CO₂.

Question 90.
Why is limestone used in the extraction of iron?
Answer:

• The ore of iron contains acidic gangue or impurity of silica, SiO₂.
• To remove silica gangue, basic flux like calcium oxide CaO, is required, which is obtained from the decomposition of limestone, CaCO₃. \(\mathrm{CaCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_{2}\)
• Silica reacts with CaO and forms a fusible slag of CaSiO₃.
  \(\mathrm{SiO}_{2}+\mathrm{CaO} \stackrel{\Delta}{\longrightarrow} \mathrm{CaSiO}_{3}\)

Therefore in the extraction of iron, lime is used.

Question 91.
Name the furnace in which iron is extracted from Haematite ore.
Answer:
Extraction of iron is carried out in Blast furnace.

Question 92.
Explain the extraction of iron from haematite.
Answer:
Iron is mainly extracted from haematite, Fe₂O₃ by reduction process.
Haematite ore contains silica (SiO₂), alumina (Al₂O₃) and phosphates as impurity or gangue.

Coke is used for the reduction of ore.

To remove acidic gangue SiO₂, a basic flux CaO is used which is obtained from lime stone CaCO₃.

The extraction process involves following steps :
(1) Concentration of an ore : The powdered ore is concentrated by gravity separation process by washing it in a current of water. The lighter impurities (gangue) are carried away leaving behind the ore.
(2) Roasting : The concentrated ore is heated strongly in a limited current of air. During this, moisture is removed and the impurities like S, As and phosphorus are oxidised to gaseous oxides which escape.

After roasting, the ore is sintered to form small lumps.
(3) Reduction (or smelting) : The roasted or calcined ore is then reduced by heating in a blast furnace.

The blast furnace is a tall cylindrical steel tower about 25 m in height and has a diameter about 5-10m lined with fire bricks inside.

Blast furnace has three parts :

• the hearth,
• the bosh and
• the stack.

At the top, there is a cup and cone arrangement to introduce the ore and at the bottom, tapping hole for withdrawing molten iron and an outlet to remove a slag.

The roasted ore is mixed with coke and limestone in the approximate ratio of 12 : 5 : 3.

A blast of hot air at about 1000 K is blown from downwards to upwards by layers arrangement. The temperature range is from bottom 2000 K to 500 K at the top. The charge of ore from top and the air blast from bottom are sent simultaneously. There are three zones of temperature in which three main chemical reactions take place.

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

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

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

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

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

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

Question 93.
Write the reaction involved in the zone of reduction in blast furnace during extraction of iron.
Answer:
Zone of reduction : At about 900 °C, CO reduces Fe₂O₃ to spongy (or porous) iron.
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Carbon also reduces Fe₂O₃ to Fe.
Fe₂O₃ + 3C → 2Fe + 3CO

Question 94
Write reactions involved at different temperatures in the blast furnace.
Answer:

Question 95.
What is the action of carbon on Fe203 in blast furnace?
Answer:
Fe₂O₃ + 3C → 2Fe + 3CO

Question 96.
What is refining of metals?
Answer:
Refining of metals : The purification of impure or crude metals by removing metallic and nonmetallic impurities is known as refining of metals. H

Question 97.
How is pure iron obtained from crude iron?
Answer:
Pure iron can be obtained by electrolytic refining.

Question 98.
Name the methods of refining of metals.
Answer:
Methods of refining of metals :

• Electrorefining
• Liquefaction
• Distillation
• Oxidation m

Question 99.
What are the factors that govern the choice of extraction technique of metals?
Answer:
The choice of extraction technique is governed by the following factors.

• Nature of ore
• Availability and cost of reducing agent. (Generally, cheap coke is used).
• Availability of hydraulic power.
• Purity of metal required.
• Value of by-products. For example. SO₂ obtained during the roasting of sulphide ores is important for the manufacture of H₂SO₄.

Question 100.
Which are the commercial forms of iron?
Answer:
Commercial forms of iron are :

• Cast iron
• wrought iron
• steel. H

Question 101.
(A) What are f-block elements?
(B) What are inner transition elements?
Answer:
(A)

• Elements in which differentiating electron enters into the pre-penultimate shell the (n – 2) f-orbital are known as f.block elements.
• They include 28 elements with atomic numbers ranging from 58-71 and atomic numbers 90 to 103 collectively.
• There are two f-series or two f-block elements, namely 4f and 5f series.
• The f-block includes two inner transition series namely the lanthanoid series. Cerium (58) to LuteUum (71) or the 4 f-block elements and the actinoid series. Thorium (90) to I.awrencium (103) or the 5f block elements.

(B) f-block elements are called inner transition elements since f-orbital lies much inside the f-orbital in relation to the transition metals, These elements have 1 to 14 electrons in their f-orbital.

Question 102.
What are fIrst inner transition elements?
Answer:

1. 4f-hlock elements are called (first) inner transition elements and have partly filled inner orbitaIs or (4f) orbitais.
2. They have general outer electronic configuration \((n-2) f^{1-14},(n-1) d^{0-1}, n s^{2}\).
3. There are two f-series, namely 4f and 5f series, called lanthanoids and acùnoids respectively.
4. They shos intermediate properties as compared to electropositive s-block elements and electronegative p-block elements. Hence they are called (first) inner transition elements.

Question 103.
What are lanthanoids (or lanthanides)?
OR
What is the lanthanoid series?
Answer:

• Lanthanoids or Lanthanoid series or Lanthanones : The series of fourteen elements from ₅₈Ce to ₇₁Lu in which a differentiating electron enters 4f sub-shell and follows lanthanum is called lanthanoid series and the elements are called lanthanoids.
• They have general electronic configuration, [Xe] 4f^1-14 ,5d^0-1, 6s².
• They follow Lanthanum (Z = 57) in 3d-series.

Question 104.
What are rare earths?
Answer:

• Lanthanoids or 4f-block elements are called rare earths.
• Lanthanoids are never found in free state, and their minerals are not pure.
• They exhibit similar chemical properties hence cannot be extracted and separated by normal metallurgical processes.
• Lanthanoid metals are available on small scale. Therefore they are called rare earths.

Question 105.
Explain the position of lanthanoids in the periodic table.
OR
How is the position of lanthanoids justified?
Answer:

1. Position of Lanthanoids in the periodic table : Group – 3; Period – 6.
2. They interrupt the third transition series of t/-block elements (i.e. 5 d series) in the sixth period.
3. They are 14 elements from ₅₈Ce to ₇₁Lu and their position is in between La and Hf. Since they follow lanthanum, they are called lanthanoids.
4. They are called 4f-series elements and for the convenience, they are placed separately below the main periodic table.
5. The actual position of lanthanoids is in between Lanthanum (Z = 57) and Hafnium (Z = 72).
6. Their position is justified due to following reasons :
   • All these elements have the same electronic configuration in ultimate and penultimate shells, one electron in 5d-orbital and two electrons in 6s-orbital.
   • Group valence of all lanthanoids is 3.
   • All lanthanoids from ₅₈Ce to ₇₁Lu have similar physical and chemical properties.

Question 106.
Explain the meaning of inner-transition series.
Answer:
A series of f-block elements having electronic configuration (n – 2)f^1-14 (n – I) d^0-1 ns² placed separately in the periodic table represents inner transition series. The f-orbitals lie much inside the e/ orbitals.

Since the last electron enters pre-penultimate shell, these elements are inner transition elements.

There are two inner transition series as follows :
4f-series ₅₈Ce → ₇₁Lu
5f-series ₉₀Th → ₁₀₃Lr

Question 107.
Draw a skeletal diagram of the periodic table to show the position of d and/- block elements.
Answer:

Question 108.
What are the properties of lanthanoids?
Answer:

• Lanthanoids are soft metak with silvery white colour, Colour and brightness reduces on exposure to air.
• They are good conductors of heat and electricity.
• Except promethium (P_m), all are non-radioactive in nature.
• The atomic and ionic radii decrease from La to Lu. (Lanthanoid contraction).
• Coordination numbers arc greater than 6.
• They are paramagnetic.
• They become ferromagnetic at lower temperature.
• Their magnetic and optical properties are independent of environment.
• They are called rare earths as their exiractioli was difficult.
• They are abundant in earth’s crust
• All lanthanoids fonn hydroxides which are ionic and basic. l3asicity decreases with atomic number,
• They react with nitrogen to give nitrides and with halogen to give halides.
  
• When heated with carbon at very high temperature give carbides
  

Question 109.
Explain the variations in ionisation enthalpy of lanthanoids.
Answer:

• The first ionisation enthalpy of lanthanoids is nearly same. It is very high for Gd and Yb.
• The ionisation enthalpy increases from first (IE₁] to third (IE₃).

First, second and third ionization enthalpies of lanthanoids in kj/mol

Question 110.
Give the general electronic configuration of 4f-series elements (OR lanthanoids).
Answer:

• The general electronic configuration of 4f-series elements is, Ln[Xe]⁵⁴ 4f^1-14 5d^0-1 6s² where Ln is a lanthanoid.
• Xenon has electronic configuration, [Xe] : Is² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶.
• In lanthanoids, the differentiating electron enters prepenultimate shell, 4f m

Question 111.
What are the important features of the electronic configuration of lanthanoids?
Answer:

1. Lanthanoids show two types of electronic configurations
   (a) an expected or idealized
   (b) an observed electronic configuration.
   In the idealized electronic configuration, the filling of the 4/-orbitals is regular but in the observed configuration, there is the shift of a single electron from 5d to 4/ sub-shell.
2. Lanthanum (57) has an electronic configuration [Xe] 4f° 5d¹6s². It does not have any f-electron.
3. The next incoming electron does not enter the 5d sub-shell but goes to the 4f sub-shell.
4. 14 electrons are progressively filled in the 4f sub-shell as the atomic number increases by one unit from La to Lu.
5. La, Gd and Lu are the only elements which possess one electron in a 5d orbital, while in all other lanthanoids the 5d sub-shell is empty.
6. La-(4f°), Gd-(4f⁷) and Lu-(4f¹⁴) posses extra stability due to their empty, half-filled and completely filled 4f-orbitals respectively.
7. The 4f-electrons in the prepenultimate shell are shielded by the outermost higher orbitals, 5s², 5p⁶, 5d¹, 6s², i.e. by eleven electrons, hence they are less effective in chemical bonding.

Electronic configuration (Idealised and observed)

[Xe]⁵⁴ ls²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶

Question 112.
Write the expected electronic configuration of (a) Nd (Z = 60) (b) Tm (Z = 69).
Answer:
Expected electronic configuration :
(a) Nd = [Xe] 4f³ 5d¹ 6s²
(b) Tm= [Xe] 4f¹⁴5d¹6s²

Question 113.
Write electronic configurations of
(i) Gd
(ii) Yb.
Answer:
(i) ₆₄Gd [Xe] 4f⁷5d¹6s² (Observed)
(ii) ₇₀Yb [Xe] 4f¹⁴5d°6s² (Observed)

Question 114.
Write expected and observed electronic configurations of
(i) Ce
(ii) Tb.
Answer:

Question 115.
Why are the expected and observed ground state electronic configurations of gadolinium and lawrencium same?
Answer:

• The degenerate orbitals like 4f and 5f acquire extra stability when they are half filled (4f⁷) or completely filled (5f¹⁴).
• The expected and observed electronic configuration of gadolinium is, ₆₄Gd [Xe] 4f⁷ 5d¹ 6s².
• The expected and observed electronic configuration of lawrencium is ₁₀₃Lr [Rn] 5f¹⁴ 6d¹ 7s².

Question 116.
Explain oxidation states of lanthanoids.
Answer:

• The common oxidation state of the Lanthanoids is 3 + due to the loss of 2 electrons from outermost 6s orbital and one electron from the penultimate 5d sub-shell.
• Gd³⁺ and Lu³⁺ show extra stability due to their half-filled and completely filled f-orbitals, Gd³⁺ = [Xe]4f⁷, Lu³⁺ = [Xe]4f¹⁴
• Ce and Tb attain the 4f° and 4f⁷configurations in the 4 + oxidation states. Eu and Yb attain the 4f⁷ and 4f¹⁴ configurations in the 2 + oxidation states. Sm and Tm also show the 2+ oxidation state although their stability can be explained based on thermodynamic factors.
• Some lanthanoids show 2 + and 4 + oxidation states even though they do not have stable electronic configuration of 4f°, 4f⁷ or 4f¹⁴. E.g. Pr⁴⁺ (4f¹), Nd²⁺ (4f⁴), Sm²⁺ (4f⁶), Dy⁴⁺ (4f⁸) etc

Question 117.
Write the. electronic configuration of the following ions :
(1) La^{3 +} ;
(2) Gd³⁺;
(3) Eu³⁺;
(4) Ce³⁺.
Answer:
(1) La^{3 +} = [Xe]
(2) Gd³⁺ = [Xe] 4f⁷
(3) Eu³⁺ = [Xe] 4f⁶
(4) Ce³⁺ = |Xe] 4f¹

Question 118.
Write the electronic configuration of
(1) Nd²⁺
(2) Nd³⁺
(3) Nd⁴⁺.
Answer:
(1) Nd²⁺ [Xe] 4f⁴
(2) Nd³⁺ [Xe] 4f³
(3) Nd⁴⁺ [Xe] 4f²

Question 119.
Among the following lathanoids, which elements show only one oxidation state 3 +? Why? Dy, Gd, Yb, Lu.
Answer:
Gd and Lu show only one oxidation state 3 +, since they acquire electronic configurations with extra stability namely 4f⁷ and 4f¹⁴ respectively.

Question 120.
Write the expected electronic configurations of :
(1) europium (Z = 63),
(2) erbium (Z = 68).
Answer:
(1) Europium (₆₃Eu) [Xe]⁵⁴4f⁶ 5d¹ 6s²
(2) Erbium (₆₈Er) [Xe]⁵⁴4f¹¹ 5d¹ 6s²

Question 121.
Why does lanthanum form La³⁺ ion, while cerium forms Ce⁴⁺ ion? (Atomic number La = 57 and Ce = 58).
Answer:

1. Electronic configuration Lanthanum is La [Xe] 4f° 5d¹ 6s². By losing three electrons, La acquires stable electronic configuration of Xe and forms La³⁺.
2. Electronic configuration of Cerium is Ce [Xe] 4f¹ 5d¹ 6s². By losing four electrons, Ce acquires stable electronic configuration of Xe and forms Ce⁴⁺.

Question 122.
₆₃EU and ₇₀Yb show 2 + oxidation state. Explain.
Answer:
₆₃EU has electronic configuration, [Xe] 4f⁷ 5d°6s². By losing 2 electrons from 6s orbital, it acquires stable configuration and 4f-orbital is half-filled.
₇₀Yb has electronic configuration, [Xe] 4f¹⁴ 5d° 6s². By losing 2 electrons from 6 s orbital, it acquires stable configuration and 4/-orbital is completely filled.
Hence Eu and Yb show 2 + oxidation states.

Question 123.
Display electronic configuration, atomic and ionic radii of lanthanoids.
Answer:
Answers are given in bold.

Electronic configuration and atomic ionic radii of lanthanoids

Question 124.
Explain the trend in atomic and ionic sizes of lanthanoids.
Answer:

• From ₅₇La (187 pm) to first element of 4f-series ₅₈Ce (183 pm), the contraction in atomic radius is very large, 4 pm.
• But from Ce onwards as atomic number increases atomic radius decreases very steadily so that total decrease in atomic radius from Ce to Lu is only 10 pm.
• In case of tripositive ions due to large pull by nucleus, the decrease in ionic radii is slightly more, i.e. 18 pm. For example, Ce³⁺ (103 pm) to Lu³⁺ (85 pm ).
• Hence all lanthanoids have similar properties. Therefore they cannot be separated from each other easily by normal metallurgical methods but require special methods.
  

Question 125.
What is meant by lanthanoid contraction?
Answer:
Lanthanoid contraction : The gradual decrease in atomic and ionic radii of lanthanoids with the increase in atomic number is called lanthanoid contraction.

Question 153.
Explain the causes of the lanthanoid contraction.
Answer:
The causes of the lanthanoid contraction are as follows :

• As the atomic number of lanthanoids or 4f-block elements increases the positive nuclear charge increases and correspondingly electrons are added to the prepenultimate 4f sub-shell.
• The attraction of nucleus on 4 f-electrons increases with the increase in atomic number.
• The outer eleven electrons namely, 5s², 5p⁶, 5d³ and 6s² do not shield inner 4 f-electrons from the nucleus.
• There is imperfect shielding of each 4f-electron from other 4 f-electrons.
• As compared to d sub-shell, the extent of shielding for 4 f-electrons is less.
• Due to these cumulative effects, 4 f-electrons experience greater nuclear attraction and hence valence shell is pulled towards the nucleus to the greater extent decreasing atomic and ionic radii appreciably.
• From 57La to 58Ce, there is a sudden contraction in atomic radius from 187 pm to 183 pm but the further decrease up to the last 4f-element, ₇₁Lu is comparatively low (about 10 pm).

Question 126.
Explain lanthanoid contraction effect with respect to (1) decrease in basicity, (2) ionic radii of post-lanthanoids.
Answer:
The lanthanoid contraction has a definite effect on the properties of lanthanoids as well as on the properties of post-lanthanoid elements.
(1) Decrease in basicity :

• In lanthanoids due to lanthanoid contraction, as the atomic number increases, the size of the lanthanoid atoms and their try positive ions decreases, i.e. from La³⁺ to Lu³⁺.
• As size of the cation decreases, according to Fajan’s rule, the polarizability increases and thus the covalent character of the M-OH bond increases, and ionic character decreases.
• Therefore the basic nature of the hydroxides decreases.
• Basicity and ionic character decrease in the order La(OH)₃ > Ce(OH)₃ > … Lu(OH)₃.

(2) Ionic radii of post-lanthanoids :

• Elements following the lanthanoids in the 6th period (third transition series, i.e. 5d-series) are known as post-lanthanoids.
• Due to lanthanoid contraction the atomic radii (size) of elements which follow lanthanum in the 6th period (3rd transition series – Hf, Ta, W, Re)-are similar to the elements of the 5th period (4d-series Zr, Nb Mo, Tc).
• Due to similarity in their size, post-lanthanoid elements (5d-series) have closely similar properties to the elements of the 2nd transition series (4d-series) which lie immediately above them.
• Pairs of elements namely Zr-Hf(Gr-4), Nb-Ta (Gr-5), Mo-W(Gr-6), Tc-Re (Gr-7) are called chemical twins since they possess almost identical sizes and similar properties.

Question 127.
Why do lanthanoids form coloured compounds?
Answer:

• The colour in lanthanoid ions is due to the presence of unpaired electrons in partially filled 4f sub-shells.
• Due to the absorption of radiations in the visible region there arises the excitations of the unpaired electrons from f-orbital of lower energy to the f-orbital of higher energy-giving f → f transitions.
• The observed colour is complementary to the colour of the light absorbed.
• The colour of try positive ions (M³⁺) depends upon the number of unpaired electrons in f-orbitals. Hence the lanthanoid ions having equal number of unpaired electrons have similar colour.
• The colours of M³⁺ ions of the first seven lanthanoids, La³⁺ to Eu³⁺ are similar to those of seven elements Lu³⁺ to Tb³⁺ in the reverse order.

Question 128.
Explain, why Ce³⁺ ion is colourless.
Answer:

• The electronic configuration of Ce³⁺ is, [Xe] 4f⁷
• Even though there is one unpaired electron in 4f sub-shell, the f → f transition involves very low energy. Hence, Ce³⁺ ion does not absorb radiation in the visible region.

Therefore Ce³⁺ ion is colourless.

Question 129.
Explain why Gd³⁺ is colourless.
Answer:

• Gd³⁺ has electronic configuration, [Xe] 4f⁷
• Due to extra stability of half filled orbital, it does not allow f → f transition, and hence does not absorb radiations in the visible region.

Hence Gd³⁺ is colourless.

Question 130.
The salts of (1) La³⁺ and (2) Lu³⁺ are colourless. Explain.
Answer:
(1) (i) La³⁺ has electronic configuration, [Xe] 4f°
(ii) Since there are no unpaired electrons in 4 f-orbital, f → f transition is not possible. Hence La³⁺ ions do not absorb radiations in visible region, and they are colourless.

(2) (i) LU³⁺ has electronic configuration [Xe] 4 f¹⁴
(ii) Since there are no unpaired electrons in 4f-orbital, f → f transition is not possible. Hence Lu³⁺ ions do not absorb radiations in visible region and they are colourless.

Question 131.
Explain giving examples, the colour of nf electrons is about the same as those having (14-n) electrons.
Answer:
(1) Consider Pr³⁺ and Tm³⁺ ions.
Tm³⁺ (4f¹²) has nf electron 12 electrons.
Pr²⁺ (4f²) has (14 – n) = (14 – 2) = 12 electrons. Both, Tm³⁺ and Pr³⁺ are green.

(2) Consider Nd³⁺ and Er³⁺ ions. Er³⁺ (4f¹¹) has nf electrons 11.
Nd³⁺ (4f³) has (14 – n) is (14 – 3) = 11 electrons. These both ions Er³⁺, Na³⁺ are pink in colour.

Question 132.
Lu³⁺ has observed magnetic moment zero. How many unpaired electrons are present?
Answer:
Since magnetic moment is zero, it has no unpaired electrons.

Question 133.
What are the application of lanthanoids?
Answer:

1. Lanthanoid compounds are used inside the colour television tubes and computer monitor. For example mixed oxide (Eu, Y)₂ O₃ releases an intense red colour when bombarded with high energy electrons.
2. Lanthanoid ions are used as active ions in luminescent materials. (Optoelectronic application)
3. Nd : YAG laser is the most notable application. (Nd : YAG = neodymium doped ytterium aluminium garnet)
4. Erbium doped fibre amplifiers are used in optical fibre communication systems.
5. Lanthanoids are used in cars, superconductors and permanent magnets.

Question 134.
What are actinoids? Give their general electronic configuration.
Answer:

• Actinoids : The series of fourteen elements from ₉₀Th to ₁₀₃Lr which follow actinium (₈₉Ac) and in which differentiating electrons are progressively filled in 5f-orbitals in prepenultimate shell are called actinoids.
• Their general electronic configuration is, [Rn]⁸⁶ 5f^1-14 6d^0-1 7s².

Question 135.
Why are actinoids called inner transition elements?
Answer:

• Actinoids are 5f-series elements in which electrons progressively enter into 5f-orbitals, which are inner orbitals.
• They have electronic configuration [Rn]⁸⁶ 5f^1-14 6d^0-1 7s².
• They show intermediate properties as compared to electropositive 5-block elements and electronegative p-block elements. Hence they are called second inner transition elements.

Question 136.
Explain the position of actinoids in the periodic table.
OR
What is the position of actinoids in the periodic table?
Answer:

• Position of actinoids in the periodic table : Group-3; Period-7.
• They interrupt the fourth transition series (6d series) in the seventh period in the periodic table.
• After Actinium, 89Ac which has electronic configuration [Rn] 6d¹7s², the electrons enter progressively 5f orbital and they have general electronic configuration, [Rn] 5f^{1 – 14} 6d^{0 – 1} 7s².
• They are fourteen elements from ₉₀Th to ₁₀₃Lr and since they follow actinium, they are called actinoids.
• They are called 5f series or second inner transition series elements and for the convenience they are placed separately below the periodic table.

Question 137.
Write idealised and observed electronic configuration of actinoids.
Answer:

Question 138.
Explain the oxidation states of actinoids.
Answer:

• Due to availability of electrons in 5f, 6d and 7s sublevels, lanthanoids show varied oxidation states.
• The most common oxidation state is + 3 due to loss of one electron from 6d and two electrons from 6s-orbitals.
• Ac, Th and Am show + 2 oxidation state.
• Th, Pa, U, Np, Pu, Am and Cm show + 4 oxidation state.
• Np and Pu show the highest oxidation state + 7.
• U, Np, Bk, Cm and Am show stable oxidation state + 4.
• In + 6 oxidation state, due to high charge density the actinoid ions form oxygenated ions, e.g. \(\mathrm{UO}_{2}^{+}, \mathrm{NpO}_{2}^{+},\) etc.

Question 139.
Why do actinoids show variable oxidation states?
Answer:

• The large number of variable oxidation states of actinoids is due to very small energy difference between 5f, 6d and 7s subshells.
• The electronic configuration of actinoids is, [Rn] 5f^1-14 6d^0-1, 7s²
• Due to the loss of three electrons from 6d¹ and 7s², the common oxidation state is + 3, but due to further loss of electrons from 5f subshell, actinoids show higher oxidation states.
• The variable oxidation states are + 2 to + 7.

Electronic configuration of actinoids and their ionic radii in + 3 oxidation state

Question 140.
What is meant by actinoid contraction?
Answer:
Actinoid contraction: The gradual decrease in atomic and ionic radii of actinoids with the increase in atomic number is called actinoid contraction.

Question 141.
The extent of actinoid contraction is greater than lanthanoid contraction. Explain Why?
Answer:

• The electronic configurations of :
  Lanthanoids [Xe] 4f^{1 – 14} 5d^{0 – 1} 6s²
  Actinoids [Rn] 5f^{1 – 14}, 6d^{0 – 1} 7s²
• The mutual screening offered in case of 5f-electrons is less than that in the 4f-electrons.
• Hence, the outer orbitals are pulled to the greater extent by nuclei in actinoids (5f-series) than in lanthanoids (4f-series).
• Therefore, actinoid contraction is greater than lanthanoid contraction.

Question 142.
Describe the important properties of actinoids.
Answer:
Properties of actinoids :

• Actinoids are silvery white ( similar to lanthanoids).
• They are highly reactive radioactive elements.
• Most of these elements are not found in nature. They are radioactive and man made.
• They experience decrease in the atomic and ionic radii from Ac to Lw, known as actinoid contraction.
• The common oxidation state is +3. Elements of the first half of the series exhibit higher oxidation states.

Question 143.
What are the applications of actinoids?
Answer:

• Thorium oxide (ThO₂) with 1% CeO₂ is used as a major source of indoor lighting, as well as for outdoor camping.
• Uranium is used in the nuclear reactors.
• The isotopes of Thorium and Uranium have very long half-life, so that we get very negligible radiation from them: Hence they can be used safely.

Question 144.
What are transuranic elements?
Answer:

• The man-made elements heavier titan Uranium (Z = 92) in the Actinoid señes are called transuranic elements.
• These are synthetically or artificially prepared (man-made) elements starting from Neptunium (Z= 93).
• Transuranic elements arc generally considered to be from Neptunium (Z = 93) to Lawrencium (Z = 103).
• Recently elements from atomic number 104 (Rf) to atomic number 118 (Og) or (Uuo) in 6 d series have also been identified as transuranic elements.
• All transuranic elements are radioactive.

Question 145.
What are post actinoid elements?
Answer:

• Elements from atomic number 104 to 118 are called postactinoid elements.
• The post actinoid elements known so far are transition metals.
• They can be synthesised in the nuclear reactions.
• As they have very short half life period, it is difficult to study their chemistry.
• Ruiherfordium forms a chloride (RfCl₄) similar to zirconium and hafnium in + 4 oxidation state.
• Dubniurn resembles niobium and protactinium.

Question 146.
Name the transuranic elements.
Answer:
Names of transuranic elements

In the transuranic elements, elements from atomic number 93 to 103 are actinoids and from atomic number 104 to 118 are called postactinoid elements.

Question 147.
What are the similarities between lanthanides and actinides.
Answer:
Lanthanides and actinides show similarities as follows :

• Both, lanthanides and actinides show+ 3 oxidation state.
• In both the series, the f-orbitals are filled gradually.
• Ionic radius of the elements in both the series decreases with increase in atomic number.
• Electronegativity in both the series is low for all the elements.
• They all are highly reactive.
• The nitrates, perchlorates and sulphates of all elements are soluble while their hydroxides, theorides and carbonates
  are insoluble.

Question 148.
Differentiate between lanthanoids and actinoids.
Answer:

Question 149.
Compare Pre-transition metals, Lanthanoid and transition metals.
Answer:

Multiple Choice Questions

Question 150.
Select and write the most appropriate answer from the given alternatives for each sub-question :

1. In transition elements, the different electron enters into
(a) ns subshell
(b) np subshell
(c) (n – 1) d subshell
(d) (n – 2)f subshell
Answer:
(c) (n – 1) d subshell

2. Chromium (Z = 24) has electronic configuration
(a) [Ar]4d^A 4s²
(b) [Ar] 4d⁵ 45¹
(c) [Ar] 3d⁵ 3s¹
(d) [Ar] 3d⁵ 4s¹
Answer:
(d) [Ar] 3d⁵ 4s¹

3. Manganese achieves the highest oxidation state in its compounds
(a) Mn₃O₄
(b) KMnO₄
(c) K₂MnO₄
(d) MnO₂
Answer:
(b) KMnO₄

4. The group which belongs to transition series is
(a) 2
(b) 7
(c) 13
(d) 15
Answer:
(b) 7

5. The last electron of transition element is called
(a) s-electron
(b) p-electron
(c) d-electron
(d) f-electron
Answer:
(c) d-electron

6. Which one of the following elements does NOT belong to first transition series?
(a) Fe
(b) V
(c) Ag
(d) Cu
Answer:
(c) Ag

7. The incomplete d-series is
(a) 3d
(b) 4d
(c) 5d
(d) 6d
Answer:
(d) 6d

8. The electronic configuration of Sc is
(a) [Ar] 3d² 4s²
(b) [Ar] 3d¹ 4s²
(c) [Kr] 3d¹ 4s²
(d) [Kr] 3d² 4s¹
Answer:
(b) [Ar] 3d¹ 4s²

9. The observed electronic configuration of copper is
(a) [Ar]¹⁸ 3d⁹ 4s²
(b) [Kr] 3d¹⁰ 45¹
(c) [Kr] 3d⁹ 4s²
(d) [Ar] 3d¹⁰ 45¹
Answer:
(d) [Ar] 3d¹⁰ 45¹

10. Fe belongs to the
(a) 3d-transition series elements
(b) 4d-transition series elements
(c) 5d-transition series elements
(d) 6d-transition series elements
Answer:
(a) 3d-transition series elements

11. Which one of the following elements does not exhibit variable oxidation states?
(a) Iron
(b) Copper
(c) Zinc
(d) Manganese
Answer:
(c) Zinc

12. In KMnO₄, oxidation number of Mn is
(a) 2+
(b) 4 +
(c) 6 +
(d) 7+
Answer:
(d) 7+

13. Which one of the following transition elements shows the highest oxidation state?
(a) Sc
(b) Ti
(c) Mn
(d) Zn
Answer:
(c) Mn

14. The colour of transition metal ions is due to
(a) s → s transition
(b) d → d transition
(c) p → p transition
(d) f → f transition
Answer:
(b) d → d transition

15. Which one of the following compounds is expected to be coloured?
(a) AgNO₃
(b) CuSO₄
(c) ZnCl₂
(d) CuCl
Answer:
(b) CuSO₄

16. The metal ion which is NOT coloured, is
(a) Fe³⁺
(b) V²⁺
(c) Zn²⁺
(d) Ti³⁺
Answer:
(c) Zn²⁺

17. A pair of coloured ion is
(a)Cu²⁺, Zn²⁺
(b)Cr³⁺ , Cu⁺
(c) Cd²⁺, Mn⁵⁺
(d) Fe²⁺, Fe³⁺
Answer:
(d) Fe²⁺, Fe³⁺

18. The highest oxidation state is shown by
(a) Fe
(b) Mn
(c) Os
(d) Cr
Answer:
(c) Os

19. Transition elements are good catalysts since
(a) they show variable oxidation states
(b) they have partially filled d-orbitals
(c) they have low I.P
(d) they have small atomic radii
Answer:
(a) they show variable oxidation states

20. Highest magnetic moment is shown by the ion
(a) V³⁺
(b) Co³⁺
(c) Fe³⁺
(d) Cr³⁺
Answer:
(c) Fe³⁺

21. The most common oxidation state of lanthanoids is
(a) +4
(b) +3
(c) +6
(d) +2
Answer:
(b) +3

22. Which one of the following elements belong to the actinoid series?
(a) Cerium
(b) Lutetium
(c) Thorium
(d) Lanthanum
Answer:
(c) Thorium

23. The total number of elements in each of f-series is
(a) 10
(b) 12
(c) 14
(d) 15
Answer:
(c) 14

24. The general electronic configuration of Lanthanoids is
(a) [Xe] 4f^{1 – 14} 5d^{0 – 1} 6s²
(b) [Xe] 4f^{2 – 14} 5d^{0 – 1} 6s²
(c) [Xe] 4f^{1 – 13} 5d^{0 – 1} 6s²
(d) [Xe] 4f^{0 – 14} 5d^{0 – 1} 6s¹
Answer:
(a) [Xe] 4f^{1 – 14} 5d^{0 – 1} 6s²

25. f-block elements are called ………………….
(a) transition elements
(b) representative elements
(c) inner transition elements
(d) alkalin earth metals
Answer:
(c) inner transition elements

26. Actinoids form coloured salts due to the transition of electrons in
(a) d – d
(b) f – f
(c) f – d
(d) s – f
Answer:
(b) f – f

27. In the periodic table, Gadolinium belongs to
(a) 4th Group 6th period
(b) 4th group 4th period
(c) 3rd group 5th period
(d) 3rd group 7th period.
Answer:
(d) 3rd group 7th period.

28. The transuranic elements are prepared by
(a) addition reaction
(b) substitution reactions
(c) decomposition reaction
(d) nuclear reactions
Answer:
(d) nuclear reactions

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

Maharashtra State Board 12th Chemistry Important Questions Chapter 7 Elements of Groups 16, 17 and 18

Question 1.
What are p-block elements?
Answer:
The elements in which the differentiating electron (last filling electron) enters the p-orbital of the outermost shell of the atoms are called p-block elements. The elements of groups 16, 17 and 18 are p-block elements.

Question 2.
To which groups in the periodic table do p-block elements belong?
Answer:
A maximum of 6 electrons can be accommodated in the p-subshell, giving rise to six groups. In the periodic table, groups 13 to 18 include p-block elements.

Question 3.
What is the general electronic configuration of p-block elements?
Answer:
The general electronic configuration of valence shell of p-block elements is ns² np^{1 – 6} (except He which has electronic configuration Is²).

Question 4.
What factors influence the properties of p-block elements?
Answer:
The properties of p-block elements are influenced by

• atomic and ionic radii
• ionisation enthalpy
• electron gain enthalpy (or electron affinity)
• electronegativity
• the presence or absence of electrons in d or d and orbitals in the p-block elements. m

Question 5.
Mention the symbols and atomic numbers of p-block elements. Also show the positions in the periods and groups.
Answer:

Question 6.
Name the elements in groups 16, 17 and 18. What is the oxygen family?
Answer:
Group 16 elements :
Oxygen (₈O); Sulphur (₁₆S); Selenium (₃₄Se); Tellurium(₅₂Te); Polonium(₈₄Po)

Group 17 elements :
Fluorine (₉F); Chlorine (₁₇Cl); Bromine (₃₅Br); Iodine (₅₃I); Astatine (₈₅At).

Group 18 elements :
Helium : (₂He) Neon (₁₀Ne), Argon (₁₈Ar), Krypton (₃₆Kr), Xenon(₅₄Xe), Radon (₈₆Rn)

The group 16 is called the oxygen family.

Question 7.
What are chalcogens?
Answer:
Group 16 elements are called chalcogens or ore-forming elements, as a large number of metal ores are oxides or sulphides.

Question 8.
Which is the most abundant element on the earth?
Answer:
(i) Oxygen is the most abundant of all the elements on the earth.
(ii) 46.6% by mass of earth’s crust contains oxygen and oxygen forms 20.95 % by volume of air.

Question 9.
How does sulphur occur?
Answer:
Sulphur forms 0.034 % by mass of the earth’s crust. In the combined form it occurs as follows :

• Sulphides : Galena (PbS), Zinc blende (ZnS), Copper pyrites (CuFeS₂)
• Sulphates : Gypsum (CaSO₄.2H₂O), Epsom salt (MgSO₄.7H₂O), Baryte (BaSO₄)

Question 10.
How do selenium, tellurium and poloniumoccur?
Answer:
Selenium and tellurium are found as metal selenides and tellurides in sulphide ores. Polonium occurs in nature as a decay product of thorium and uranium metals.

Question 11.
What are group 17 elements called?
Answer:
Group 17 elements are collectively called halogens. In Greek, halo means salt and gene means born, so halogens are salt produClng elements.

Question 12.
Why are halogens not found in the free state?
Answer:
Halogens are highly reactive due to high elec-tronegativities, hence do not occur in the free state. They mostly occur in the form of compounds.

Question 13.
How do group 17 elements occur 1
Answer:
Group 17 elements occur mostly in the form of compounds.

• Fluorine is available as insoluble fluorides in the earth’s crust.
• The important minerals are fluorspar, CaF₂, cryolite, Na₃AlF₆, fluorapatite 3Ca₃(PO₄)₂.CaF₂.
• Halides (chloride, bromides, iodides) of Na, K, Mg and Ca are present in sea water.
• The dried up sea beds contain sodium chloride and the mineral carnallite (KCl.MgCl₂.6H₂0).
• Seaweeds contain up to 0.5 % of iodine. The compound chile saltpetre contains 0.2% of sodium iodate.
• Astatine is the last member of the halogen family. It is radioactive and has a half life of 8.1 hours.

Question 14.
How do noble gases occur?
Answer:

• The most important source of noble gases is the atmosphere where they make up about 1% by volume of air. Argon is the major constituent.
• All noble gases occur in nature except Radon. Radon is a decay product of radioactive ²²⁶Ra.
• Helium on a large scale is obtained from natural gas. Helium and neon are found in the minerals pitchblende, monazite and cleveite.
• Xenon and Radon are the rarest noble gases.

Question 15.
How does the general electronic configuration of groups 16, 17 and 18 vary?
Answer:
The general electronic configuration of group 16 elements is ns²np⁴, group 17 is ns²np⁵ and group 18 is ns²np⁶. Group 16 has two electrons less than the stable electronic configuration of the nearest noble gas. while group 17 has one electron less than the stable electronic configuration of the nearest noble gas.

Question 16.
Give the names, atomic numbers and electronic configuration of the group 16 elements.
Answer:

Question 17.
Write the names and electronic configuration of group 17 elements.
Answer:
Electronic configuration of group 17 elements

Question 18.
Write the names of group 18 elements and their electronic configuration.
Answer:
Electronic configuration of group 18 elements

Question 19.
Mention the different atomic and physical properties of group 16 elements.
Answer:
Atomic and physical properties of group 16 elements

Question 20.
Mention the different atomic and physical properties of group 17 elements.
Answer:
Atomic and physical properties of group 17 elements

Question 21.
Mention the different atomic and physical properties of group 18 elements.
Answer:
Atomic and physical properties of group 18 elements

Question 22.
Discuss the trends in the following in case of groups 16, 17 and 18 elements.
(1) Atomic and ionic radii
(2) Ionisation enthalpy
(3) Electronegativity
(4) Electron gain enthalpy
Answer:

1. Atomic and ionic radii :In groups 16, 17, 18 the atomic and ionic radii increase down the group, due to increase in the number of quantum shells. Across a period atomic or ionic radii decrease due to increase in effective nuclear charge.
2. Ionisation enthalpy :
   • The elements of groups 16, 17 and 18 have a high ionisation enthalpy
   • In groups 16, 17, 18 the ionisation enthalpy decreases down the group, due to increase in atomic size.
3. Electronegativity : In groups 16, 17, 18 the elec-tronegativity decreases down the group.
4. Electron gain enthalpy :
   • In groups 16 and 17 the electron gain enthalpy becomes less negative down the group.
   • Group 18 elements have large positive electron gain enthalpy.

Question 23.
Explain the electron gain enthalpy of group 16 elements.
Answer:

1. Electron gain enthalpy (or electron affinity) is the energy released when an electron is added to the valence shell of a gaseous atom forming gaseous ion.
   \(\mathrm{M}_{(\mathrm{g})}+\mathrm{e}^{-} \longrightarrow \mathrm{M}^{+}{ }_{(\mathrm{g})}+\text { energy }\)
   More the energy released more is electron gain enthalpy or electron affinity.
2. Group 16 elements have high values for electron gain enthalpy. On moving down the group, electron gain enthalpy decreases from S to Po.
3. Oxygen has less negative electron gain enthalpy, due to high electronegativity, low atomic size and high electron density so that the incoming electron is repelled.
4. Electron gain enthalpy of the elements decreases down the group due to successive decrease in electronegativity and nuclear attraction and increase in atomic size.

Question 24.
Oxygen has low electron gain enthalpy in group 16 elements. Explain.
Answer:

• Oxygen has low atomic size.
• It has high electronegativity (3.5).
• It has high electron density.
• The incoming electron is repelled due to high electron density hence oxygen has less negative value of electron gain enthalpy as compared to other group 16 elements. O_(g) + e^– → O^–_(g) Δ_egH = – 141 kJ mol^-1

Question 25.
Bromine has the highest negative electron gain enthalpy as compared to other elements of 4th period. Explain.
Answer:

• As compared to other elements (K to Kr) of 4th period, bromine (Br) has the lowest atomic size.
• It has electronic configuration, ₃₅Br [Ar]¹⁸ 3d¹⁰ 4s² 4p⁵.
• It has seven valence electrons and it needs one electron to complete octet.
• Hence Br has the strongest tendency to gain an electron and energy released is maximum in the period. Br_(g) + e^– → Br^–_(g) Δ_egH = -325 kJ mol^-1

Therefore Br has the highest electron gain enthalpy as compared to other elements of 4th period.

Question 26.
Fluorine has higher electronegativity but less electron gain enthalpy. Explain.
Answer:

• Halogens have the highest values for electronegativity due to their small atomic radii and high nuclear charge.
• Fluorine has the highest electronegativity due to its small size.
• In fluorine due to strong inter electronic repulsions in the relatively small 2p orbitals and higher electron density, there is not much attraction of the nucleus for the incoming electron.
• Thus the electron gain enthalpy of fluorine is less in spite of higher electronegativity.

Question 27.
Explain the trend in atomic radius of group 18 elements.
Answer:

• The atomic radii of group 18 elements is larger than the atomic radii of group 17 elements.
• Down the group from He to Xe, atomic radii increases due to increase in the number of quantum shells.
• The atomic radii increase in the order He < Ne < Ar < Kr < Xe

Question 28.
Explain the variation in (1) ionisation enthalpy and (2) electron gain enthalpy in group 18 elements.
Answer:
(1) Ionisation enthalpy :

• In general, group 18 elements have high values of ionisation enthalpy.
• In a period, each noble gas has the highest ionisation enthalpy.
• The noble gases have electronic configuration, ns² np⁶, they have complete octet with paired electrons and very stable closed-shell electronic configuration. Therefore high energy is required to remove the electron from valence shell.
• Down the group ionisation enthalpy decreases.

(2) Electron gain enthalpy :

• Group 18 elements have electronic configuration ns² np⁶ and complete octet of electrons, hence they have no tendency to accept electrons.
• Therefore, they have a large positive electron gain enthalpy.

Question 29.
How does the atomic radii and ionisation enthalpy vary across a period?
Answer:

• The atomic or ionic radii decreases across a period with increase in atomic number and increase in the effective nuclear charge (Z_eff).
• The ionisation enthalpy also increases across a period with increase in atomic number. This is due to addition of electrons to the same shell, while moving across a period.
• Inert element at the end of period has the highest value of ionisation enthalpy.

Question 30.
Why are atomic radii of noble gases larger than those of halogens?
Answer:

• Halogens (F to At) are in group 17 while noble gas (He to Rn) are in group 18.
• Generally, atomic radii decrease on moving from left to right in a period, hence atomic radii of noble gases are expected to be smaller than those of halogens.
• Due to the crowding of eight electrons in the valence shell, electron density increases and there is an appreciable electronic repulsion between them.
• To decrease this repulsion and electronic density, the volume of valence shell increases.

Therefore atomic radii of noble gases are larger than those of halogens.

Question 31.
Which element has higher electron affinity among O and S? Why?
Answer:

1. Oxygen has less negative electron gain enthalpy, due to high electronegativity, low atomic size and high electron density so that the incoming electron is repelled.
2. Electron gain enthalpy of the elements decreases down the group due to successive de­crease in electronegativity and nuclear attraction and increase in atomic size.
   

   Element	S	Se	Te	Po	O
   Electron gain enthalpy kJ/mol	-200	-195	– 190	– 174	– 141

Question 32.
Why is the first ionisation energy of oxygen lower than that of nitrogen?
Answer:

1. Electronic configuration :
   ₇N Is² 2s² 2p_x¹ 2p_y¹ 2p_z¹
   ₈O Is² Is² 2p_x² 2p_y¹ 2p_z¹
   
2. Nitrogen has extra stability due to half-filled p-orbitals. Therefore it has higher first ionisation enthalpy.
3. Oxygen by losing one electron from 2px orbital acquires extra stability of half-filled 7-orbitals and hence has lower ionisation enthalpy.
4. Therefore oxygen has lower ionisation enthalpy than nitrogen.

Question 33.
Why do halogens possess very high values of electronegativity? Arrange the halogens in the order of decreasing electronegativity.
Answer:

• Since halogens have eletronic configuration, ns² np⁵, they have tendency to accept one electron to complete an octet.
• Group 17 elements have the lowest atomic radii and high effective nuclear charge.
• Therefore halogens have high values of electronegativity.
• The decreasing order of electronegativity is, F > Cl > Br > I

Question 34.
Discuss the physical states and the types of elements of Groups 16,17 and 18.
Answer:
Physical states :

• Oxygen is a gas, while the other elements of Group 16 are solids at room temperature.
• All the elements of group 16 show allotropy and exist in several allotropic modifications.
• In group 17, fluorine and chlorine are gases, bromine is a liquid and iodine is a solid at room temperature.
• All halogens are coloured. Fluorine is a yellow coloured gas, chlorine is a greenish-yellow coloured gas, bromine is a red coloured liquid and iodine is a violet coloured solid.
• All the elements of group 18 are monoatomic gases.

Types of elements :

• In group 16 elements, oxygen and sulphur are non-metals, selenium and tellurium are metalloids, while polonium is a radioactive metal with half-life 138 days.
• All halogens are non-metals.
• Group 18 elements are chemically inert and hence called inert elements.

Question 35.
Explain the colour of halogens.
Answer:

• All halogens are coloured.

• The colour arises due to absorption of radiation in visible region and exCltation of electrons.
• From F to I atomic size increases, hence the valence electrons are loosely bound.
• For example, fluorine absorbs violet radiation of higher energy and transmits yellow light of lower energy while iodine absorbs yellow light of lower energy and transmits violet radiation of higher energy.

Question 36.
Explain the trend in the density, melting and boiling points of Group 16 elements.
Answer:
Density :

• The density of group 16 elements increases down the group.
• On moving down the group, the increase in atomic mass is more than the increase in atomic size.
• Down the group the magnitude of van der Waals forces of attraction increases resulting in compact lattice formation of elements.

Therefore the density increases down the group.

The density increases in the order O < S < Se < Te

Melting and boiling point :

• The melting and boiling points of the elements increase regularly on moving down the group.
• However, the melting and boiling points of polonium are lower than that of tellurium.
• This is because the intermolecular van der Waals forces are weaker in polonium.
• There is also a large difference in the melting and boiling points of oxygen and sulphur as oxygen has a smaller size, while sulphur has a larger size and stronger van der Waals forces.

Question 37.
Explain the following in the case of group 17 elements :
(i) Density
(ii) Melting and boiling point
Answer:
Density :

• Down the group, density of halogens increases.
• This is, because down the group, van der Waals forces of intermolecular attraction increase, and hence tendency for agglomerisation increases. Therefore density increases.

Melting point and boiling point :

• Halogens have low melting points and boiling points.
• Melting points and boiling points increase down the group.

Question 38.
In case of group 18 elements, explain the variation in melting and boiling points.
Answer:
Melting points and boiling points :

• Group 18 elements have very low melting points and boiling points.
• The melting points and boiling points increase down the group from He to Rn.
• As the size of the atoms increases on moving down the group, the magnitude of the van der Waals forces increase from He to Rn.
• Thus, the melting and boiling points increase from He to Rn. Helium has the lowest boiling point (4.2 K) of any known substance.
• The boiling points increase in the order He < Ne < Ar < Kr < Xe < Rn

Question 39.
Explain reactions of halogens with water.
Answer:

• Halogens react with water forming halo acids and haloxyaClds.
• The reactivity decreases from fluorine to iodine.
• 2F₂ + 2H₂O → 4HF + O₂
  3F₂ + 3H₂O → 6HF + O₃
  X_2(g) + H₂O₍₁₎ → HX_(aq) + HOX (X = Cl, Br)

Question 40.
How do halogens react with other solvents and with hydrocarbons?
Answer:

• Halogens are soluble in various organic solvents such as chloroform, carbon disulphide and carbon tetrachloride
• All halogens react with hydrocarbons and form mono or multisubstituted compounds.
• The reactivity decreases down the group from fluorine to iodine.
  CH₄ + 2F₂ → C + 4HF
  C + 2F₂ → CF₄
  CH₄ + Cl₂ → CCl₄ + 4HCl

Question 41.
Explain the trend in the bond dissociation enthalpies of halogen molecules.
Answer:

• The bond dissociation enthalpies of halogen molecules decrease down the group with an increase in atomic size.
• However, the bond dissociation enthalpy of fluorine is lower than that of chlorine and bromine because the F – F bond is weak.
• The bond dissociation enthalpies of the halogen molecules decreases in the order,
  Cl – Cl > Br – Br > F – F > I -1.

Question 42.
Why does the density of halogens increase on moving down the group?
Answer:
The bond dissociation enthalpies of halogen molecules decrease down the group with an increase in atomic size.

Question 43.
Why do melting and boiling points of halogens increase on moving down the group?
Answer:

• The atomic size of halogens increases on moving down the group and hence the strength of van der Waals forces between the molecules also increases.
• In the lighter elements, F and Cl, these forces are weak, hence they are gases at room temperature and possess low melting and boiling points.
• In the heavier elements of the group, the van der Waals forces become stronger. Thus, bromine is a liquid and iodine is a solid and they possess higher melting and boiling points.

Question 44.
Why does oxygen show anomalous behaviour?
OR
Oxygen differs from the rest of the members of the family. Explain
Answer:
Oxygen is the first element of group 16.
Reasons for anomalous behaviour of oxygen :

• It has the smallest size in the group.
• It has the highest electronegativity (3.5).
• It does not have vacant d-orbitals like other elements in group 16.

Question 45.
Explain the anomalous behaviour of oxygen in relation to the properties of the Group 16 elements.
Answer:
Oxygen shows following anomalous behaviour :

• Physical state : Oxygen is a gas while other elements in the group are solids at ordinary temperature.
• AtomiClty : Oxygen exists as a diatomic molecule O₂ while other elements are polyatomic molecules like S₈, Se₈ have puckered ring structure.
• Magnetic behaviour : Molecular oxygen O₂ is paramagnetic while other elements are diamagnetic. Molecular O₂ has two unpaired electrons in the antibonding molecular orbitals.
• Oxidation states : Oxygen shows oxidation state-2 in oxides, – 1 in peroxides while + 2 in oxygen difluoride, OF₂.
• Since it does not have vacant d-orbital it doesn’t show higher oxidation states while other elements of group 16 show + 2, + 4 and + 6 oxidation states.
• Hydrogen bonding : Since oxygen has high elec-tronegativity (3.5), it forms hydrogen bonding in its compounds like H₂O, alcohols, etc. Other elements in the group do not show this property.
• Hydrides : The hydride of oxygen, H₂O is a liquid while the hydrides of all other elements in group 16 are gases.
• Covalency : Oxygen shows a common covalency 2 since it has only two unpaired electrons and no d -orbitals in its valence shell. In rare cases, it shows covalency 4.
• The other members of Group 16 can show covalency, more than 4, due to the presence of J-orbitals in their valence shell.

Question 46.
Oxygen generally exhibits an oxidation state of – 2, whereas the other members of its family show oxidation states of +2, +4 and +6 as well. Explain.
Answer:

• The electronic configuration of oxygen is ls²2s²2p⁴.
• It has two half filled p-orbitals and no d-orbitals for exCltation of electrons. Hence it cannot show higher oxidation states.
• Oxygen being highly electronegative, it mostly shows an oxidation state of – 2 only.
• Other members of the family like sulphur, have vacant d-orbitals, thereby giving four and six half- filled orbitals for bonding.
• Furthermore, they can combine with more electronegative elements.
• Hence, they show oxidation states of +2, +4 and 4- 6 also.

Question 47.
Why is H₂O a liquid and H₂S a gas?
OR
Water is a liquid and other hydrides of the group, are gases. Explain.
Answer:

• The boiling point of water (373 K) is very high hence it is less volatile while that of H₂S is low (213 K) and hence it is volatile.
• Since oxygen is more electronegative than sulphur, O-H bond is more polar and there arises associ-
  
• Hydrogen bonding and molecular assoClation are not present in H₂S and other hydrides.
• In H₂S, there are weak van der Waals forces.
• Therefore to separate H₂O molecules in H₂O liquid, energy required is higher than that required to separate H₂S molecules.

Hence H₂O is a liquid and H₂S (and other hydrides) is a gas.

Question 48.
Why does fluorine show anomalous behaviour?
Answer:
Fluorine exhibits anomalous behaviour as compared to other halogens in the group.

The reasons for anomalous behaviour of fluorine are as follows :

• the smallest size of fluorine
• the highest electronegativity
• low bond dissoClation enthalpy of F-F bond
• non-availability of d-orbitals in its valence shell.

Question 49.
Explain the anomalous properties of fluorine.
Answer:
The anomalous properties of fluorine are as follows :

• Fluorine has the highest reactivity among other halogens.
• Fluorine forms strong hydrogen bonding in its hydrides unlike other halogens.
• HF is a liquid while other hydrogen halides are gases at room temperature.
• HF is a weak aCld while other haloaClds are strong aClds.
• Fluorine shows only one oxidation state – 1 while all other halogens show variable oxidation states like -1, +1, +3, +5 and + 7.
• Fluorine has the highest electronegativity but less negative electron gain enthalpy than chlorine.
• The compounds of fluorine have higher ionic character than other halogens.
• Fluorine has no tendency to form polyhalide ion whereas other halogens form polyhalide ions like, Cl₃^–, Br₃^– and I₃^–.
• Fluorine unlike other halogens when reacts with water and produces O₂ and O₃.
  2F₂ + 2H₂O → 4HF + O₂;
  3F₂ + 3H₂O → 6HF + O₃
• Fluorine shows much higher values of ionisation enthalpy, electronegativity and standard electrode potentials compared to the other halogens.
• Fluorine shows much lower values of ionic and covalent radii, melting and boiling points and electron gain enthalpy than expected.
• Fluorine forms only one oxoacid HOF, while the other halogens form a number of oxoacids.

Question 50.
Describe anomalous behaviour of fluorine with the other elements of group 17 with reference to :
(a) Hydrogen bonding
(b) Oxidation state
(c) Polyhalide ions.
Answer:
Anomalous behaviour of fluorine with other 17 group elements (Cl, Br, I) :

(a) Hydrogen bonding : Hydrogen bonding is present in HF only, while it is absent in other haloaClds (HCl, HBr and HI).
(b) Oxidation state : Fluorine shows only one oxidation state namely – 1 due to absence of d-orbital. Other halogens exhibit along with – 1 other oxidation states namely +1, +3, + 5 and +7.
(c) Polyhalide ion : Fluorine does not form polyhalide ion but other halogens form polyhalide ions like Cl^–₃, Br^–₃ and I^–₃.

Question 51.
At room temperature hydrogen fluoride is a liquid while all other hydrogen halides are gases. Explain.
Answer:

• Fluorine has the highest electronegativity compared to remaining halogens.
• H-F bond has the highest bond polarity compared to other halides.
• The presence of strong intermolecular hydrogen bonding leads to the association of HF molecules as follows :
  
• Other hydrogen halides do not show hydrogen bonding due to their larger atomic size and less electronegativity.

Hence HF is a liquid while other hydrogen halides (HCl to HI) are gases.

Question 52.
Explain the oxidation states of Group 16 elements.
Answer:

• Group 16 elements have general electronic configuration, ns² np⁴,
• They show variable oxidation states, – 2, + 2, + 4 and -6.
• Since all elements have 6 valence electrons, they tend to gain or share 2 electrons to complete an octet, and show common oxidation state – 2.
• Oxygen being highly electronegative, it shows the common oxidation state -2 in oxides (H₂O, MgO). It shows – 1 oxidation state in peroxides (H₂O₂, Na₂O₂) and + 2 oxidation states in OF₂.
• Except oxygen all other elements have vacant d-orbitals, hence they show higher oxidation states + 4 and + 6. For example
  
• The stability of the + 6 oxidation state decreases but the stability of the +4 oxidation state increases down the group due to the inert pair effect.
• Bonding in +4 and +6 oxidation states is covalent in nature.

Question 53.
What is the oxidation state of S in H₂S₂O₆?
Answer:

∴ The oxidation state of S atom = + 5

Question 54.
What is the oxidation state of oxygen in OF₂?
Answer:
The oxidation state of oxygen in OF₂ is +2.

Question 55.
What is the oxidation state of oxygen in compounds
(i) O₂F₂ and
(ii) H₂O₂?
Answer:
(i) In O₂F₂, the oxidation state of oxygen is + 1.

(ii) In H₂O₂, the oxidation state of oxygen is – 1.

Question 56.
Oxygen usually exhibits – 2 oxidation state, but it exhibits + 2 oxidation state in OF₂?
Answer:

• Oxygen being highly electronegative, it shows the common oxidation state – 2 in oxides.
• In the presence of a highly electronegative fluorine atom it exhibits + 2 oxidation state in OF₂.

Question 57.
Why do sulphur and other heavier elements of group 16 exhibit higher oxidation states?
Answer:
Since sulphur and heavier elements of group 16 have large atomic radii and available d-orbitals, they exhibit higher oxidation states.

Question 58.
Why does the tendency of group 16 elements to exist in -2 oxidation state decrease on moving down the group?
Answer:

• The electronegativity of the elements of Group 16 elements decreases on moving down the group in the order O > S > Se > Te > Po.
• Thus, the tendency to show -2 oxidation state also decreases.

Question 59.
Discuss the oxidation states in Group 18 eiements.
Answer:

• The group 18 elements have a stable electronic configuration ns2np6 with completely filled orbitals.
• Due to completely filled orbitals and complete octets these elements do not show a tendency to lose, gain or share electrons.
• They have zero valency and mostly exist as mono- atomic gases.
• Xenon exhibits higher oxidation states, as the paired electrons of the valence shell can be unpaired and promoted to the empty d-orbitals.
• The unpaired electrons are shared with fluorine or oxygen atoms to form covalent compounds with higher oxidation states such as XeF₂, XeF₄, XeF₆, XeO₃ and XeOF₆.

Question 60.
Why does xenon being a noble gas form compounds with other elements?
Answer:
Xenon forms compounds with other elements due to the following reasons :

• Xenon has large atomic size and lower ionisation enthalpy.
• The paired electrons of the valence shell can be unpaired and promoted to the empty d-orbitals.
• The unpaired electrons are shared with fluorine or oxygen atoms to form covalent compounds with higher oxidation states.

Question 61.
Explain the reactivity of group 16 elements with hydrogen.
Answer:

• All the elements of group 16 react with hydrogen and form hydrides of the type H₂M where M = O, S, Se, Te and Po. For example H₂O, H₂S, H₂Se, H₂Te and H₂Po.
• The hydrides of these elements show regular trends in their physical and chemical properties.

Question 62.
Explain the general structures of hydrides of group 16 elements.
OR
Give reason : Bond angle decreases from H₂O to H₂S.
Answer:

• Group 16 elements form hydrides of the type H₂M where M = O, S, Se, Te and Po.
• All these hydrides are formed by sp³ hybridisation of the central atom and have angular shape.
• All hydrides have similar structure but differ in H-M-H bond angle. This bond angle decreases from H₂O to H₂Te.
• All these hydrides have two M-H bond pairs and two lone pairs of electrons.
• Due to repulsion between lone pairs-lone pairs and lone pairs-bond pairs, and bond pairs-bond pairs the bond angle is reduced from regular 109° 28′.
• On moving down the group, atomic size increases, electronegativity decreases, electron density decreases.
• Since among group 16 elements, oxygen has the highest electronegativity, lowest atomic size and hence the highest electron density, the repulsion between electron pairs is maximum in H₂O, hence H-O-H bond angle is maximum.
• The electron pair repulsion decreases down the group, hence bond angle also decreases down the group.

Question 63.
Explain the physical states, volatility and chemical properties of hydrides of group 16 elements.
Answer:
(1) Physical state : Hydride of oxygen, H₂O is a colourless odourless liquid while the hydrides of other group 16 elements are colourless poisonous gases with unpleasant odours.

(2) Volatility : Volatility of hydrides increase from H₂O to H₂S and then decreases. H₂O < H₂S > H₂Se > H₂Te
At ordinary temperature H₂O is a liquid while all other hydrides are gases.

(3) Thermal stability :

• The thermal stability of hydrides decreases in the order of H₂O > H₂S > H₂Se > H₂Te.
• Since atomic size increases from O to Te, the tendency to form hydride bond, M-H decreases. Hence M-H bond in O-H is the strongest and in Te-H the weakest. Therefore thermal stability decreases from H₂O to H₂Te.

(4) ACldic character :

• The hydrides of group 16 elements are weakly acidic and acidic strength increases from H₂O to H₂Te.
  Since M-H bond strength decreases, bond enthalpy decreases, acidic character increases.

(5) Reducing power :

• Except H₂O, all hydrides of group 16 elements are reducing agents.
• Reducing power increases from H₂S to H₂Te.
• Reducing power of the hydrides is due to their less stability and tendency to dissociate, which increases from H₂S to H₂Te.

Question 64.
Giving suitable reasons, arrange the hydrides of group 16 elements in the decreasing order of their thermal stability.
Answer:

• The thermal stability of hydrides decreases in the order of H₂O > H₂S > H₂Se > H₂Te.
• Since atomic size increases from O to Te, the tendency to form hydride bond, M-H decreases.
• Hence M-H bond in O-H is the strongest and in Te-H the weakest. Therefore thermal stability decreases from H₂O to H₂Te.

Question 65.
What is the oxidation state of S in the following :
(a) S₈
(b) \(\mathrm{HSO}_{4}^{-}\)
(c) K₂S₂Og?
Answer:
(a) Oxidation state of S in, S₈ is zero.
(b) In \(\mathrm{HSO}_{4}^{-}\) it is +6
(c) in K₂S₂O₈ it is +6

Question 66.
H₂S is less acidic than H₂Te, why?
Answer:

• Sulphur is more electronegative than Tellurium.
• Bond energy of S-H in H₂S is (347 kJ mol^-1) more than the bond energy of Te-H (238 kJ mol^-1).
• Hence H₂Te dissociates more giving H + (or H₃O⁺) than H₂S in the solution. Therefore H₂S is less acidic than H₂Te.

Question 67.
“The reducing power of the hydrides of group 16 elements increases from H₂S to H₂Te”. Explain.
Answer:

• Except H₂O, all hydrides of group 16 elements are reducing agents.
• reducing power increases from H₂S to H2Te.
• reducing power of the hydrides is due to their less stability and tendency to dissociate, which increases from H₂S to H₂Te.

Question 68.
Explain the volatility of hydrides from H₂S to H₂Te.
OR
“The boiling points of the hydrides of Group 16 elements increases from H₂S to H₂Te.” Explain.
Answer:

• In group 16 elements, except H₂O which is a liquid, all other hydrides are gases.
• From H₂S to H₂Te boiling point increases due to an increase in atomic size from S to Te.
• The magnitude of van der Waals forces increases as atomic size increases, from S to Te, therefore boiling points increase from H₂S to H₂Te.

Question 69.
Give reasons : H₂S has a lower boiling point than H₂O.
Answer:

• Hydrogen bonding and molecular association are not present in H₂S.
• The H₂O molecules are associated with each other through intermolecular hydrogen bonding.
• Hence, H₂S has a lower boiling point than H₂O.

Question 70.
Among the hydrides of group 16, water shows unusual properties. Why?
Answer:

• Oxygen being more electronegative, the O-H bond is more polar and there arises association of H₂O molecules through intermolecular hydrogen bonding.
• The other hydrides of group 16 do not form H bonds and hence exist as discrete molecules.
• As a result, water shows unusual properties like high B.P, high thermal stability and weaker acidic character as compared to the other hydrides of group 16.

Question 71.
Explain the reactivity of halogens towards hydrogen.
Answer:

• All halogens react with hydrogen and form hydrogen halides HX where X = F, Cl, Br, I.
• The affinity and reactivity decrease down the group from F to I.
• 

Question 72.
Explain the acidic strength of halo acids.
OR
Write a short note on the acidic strength of hydrogen halides.
Answer:

• The aCldic strength of haloaClds vary in the following order, HF < HCl < BHr < HI
• The stability of hydrogen halides decreases down the group. HF > HCl > HBr > HI.
  This is because from F to I, atomic size increases and bond dissociation enthalpy of H-X decreases.

Question 73.
HF is the most stable among all the hydrogen halides. Explain.
OR
Explain the thermal stability of hydrogen halides.
Answer:

• HF is the most stable among all the hydrogen halides.
• The thermal stability decreases from HF to HI.
• This is due to a decrease in bond dissociation enthalpy and bond strength of H-X bond from F to I.
• Down the group, from F to I, atomic size increases, bond length of H-X bond increases, bond polarity decreases and hence bond strength decreases.
• F being of the lowest atomic size and the highest electronegativity, HF bond is stronger and highly thermally stable.

Question 74.
Explain the reducing character of hydrogen halides.
Answer:

• The reducing character of hydrogen halides increases from HF to HI.
• HF does not show any reducing property while HI is a strong reducing agent.
• H-X bond strength and thermal stability of hydrogen halides decrease from HF to HI.

Unlike other hydrogen halides, HF does not dissociate releasing hydrogen, hence HF is not a reducing agent.

Question 75.
Why is HCl a weaker acid than HI?
Answer:

• Bond length of HCl is less than HI.
• Cl is more electronegative than I.
• Hence bond strength of HCl is more than HI.
• Therefore HCl dissociates less than HI, which makes HCl a weaker acid than HI.

Question 76.
Why is HF a weaker acid than HCl₂
Answer:

• The H-F bond is stronger than H-Cl bond.
• Hence, HF ionises less readily than HCl in H₂O, to give H⁺ ions.
  Therefore, HF is a weaker acid than HCl.

Question 77.
Arrange the halogen acids in the decreasing order of :
(1) thermal stability
(2) acidic strength.
Answer:
(1) Thermal stability : As the atomic size increases from F to I, the bond dissociation enthalpy of H-X decreases.
Thus, the thermal stability of hydrogen halides decreases down the group in the order HF > HCl > HBr > HI

(2) ACldic strength : Since, the bond dissociation enthalpy of H-X decreases down the group, the acidic strength varies in the order HI > HBr > HCl > HF

Question 78.
Noble gases do not form compounds with hydrogen. Why?
Answer:
Noble gases have a stable electronic configuration, and so are unreactive. Since, they are chemically inert, they do not form compounds with hydrogen.

Question 79.
Explain the reactivity of group 16 elements with oxygen.
OR
Give brief account of the oxides of group 16 elements.
Answer:

• All elements of group 16 react with oxygen and form oxides of the types EO₂ and EO₃ where E = S, Se, Te or Po. Examples are SO₂, SeO₂, TeO₂, SO₃, TeO₃, etc.
• SO₂ and SeO₂ are acidic in nature and react with water to form acids.
  
• The reducing power reduces from SO₂ to TeO₂. SO₂ is an oxidising agent, while TeO₂ is a reducing agent.
• SO₃, SeO₃ and TeO₃ are also acidic in nature. They react with water to form acids.
  

Question 80.
Complete the following reactions :
(i) S + O_2(air)
(ii) Se + O_2(air)
Answer:

Question 81.
Explain the reactivity of halogens towards oxygen.
Answer:

• Halogens with oxygen form many oxides with different oxidation states of halogens.
• Fluorine forms O₂F₂ and OF₂.
• Chlorine forms oxides like  All oxides are oxidising agents.
• Bromine forms oxides like Br₂O, BrO₂, BrO₃ and iodine forms I₂O₄, I₂O₅, I₂O₇.

Question 82.
Discuss the properties of the oxides of halogens.
Answer:
Properties of oxides of halogens :

• Both OF₂ and O₂F₂ are strong fluorinating agents. Only OF₂ is thermally stable at 298 K. OF₂ oxidises plutonium to PuF₆. This reaction is used to remove Pu as PUF₆ from spent nuclear fuel.
• The chlorine oxides Cl₂O, ClO₂, Cl₂O₆ and Cl₂O₇ are highly reactive oxidizing agents and tend to explode. ClO₂ is used as a bleaching agent in the paper industry and textiles and in water treatment.
• Bromine forms oxides like Br₂O, BrO₂, BrO₃, which are the least stable halogen oxides. They are powerful oxidising agents.
• Iodine forms oxides like I₂O₄, I₂O₅ and I₂O₇. These solids are insoluble in water and decompose on heating. I₂O₅ is a powerful oxidising agent and is used to determine the amount of carbon monoxide.

For a particular halogen, higher oxides are more stable than the lower ones.

Question 83.
Explain the reactivity of group 16 elements towards halogens.
Answer:

1. The group 16 elements with halogens form a large number of halides of the type EX₂, EX₄ and EX₆ where E is an element of the group and X is a halogen.
2. The stability of the halides decreases in the order fluoride > chloride > bromide > iodide.
   
3. The hexahalides SF₆, SeF₆, TeF₆ are formed by direct combination. They are colourless gases. The hexahalides have sp³d² hybridisation and have an octahedral structure. SF₆ is exceptionally stable due to steric reasons.
4. The tetrahalides SF₄, SeF₄, TeF₄, TeCl₄ have sp³d hybridisation and thus have a trigonal bipyramidal structure in which one of the equatorial positions is occupied by a lone pair of electrons. This geometry is also called see-saw geometry.
5. The dihalides SCl₂, SeCl₂, TeCl₂ have sp³ and thus have a tetrahedral structures with two equatorial positions occupied by lone pairs.
6. The monohalides S₂F₂, S₂Cl₂, Se₂Cl₂ and Se₂Br₂ are dimeric in nature. These halides have a tendency to undergo disproportionation. For example, 2Se₂Cl₂ → SeCl₄ + 3Se

Question 84.
Complete the following reaction :
(i) S + 3F₂
(ii) Se + 2Cl₂
Answer:
(i) S + 3F₂ → SF₆
(ii) Se + 2Cl₂ → SeCl₄

Question 85.
Explain the reactivity of halogens towards other halogens.
OR
What are interhalogen compounds?
Answer:

• The halogens have a tendency to combine amongst themselves forming different compounds called interhalogen compounds.
• Interhalogens are of the type XX, XX’₃, XX’₅ and XX’₇ where X is larger size halogen than X’.
• They are covalent, diamagnetic, reactive and good oxidising agents.
• Preparation :
  

Question 86.
Explain the reactivity of Group 18 towards halogens.
Answer:

• Group 18 elements are chemically inert.
• However, inert elements like krypton and xenon react directly with fluorine under appropriate conditions to give their fluorides. For example,
  
• The xenon fluorides XeF₂, XeF₄ and XeF₆ are colourless crystalline solids which sublime at 298 K.
• These fluorides are strong fluorinating agents.

Question 87.
What is the action of metals on group 16 elements?
Answer:
Metals react with O, S, Se to form oxides, sulphides and selenides respectively.

Question 88.
Complete the following reactions :
(i) Cu + S
(ii) Cd + Se
Answer:
(i) Cu + S → CuS
(ii) Cd + Se → CdSe

Question 89.
Explain the reactivity of halogens with metals.
Answer:

• All halogens react with metals instantly to give metal halides.
• Down the group reactivity decreases from fluorine to iodine.
• 2Na_(s) + Cl₂₍₁₎ → 2NaCl_(s), Mg_(s) + Br_2(s) → MgBr_2(s)
• Halogens being highly electronegative, the metal-halogen bonds are ionic and ionic character decreases down the group. For example M-F > M-Cl > M-Br > M-I.
• The metal halides with higher oxidation state of the metal are more covalent than the halides with lower oxidation state of metal. For example SnCl₄, PbCl₄, SbCl₅ and UF₆ are more covalent than

Question 90.
Explain the property of allotropy in Group 16 elements.
Answer:

• All the elements of group 16 exhibit allotropy.
• These elements exist in different allotropic modifications.
• Oxygen exists as O₂ and ozone O₃.
• Sulphur exists as α-sulphur, β-sulphur, γ-sulphur, homocyclic sulphur, plastic sulphur, etc. Rhombic sulphur (α sulphur) and mono-clinic sulphur (β sulphur) are the important allotropes. Both are non-metallic in nature.
• Selenium exists in two allotropic forms red (non-metallic) and grey (metallic).
• Tellurium exists in two allotropic forms crystalline form and the amorphous form.
• Po has two forms namely α-form and β-form both being metallic.

Question 91.
What are the allotropes of oxygen?
Answer:
Allotropes of oxygen :

• Oxygen O₂
• Ozone O₃

Question 92.
Which are the most important allotropic forms of sulphur?
Answer:
The most important allotropes of sulphur are :

• Rhombic (α-sulphur)
• Monoclinic (β-sulphur)

Question 93.
Write a note on the following :
(1) Rhombic sulphur
(2) Monoclinic sulphur
Answer:
(1) Rhombic sulphur :

• Rhombic sulphur are orthorhombic crystals.
• This is the most stable form and common form of sulphur.
• It is pale yellow., having density 2.069 g/cm³ and melting point 385.8 K
• It is insoluble in water, but soluble in CS₂
• It is stable below 369 K and transforms to β-sulphur above this temperature.
• It exists as S₈ molecules with a structure of a puckered ring.
• It is obtained by the evaporation of roll sulphur is CS₂.
  

(2) Monoclinic sulphur :

• Monoclinic sulphur (β sulphur or prismatic sulphur) are needle-shaped monoclinic crystals.
• It is bright yellow, having a density 1.989 gcm³ and melting point 393 K.
• It is soluble in CS₂.
• It is stable above 369 K and transforms into α-sulphur below this temperature.
• It exists as, S₈ molecules with a structure of a puckered ring.
• It is prepared by melting rhombic sulphur and cooling it till a crust is formed. Two holes are pierced in the crust and the remaining liquid is poured to obtain needle-shaped crystals of monoclinic sulphur.

Question 94.
What are the different oxyacids of sulphur?
Answer:
Sulphurous acid : H₂SO₃
Sulphuric acid (Oil of vitriol): H₂SO₄
Di or pyrosulphuric acid or oleum H₂S₂O₇
Peroxy monosulphuric acid or Caro’s acid H₂SO₅
Peroxy sulphuric acid (or Marshall’s acid : H₂S₂O₈
Thiosulphuric acid H₂S₂O₃

Question 95.
Write the structures of oxyacids of sulphur. Write the oxidation number of sulphur.
Answer:

Question 95.
Write molecular formulae and structure of the following compounds :
(1) Peroxy monosulphuric acid
(2) Pyrosulphuric aCld
Answer:
(1)

(2)

Question 96.
What is oil of vitriol?
Answer:
In ancient days, sulphuric acid, H₂SO₄ was called, oil of vitriol.

Question 97.
What is the oxidation state of S in the following compounds?
(i) H₂SO₂
(ii) H₂S₂O₄
(iii) H₂S₂O₆
(iv) H₂S₂O₅
Answer:

Question 98.
Which are different oxyacids (or oxoacids) of halogens?
Answer:
Fluorine forms only one oxyacid namely, hypofluorous acid or fluoric acid, HOF while other halogens form several oxyacids.

Question 99.
How do
(i) oxidising power and
(ii) thermal stability of oxyacids (or oxoacids) of halogens vary?
Answer:
(i) Oxidising power of oxyacids of halogens decreases as the oxidation number of halogens increases.

(ii) The thermal stability of oxyacids of halogens increases with the increase in the oxidation state of halogen. Hence the increasing order of thermal stability is,

Question 100.
How does the acid strength of the halogen oxoacids vary?
Answer:
The acid strength of the halogen oxoacids increases with increase in the oxidation state of the halogen.

Question 101.
Explain the laboratory methods for the preparation of dioxygen.
Answer:
Laboratory methods :
(i) By heating chlorates, nitrates and permanganates.
Potassium chlorate in the presence of manganese dioxide on heating decomposes to form potassium chloride and oxygen.

(ii) By the thermal decomposition of metal oxides.

(iii) By the thermal decomposition of hydrogen peroxide :
The thermal decomposition of hydrogen peroxide in the presence of finely divided metal and manganese dioxide used as catalyst gives oxygen.

Question 102.
How is oxygen prepared by thermal decomposition of certain metallic oxides?
Answer:

Question 103.
Write the reactions for the decomposition of following oxides on heating :
(i) HgO
(ii) Ag₂O
(iii) PbO₂
(iv) H₂O₂.
Answer:

Question 104.
How is dioxygen manufactured on a large scale?
Answer:
Dioxygen on a large scale or commercial scale is obtained by two following methods :
(1) From water : By electrolysis of acidified water, H₂ gas is obtained at the cathode and O₂ is obtained at anode.

(2) From air :

• Carbon dioxide and water vapour is removed from air and the remaining gases are liquefied.
• O₂ on large scale is obtained by fractional distillation of liquid air.
• On distillation, liquid dinitrogen having low boiling point distils out first leaving behind liquid dioxygen. Then liquid O₂ is distilled out and separated.

Question 105.
What is the action of heat on KClO₃?
Answer:
Laboratory methods :
(i) By heating chlorates, nitrates and permanga­nates.
Potassium chlorate in the presence of manganese dioxide on heating decomposes to form potassium chloride and oxygen.
\(2 \mathrm{KClO}_{3} \frac{\Delta}{\mathrm{MnO}_{2}} 2 \mathrm{KCl}+3 \mathrm{O}_{2(\mathrm{~g})}\)

(ii) By the thermal decomposition of metal oxides.
\(2 \mathrm{Ag}_{2} \mathrm{O}_{(\mathrm{s})} \stackrel{\text { heat }}{\longrightarrow} 4 \mathrm{Ag}+\mathrm{O}_{2(\mathrm{~g})}\)
\(\begin{aligned}
&2 \mathrm{HgO}_{(\mathrm{s})} \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{Hg}+\mathrm{O}_{2(\mathrm{~g})} \\
&2 \mathrm{PbO}_{2(\mathrm{~s})} \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{PbO}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{~g})}
\end{aligned}\)

(iii) By the thermal decomposition of hydrogen per­oxide :
The thermal decomposition of hydrogen peroxide in the presence of finely divided metal and manganese dioxide used as catalyst gives oxygen.
\(2 \mathrm{H}_{2} \mathrm{O}_{2 \text { (aq) }} \frac{\text { heat }}{\mathrm{MnO}_{2}} 2 \mathrm{H}_{2} \mathrm{O}_{(1)}+\mathrm{O}_{2(\mathrm{~g})}\)

Question 106.
Write the reactions of dioxygen with
(i) ZnS
(ii) CH₄
Answer:

Question 107.
Complete the following reactions :

Answer:

Question 108.
What are the uses of dioxygen?
Answer:
The uses of dioxygen are as follows :

• Dioxygen is essential for sustaining life. It is used in hospitals for artificial respiration.
• It is used in oxy-hydrogen (2800 °C) and oxy-acetylene (3200 °C) torches used for cutting and welding metals.
• Oxygen cylinders are used in hospitals, for high altitude flying and in mountaineering.
• It is used in the combustion of fuels. In rockets, hydrazine in oxygen is used as a fuel as it provides tremendous thrust.

Question 109.
What are oxides? How are they classified? Give examples.
OR
What are the different types of oxides? Give example.
Answer:
The oxides are binary compounds in which one element is oxygen and another may be a metal or a non-metal.
They are classified as follows :

• acidic oxides, CO₂, SO₂, etc.
• Basic oxides, CaO, BaO, etc.
• Amphoteric oxides, Al₂O₃, ZnO, etc.
• Neutral oxides, NO, N₂O, CO, etc.

Question 110.
Explain different types of oxides.
Answer:
Oxides are of four types as follows :
(1) acidic oxides :

• The oxide, which on reaction with water forms an acid or reacts with a base to give a salt is called an acidic oxide.
• It is formed by the combination of oxygen with non-metals. For example CO₂, SO₃, etc.
  

(2) Basic oxide :

• The oxide, which on reaction with water forms a base or reacts with an acid to give a salt is called basic oxide.
• It is formed by the reaction of oxygen with highly electropositive metals.
  For example, Na₂O, CaO, etc.
  Na₂O_(s) + H₂O₍₁₎ → 2NaOH_(aq)
  CaO_(s) + H₂O₍₁₎ → Ca(OH)_2(aq)
  BaO_(s) + 2HCl_(aq) → BaCl₂ + H₂O
• Basic oxides are generally ionic in nature.

(3) Amphoteric oxide :

• The oxide, which shows both acidic and basic I characteristics is called an amphoteric oxide.
• They are formed by the reaction of oxygen with elements which lie on the border of electroposi- five (metals) and electronegative (non-metals) nature.

Along the period, from left to right the nature of oxides changes from basic to amphoteric to acidic.

(4) Neutral oxide : The oxide which behaves neither acidic nor basic is called a neutral oxide.
For example, CO, NO, N₂O.

Question 111.
What is the nature of the following oxides :
(i) N₂O_s
(ii) P₄O₁₀
(iii) Cl₂O₇
(iv) ZnO
(v) Al₂O₃
(vi) K₂O
(vii) BaO
(viii) CO
(ix) N₂O?
Answer:

Question 112.
Complete the following reactions :
(i) P₄O₁₀ + H₂O
(ii) C1₂O₇ + H₂O
(iii) ZnO + HCl
(iv) ZnO + NaOH
(v) Al₂O₃ + NaOH.
Answer:

Question 113.
Explain the nature of zinc oxide with the help of the reactions.
OR
Explain with chemical reactions, why is zinc oxide amphoteric in nature.
Answer:

Question 114.
What is ozone umbrella?
OR
Explain Ozone as a protective umbrella for UV from sun.
Answer:
The stratospheric pool of ozone which is a layer above earth’s surface and protects from harmful high energetic ultraviolet (UV) rays is called the ozone umbrella or ozonosphere.

Question 115.
How is ozone formed naturally?
Answer:

• In the atmosphere, ozone is naturally formed through photochemical reactions.
• Oxygen present in the lower mesosphere on the absorption of solar radiations, is dissociated into two oxygen atoms which oxidise oxygen to ozone.
• One atomic oxygen combines with molecular oxygen to form O₃.
  

Question 116.
Explain laboratory preparation of ozone.
Answer:

• When a slow dry stream of oxygen is passed through a silent electric discharge, oxygen is converted into ozone (about 10%). The mixture is called ozonised oxygen.
  
• It is an endothermic reaction.
• Silent electric discharge prevents the decomposition of ozone.

Question 117.
What are the physical properties of ozone?
Answer:
The physical properties of ozone are as follows :

• Gaseous ozone is blue, liquid ozone is dark blue and solid ozone is violet-black.
• It has a pungent odour hence the name is ozone.
• At higher concentrations (about 100 ppm), it is harmful and results into nausea and headache while in small concentrations it is harmless.
• Ozone is thermodynamically less stable than oxygen.
• The decomposition of ozone (2O₃ → 3O₂) is exothermic (ΔH < O) and has ΔS > 0.
• Therefore ΔG < 0 and hence decomposition of O₃ is spontaneous.
• Ozone is diamagnetic in nature.

Question 118.
Ozone acts as an oxidising agent and a reducing agent. Explain with examples.
Answer:
(A) Ozone as an oxidising agent : Since ozone decomposes to liberate nascent oxygen, it is a powerful oxidising agent next to fluorine.
O_3(g) → O_2(g) + O
For example :

• It oxidises lead sulphide (PbS) to lead sulphate (PbSO₄) changing the oxidation state of S from – 2 to +6.
  PbS_(s) + 4O_3(g) → PbSO_(s) + 4O_2(g)
• Potassium iodide, KI is oxidised to iodine, I₃ in the solution.
  2KI_(aq) + H₃O₍₁₎ + O_3(g) → 2KOH_(aq) + I_2(s) + O_2(g)
• Ozone oxidises nitrogen oxide to nitrogen dioxide.
  NO_(g) + O_3(g) → NO_2(g) + O_2(g)

(B) Ozone as a reducing agent : Ozone reduces peroxides (O^1-) to oxides (O^2-).

• Ozone reduces barium peroxide, BaO₂ to barium oxide, BaO.
  BaO₂ + O₃ → BaO + 2O₂
• Ozone reduces hydrogen peroxide, H₂O₂ to water, H₂O.
  H₂O₂ + O₃ → H₂O + 2O₂

Question 119.
What is meant by ozone depletion?
Answer:
The thinning of the ozone layer in the upper atmosphere is called ozone depletion. This thinning has been more pronounced in the polar regions, especially over the Antarctica.

Question 120.
How does ozone protect the people of the earth?
Answer:
Ozone layer in the upper atmosphere absorbs the harmful high energetic UV radiations of the Sun and protects the earth. The ultraviolet radiations damage plant and animal life on earth.

Question 121.
How do exhaust systems of cars and supersonic jet aeroplanes deplete the concentration of ozone layer?
Answer:
The exhaust gases from cars and supersonic jet aeroplanes contain nitric oxide (NO), which combines with ozone in the atmosphere forming oxygen.
NO + O₃ → NO₂ + O₂
Hence the concentration of ozone is depleted in the upper layer of the atmosphere.

Question 122.
How do the coolents in refrigerants deplete the concentration of ozone?
Answer:

• The coolents used in refrigerants are generally chlorofluorocarbons, CFCs also known as Freon, for example CF₂Cl₂.
• CFCs have very long lifetime about 20 to 100 years.
• CF₂Cl₂ undergoes photochemical decomposition giving free radicals which react and destroy ozone.
  

Thus Cl- propagates the destruction of ozone in the atmosphere.

Question 123.
What is the action of ozone on unsaturated hydrocarbons? Give reaction.
Answer:
Ozone with unsaturated hydrocarbons form ozonides.

Question 124.
Ozone depletion is a major environmental problem. Explain.
Answer:

• The depletion of ozone layer increases the amount of ultraviolet radiation reaching the Earth.
• This has caused an increase in the rate of skin cancer, eye cataracts and damage to the genetic as well as immune system.
• Thus, ozone depletion has become a major environ-mental problem.

Question 125.
What are the uses of ozone?
Answer:
Uses of Ozone :

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

Question 126.
How is sulphur dioxide prepared?
Answer:
Sulphur dioxide, SO₂ is prepared by following methods :
(1) From sulphur : When sulphur is burnt in air or oxygen, SO₂ is formed along with SO₃.

(2) Laboratory method (From sulphite) :

• Sodium sulphite on treating with dilute H₂SO₄ forms SO₂.
  Na₂SO₃ + H₂SO_4(aq) → Na₂SO₄ + H₂O₍₁₎ + SO_2(g)
• Sodium sulphite, Na₂SO₃ on reaction with dilute hydrochloric acid solution forms SO₂.
  Na₂SO_3(aq) + 2HCl_(aq) → 2NaCl_9aq0 + H₂O₍₁₎ + SO_2(g)

(3) Industrial method (From sulphides): It is obtained as a by-product in the roasting of ores of pyrites and blendes.

The gas is dried and liquefied under pressure and stored.

Question 127.
What happens when an excess of SO₂ is passed through sodium hydroxide solution?
Answer:
When SO₂ gas is passed through sodium hydroxide solution (NaOH), it forms sodium sulphite, Na₂SO₃ which further with excess of SO₂ forms sodium hydrogen sulphite, NaHSO₃.
2NaOH + SO₂ → Na₂SO₃ + H₂O
Na₂SO₃ + H₂O + SO₂ → 2NaHSO₃

Question 128.
Give reactions to show that SO₂ gas acts as a reducing agent.
Answer:
Sulphur dioxide (SO₂) acts as a reducing agent in the presence of moisture.

• Sulphur dioxide, SO₂ reduces halogens (X°) to haloacids (X^1-).
  I₂ + 2H₂O + SO₂ → 2HI + H₂SO₄
• SO₂ gas when passed through an acidified solution of KMnO₄ it is decolourised due to reduction (Mn⁷⁺ to Mn²⁺).
  2KMnO₄ + 2H₂O + 5SO₂ → K₂SO₄ + 2MnSO₄ + 2H₂SO₄
• Sulphur dioxide (SO₂) reduces Fe³⁺ to Fe²⁺
  2FeCl₃ + SO₂ + 2H₂O → 2FeCl₂ + H₂SO₄ + 2HCl

Question 129.
Complete the following reactions :

Answer:

Question 130.
What are the uses of sulphur dioxide?
Answer:
The uses of sulphur dioxide are as follows :

• SO₂ is used in the manufacture of H₂SO₄.
• In refining petroleum and also in sugar industry.
• In the manufacture of NaHSO₃.
• As an antichlor, as a disinfectant and preservative.
• Liquid SO₂ is used as a selective solvent to dissolve many inorganic and organic compounds.
• Sulphur dioxide is used for bleaching wool and silk. As a bleaching agent in moist condition due to reduction reaction (SO₂ + 2HaO → H₂SO₄ + 2[H]) colouring matter + [H] → colourless matter.
• On exposing the bleached matter, the colour is restored due to oxidation. Colourless matter + [O] → coloured matter. Hence the bleaching action is temporary.

Question 131.
Explain, the nature of S-O bond in SO₂.
Answer:

• In SO₂, sulphur atom is sp² hybridised forming three hybrid orbitals.
• In SO₂, each oxygen atom is bonded to sulphur by a σ and a π bond. Hence in SO₂ there are two σ and two π bonds.
• a bonds are formed by sp²-p overlap while one of n bonds is due to pπ-pπ overlaps and other is due to pπ-pπ overlap.
• But both S-O bonds are identical due to resonance and has bond length 143 pm.

Question 132.
What are the physical properties of H₂SO₄?
Answer:

• It is a colourless, dense, oily liquid having speClfic gravity 1.84 at 298 K.
• It has freezing point 283 K and boiling point 611 K.
• It is a strong dehydrating agent and dissolves in water with the evolution of a large amount of heat.
• It is a strong dibasic acid and acts as an oxidising agent.
• Due to hydrogen bonding, it is viscous.
• It is highly corrosive and produces severe bums on skin.

Question 133.
How is a solution of H₂SO₄ prepared from concentrated sulphuric acid solution?
Answer:

• Concentrated H₂SO₄ dissolves in water with the evolution of a large amount of heat.
• Generally a dilute solution is prepared by adding water to the concentrated solution.
• In case of concentrated H₂SO₄, since its specific gravity is higher, the added water remains on the surface of it and due to evolution of heat, a glass vessel cracks at the interface.
• Hence concentrated H₂SO₄ must be added slowly into fine stream of water with constant stirring to obtain a dilute H₂SO₄ solution.

Question 134.
Give two chemical reactions to explain oxidizing property of concentrated H₂SO₄.

Question 135.
Why is sulphuric acid a strong acid?
Answer:
Sulphuric acid ionises in an aqueous solution in two steps as follows :

The larger value of K_a (K_a >10) shows more dissociation of the acid into H₃O⁺ and H₂SO₄. Thus H₂SO₄ is a strong acid.

Question 136.
What is the action of hot and concentrated sulphuric acid on the following :
(i) Carbon
(ii) Sulphur
(iii) Phosphorus
(iv) Zn
(v) Cu
(vi) Haloacid (HX)?
Answer:

Question 137.
Write the reactions of the following with concentrated H₂SO₄ :
(i) NaCl
(ii) KNO₃
(iii) CaF₂.
Answer:
(i) NaCl + H₂SO_4conc. → NaHSO₄ + HCl
(ii) 2KNO₃ + H₂SO_4conc. → K₂SO₄ + 2HNO₃
(iii) CaF₂ + H₂SO_4conc. → CaSO₄ + 2HF.

Question 138.
What is the action of hot and concentrated H₂SO₄ on
(i) FeSO₄ and
(ii) HI?
Answer:

• FeSO₄ is oxidised to Fe₂(SO₄)₃ by hot and concentrated H₂SO₄.
  \(2 \mathrm{FeSO}_{4}+2 \mathrm{H}_{2} \mathrm{SO}_{4} \stackrel{\Delta}{\longrightarrow} \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\)
• HI is oxidised to I2 by hot and concentrated H2S04.
  2HI + H₂SO₄ → 2H₂O + SO₂ + I₂

Question 139.
Complete the following reactions :
(i) Fe + dil. H₂SO₄
(ii) Benzene + H₂SO_2(conc.)
(iii) PCl₅ + H₂SO₄
(iv) K₄[Fe(CN)₆]
(v) KClO₃
(vi) Fluorspar, CaF₂ + dil. H₂SO₄.
Answer:

Question 140.
In the dissociation of H₂SO₄ in water, why is the second dissociation constant smaller than the first?
Answer:

• H₂SO₄ is a dibasic acid. In aqueous, solution it dissociates in two steps as follows :
  H₂SO_4(aq) ⇌ H⁺_(aq) + HSO^–_4(aq) Ka₁ > 10
  HSO; _(aq) ⇌ H⁺_(aq) + SO^2-_4(aq) Ka₂ = 1.2 x 10^-2
  
• Neutral H₂SO₄ molecule has more tendency to lose proton (H⁺) than anionic Lowry-Bronsted base \(\mathrm{HSO}_{4}^{-} \text {. }\)
• Therefore second dissociation constant Ka₂ is smaller than first dissociation constant Ka₁.

Question 141.
Mention the conditions to maximize the yield of H₂SO₄ by contact process.
Answer:

• In the contact process of the manufacture of sulphuric acid, SO₂ is oxidised to SO₃ by heating the mixture on the heterogeneous catalyst V₂O₅.
  2SO_2(g) + O_2(g) ⇌ 2SO_3(g) ΔH= – 196.6 kJ
• The forward reaction is exothermic and there is a decrease in number of moles and volume.
• The optimum conditions to maximise the yield of H₂SO₄ are pressure of 2 bar and temperature around 720 K.

Question 142.
What are the uses of H₂SO₄?
Answer:
H₂SO₄ has following uses :

• In the preparation of HNO₃, HCl, H₂PO₄, Na₂CO₃ sulphates, alums, alcohols, ethers, etc.
• In the manufacture of dyes, fertilizers like ammonium sulphate, super phosphate, detergents.
• As an electrolyte in lead storage battery.
• As a dehydrating agent.
• As an oxidising agent.
• For refining petroleum.
• As a pickling agent for removing layers of basic oxides from the metal surfaces like Fe, Cu, etc. before the metals are galvanized, electroplated, etc.
• It is used as a laboratory reagent.
• It is used in the manufacture of nitrocellulose products.

Question 143.
What happens when,
(i) SO₃ gas is passed through water
(ii) Concentrated sulphuric acid is added to sugar?
Answer:
(i) SO₃ dissolves in water and forms sulphuric acid, H₂SO₄.
SO_3(g) + H₂O₍₁₎ → H₂SO_4(aq)
(ii) Concentrated sulphuric acid when added to sugar, it is dehydrated giving carbon.

Question 144.
How is chlorine obtained from HCl?
Answer:
Chlorine is obtained by the oxidation of hydrochloric acid by various oxidising agents:

Question 155.
How is chlorine obtained from rock salt or NaCl?
Answer:
Rock salt or NaCl in the presence of MnO₂ and concentrated H₂SO₄ forms chlorine.

The reaction takes place in two steps.

Question 156.
How is chlorine manufactured by Deacon’s process?
Answer:
In Deacon’s process, chlorine is manufactured by oxidising hydrogen chloride gas by atmospheric oxygen in the presence of CuCl₂ as a catalyst at 723 K.

Question 147.
How is chlorine manufactured by the electrolytic process?
Answer:

• On a large scale chlorine is obtained by an electrolytic process.
• The electrolyte is brine, a concentrated solution of NaCl.
• Nelson’s two compartments diaphragm cell is used for electrolysis in which stout graphite rod is used as an anode while U shaped steel vessel is used as a cathode.
• Reactions :
  NaCl → Na⁺ + Cl^–
  H₂O ⇌ H⁺+ OHAt cathode :
  2H₂O + 2e^– → H₂ + 20H^–
  Na⁺ +OH^– → NaOHAt anode :
  2Cl → 2Cl + 2e^–
  2Cl → Cl_2(g)

Question 148.
What are the physical properties of chlorine?
Answer:
The physical properties of chlorine are as follows :

• It is greenish-yellow gas with a suffocating and pungent smell and is heavier than air.
• It is a poisonous gas.
• When liquefied, chlorine forms a greenish-yellow liquid which has boiling point 239 K.
• Chlorine is soluble in water and the solution is called chlorine water.

Question 149.
What is the action of chlorine on the following :
(i) Na
(ii) K
(iii) Ca
(iv) Fe
(v) A1
(vi) Cu?
Answer:
(i) 2Na + Cl₂ → 2NaCl
(ii) 2K + Cl₂ → 2KCl
(iii) Ca + Cl₂ → CaCl₂
(iv) 2Fe + 3Cl₂ \(\stackrel{\Delta}{\longrightarrow}\) 2FeCl₃
(v) 2A1 + 3Cl₂ \(\stackrel{\Delta}{\longrightarrow}\) 2A1Cl₃
(vi) Cu + Cl₂ → CuCl₂

Question 150.
What is the action of Cl₂ on the following :
(i) P₄
(ii) As
(iii) Sb
(iv) B
(v) S.
Answer:
(i) P₄ + 6Cl₂ → 4PCl₃ and P₄ + 10Cl₂ → 4PCl₅
(ii) 2As + 3Cl₂ → 2AsCl₃
(iii) 2Sb + 3Cl₂ → 2SbCl₃
(iv) 2B + 3Cl₂ → 2BCl₃
(v) S₈ + 4Cl₂ → 4S2Cl₂
Sulphur monochloride

Question 151.
What is the action of chlorine on
(i) hydrogen
(ii) hydrogen sulphide?
OR
Give two reactions to show the affinity of hydrogen towards chlorine.
Answer:
(i) Since chlorine has high affinity for hydrogen, it forms hydrogen chloride in the presence of sunlight.
\(\mathrm{H}_{2}+\mathrm{Cl}_{2} \stackrel{h v}{\longrightarrow} 2 \mathrm{HCl}\)
(ii) Chlorine reacts with hydrogen sulphide to form hydrogen chloride and sulphur.
H₂S + Cl₂ → 2HCl + S

Question 152.
What is the reaction of chlorine with ammonia?
Answer:

• Chlorine reacts with the excess of ammonia to form ammonium chloride, NH₄Cl and nitrogen.
  
• When chlorine is in excess, then with ammonia it forms explosive nitrogen trichloride, NCl₃.
  

Question 153.
Explain the reactions of chlorine with alkalies.
Answer:

• Chlorine with cold and dilute caustic soda, NaOH forms sodium chloride and sodium hypochlorite, NaOCl.
  
• Chlorine with hot and concentrated NaOH forms sodium chloride and sodium chlorate, NaClO₃.
  
• When chlorine is passed over dry slaked lime Ca(OH)₂, bleaching powder, CaOCl₂ is obtained.
  Ca(OH)₂ + Cl₂ → CaOCl₂ +H₂O

Question 154.
What is the action of chlorine on
(i) methane and
(ii) ethylene?
Answer:
(i) With methane, chlorine forms substituted products.

(ii) With unsaturated hydrocarbons, chlorine forms addition products.

Question 155.
Explain oxidising nature of chlorine with chemical reactions.
Answer:

Question 156.
Explain the bleaching action of chlorine.
Answer:
(i) Chlorine acts as a powerful bleaching agent due to its oxidising nature.
(ii) In moist conditions or in the presence of water it forms unstable hypochlorous acid, HOCl which decomposes giving nascent oxygen which oxidises the vegetable colouring matter of green leaves, flowers, litmus, indigo, etc.
Cl₂ + H₂O → HCl + HOCl
HOCl → HCl + [O]
Vegetable coloured matter + [O] → colourless matter.

Question 157.
Explain disinfecting action of chlorine.
OR
Chlorine is used for the sterilization of water. Explain.
Answer:
(i) Since chlorine has an ability of killing harmful micro-organisms, it acts as a good disinfecting agent.
(ii) This ability is due to the oxidising nature of chlorine which produces nascent oxygen in aqueous solutions.
H₂O + Cl₂ → HCl + HOCl
HOCl → HCl + [O]

Question 158.
What is the action of Cl₂ on
(a) Cold and dil. NaOH
(b) Hot and cone. NaOH
(c) P₄ molecule
(d) Iron (II)
Answer:

Question 159.
Write the formulae of tear gas, phosgene and mustard gas.
Answer:

Question 160.
Who prepared hydrochloric acid first?
Answer:
Glauber in 1648, first prepared hydrochloric acid by heating common salt NaCl with concentrated H₂SO₄.
\(2 \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{SO}_{4}(\text { conc. }) \stackrel{\Delta}{\longrightarrow} \mathrm{Na}_{2} \mathrm{SO}_{4}+2 \mathrm{HCl}\)

Question 161.
What are the physical properties of hydrogen chloride gas?
Answer:

• Hydrogen chloride is a colourless gas with a pungent odour.
• It is heavier than air.
• Hydrogen chloride on liquefication forms a colourless liquid having boiling point 189 K and on freezing it forms white solid crystals having melting point 159 K.
• Hydrogen chloride is highly soluble in water.

Question 162.
What is hydrochloric acid?
Answer:
An aqueous solution of hydrogen chloride is called hydrochloric acid.

It is acidic in nature. It is a strong acid and dissociates almost completely.
HCl_(aq) + H₂O → H₃O⁺_(aq) + Cl^–_(aq) K_a = 10⁷

Question 163.
What is aquaria? What is the action of aquaria on (i) Au and (ii) Pt?
Answer:
A mixture of three parts of concentrated HCl solution and one part of concentrated HNO₃ solution is known as aquaria.

Aquaregia dissolves almost all substances including noble metals like gold, platinum.

This high solubility is due to the formation of nascent chlorine.

Question 164.
Write the reactions of hydrochloric acid with the following :
(a) NaHSO₃
(b) NaHCO₃
(c) CaO
(d) Mg(OH)₂
(e) NaOH.
Answer:
(a) NaHSO₃ + HCl → NaCl + H₂O + SO₂
(b) NaHCO₃ + HCl → NaCl + H₂O + CO₂
(c) CaO + 2HCl → CaCl₂ + H₂O
(d) Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O
(e) NaOH + HCl → NaCl + H₂O

Question 165.
What are the uses of hydrogen chloride (hydrochloric acid)?
Answer:
Hydrogen chloride (OR hydrochloric acid) is used :

• in the manufacture of chlorine and ammonium chloride,
• to manufacture glucose from com, starch
• to manufacture dye
• in mediClne and galvanising
• as an important reagent in the laboratory
• to extract glue from bones and for the purification of bone black.
• for dissolving metals, Fe + 2HCl_(aq) → FeCl₂ + H_2(g)

Question 166.
What are the general compositions of interhalogen compounds?
Answer:
(i) The general compositions of interhalogen compounds are XX’, XX’₃, XX’₅ and XX’₇ where halogen X is more electropositive and has larger size than another halogen X’. For example,

(ii) As the ratio of radii of X and X’ increases, the number of atoms of X’ per molecule of interhalogen compound increases. For example, iodine having the largest size with fluorine of the smallest size forms stable IF₇.

Question 167.
Which are the different types of interhalogen compounds?
Answer:
(i) The general compositions of interhalogen compounds are XX’, XX’₃, XX’₅ and XX’₇ where halogen X is more electropositive and has larger size than another halogen X’. For example,

(ii) As the ratio of radii of X and X’ increases, the number of atoms of X’ per molecule of interhalogen compound increases. For example, iodine having the largest size with fluorine of the smallest size forms stable IF₇.

Question 168.
What are general characteristics of interhalogen compounds?
Answer:
The general characteristics of interhalogen compounds are as follows :

• In XX’_n, X is the halogen which has larger size and is more electropositive, while X’ is the halogen having smaller size and is less electropositive, n is the number of atoms of X’ attached to X.
• Interhalogen compounds are named as halogen halides. The more electropositive halogen is named as such and the less electropositive halogen is named as the halide. In ClF, since Cl is larger and more electropositive than F, the interhalogen compound is named as Chlorine monofluoride,
• As the ratio of radii (radius of X : radius of X’ ) between the atoms X and X’ increases, the number of halogen atoms (n) per interhalogen compound also increases.
• The interhalogen compounds have even number of atoms i.e. 2,4,6,8. For example, ClF₃ has 4 atoms and BrF₅ has 6 atoms.
• The number of X’ atoms in the interhalogen compounds are always odd.
• The properties of interhalogen compounds are generally intermediate between those of the halogens from which they are made.
• The oxidation state of the atom X in XX’_n, is equal to +1, +3, +5, +7 and that of X’ is -1.
• Since the electronegativity difference between two different halogens is low, the interhalogen compounds are covalent in nature.
• Interhalogen compounds exist as gases, liquids and solids depending upon their composition. They are volatile and less stable but not explosive.
• They are diamagnetic in nature.
• Since the interhalogen bond (X – X’) is weaker than parent halogens, they are more reactive than halogens.
  x- l so >oui bruin Timor! N

Question 169.
How are interhalogen compounds prepared?
Answer:
All interhalogen compounds are prepared by direct reaction of halogens or reaction of halogen on lower interhalogen compounds. The compound formed depends upon specific conditions of the reactions,

(a) By direct combination :

• Chlorine monofluoride, ClF : It is a colourless gas obtained by reacting equal volumes of Cl₂ and F₂.
  
• Chlorine trifluoride, ClF₃ : It is a colourless gas, obtained by reacting Cl₂ with excess of F₂.
  
• Bromine trifluoride, BrF₃ : It is a yellow green liquid obtained by reacting Br₂.
  
• Iodine trichloride, ICl₃ : It is an orange solid obtained by reacting I₂ with excess of Cl. This orange solid dimerises to form I₂Cl₆ having Cl-bridges.
  
• Iodine chloride, ICl : It is obtained by reaction between equimolar amounts of I₂ and Cl₂. a- form of it is a ruby red solid while 1-form is a brown-red solid.
  
• Bromine pentafluoride, BrF₅ : It is a colourless liquid and obtained by the reaction of bromine with the excess of fluorine.
  
• Bromine chloride BrCl : It is a gas, obtained by the reaction of bromine with chlorine taken in equal volumes.
  

(b) By the reaction of halogen with interhalogen compound :

• Bromine fluoride (BrF) : It is a pale brown coloured gas prepared by the action of Br₂ on BrF₃. Br₂ + BrF₃ → 3BrF
• Bromine trifluoride ( BrF₃) : BrF₃ can also be prepared by the action of Br₂ on ClF’₃.
  Br₂ + ClF₃ → 2BrF₃ + BrCl

(c) SpeClal reaction for the preparation of ICl :

• ICl( Wij’s solution) can be prepared by the action of Iodinef (I₂) on potassium chlorate (KClO₃)
  I₂ + KClO₃ → ICl + KlO₃

Question 170.
Give the physical states of the following molecules :
BrCl, IBr, IF₅, BrF₅, ClF₃, IF₇
Answer:

Question 171.
Discuss the hydrolysis reactions given by interhalogen compounds.
Answer:
Interhalogen compounds undergo hydrolysis and form haloacids of lower halogen and oxyacid of higher halogen in interhalogen compounds.

BrCl + H₂O → HOBr + HCl
5ICl + 3H₂O → HIO₃ + 5HCl + 2I₂
2ICI₃ + 3H₂O → HIO₃ + 5HCl + ICl

Question 172.
Give the addition reaction given by interhalogen compounds.
Answer:
Interhalogen compounds give addition reactions with unsaturated hydrocarbons.
CH₂ = CH₂ + XX’ → CH₂X – CH₂X’
CH₂ = CH₂ + ICl → CH₂I – CH₂Cl

Question 173.
Arrange the given compounds in the order of their thermal stability. ClF, IBr, BrCl, ICl.
Answer:
The thermal stability of interhalogen compounds depends upon the electronegativity of the combining atoms.

More the difference between the electronegativity the combining atoms, more is the thermal stability. Hence the order of thermal stability is ClF > ICl > IBr > BrCl

Question 174.
Give one fluorinating reaction given by inter-halogen compounds.
Answer:
Interhalogen compounds like ClF₃, BrF₃, etc. are very reactive and they are strong fluorinating agents.
U_(S) + 3ClF₃₍₁₎ → UF_6(g) + 3ClF_(g)

Question 175.
What are the uses of interhalogen compounds?
Answer:
The uses of interhalogen compounds are as follows :

• ICl is used as a halogenating agent and also in the estimation of iodine number of fats and oils.
• In the preparation of polyhalides, interhalogen compounds are used.
• ClF₃ and BrF₃ are widely used as fluorinating agent. They are used in the separation of ²³⁵U isotope by fluorinating.
  ²³⁵U_(S) + 3ClF_3(l) – ²³⁵UF_6(g) + 3ClF_(g)
• In propellants ClF₃ and IF₃ are used as oxidisers.
• They are used as non-aqueous solvents.
• The XX’ interhalogen compounds are used as a catalyst for the oxidation of As(III)

Question 176.
Give the disproportionation reactions given by the following :
(i) BrF
(ii) ClF₃
(iii) ICl₃
Answer:

Question 177.
Give hybridisation and geometry of the following molecules :
(1) C1F₃
(2) IF₅
(3) IF₇.
Answer:

Question 178.
Draw structure of :
(a) Chlorine trifluoride
(b) Chlorine pentafluoride
(c) Bromine trifluoride
(d) Iodine dichloride
Answer:

Question 179.
Complete the given table :

Answer:

Question 180.
Explain IC1 is more reactive than I2.
Answer:

• in interhalogen compound ICl and I₂ the atoms are bonded by covalent bonds.
• The covalent bond between dissimilar atoms, I and Cl atoms in ICl is weaker than between similar atoms in I₂.
• Therefore bond dissociation enthalpy of ICl bond is less than that of I₂ bond.
  Hence ICl is more reactive than I₂.

Question 181.
What is the action of
(i) H₂O and
(ii) PF₅ on XeF₂?
Answer:
(i) Action of H₂O ( Hydrolysis) :
XeF₂ undergoes hydrolysis to form HF.
2XeF₂ + 2H₂O → 4HF + 2Xe + O₂

(ii) Reaction with PF₅ :
XeF₂ forms adducts on reaction with PF₅.
XeF₂ + PF₅ → XeF₂.PF₅

Question 182.
Give the method of preparation of XeF₄ and its structure.
Answer:
For the preparation of XeF₄ :
The covalent bond between dissimilar atoms, I and Cl atoms in ICl is weaker than between similar atoms in I₂.

Question 183.
Complete the following :
(1) XeF₄ + H₂O → ……………… + HF
Answer:
XeF₄ + H₂O → XeOF₂ + 2HF

Question 184.
What are the uses of :
(1) Helium
(2) Neon
(3) Argon.
Answer:
(1) Uses of helium (He) :

• A mixture of helium (85%) and oxygen (15%) is used for filling balloons.
• A mixture of helium and oxygen is also used for respiration by sea divers instead of air because helium is less soluble in blood than nitrogen under high pressure. It is also used for treatment of asthma.
• Helium is used in produClng inert atmosphere in metallurgical operations and welding of metals.
• Liquid helium is used in produClng low temperature required for research.
• Helium is also used in produClng lasers in low temperature gas thermometry.
• It is used in magnetic resonance imaging.
• It is used as a shielding gas for arc welding, (viii) It is used in supersonic wind tunnels.
• Helium nucleus is used as a bombarding particle for disintegration of atoms.

(2) Uses of neon (Ne) :

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

(3) Uses of argon (Ar) :

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

Question 185.
Helium and neon do not form compounds with fluorine. Why?
Answer:
Helium and Neon do not contain d-orbitals in their respective valence shells and hence their electrons cannot be promoted to higher energy levels like other elements in group 18. Therefore, helium and neon do not form compounds with fluorine.

Multiple Choice Questions

Question 223.
Select and write the most appropriate answer from the given alternatives for each sub¬question :

1. Oxygen molecule shows :
(a) Paramagnetism
(b) Diamagnetism
(c) Ferromagnetism
(d) ferrimagnetism
Answer:
(a) paramagnetism

2. Among group 16 elements, the radioactive element is
(a) Ra
(b) Rn
(c) Bi
(d) Po
Answer:
(d) Po

3. Hybridization involved in H₂O is
(a) sp
(b) sp³
(c) sp³d²
(d) sp²
Answer:
(b) sp³

4. The structure of SF₄ is
(a) Octahedral
(b) Bipyramidal
(c) Square planar
(d) Tetrahedral
Answer:
(b) Bipyramidal

5. Which one has the highest bond energy?
(a) 0-0
(b) S-S
(c) Se-Se
(d) Te-Te
Answer:
(b) S – S

6. Which of the following compounds contains S = O as well as S = S bonds?
(a) Sulphuric acid
(b) Thiosulphuric acid
(c) Sulphurous acid
(d) Thiosulphurous acid
Answer:
(b) Thiosulphuric acid

7. Ozone is :
(a) A compound of oxygen
(b) An allotropic oxygen
(c) An isotope of oxygen
(d) An isobar of oxygen
Answer:
(b) An allotropic oxyen

8. Which one has lowest boiling point?
(a) H₂O
(b) H₂S
(C) H₂Se
(d) H₂Te
Answer:
(b) H₂S

9. Maximum covalency of sulphur is :
(a) Four
(b) Six
(c) Three
(d) Two
Answer:
(b) Six

10. Oxygen exhibits-1 oxidation state in :
(a) OF₂
(b) H₂O₂
(c) HClO
(d) H₂O
Answer:
(b) H₂O₂

11. The oxidation state of oxygen in OF₂ is
(a) -1
(b) -2
(c) +1
(d) +2
Answer:
(d) +2

12. In the hydrides of group 16 elements, the minimum bond angle is in
(a) H₂O
(b) H₂S
(C) H₂Se
(d) H₂Te
Answer:
(d) H₂Te

13. In SF₆, sulphur atom undergoes hybridisation
(a) sp³
(b) sp³d
(c) sp³d²
(d) sp³d³
Answer:
(c) sp³d²

14. Al₂O₃ is
(a) acidic oxide
(b) basic oxide
(c) amphoteric oxide
(d) neutral oxide
Answer:
(c) amphoteric oxide

15. Which of the following is a basic oxide?
(a) SiO₂
(b) P₄O₁₀
(c) MgO
(d) Al₂O₃
Answer:
(c) MgO

16. The most stable allotropic form of sulphur is :
(a) Rhombic sulphur
(b) Flowers of sulphur
(c) Platic sulphur
(d) Mono clinic sulphur
Answer:
(a) Rhombic sulphur

17. The atomicity of sulphur in orthorhombic sulphur is –
(a) 8
(b) 6
(c) 4
(d) 2
Answer:
(a) 8

18. The oxidation state of sulphur in peroxy disul- phuric acid is
(a) +2
(b) +3
(c) +4
(d) +6
Answer:
(d) +6

19. In the contact process, the catalyst used is
(a) TiO₂
(b) Pt
(c) V₂O₅
(d) Fe₂O₃
Answer:
(c) V₂O₅

20. What is the molecular formula of the oleum?
(a) H₂SO₃
(b) H₂SO₄
(c) H₂S₂O₇
(d) H₂S₂O₈
Answer:
(c) H₂S₂O₇

21. The electronic configuration of bromine is
(a) [Ne] 3s²3p⁵
(b) [Kr] 4d¹⁰5s²5p^s
(c) [Ar] 3d¹⁰4s²4p⁵
(d) [Ar]4d¹⁰5s²5p⁵
Answer:
(c) [Ar] 3d¹⁰4s²4p⁵

22. The decreasing order for negative values for electron gain enthalpies is
(a) F > Cl > Br
(b) Cl > F > Br
(c) Br > Cl > F
(d) F > Br > Cl
Answer:
(b) Cl > F > Br

23. The molecular formula H₂S₂O₂ represents which oxoacid among the following?
(a) Hydrosulphurous acid
(b) Thiosulphurous acid
(c) Sulphuric acid
(d) Pyrosulphurous acid
Answer:
(b) Thiosulphurous acid

24. General electronic configuration of group 16 elements is
(a) ns²np³
(b) ns²np⁴
(c) ns²np²
(d) ns²np⁵
Answer:
(b) ns²np⁴

25. The decreasing order of electronegativity is
(a) Cl > I > Br
(b) Br > I > F
(c) F > Br > I
(d) Cl > F > Br
Answer:
(c) F > Br > I

26. The increasing order of acidic strength of haloacids is
(a) HF > HCl > HBr > HI
(b) HCl > HF > HBr > HI
(c) HCl < HF < HBr < HI
(d) HF < HCl < HBr < HI
Answer:
(d) HF < HCl < HBr < HI

27. Amongst haloacids, the strongest reducing agent is
(a) HCl
(b) HF
(c) HI
(d) HBr
nswer:
(c) HI

28. Haloacid with the maximum thermal stability is
(a) HI
(b) HCl
(c) HF
(d) HBr
Answer:
(c) HF

29. The bleaching action of chlorine is due to
(a) oxidation
(b) reduction
(c) neutralisation
(d) double decomposition
Answer:
(a) oxidation

30. The halogen which does not form polyhalide ion is
(a) F
(b) Br
(c) I
(d) Cl
Answer:
(a) F

31. Camalite is a natural source of
(a) F
(b) Cl
(c) Br
(d) I
Answer:
(b) Cl

32. In Deacon’s process the catalyst used is
(a) V₂O₅
(b) CuCl₂
(c) PtCl₄
(d) FeCl₃
Answer:
(b) CuCl₂

33. Chlorine on reaction with carbon disulphide forms
(a) CCl₄
(b) SO₂Cl₂
(c) S₂Cl₂
(d) CSCl₂
Answer:
(c) S₂Cl₂

34. When chlorine is passed over dry slaked lime the product obtained is
(a) CaCl₂
(b) HCl
(c) CaOCl₂
(d) HOCl
Answer:
(c) CaOCl₂

35. Hybridisation in ClF₃ is
(a) sp³
(b) sp³d
(c) dsp³
(d) sp³d²
Answer:
(b) sp³d

36. The geometry of ClF₃ is
(a) tetrahedral
(b) trigonal bipyramidal
(c) square planar
(d) triangular
Answer:
(b) trigonal bipyramidal

37. The structure of interhalogen compound XX’₅ is
(a) square pyramidal
(b) trigonal bipyramidal
(c) tetrahedral
(d) pentagonal bipyramidal
Answer:
(a) square pyramidal

38. Interhalogen compound used as a fluorinating agent is
(a) IF₇
(b) BrF₅
(c) IF₅
(d) ClF₃
Answer:
(d) ClF₃

39. The increasing order of thermal stability of oxoacids of halogens is
(a) HClO₂ < HCIO₃ < HClO₄
(b) HClO₄ < HCIO3 < HClO₂
(c) HClO₄ < HClO₂ < HCIO3
(d) HCIO3 < HClO₂ < HClO₄
Answer:
(a) HClO₂ < HCIO₃ < HClO₄

40. The shape of IF₇ molecules is :
(a) Tetrahedral
(b) Octahedral
(c) Trigonal bipyramidal
(d) Pentagonal bipyramidal
Answer:
(d) Pentagonal bipyramidal

41. The strongest oxidizing agent is :
(a) HOCl
(b) HClO₄
(c) HClO₃
(d) HClO₂
Answer:
(b) HClO₄

42. Iodine can exist in the oxidation states
(a) +1, +3, +5
(b) -1, +1, +3
(c) +3, +5, +7
(d) -1, +1, +3, +5, +7
Answer:
(d) -1, +1, +3, +5, +7

43. Fluorine does not show positive oxidation state due to absence of:
(a) d-orbitals
(b) s-orbitals
(c) p-orbitals
(d) f-orbitals
Answer:
(a) d-orbitals

44. The most powerful oxidising agent is :
(a) Fluorine
(b) Chlorine
(c) Iodine
(d) Bromine
Answer:
(a) Fluorine

45. Which one is the strongest reducing agent?
(a) HF
(b) HCl
(c) HBr
(d) HI
Answer:
(d) HI

46. Fluorine can exist in the oxidation state :
(a) – 1 only
(b) – 1 and + 1
(c) 1-, +1, +3 only
(d) -1, +l,. + 3, +7
Answer:
(a) – 1 only

47. Which of the following halogen is radioactive in nature?
(a) Chlorine
(b) Bromine
(c) Iodine
(d) Astatine
Answer:
(d) Astatine

48. The electronic configuration of krypton is
(a) [Ar] 3d¹⁰ As² Ap⁶
(b) [Ne] 3s² 3p⁶
(c) [Ai] 4d¹⁰ 5s² 5p⁶
(d) [Xe] 6s² 6p⁶
Answer:
(a) [Ar] 3d¹⁰ As² Ap⁶

49. Which one of the following does not exist?
(a) HeF₄
(b) XeF₄
(c) CF₄
(d) SF₆
Answer:
(a) HeF₄

50. Name the noble gas not present in the air.
(a) Radon
(b) Argon
(c) Krypton
(d) Helium
Answer:
(a) Radon

51. Which is the most abundant noble gas?
(a) Argon
(b) Helium
(c) Neon
(d) Krypton
Answer:
(a) Argon

52. A mixture used for respiration by the sea divers is OR Which mixture is used for respiration by deep sea divers?
(a) He + O₂
(b) Ne + O₂
(c) Ar + O₂
(d) Kr + O₂
Answer:
(a) He + O₂

53. Iodine exists as-
(a) polar molecular solid
(b) ionic solid
(c) nonpolar molecular solid
(d) hydrogen-bonded molecular solid
Answer:
(c) nonpolar molecular solid

54. Hybridisation in H₂S is
(a) sp
(b) sp³
(c) sp³d²
(d) sp²
Answer:
(b) sp³

55. Noble gas which finds the use as a cryogenic agent is
(a) He
(b) Ne
(c) Ar
(d) Kr
Answer:
(a) He

56 Which of the following compounds of chlorine is used as refrigerant?
(a) CCl₃NO₂
(b) CCl₂F₂
(c) COCl₂
(d) CCl₄
Answer:
(b) CCl₂F₂

57. The increasing order of oxidising power of oxyacids of chlorine is
(a) HClO₂ > HClO₃ > HClO₄
(b) HOCl > HCl0₂ > HClO₄
(c) HClO₂ < HClO₃ < HClO₄
(d) HClO₄ < HOCl < HCO₂
Answer:
(c) HClO₂ < HClO₃ < HClO₄

58. S-O-O-S bonds are present in
(a) H₂S₂O₆
(b) H₂SO₅
(c) H₂S₂O₅
(d) H₂S₂O₈
Answer:
(d) H₂S₂O₈

59. The increasing order of thermal stability of hydrides of sixteenth group elements is
(a) H₂O < H₂S < H₂Se < H₂Te
(b) H₂S < H₂O < H₂Se < H₂Te
(c) H₂Te < H₂S < H₂Se < H₂O
(d) H₂Te < H₂Se < H₂S < H₂O
Answer:
(d) H₂Te < H₂Se < H₂S < H₂O

60. The bond angle F-Cl-F in ClF₃ is
(a) 89°
(b) 109°29
(c) 95°
(d) 120°
Answer:
(a) 89°

61. The oxidation number of Xe in XeOF₂ is :
(a) O
(b) +2
(c) +4
(d) +3
Answer:
(c) +4

62 XeF₄ on partial hydrolysis produces
(a) XeF₂
(b) XeOF₂
(c) XeOF₄
(d) XeO₃
Answer:
(c) XeOF₄

63. Shape of XeO₄ is
(a) Octahedral
(b) Square pyramidal
(c) Pyramidal
(d) T-shaped
Answer:
(b) Square pyramidal

64. In XeF₂, XeF₄ and XeF₆ the number of lone pairs on Xe is respectively
(a) 2, 3, 1
(b) 1, 2, 3
(c) 4, 1, 2
(d) 3, 2, 1
Answer:
(d) 3, 2, 1

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 6 Chemical Kinetics Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 6 Chemical Kinetics

Question 1.
What is chemical kinetics?
Answer:
Chemical kinetics is a branch of physical chemistry that involves the study of the rates and mechanisms of chemical reactions and the influence of various factors like temperature, pressure, catalyst, etc., on the rates of reactions.

Question 2.
What is the importance of chemical kinetics?
Answer:

• It deals with the study of the rates and mechanism of reactions.
• The effect of temperature on the reaction rates can be studied.
• The influence of catalysts can be studied.
• The conditions for altering the rates and mechanisms of chemical reactions can be predicted.
• Thermodynamic parameters like energy, enthalpy changes, Δ5, ΔG of the reactions can be calculated.

Question 3.
How are reactions classified according to their rates? Give one example of each.
Answer:
According to the rates of the reactions, they can be classified as :
(1) Fast reactions,
(2) Very slow reactions,
(3) Moderately slow reactions.

(1) Fitst actions : In this, reactants react almost instantaneously, e.g., neutralisation reaction between H+ and OH-, forming water.
\(\mathrm{H}_{(\mathrm{xa})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{0 \mathrm{D}}\)

(2) Very slow reactions : In this, the reactants react extremely slow, so that there is no appreciable change in the concentrations of the reactants over a long period of time. E.g., reaction of silica with mineral acids, rusting of iron, etc.

(3) Moderately slow reactions : In this, the reactants react moderately slow with a measurable velocity, e.g., the hydrolysis of the esters.
\(\begin{aligned}
\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{COOH} \\
&+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}
\end{aligned}\)

Question 4.
Define rate of a reaction.
Answer:
Definition : The rate of a chemical reaction is defined as the change in the concentration of the reactants or products per unit time.

It is often expressed in mol dm^-3s^-1.

Question 5.
Explain the following :
(A) Rate of the reaction in terms of the concentration of the reactants.
(B) Rate of reaction in terms of the concentration of the products.
Answer:
(A) Rate of the reaction in terms of the concentration of the reactants :
If c₁ and c₂ are the concentrations of the reactant A at time t₁ and t₂ respectively, then, the change in concentration, Δc = c₂ – c₁
Since c₂ < c₁, the term Δc is negative often written as – Δc.
The time interval is, Δt – t₂ – t₁
If Δ [A] is the change in concentration of A, then A[A] = C₂ – C₁
∴ Rate of the reaction = \(\mathrm{A}=\frac{-\Delta[\mathrm{A}]}{\Delta t}\)
∴ Rate of the reaction = \(\frac{-\Delta c}{\Delta t}\)

(B) Rate of the reaction in terms of the concentration of the products :
If x₁ and x₂ are the concentrations of the product B at time t₁ and t₂ respectively, then the change in concentration, Δx = x₂ – x₁.

∴ x₂ > x₁, the term Δx is positive.
The time interval is, Δt = t₂ – t₁

If Δ B is the change in concentration of product B, then Δ[B] = x₂ – x₁ = Δx
∴ Rate of formation of \(\mathrm{B}=+\frac{\Delta[\mathrm{B}]}{\Delta t}\)
∴ Rate of the reaction \(=\frac{\Delta x}{\Delta t}\)

Question 6.
What are the units of rate of a chemical reaction?
Answer:

∴ The unit of the rate of a chemical reaction : mol L^-1 3t^-1 or mol dm^-3s^-1 (According to IUPAC, the rate of a chemical reaction should be expressed in mol m^-3s^-1 [SI unit]).

Question 7.
Mention the factors that affect the rate of a chemical reaction.
Answer:
The rate of a chemical reaction depends on the following factors :

• Nature of the reactants.
• The concentration of the reactants. In case of a gaseous reaction the rate depends on the pressures of the reactants.
• Temperature of the reaction.
• The presence of a catalyst and its nature.

Question 8.
Explain the term Average rate of a reaction.
Answer:
In chemical kinetics the rate of a reaction is measured in terms of the changes in the concentrations of the reactants or the products per unit time. Average rate of a chemical reaction : It is expressed as a finite change in concentration (- Δc) of the reactant divided by the time interval (Δt) for the change in concentration.

Consider a reaction,
A → B
The rate of a reaction, \(R=\frac{-\Delta[\mathrm{A}]}{\Delta t}=\frac{-\Delta c}{\Delta t}=\frac{c_{2}-c_{1}}{t_{2}-t_{1}}\)


∴ Average rate \(=\frac{-\Delta[\mathrm{A}]}{\Delta t}\) (in mol dm^-3s^-1)

Δc is negative, since the concentrartion of the reactant decreases with the time.

The rate of a reaction is also measured in terms of a finite change in the concentration (Δx) of the product divided by the time interval (Δt), for the change.

For the reaction,

Question 9.
Explain the term Instantaneous rate of a reaction.
Answer:
Instantaneous rate of a reaction : It is defined as a rate of a reaction at a specific instant during a course of the reaction.

If the average reaction rate is calculated over shorter and shorter intervals (making Δt very small) then instantaneous rate is obtained.

In case of reactant A, the instantaneous rate is represented as, \(R=\frac{-d[\mathrm{~A}]}{d t}\) and in case of product B, it is represented as \(R=\frac{+d[B]}{d t}\)

Question 10.
Define :
(a) Average rate of reaction.
(b) Instantaneous rate of reaction.
Answer:
(a) Average rate of a chemical reaction : It is expressed as a finite change in concentration (- Δc) of the reactant divided by the time interval (Δt) for the change in concentration.

∴ Average rate, \(R=\frac{-\Delta c}{\Delta t}\)

(b) Instantaneous rate of reaction : It is defined as a rate of a reaction at a specific instant during a course of the reaction.

Instantaneous rate \(=\frac{-d c}{d t}\)

Question 11.
Represent the average rates of the following reaction. N_2(g) + 3H_2(g) → 2NH_3(g).
Answer:
For the reation,

This is because the rate of consumption of H₂ is thrice the rate of consumption of N₂ while the rate of formation of NH₃ will be twice the rate of consumption of N₂.

Question 12.
Express the rate of a reaction in terms of change in concentration of each constituent in the following reaction : aA+bB → cC+ dD
Answer:
The rate of a reaction may be expressed in terms of decrease in the concentration of the reactants or in-crease in the concentration of the product per unit time,

∴ For the given reaction, aA T bB → cC +dD

Question 13.
For a hypothetical reaction, A + 2B → products, the concentration of A and B at different intervals of time are given in the following table. Find the rates of the reaction in terms of concentration changes in A and B.

The equilibrium concentration of A and B at different time intervals :

Answer:
Rate of a reaction = \(\frac{-\Delta[\mathrm{A}]}{\Delta t}=-\frac{1}{2} \frac{\Delta[\mathrm{B}]}{\Delta t}\)
(1) Over time interval from O to 10 minutes

(Note that the rate of a reaction in terms of changes in concentration of any reactant or product at the given time remains the same.)

(2) Over the time interval from 10 to 20 minutes,

Question 14.
Show that the rate of reaction is the same whether expressed in terms of the rate of consumption of any reactant or of the formation of any product.
2N₂O_5(g) → 4NO_2(g) + O_2(g)
The concentrations of reactants and products at different time intervals are given in the following table :
Concentrations of various species at different times for the reaction N₂O_5(g) → 4NO_2(g) + O_2(g) :

Answer:
The rate of the reaction can be expressed in terms of rate of consumption of reactants or rate of formation of products.

Consider concentrations at time t₁ = 200 seconds and t₂ = 400 seconds

The constant values of rate of reaction proves that the rate of the reaction may be measured in terms of concentration changes of reactants or products per unit time.

Question 15.
Define Rate law (or differential rate law).
Answer:
Rate law (or differential rate law) : It is defined as an experimentally determined mathematical equation which expresses the rate of a chemical reaction in terms of molar concentrations of the reactants which influence the rate of the reaction. For example, for a reaction, A + B → Products By rate law, Rate = R = k[A] x [B] where k is a rate constant and [Al and [B] are molar concentrations of the reactants A and B respectively.

Question 16.
Give examples of rate law with illustrations.
Answer:
Consider following examples :
(i) H_2(g) + I_2(g) → 2HI_(g)
R = k[H₂] [I₂]

(ii) 2H₂O_2(g) → 2H₂O_(I) + O_2(g)
Experimentally it is observed that the rate of the reaction is proportional to the concentration of H₂O₂.
∴ R = k [H₂O₂]

(iii) NO_2(g) + CO_(g) → NO_(g) + CO_2(g)
Experimentally it is observed that rate of the reaction does not depend on the concentration of CO but it is proportional to [NO₂]².
∴ R = k[NO₂]²

Question 17.
What are the applications of the rate law?
Answer:

• The rate of any reaction at the given concentration can be measured by knowing the rate law and the rate constant.
• The concentration of the reactants or the products at any instant during the progress of a reaction can be estimated with the help of rate law and the rate constant.
• The mechanisms of simple or complex chemical reactions can be predicted and studied.

Question 18.
Define the rate constant. What are the factors which influence the rate constant of a chemical reaction?
Answer:
(A) Rate constant : The rate constant of a chemical reaction is defined as the rate of the chemical reaction when the concentration (or active masses) of each reactant has unit value, i.e., 1 mol dm^-3 in the case of solution and the pressure is 1 atm in case of gases, e.g., for a reaction, A → products, Rate R = k[A].

If [A] = 1 mol dm^-3, then k = R.

(B) The rate constant of a reaction depends on the following factors:

• Nature of the reactants.
• Temperature of the reaction. As the temperature increases, the velocity constant (rate constant) increases.
• The conditions of the reactions like the presence of the catalyst, solvent, pH, etc.
• It does not depend on the concentration of the reactants. But if one or more substances are in excess concentration, then the order of the reaction is independent of them.

Question 19.
What are the characteristics of rate constant?
Answer:
The characteristics of rate constant are as follows :

• The rate constant depends upon the nature of the reaction.
• Higher the value of the rate constant, faster is the reaction.
• Lower the value of the rate constant, slower is the reaction.
• By increasing the temperature, the magnitude of the rate constant increases.
• For the given reaction, the rate constant has higher value in the presence of a catalyst than in the absence of the catalyst.
• The reactions having lower activation energy have higher values for rate constants.

Solved Examples 6.2 – 6.3.2

Question 20.
Solve the following :

(1) Write the rate expressions for the following reactions in terms of rate of consumption of the reactants and the rate of formation of the products.
(i) 2NO_(g) + O_2(g) → 2NO_2(g)
(ii) H_2(g) + I_2(g) → 2HI_(g)
Solution :
(i) Given : 2NO_(g) + O_2(g) → 2NO_2(g)
Rate of consumption of NO at time \(t=\frac{-d[\mathrm{NO}]}{d t}\)
Rate of consumption of O₂ at time \(t=\frac{-d\left[\mathrm{O}_{2}\right]}{d t}\)
Rate of formation of NO₂ at time \(t=\frac{d\left[\mathrm{NO}_{2}\right]}{d t}\)
Rate of the reaction \(=-\frac{1}{2} \frac{d[\mathrm{NO}]}{d t}=\frac{-d\left[\mathrm{O}_{2}\right]}{d t}\)
\(=\frac{1}{2} \frac{d\left[\mathrm{NO}_{2}\right]}{d t}\)

(ii) Given : H_2(g) + I_2(g) → 2HI_(g)
Rate of consumption of H₂ at time \(t=\frac{-d\left[\mathrm{H}_{2}\right]}{d t}\)
Rate of consumption of I₂ at time \(t=\frac{-d\left[\mathrm{I}_{2}\right]}{d t}\)
Rate of formation of HI at time \(t=\frac{d[\mathrm{HI}]}{d t}\)
∴ Rate of reaction at any time t \(=-\frac{d\left[\mathrm{H}_{2}\right]}{d t}=-\frac{d\left[\mathrm{I}_{2}\right]}{d t}=\frac{1}{2} \frac{d[\mathrm{HI}]}{d t}\)

(2) The gas-phase reaction between NO and Br₂ is represented by the equation. 2NO_(g) + Br_2(g) → 2NOBr_(g)
(a) Write the expressions for the rate of consumption of reactants and formation of products.
(b) Write the expression for the rate of overall reaction in terms of rates of consumption of reactants and formation of products.
Solution :
Given : 2NO_(g) + Br_2(g) → 2NOBr_(g)
(a) Rate of consumption of NO at time t \(=-\frac{d[\mathrm{NO}]}{d t}\)
Rate of consumption of Br₂ at time t \(=\frac{-d\left[\mathrm{Br}_{2}\right]}{d t}\)
Rate of formation of NOBr at time \(t=\frac{d[\mathrm{NOBr}]}{d t}\)
(b) Rate of reaction \(=-\frac{1}{2} \frac{d[\mathrm{NO}]}{d t}=\frac{-d\left[\mathrm{Br}_{2}\right]}{d t}\)
\(=\frac{1}{2} \frac{d[\mathrm{NOBr}]}{d t}\)

(3) The decomposition of N2Os is represented by the equation
2N₂O_5(g) → 4NO_2(g) + O_2(g)
(a) How is the rate of formation of NO₂ related to the rate of formation of O₂?
(b) How is the rate of formation of O₂ related to the rate of consumption of N₂O₅?
Solution :
Given : 2N₂O_5(g) → 4NO_2(g) + O_2(g)
(a) Rate of formation of NO₂ at time \(t=\frac{d\left[\mathrm{NO}_{2}\right]}{d t}\)
Rate of formation of O₂ at time \(t=\frac{d\left[\mathrm{O}_{2}\right]}{d t}\)

They are related to each other through rate of reaction.
∴ Rate of reaction \(=\frac{1}{4} \frac{d\left[\mathrm{NO}_{2}\right]}{d t}=\frac{d\left[\mathrm{O}_{2}\right]}{d t}\)

(b) Rate of consumption of N₂O₅ at time t \(=-\frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}\)

Rate of reaction \(=-\frac{1}{2} \frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}=\frac{d\left[\mathrm{O}_{2}\right]}{d t}\)

In general,
Rate of reaction \(=-\frac{1}{2} \frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}=\frac{1}{4} \frac{d\left[\mathrm{NO}_{2}\right]}{d t}=\frac{d\left[\mathrm{O}_{2}\right]}{d t}\)

(4) Nitric oxide reacts with H₂ according to the reaction. 2NO_(g) + 2H_2(g) → N_2(g) + 2H₂O_(g)
What is the relationship among \(\frac{d[\mathrm{NO}]}{d t}=\frac{d\left[\mathrm{H}_{2}\right]}{d t}=\frac{d\left[\mathrm{~N}_{2}\right]}{d t} \text { and } \frac{d\left[\mathrm{H}_{2} \mathrm{O}\right]}{d t} ?\)
Solution :
Given : 2NO_(g) + 2H_2(g) → N_2(g) + 2H₂O_(g)
The relationship among the rate of consumption of the reactants and the rate of formation of products is as follows :

Rate of reaction :
\(R=-\frac{1}{2} \frac{d[\mathrm{NO}]}{d t}=-\frac{1}{2} \frac{d\left[\mathrm{H}_{2}\right]}{d t}=\frac{d\left[\mathrm{~N}_{2}\right]}{d t}=\frac{1}{2} \frac{d\left[\mathrm{H}_{2} \mathrm{O}\right]}{d t}\)

(5) The rate of decomposition of N2Os was studied in liquid bromine,
2N₂O_5(g) → 4NO_2(g) + O_2(g)
If at a certain time, the rate of disappearance of N₂O₅ is 0.015 Ms^-1 find the rates of formation of NO₂ and O₂. What is the rate of the reaction at this instant?
Solution :
Given : 2N₂O_5(g) → 4NO_2(g) + O_2(g)
Rate of disappearance of N₂O₅ = 0.015 M s^-1
Rate of formation of NO₂ =?
Rate of formation of O₂ =?
Rate of reaction = ?
Rate of disappearance of \(\mathrm{N}_{2} \mathrm{O}_{5}=\frac{-d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}\)
= 0.015 M s^-1

Since 4 moles of NO₂ are formed from 2 moles of N₂O₅ Rate of formation of NO₂
Answer:
Rate of formation of NO₂ = 0.03 Ms^-1
Rate of formation of O₂ = 0.0075 M s^-1
Rate of reaction = 0.0075 Ms^-1.

(6) In the reaction, PCl_5(g) → PCl_3(g) + CI_2(g), at a particular moment, the rate of disappearance of PCl₅ is 0.015 Ms^-1. What are the rates of formation of PCI₃ and Cl₂?
Solution :
Given : PCl_5(g) → PCl_3(g) + Cl_2(g)


Answer:
Rate of formation of PCl₃ = 0.015 Ms^-1
Rate of formation of Cl₂ = 0.015 Ms^-1

(7) In the reaction, 2N₃O_5(g) → 4NO_2(g) + O_2(g), at a certain time, the rate of formation of NO₂ is 0. 04 Ms^-1. Find the rate of consumption of N₂O₅, rate of formation of O₂ and the rate of the reaction.
Solution :
Given : 2N₂O_5(g) → 4NO_2(g) + O_2(g)
Rate of formation of NO₂ = \(\frac{d\left[\mathrm{NO}_{2}\right]}{d t}\) = 0.04 Ms^-1

From the reaction, rate of consumption of N₂O₅ is half the rate of formation of NO₂ since when 2 moles of N₂O₅ are consumed, 4 moles of NO₂ are formed.

Rate of formation of O₂ is one-fourth rate of formation of NO₂.

Answer:
(i) Rate of consumption of N₂O₅
(ii) Rate of formation of O₂ = 0.01 Ms^-1
(iii) Rate of reaction = 0.01 Ms^-1

(8) Consider the reaction 2A + B → 2C. Suppose that at a particular moment during the reaction, rate of disappearance of A is 0.076 M/s,
(a) What is the rate of formation of C?
(b) What is the rate of consumption of B?
(c) What is the rate of the reaction?
Solution :
Given : 2A + B → 2C
Rate of disappearance of A = 0.076 Ms^-1
(a) Rate of formation of C =?
(b) Rate of consumption of B =?
(c) Rate of reaction = ?

Answer:
(a) Rate of formation of C = 0.076 Ms^-1
(b) Rate of consumption of B = 0.038 M s^-1
(c) Rate of reaction = 0.038 Ms^-1

(9) Consider the reation \(\mathbf{3 I}_{(\mathbf{a q})}^{-}+\mathbf{S}_{2} \mathbf{O}_{8(u q)}^{2-} \longrightarrow \mathbf{I}_{3(\mathrm{aq})}^{-}+2 \mathrm{SO}_{4}^{2-}\) At a particular time t, \(t, \frac{d\left[\mathrm{SO}_{4}^{2-}\right]}{d t}=2.2 \times 10^{-2} \mathrm{M} / \mathrm{s}\) What are the values of \(\text { (a) }-\frac{d\left[\mathrm{I}^{-}\right]}{d t}\) \(-\frac{d\left[\mathrm{~S}_{2} \mathrm{O}_{8}^{2-}\right]}{d t}\) \(\text { (c) } \frac{d\left[\mathbf{I}_{3}^{-}\right]}{d t}\) at the same time?
Solution :

(a) Rate of consumption of \(\mathrm{I}^{-}=-\frac{d\left[\mathrm{I}^{-}\right]}{d t}\)
When 2 moIes of \(\mathrm{SO}_{4}^{2-}\) are formed, 3 moves of I^– are consumed in the same time.

(b) In the formation of 2 moles of \(\mathrm{SO}_{4}^{2-}\), 1 mole of \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) is consumed in the same time.

Answer:

(10) Ammonia and oxygen react at high temperature as :
4NH_3(g) + 5O_2(g) → 4NO_(g) + 6H₂O_(g)
In an experiment, rate of formation of NO_(g) is 3.6 x 10^-3 mol L^-1s^-1.
Calculate-
(a) Rate of disappearance of ammonia
(b) Rate of formation of water.
Solution :

Answer:
(a) Rate of disappearance of NH₃
= 3.6 x 10^-3 mol L^-1s^-1
(b) Rate of formation of H₂O
= 5.4 x 10^-3 mol L^-1s^-1

(11) The rate law for the reaction
C₂H₄Br₂ + 3I^– → C₂H₄ + 2Br^– +I^–₃ is Rate = k [C₂H₄Br₂][I^–]. The rate of the reac-tion is found to be 1.1 x 10^-4 M/s when the concentrations of C₂H₄Br₂ and I^–– are 0.12M and 0.18 M respectively. Calculate the rate constant of the reaction.
Solution :
Given : C₂H₄Br₂ + 3I^– → C₂H₄ + 2Br^– +I^–₃
By rate law, Rate of reaction = R = k x [C₂H₄Br₂][I^–]
R = 1.1 x 10^-4 Ms^-1
[C₂H₄Br₂] = 0.12 M; [I^–] =0.18 M
Rate constant = k =?
R = k x [C₂H₄Br₂] x [I^–]

Answer:
Rate constant = k = 5.1 x 10^-3 M^-1s^-1

(12) For a reaction, 2A + B → C, the rate law is, rate =k x [A]² x [B]. If the rate constant of the reaction is 3.74 x 10^-2M^-2s^-1, calculate the rate of the reaction when the concentrations of A, B and C are 0.108 M, 0.132 M and 0.124 M respectively.
Solution :
Given : Rate constant of the reaction = k
= 3.74 x 10^-2M^-2s^-1
[A] =0.108 M, [B] = 0.132M, [C] = 0.124 M
Rate of the reaction = R = ?
By rate law,
R = k [A]² x [B] = (0.108)² x 0.132 = 1.54 x 10^-3 Ms^-1
(Concentration of C need not be considered since it is a product.)
Answer:
Rate of reaction = 1.54 x10^-3 Ms^-1

(13) For a reaction, A + B → C, if the concentration of A doubles, the rate of the reaction doubles. While if the concentration of B doubles the rate of the reaction increases by four fold. Write rate law. .
Solution :
Let x moles of A react with y moles of B. xA + yB → C
To write rate law, it is necessary to find x and y values.

(i) Initial rate \(=R_{1}=k[\mathrm{~A}]_{1}^{x}[\mathrm{~B}]_{1}^{y}\)
Final rate R₂ is doubled when the concentration of A is doubled, i.e., R₂ = 2R₁ when final concentration,

(It is assumed that the concentration of B remains same.)

(ii) Initial rate \(=R_{1}=k[\mathrm{~A}]_{1}^{x}[\mathrm{~B}]^{y}\)
If the concentration of B is doubled keeping of A constant, rate becomes four times, i.e.,

Hence the rate law is represented by an expression.
Rate = k[A] [B]²
Answer:
Rate law is. Rate = k [A] [B]²

(14) For the reaction, A2 + B + C → AC + AB, it is found that tripling the concentration of A₂ triples the rate, doubling the concentration of C doubles the rate and doubling the concentration of B has no effect,
(a) What is the rate law?
(b) Why the change in concentration of B has no effect?
Solution :
Given : A2 + B + C → AC + AB
(a) The rate law may be represented as,
Rate = k [A₂]^x [B]^y [C]^z
Let [A]₁, [B]₁ and [C]₁ represent initial concentration and [A]₂, [B]₂ and [C]₂ represent final concentrations, and let R₁ and R₂ be initial and final rates of the reaction when the concentrations are changed.

(i) If [A]₂ = 3[A]₁, R₂ = 3R₁

If the concentrations of B and C remain constant, then

(b) In the rate determining step, B may not be involved as the reactant, hence rate is independent of changes in concentration of B. (OR B may be in large excess as compared to the concentrations of A and C.)
Answer:
(a) Rate law : Rate = k [A] [C]

Question 21.
Define and explain the term order of a chemical reaction.
Answer:
Order of a chemical reaction : The order of a chemical reaction is defined as the number of molecules (or atoms) whose concentrations influence the rate of the chemical reaction.
OR
The order of a chemical reaction is defined as the sum of the powers (or exponents) to which the concentration terms of the reactants are raised in the rate law expression for the given reaction.

Explanation :
Consider a reaction,
n₁A + n₂B → Products
where n₁ moles of A react with n₂ moles of B.

The rate of this reaction can be expressed by the rate law equation as,
R = k [A]^n₁ [B]^n₂
where k is the rate constant of the reaction, hence, the order of the reaction is n – n₁ + n₂, (observed, experimentally).

If n = 1, the reaction is called the first order reaction, if n = 2, it is called the second order reaction, etc.

If n = 0, it is called the zero order reaction, e.g., photochemical reaction of H_2(g) and Cl_2(g).

Question 22.
What are the features (or key points) of order of a reaction?
Answer:
The features of order of reaction are as follows :

• It represents the number of atoms, ions or molecules whose concentrations influence the rate of the reaction.
• It is not related to the stoichiometric equation of the reaction, hence it cannot be predicted from stoichiometric balanced equation.
• It is experimentally determined quantity.
• It is defined only in terms of the concentrations of the reactants and not of products.
• It may have values which are integers, fractional or zero.
• Higher values are rare. Reactions of first and second order are in large number. Third order reactions are very few like,
  2NO_(g) + Cl_2(g) → 2NOCl_(g).

Solved Examples 6.3.3

Question 23.
Solve the following :
(1) From the rate expressions for the following reactions, determine their order :
(a) 2N₂O_5(g) → 4NO_2(g) + O_2(g) : Rate = k [N₂O₅]
(b) CHCl_3(g) + Cl_2(g) → CCl_4(g) + HCl_(g) : Rate = k [CHL₃] [Cl₂]^1/2
(c) C₂H₅Cl_(g) → C₂H_4(g) + HCl_(g): Rate = k [C₂H₅Cl]
(d) 2NO_2(g) + F_2(g) → 2NO₂F_(g) → : Rate = k (NO₂] [F₂]
Solution :
(a) 2N₂O_5(g) → 4NO_2(g) + O_2(g)
The rate law expression given for the reaction is Rate = k x [N₂O₅]
Hence the reaction is of first order.

(b) CHCl_3(g) + Cl_2(g) → CCl_4(g) + HCl_(g)
The given rate law expression is, R = k [CHCl₃] x [Cl₂]^1/2 Here the order of a reaction is one with respect to CHCl_3(g) and half with respect to Cl_2(g). Therefore the overall order of the reaction is 1 + 1/2 = 1.5.

(c) C₂H₅Cl_(g) → C₂H_4(g) + HCl_(g)
The given rate law expression is, Rate = k [C₂H₅Cl]
Hence the reaction has order equal to one.

(d) 2NO_2(g) + F_2(g) → 2NO₂F_(g)
The given rate law expression for the reaction is Rate = k [NO₂] x [F₂]
Hence the reaction is first order with respect to NO2 and first order with respect to F₂. The overall order of the reaction is, n = n_NO2 + nF₁ = 1 + 1 = 2.

(2) Determine the order of following reactions from their rate expressions :
(a) 2H₂O₂ → 2H₂O + O₂ Rate = k [H₂O₂]
(b) NO₂ + CO → NO + CO₂ Rate = k [NO₂]²
(c) 2NO + O₂ → 2NO₂ Rate = k [NO]² x [O₂]
(d) CHCl_3(g) + Cl_2(g) → CCl_4(g) + HCl_(g)
Rate = k [CHCl₃] [Cl₂]
Solution :
(a) For the reaction,
2H₂O₂ → 2H₂O + O₂
Since the rate law expression given is,
Rate = k [H₂O₂]
Hence the reaction is of first order.

(b) For the reaction,
NO₂ + CO → NO + CO₂
Since the rate law given is Rate = k [NO₂]², the reaction is second order with respect to NO₂ and zero order with respect to CO. Hence the net order of the reaction is, n = nNO₂ + nco = 2 + 0 = 2

(c) For the reaction,
2NO + O₂ → 2NO₂
Since the rate law expression given is, Rate = k [NO]² x [O₂] the reaction is second order with respect to NO and first order with respect to O₂. Hence the overall order of reaction is n = nNO₂ + no₂ = 2 + 1 = 3.

(d) For the reaction, by rate law,
Rate = k [CHCl₃] x [Cl₂] reaction is first order with respect to CHCl₃ and first order with respect to Cl₂. Hence the overall order is, n = ncHcl₃ + ncl₂ = 1 + 1 = 2.

(3) Write the rate law expressions for the following reactions:
(1) 2N₂O_5(g) → 4NO₂ + O₂; order of the reaction is 1.
(2) CH₃CHO → CH₄ + CO; order of the reaction Is ³/₂.
Solution :
(1) For the given reaction, order is one hence the rate law expression is, Rate = k [N₂O₅].
(2) For the given reaction, order is ³/₂, hence the rate law expression is Rate = k x [CH₂CHO]^3/₂.

(4) The reaction \(\mathbf{H}_{2} \mathbf{O}_{2(\mathbf{a q})}+3 \mathbf{I}_{(\mathbf{a q})}^{-}+2 \mathbf{H}_{(\mathrm{aq})}^{+} \longrightarrow 2 \mathbf{H}_{2} \mathbf{O}_{(0)}+\mathbf{I}_{3(a q)}^{-}\) is first order in H₂O₂ and I^–, zero order in H⁺. Write the rate law.
Solution:
Given :
\(\mathrm{H}_{2} \mathrm{O}_{2(\mathrm{~g})}+3 \mathrm{I}_{(\mathrm{aq})}^{-}+2 \mathrm{H}^{+}{ }_{(\mathrm{aq})} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{i})}+\mathrm{I}_{3(\mathrm{aq})}^{-}\)
Since the reaction is first order in H₂O₂ and F and zero order in H⁺, the expression for rate law will be,
Rate =k [H₂O₂]¹ [I^–]¹ [H⁺]⁰
∴ Rate = k [H₂O₂] [I^–]
Answer:
Rate = k [H₂O₂] [I^–]

(5) The rate law for the gas-phase reaction
2NO_(g) + O_2(g) → 2NO_2(g) is rate = k [NO₂]² [O₂]. What is the order of the reaction with respect to each of the reactants and what is the overall order of the reaction?
Solution :
Given : 2NO_(g) + O_2(g) → 2NO_2(g)
Rate = k [NO]²[O₂]
Order of the reaction with respect to NO = n_No = 2
Order with respect to O₂ = n_O2 = 1
Overall order of the reaction = n = n_NO + n_O2
= 2 + 1
= 3
Answer:
Order with respect to NO = 2
Order with respect to O₂ = 1
Overall order = 3

(6) What is the order for the following reactions?
(a) 2NO_2(g) + F_2(g) → 2NO₂F_(g), rate = k [NO₂][F₂]
(b) CHCl_3(g) + Cl_2(g) → CCl_4(g) + HCl_(g), rate = k[CHCl₃][Cl₂]^1/₂
Solution :
(a) Given : 2NO_2(g) + F_2(g) → 2NO₂F
Rate = k [NO₂][F₂]
Hence the reaction is first order with respect to NO₂ and first order with respect to F₂
∴ Order of reaction = n_NO2 + n_F2 = 1 + 1 = 2

(b) Given :
CHCl_3(g) + Cl_2(g) → CCl_4(g) + HCl_(g),
Rate = k [ CHCl₃] [Cl₂]^1/₂
Hence the reaction is first order in CHCl₃ and half order in Cl₂.
∴ Order of reaction
= n_CHCl3 + n_Cl2 = 1 + \(\frac{1}{2}\) = \(\frac{3}{2}\)
Answer:
(a) Order of the reaction = 2
(b) The order of the reaction = \(\frac{3}{2}\)

(7) Write the rate law for the following reactions :
(a) A reaction that is zero order in A and second order in B.
(b) A reaction that is second order in NO and first order in Br₂.
Solution :
(a) Given : A + B → Products
The reaction is zero order in A and second order in B. Hence the rate law is represented as, Rate = k [A]^O[B]²
Rate = k[B]²

(b) Given : 2NO_(g) + Br_2(g) → 2NOBr_(g)
The reaction is second order in NO and first in Br₂. Hence the rate law is,
∴ Rate = k [NO]²[Br₂]
Answer: (a) Rate law : Rate = k[B]²
(b) Rate law : Rate = k [NO]²[Br₂]

(8) The reaction A + B → Products, is first order in each of the reactants, (a) Write the rate law.
(b) How does the reaction rate change if the concentration of B is decreased by a factor 3?
(c) What is the change in the rate if the concentration of each reactant is tripled? (d) What is the change in the rate, if the concentration of A is doubled and that of B is halved?
Solution :
(a) The reaction is first order in A and B. Hence the equation for rate law is,
Rate = k [A] [B]
(b) Before changing the concentration of B, Initial rate = R₁ – k [A]₁ [B]₁
After change in concentration of B,

Hence the rate of the reaction will be decreased by a factor 3.

(c) When the concentration of each reactant is tripled, then the final concentrations will be, [A]₂ = 3[A]₁ and [B]₂ = 3[B₁]
∴ R₂ = k x 3[A]₁ x 3 [B]₁
∴ R₂ = k x 3[A]₁ x 3 [B]₁

Hence the rate of the reaction will be increased by 9 times.

(d) When the concentration A is doubled and that of B is halved then the final concentrations will be,

Rate of the reaction will remain unchanged.
Answer:
(a) Rate law is, Rate = k [A] [B],
(b) Rate is decreased by a factor 3,
(c) Rate is increased by 9 times,
(d) Rate remains unchanged.

(9) Consider the reaction A₂ + B → products. If the concentration of A₂ and B are halved, the rate of the reaction decreases by a factor of 8. If the concentration of A₂ is increased by a factor of 2.5, the rate increases by the factor of 2.5. What is the order of the reaction? Write the rate law.
Solution :
Given : A₂ + B → Products
(i) When concentration of A₂ and B are halved :
[A₂]_2(final) = 1/2 [A₂]_1(final) and [B]₂ = 1/2 [B]₁ then, R_2(final) = 1/8R_1(intial).

(ii) When concentration of A₂ is increased by the factor 2.5,
[A₂]₂ = 2.5 [A₂]₁ (concentration of B is same) then, R₂ = 2.5 R₁
Now let the reaction be, XA₂ + yB → Products

From data in (ii),

Hence the reaction is of third order. The rate law can be represented as,
Rate = k [A₂] [B]²
Answer:
(i) Order of the reaction = 3
(ii) Rate law : Rate = k [A₂] [B]³

(10) Consider the reaction C + D → Products. The rate of the reaction increases by a factor of 4 when the concentration of C is doubled. The rate of the reaction is tripled when concentration of D is tripled. What is the order of the reaction? Write the rate law.
Solution :
Given : C + D → Products OR xC + yD → Products
(i) When the concentration of C is doubled, the rate of the reaction increases by 4.

[C]_2(final) = 2[C]_1(initial) then R_2(final) = 4R_1(initial)
(In this, the concentration of D is assumed to be constant.)

Hence, the reaction is second order in C.
∴ n_C = 2
(ii) When the concentration of D is tripled, rate is tripled. The concentration of C is assumed to be constant.

Rate law : Rate = A[C]²[D]
Answer:
(i) Order of the reaction = 3
(ii) Rate law : Rate = A[C]²[D]

(11) The reaction F_2(g) + 2ClO_2(g) → 2FClO_2(g) is first order in each of the reactants. The rate of the reaction is 4.88 x 10^-4 M/s when [F₂] = 0.015 M and [ClO₂]= 0.025 M. Calculate the rate constant of the reaction.
Solution :
Given :
F_2(g) + 2ClO_2(g) → 2FClO_2(g)
Order of reaction in F₂ = nF₂ = 1
Order of reaction in CIO₂ = n_ClO2 = 1
Rate = R = 4.88 x 10^-4 Ms^-1
[F₂] = 0.015 M; [ClO₂] = 0.025 M
Rate = k = ?
By rate law,

Answer:
Rate constant = 1 = 1.3 M^-2s^-1

(12) The reaction 2H_2(g) + 2NO_(g) → 2H₂O_(g) + N_2(g) is first order in H₂ and second order in NO. The rate constant of the reaction at a certain tem­perature is 0.42M^-2s^-1. Calculate the rate when [H₂] = 0.015 M and [NO] = 0.025 M.
Solution :
Given : 2H_2(g) + 2NO_(g) → 2H₂O_(g) + N_2(g)
Order of reaction in H₂ = n_H1 = 1
Order of reaction in NO = n_NO = 2
Rate constant = k = 0.42 M^-2s^-1
[H₂] = 0.015 M; [NO] = 0.025 M
Rate of reaction = R = ?
By rate law,
Rate = R = k [H₂] [NO]²
= 0.42 x 0.015 x (0.025)² M^-2s^-1 M M
= 3.94 x 10^-6 Ms^-1
Answer:
Rate of reaction = R = 3.94 x 10^-6 Ms^-1

(13) Find the order of following reactions whose rate laws are expressed as follows. C_A and C_B are the concentrations of reactants A and B respectively :

Solution :
Given :
(1) For, – \(\frac{d c}{d t}\) = k x \(\mathrm{C}_{A}^{0}\) the order of the reaction, n = 0. Hence it is a zero order reaction.

(2) For, – \(\frac{d c}{d t}\) = k x \(\mathrm{C}_{A}^{3 / 2}\), the overall order of the reaction is 3/2.

(3) For, –\(\frac{d c}{d t}\) = k x \(\mathrm{C}_{A}^{1 / 2} \mathrm{C}_{B}^{2}\), the reaction has order 1/2 with respect to A and 2 with respect to B.
∴ n = n_A + n_B = \(\frac{1}{2}\) + 2 = \(\frac{5}{2}\).
Hence the (overall) order of the reaction is \(\frac{5}{2}\).

(4) For, \(-\frac{d c}{d t}=k \mathrm{C}_{A}^{5 / 2} \times \mathrm{C}_{B}^{0}\)
The reaction has order \(\frac{5}{2}\) with respect to A and zero with respect to B.
∴ n = n_A + n_B = \(\frac{5}{2}\) + 0 = \(\frac{5}{2}\)
Hence the order of the reaction is \(\frac{5}{2}\).

(5) For, \(-\frac{d c}{d t}=k \times \mathrm{C}_{A}^{1 / 3} \times \mathrm{C}_{B}^{2 / 3}\). The reaction has order \(\frac{1}{3}\) with respect to A and \(\frac{2}{3}\) with respect to B.
∴ n = n_A + n_B = \(\frac{1}{3}\) + \(\frac{2}{3}\) = 1
Hence the order of the reaction is 1.

(14) The rate of a reaction, 2A + B → Products is 3.78 x 10^-4 M s^-1 when the concentrations of A and B are 0.3 M each. If the rate constant of the reaction is 4.2 x 10^-3s^-1 find the order of the reaction.
Solution :
Given : 2A + B → Products
Rate = R = 3.78 x 10^-4Ms^-1
[A] = [B] = 0.3 M
Rate constant = 1 = 4.2 x 10^-3 s^-1
Let the order of the reaction in A be x and in B be y.

Then, by rate law,
Rate = R = k [A]^x [B]^y 3.78 x 10^-4
= 4.2 x 10^-3(0.3)^x(0.3)^y
= 4.2 x 10^-3 (0.3)^x+y
∴ \(\frac{3.78 \times 10^{-4}}{4.2 \times 10^{-3}}\) = (0.3)^x+y
0.09 = (0.3)^x+y
(0.3)² = (0.3)^x+y                        .
∴ x + y = 2
Hence the order of overall reaction is 2.
Answer:
The order of the reaction is 2.

(15) The rate of the reaction, A → Products is 1.25 x 10^-2 M/s when concentration of A is 0. 45 M. Determine the rate constant if the reaction is
(a) first order in A
(b) second order in A.
Solution :
Given : A → Products
Rate = R = 1.25 x 10^-2 M/s
[A] = 0.45 M

(a) Rate constant, k = ? if order is one.
For first order, rate law is, R = k [A]
∴ \(k=\frac{R}{[\mathrm{~A}]}=\frac{1.25 \times 10^{-2}}{0.45}\)
= 2.78 x 10^-2s^-1

(b) Rate constant, k =? if order is two. For second order, rate law is, R = k [A]²

Answer:
(a) Rate constant, k = 2.78 x 10^-2
(b) Rate constant, k = 6.173 x 10^-2

Question 24.
Define and explain the term elementary reaction.
Answer:
Many reactions take place in a series of steps. Such reactions are called complex reactions. Each step taking place in a complex reaction is called an elementary reaction. This shows that a complex reaction is broken down in a series of elementary chemical reactions.

By adding all the elementary steps of a complex reaction we get the overall reaction.

The mechanism of a reaction is decided from the sequence of the elementary steps that are added to give overall reaction.

Elementary reaction : It is defined as the reac­tion which takes place in a single step and cannot be divided further into simpler chemical reactions.

The order and molecularity of the elementary reaction are same.

Some reactions take place in one step and cannot be broken down into simpler reactions. For example,

C₂H₅I_(g) → C₂H_4(g) + HI_(g)
O_3(g) → O_2(g) + O_(g)

Question 25.
Define and explain the term molecularity of a reaction. Give examples.
OR
Define the molecularity of a chemical reaction.
Answer:
Molecularity : The molecularity of an elementary reaction is defined as the number of molecules (or atoms or ions) which take part in a chemical reaction.

Explanation :

• The molecularity of a reaction is always integral.
• It cannot be determined experimentally.
• The minimum value of the molecularity is one.
• It cannot have fractional or zero values.
• The reactions are classified according to the mole­cularity as follows :

(a) Unimolecular reaction (OR First order reac­tion) : In this only one molecule takes part in the reaction, e.g., N₂O₅₍g) → 2NO_2(g) + \(\frac{1}{2}\)O_2(g)

The rate law expression for this reaction is, Rate = k [N₂O₅]. Hence it is unimolecular and first order.

Other unimolecular reactions are,
O_3(g) → O_2(g) + O_(g)
C₂H₅I_(g) → C₂H_2(g) + HI_(g)

(B) Bimolecular reaction In this two molecules take part in the reaction,
e.g., 2HI_(g) → H_2(g) + I_2(g)
O_3(g) + O_(g) → 2O_2(g)
2NO_2(g) → 2NO_(g) + O_2(g)