NCERT Solutions Class 12 Chemistry Chapter 7

Q.1 Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionization enthalpy and electronegativity.

Ans.

The general characteristics of Group 15 elements:

(i)Electronic configuration:

All the elements in group 15(N, P, As, Sb, Bi) have 5 valence electrons.

The general electronic configuration: ns² np³

(ii) Oxidation states: All these elements have 5 valence electrons and require three more electrons to complete their octets. The gain of electron is very difficult as the nucleus have to attract three more electrons. Only nitrogen accepts electrons in its valance shell due to its small size and the distance between the nucleus and the valence shell is relatively small. The remaining elements of this group show a formal oxidation state of −3 in their covalent compounds. In addition to the −3 state, N and P also have −2 oxidation states (in NH₂NH₂ and P₂H₄ respectively). N only shows oxidation state of –1 in NH₂OH.

All the elements present in this group show +3 and +5 oxidation states. The stability of +5 oxidation state decreases down a group, whereas the stability of +3 oxidation state increases due to the inert pair effect.

(iii) Ionization energy and electronegativity:

First ionization energy decreases on moving down a group due to the increase in atomic size. As we move down a group, electronegativity decreases, owing to an increase in atomic radii.

(iv)Atomic size: On moving down a group, the atomic size increases. As the number of shells increases down the group, atomic size also increases.

Q.2 Why does the reactivity of nitrogen differ from phosphorus?

Ans.

Nitrogen is less reactive than phosphorus. Nitrogen exists as a diatomic molecule. In N₂, the two nitrogen atoms are attached via triple bond (N≡N). This triple bond has very high bond strength, which is very difficult to break. It is because of small size of nitrogen, it is able to form pπ−pπ bonds with itself. As a result, nitrogen is inert and unreactive in its elemental state.

On the other hand, phosphorus exists as a tetra-atomic molecule (P₄). Since the P—P single bond is much weaker than N≡N triple bond; therefore, phosphorus is much more reactive than nitrogen in its elemental state.

Q.3 Discuss the trends in chemical reactivity of group 15 elements.

Ans.

General trends in chemical reactivity of group-15 are as follows:

(i) Reactivity towards metals: The group 15 elements react with metals to form binary compounds in which metals exhibit −3 oxidation states.

(ii) Reactivity towards hydrogen: The elements of group 15 react with hydrogen to form hydrides of type EH₃, where E = N, P, As, Sb, or Bi. The stability of hydrides decreases on moving down from NH₃ to BiH₃.

(iii) Reactivity towards halogens: The group 15 elements react with halogens (Cl, Br, I) to form two series of salts: EX₃ and EX₅. However, nitrogen does not form NX₅ as it lacks the d-orbital. The trihalides of N are unstable, however NF₃ is stable. Other trihalides are known only as their unstable ammoniates, i.e., NBr₃.NH₃ and NI₃.NH₃. The instability of NCl₃, NBr₃ and NI₃ is because of weak N–X bond due to the large difference in size.

(iv) Reactivity towards oxygen: The elements of group 15 form two types of oxides: E₂O₃ and E₂O₅, where E = N, P, As, Sb, or Bi. The acidic character decreases on moving down a group. Oxides of N are acidic in nature; Oxides of As and Sb are amphoteric, while oxide of Bi is basic in nature.

Q.4 Why does NH₃ form hydrogen bond but PH₃ does not?

Ans.

The electronegativity of N (3.0) is much higher than that of H (2.1). Thus N–H bond is polar and NH₃ undergoes intermolecular H-bonding.

In contrast, both P and H both have an electronegativity of 2.1. Thus P–H bond is not-polar and hence PH₃ does not undergo H-bonding because one main condition of H-bonding is H atom should be attached to an electronegative atom.

Q.5 How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.

Ans.

In the laboratory, dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite.

${\mathrm{NH}}_4\mathrm{Cl}\left(\mathrm{aq}\right)+{\mathrm{NaNO}}_2\left(\mathrm{aq}\right)\stackrel{\mathrm{\Delta }}{\to }{\mathrm{N}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+\mathrm{NaCl}\left(\mathrm{aq}\right)$

Small amounts of NO and HNO₃ are also formed in this reaction; these impurities can be removed by passing the gas through aqueous sulphuric acid containing potassium dichromate.

Q.6 How is ammonia manufactured industrially?

Ans.

On a large scale, ammonia is manufactured by Haber’s process.

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3\left(\mathrm{g}\right);$

The optimum conditions for manufacturing ammonia are:

1. Pressure: 200 × 10⁵ Pa.
2. Temperature: ~700 K.
3. Catalyst is used such as iron oxide with small amounts of A₂O₃ and K₂O.

Q.7 Illustrate how copper metal can give different products on reaction with HNO₃.

Ans.

Concentrated nitric acid is a strong oxidising agent and attacks most metals except noble metals such as gold and platinum.

Cu gives different products on reaction with HNO₃ depending upon the conc. of nitric acid.

\begin{array}{l}{\text{3Cu+8HNO}}_{\text{3}}\left(\text{dilute}\right)\to \text{3Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}{\text{+2NO+4H}}_{\text{2}}\text{O}\\ {\text{Cu+4HNO}}_{\text{3}}\left(\text{conc}\text{.}\right)\to \text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}{\text{+2NO}}_{\text{2}}{\text{+2H}}_{\text{2}}\text{O}\end{array}

Q.8 Give the resonating structures of NO₂ and N₂O₅.

Ans.

The resonating structure of NO₂ is:

The resonating structure of N₂O₅ are as follows:

Q.9 The H-N-H angle value is higher than H-P-H, H-As-H and H-Sb-H angles. Why?

Ans.

Since nitrogen is highly electronegative, there is high electron density around nitrogen. This causes greater repulsion between the electron pairs around nitrogen, resulting in maximum bond angle. As we move down the group size of the central atom goes on increasing and its electronegativity goes on decreasing. Consequently, the repulsive interactions between the electron pairs decrease, thereby decreasing the H−E−H bond angle.

Q.10 Why does R₃P=O exist but R₃N=O does not (R = alkyl group)?

Ans.

Due to the absence of d-orbital N cannot participate in pπ-dπ bonding. For this reason nitrogen cannot expand its coordination number beyond four. In R₃N=O, N has a covalency of 5.

The compound R₃N=O does not exist.

In contrast, P due to presence of d-orbitals form pπ-dπ bonds and hence can expand its covalency beyond 4.

P form R₃P=O in which the covalency of P is 5.

Q.11 Explain why NH₃ is basic while BiH₃ is only feebly basic.

Ans.

Among the group 15 elements, N has small size due to which the lone pair of electrons is concentrated in a small region. This means that the charge density per unit volume is high. On moving down a group, the size of the central atom increases and the charge gets distributed over a large area decreasing the electron density. Hence, the electron donating capacity of group 15 element hydrides decreases on moving down the group. So, NH₃ is highly basic but BiH₃ is feebly basic.

Q.12 Nitrogen exists as diatomic molecule and phosphorus as P₄. Why?

Ans.

Nitrogen owing to its small size has a tendency to form pπ−pπ multiple bonds with itself. Nitrogen thus forms a very stable diatomic molecule (NºN).

On the other hand, due to the larger size of P and low electronegativity, its pπ−pπ bonding is not feasible. Instead it prefers to form P–P single bonds; hence it exists as P₄ molecules.

Q.13 Write main differences between the properties of white phosphorus and red phosphorus.

Ans.

Q.14 Why does nitrogen show catenation properties less than phosphorus?

Ans.

Catenation property depends upon the strength of the element-element bond (same elements). Since the N–N bond strength is much weaker than P–P bond strength, N does not show catenation property. Again, due to the smaller size of nitrogen atom, there is greater repulsion of electron density of two nitrogen atoms, thereby weakening the N−N single bond.

Q.15 Give the disproportionation reaction of H₃PO₃.

Ans.

Orthophosphorus acid (H₃PO₃) on heating undergoes self-oxidation reduction to give orthophosphoric acid (H₃PO₄) and phosphine (PH₃). This kind of reaction is called disproportionation reaction. The oxidation states of P in various species involved in the reaction are mentioned below.

$\stackrel{+3}{4{\mathrm{H}}_3{\mathrm{PO}}_3}\stackrel{\mathrm{\Delta }}{\to }\stackrel{+5}{3{\mathrm{H}}_3{\mathrm{PO}}_4}+\stackrel{-3}{{\mathrm{PH}}_3}$

Q.16 Can PCl₅ act as an oxidising as well as a reducing agent? Justify.

Ans.

The oxidation state of P in PCl₅ is +5. Since P has five electrons in its valence shell thus it can donate its 5 valence electrons and cannot increase its oxidation state beyond +5. Therefore, PCl₅ can only act as an oxidising agent (which can donate electrons). It can decrease its oxidation state from +5 to +3 and act as an oxidising agent. For example,

$\begin{array}{l}\stackrel{0}{2\mathrm{Ag}}+\stackrel{+5}{{\mathrm{PCl}}_5}\to \stackrel{+1}{2\mathrm{AgCl}}+\stackrel{+3}{{\mathrm{PCl}}_3}\\ \stackrel{+5}{{\mathrm{PCl}}_5}+\stackrel{0}{{\mathrm{H}}_2}\to \stackrel{+3}{{\mathrm{PCl}}_3}+\stackrel{+1}{2\mathrm{HCl}}\end{array}$

In the above reactions, oxidation number of P decreases from +5 in PCl₅ to +3 in PCl₃ and that of Ag and H increases from 0 to +1. Thus PCl₅ acts as an oxidising agent.

Q.17 Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.

Ans.

The elements of group 16 are collectively called chalcogens.

(i) Electronic configuration: All these elements have the same ns²np⁴ (n=2 to 6) valence shell electronic configuration and hence are justified to place in group 16 of the periodic table.

$\begin{array}{l}{}_8\mathrm{O}=\left[\mathrm{He}\right]2{\mathrm{s}}^{2}2{\mathrm{p}}^4{}_{16}\mathrm{S}=\left[\mathrm{Ne}\right]3{\mathrm{s}}^{2}3{\mathrm{p}}^4\\ {}_{34}\mathrm{Se}=\left[\mathrm{Ar}\right]3{\mathrm{d}}^{10}4{\mathrm{s}}^{2}4{\mathrm{p}}^4{}_{52}\mathrm{Te}=\left[\mathrm{Kr}\right]4{\mathrm{d}}^{10}5{\mathrm{s}}^{2}5{\mathrm{p}}^4\\ {}_{84}\mathrm{Po}=\left[\mathrm{Xe}\right]4{\mathrm{f}}^{14}5{\mathrm{d}}^{10}6{\mathrm{s}}^{2}6{\mathrm{p}}^4\end{array}$

(ii)Oxidation states: As these elements have six valence electrons (ns²np⁴), they accept 2 electrons to complete their octets and they can display an oxidation state of −2. However, only oxygen predominantly shows the oxidation state of −2 owing to its high electronegativity. It also exhibits the oxidation state of −1 (H₂O₂), zero (O₂), and +2 (OF₂). However, the stability of the −2 oxidation state decreases on moving down a group due to a decrease in the electronegativity of the elements. The heavier elements of the group show an oxidation state of +2, +4, and +6 due to the availability of d-orbitals.

(iii) Formation of hydrides: These elements form hydrides of formula H₂E, where E = O, S, Se, Te, Po (Like H₂O, H₂S, H₂Se, H₂Te and H₂Po).

Q.18 Why is di-oxygen a gas but sulphur a solid?

Ans.

Due to small size and high electronegativity, oxygen forms pπ-pπ multiple bonds. As a result oxygen can exist as diatomic (O₂) molecules. These molecules are held together by weak van der Waals forces of attraction which can easily overcome by the collision of the molecule at room temperature. So O₂ exist as gas at room temperature.

On the other hand, due to the bigger size and lower electronegativity, S does not form multiple bonds. Instead, it prefers S–S single bond. Again, S–S single bond is stronger than O–O single bond; more energy is required to break this bond. S has a much greater tendency to undergo catenation and lower tendency for pπ–pπ multiple bonds. It exists as a puckered structure held together by strong covalent bonds. Hence, it is a solid at room temperature.

Q.19 Knowing the electron gain enthalpy values for

O →O⁻ and O →O²⁻ as −141 and 702 kJmol⁻¹ respectively, how can you account for the formation of a large number of oxides having O²⁻ species and not O⁻?

Ans.

Stability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable will be the ionic compound.

Consider the following reaction of a divalent metal (M) with oxygen. The formation of MO₂ and MO involves the following steps:

$\begin{array}{l}\mathrm{M}\left(\mathrm{g}\right)\stackrel{{\mathrm{\Delta }}_{\mathrm{i}}{\mathrm{H}}_1}{\to }{\mathrm{M}}^{+}\left(\mathrm{g}\right)\stackrel{{\mathrm{\Delta }}_{\mathrm{i}}{\mathrm{H}}_2}{\to }{\mathrm{M}}^{2+}\left(\mathrm{g}\right)\\ \mathrm{O}\left(\mathrm{g}\right)\stackrel{{\mathrm{\Delta }}_{\mathrm{eg}}{\mathrm{H}}_1}{\to }{\mathrm{O}}^{-}\left(\mathrm{g}\right)\\ {\mathrm{M}}^{2+}\left(\mathrm{g}\right)+2{\mathrm{O}}^{-}\left(\mathrm{g}\right)\stackrel{\begin{array}{l}\mathrm{Lattice}\\ \mathrm{energy}\end{array}}{\to }{\mathrm{M}}^{2+}{\left({\mathrm{O}}^{-}\right)}_2\left(\mathrm{S}\right)\\ {\mathrm{M}}^{2+}\left(\mathrm{g}\right)+2{\mathrm{O}}^{2-}\left(\mathrm{g}\right)\stackrel{\begin{array}{l}\mathrm{Lattice}\\ \mathrm{energy}\end{array}}{\to }{\mathrm{M}}^{2+}{\mathrm{O}}^{2-}\left(\mathrm{S}\right)\end{array}$

If electron gain-enthalpy were the only factor involved in the formation of O^2– ions, we would expect that O will prefer to form O^– rather than O^2– ions. But a large number of oxides have O^2– species and not O^– is present. The reasons behind it are

1. O^2– is more stable than O^– because it has stable noble gas configuration.
2. Due to the higher charge on O²⁻ than on O^–ion, the lattice energy released during the formation of oxides containing O²⁻ species in the solid state is much higher than lattice energy produced during the formation of oxides containing O^– species. The lattice energy of formation of MO type of species is quite higher that can compensates the higher electron gain enthalpy (Δ_egH₂) needed during the formation of O^2– to O^– ions. Thus the formation of MO variety is energetically more feasible than the compound of O^–ions. For that reason, oxygen forms a large number of oxides having O^2– species and not O^–.

Q.20 Which aerosols deplete ozone?

Ans.

Aerosols such as chlorofluorocarbons (CFCs) and Freon (CCl₂F₂) accelerate the depletion of ozone. In the presence of ultraviolet radiations, molecules of CFCs break down to form chlorine free radicals that combine with ozone to form oxygen. The ozone layer depletes By the free radical mechanism, initiated by chlorine free radicals. This causes hole in the layer.

Q.21 Describe the manufacture of H₂SO₄ by contact process?

Ans.

Sulphuric acid is manufactured by the Contact Process which involves three steps:

(i) Production of SO₂ by burning sulphur or roasting of sulphur pyrites.

$\begin{array}{l}{\mathrm{S}}_8+8{\mathrm{O}}_2\to 8{\mathrm{SO}}_2\\ \underset{\begin{array}{l}\mathrm{Iron}\\ \mathrm{Pyrites}\end{array}}{4{\mathrm{FeS}}_2}+{110}_2\to 2{\mathrm{Fe}}_2{\mathrm{O}}_3+8{\mathrm{SO}}_2\\ \left(\text{ii}\right)\mathrm{}{\text{Catalytic conversion of SO}}_{\text{2}}{\text{to SO}}_{\text{3}}\text{by the reaction with oxygen in the presence of a catalyst}\left({\text{V}}_{\text{2}}{\text{O}}_{\text{5}}\right),\\ 2{\mathrm{SO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{So}}_3\left(\mathrm{g}\right){\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{H}}^{\mathrm{o}}=-196.6\mathrm{Kj}/\mathrm{mol}\\ {\mathrm{SO}}_3\text{produced is absorbed on 98}\mathrm{\%}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{to give H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{oleum}\right).\\ {\mathrm{SO}}_3+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_7\end{array}$

This oleum is then diluted to obtain H₂SO₄ of the desired concentration. In practice, the plant is operated at 2 bar (pressure) and 720 K (temperature). The sulphuric acid thus obtained is 96-98% pure.

Q.22 How is SO₂ an air pollutant?

Ans.

SO₂ can act as an air pollutant because of the following reasons:

(i) Even in very low concentrations (5 ppm), SO₂ causes irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness.
(ii) It combines with rain water present to form sulphurous acid. This causes acid rain. Acid rain damages soil, plants, and buildings, especially those made of marble (CaCO₃).

${\mathrm{CaCO}}_3+{\mathrm{H}}_2{\mathrm{SO}}_3\to {\mathrm{CaSO}}_3+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$

(iii) Even in low concentration (0.03 ppm) of SO₂ is extremely harmful to plants. Plants exposed to sulphur dioxide for a long time lose colour from their leaves (loss of green colour). This condition is known as chlorosis. This happens because the formation of chlorophyll is affected by the presence of sulphur dioxide.

Q.23 Why are halogens strong oxidising agents?

Ans.

The general electronic configuration of halogens is np⁵, where n =2–6. Thus, halogens need only one more electron to complete their octet and to attain the stable noble gas configuration.

X_2+2e^{-}\to 2X^{-}

Due to low bond dissociation enthalpy, high electronegetivity and large electron gain enthalpy halogens have a strong tendency to accept electrons and get reduced. They only can act as strong oxidising agents.

According to the electrode potentials of halogen atoms it is evident that F₂ is the strongest oxidising agent and I₂ is the weakest oxidising agent.

Q.24 Explain why fluorine forms only one oxoacid, HOF.

Ans.

Fluorine forms only one oxoacid (HOF), because of its high electronegativity and small size and in this compound oxidation state of F is +1.

Again, due to the absence of d-orbitals, F can’t expand its valency beyond +1 (i.e. +3, +5, and +7); hence F does not form higher oxoacids such as HOFO, HOFO₂, HOFO₃.

{\text{F}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\left(\text{ice}\right)\rightleftharpoons \stackrel{\text{+1}}{\text{HOF}}\text{+}\stackrel{\text{-1}}{\text{HF}}

Q.25 Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.

Ans.

Both chlorine and oxygen have almost the same electronegativity values, but the size of chlorine is larger than oxygen atom. Thus electron density per unit volume on oxygen is higher than that of chlorine. So oxygen is able to form hydrogen bond but chlorine cannot.

Q.26 Write two uses of ClO₂.

Ans.

The uses of ClO₂:

(i) It is an excellent bleaching agent.
(ii) It is a powerful oxidizing agent and chlorinating agent.

Q.27 Why are halogens coloured?

Ans.

The halogens are coloured because halogens absorb radiations in the visible region. This results in the excitation of valence electrons to a higher energy region while the remaining light is transmitted. Since the amount of energy required for excitation differs for each halogen, each halogen displays a different colour.

For example, F₂ absorbs violet light (higher excitation energy) and hence appears pale yellow while I₂ absorbs yellow and green light (lower excitation energy), so it appears deep violet. For the same reason Br₂ shows orange red colour and Cl₂ shows greenish yellow colour.

Q.28 Write the reactions of F₂ and Cl₂ with water.

Ans.

Reactions are as follows:

\begin{array}{l}{\text{2F}}_{\text{2}}\left(\text{g}\right){\text{+2H}}_{\text{2}}\text{O}\left(\text{I}\right)\to {\text{4H}}^{\text{+}}\left(\text{aq}\right){\text{+4F}}^{\text{–}}\left(\text{aq}\right){\text{+O}}_{\text{2}}\left(\text{g}\right)\\ {\text{3F}}_{\text{2}}\left(\text{g}\right){\text{+3H}}_{\text{2}}\text{O}\left(\text{I}\right)\to {\text{6H}}^{\text{+}}\left(\text{aq}\right){\text{+6F}}^{\text{–}}\left(\text{aq}\right){\text{+O}}_{\text{3}}\left(\text{g}\right)\\ {\text{Cl}}_{\text{2}}\left(\text{g}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{I}\right)\to \text{HCl}\left(\text{qa}\right)\text{+}\underset{\text{Hypochlorous}\text{acid}}{\text{HOCl}\left(\text{aq}\right)}\end{array}

Q.29 How can you prepare Cl₂ from HCl and HCl from Cl₂? Write reactions only.

Ans.

(a) HCl can be oxidized to Cl₂ by a number of oxidizing agents such as MnO₂, KMnO₄, K₂Cr₂O₇ etc.

MnO₂ + 4 HCl → MnCl₂ + Cl₂ 2H₂O

(b) HCl can be prepared from Cl₂ by dissolving it in water.

{\text{Cl}}_{\text{2}}\left(\text{g}\right){\text{H}}_{\text{2}}\text{O}\left(\text{I}\right)\to \text{HCl}\left(\text{aq}\right)\text{+}\underset{\text{Hypochlorous}\text{acid}}{\text{HOCl}\left(\text{aq}\right)}

Q.30 What inspired N. Bartlett for carrying out reaction between Xe and PtF₆?

Ans.

Neil Bartlett initially carried out a reaction in which oxygen reacts with PtF₆ to yield an ionic solid, O₂⁺PtF₆^–.

In this reaction, O₂ gets oxidized to O²⁺ by PtF₆.

O₂(g) + PtF₆(g) → O₂⁺[PtF₆]^–

Later, he realized that the first ionization energy of oxygen (1175 kJ/mol) and Xe (1170 kJ/mol) is almost the same. He thought that PtF₆ should oxidize Xe to Xe⁺. Thus, he tried to prepare a compound with Xe and PtF₆. When Xe and PtF₆ were mixed, a rapid reaction took place and a red solid with formula, Xe⁺PtF₆^– was formed.

Xe+Pt{F}_6\to X{e}^{+}{\left[Pt{F}_6\right]}^{-}

Q.31 What are the oxidation states of phosphorus in the following:
(i)H₃PO₃, (ii) PCl₃, (iii) Ca₃P_2, (iv)Na₃PO_{4 ,} (v)POF₃?

Ans.

Let the oxidation state of P be x.

(i) H₃PO₃ :
3(+1) + x +3(–2) = 0
Or, x = +3

(ii) PCl₃ :
x + 3(–1) = 0
Or, x = +3

(iii) Ca₃P₂:
3(+2) + 2 x = 0Or, x = –3

(iv) Na₃PO₄ :
3(+1) + x + 4 (–2) = 0
Or, x = +5

(v) POF₃ :
x + 1(–2) + 3(–1) = 0
Or, x = +5

Q.32 Write balanced equations for the following:
(i) NaCl is heated with sulphuric acid in the presence of MnO₂.
(ii) Chlorine gas is passed into a solution of NaI in water.

Ans.

The required reactions are as follows:

$\begin{array}{l}\left(\text{i}\right){\text{4NaCl +MnO}}_2+4{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{MnCl}}_2+4{\mathrm{NaHSO}}_4+{\mathrm{Cl}}_2+2{\mathrm{H}}_2\mathrm{O}\\ \left(\mathrm{ii}\right){\text{Cl}}_2\left(\mathrm{g}\right)+2\mathrm{NaI}\left(\mathrm{aq}\right)\to 2\mathrm{NaCl}\left(\mathrm{aq}\right)+{\mathrm{I}}_2\left(\mathrm{s}\right)\end{array}$

Q.33 How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained?

Ans.

XeF₂, XeF₄, and XeF₆ are obtained by a direct reaction between Xe and F₂. The products obtained are dependent on the reaction conditions.

$\begin{array}{l}\underset{\mathrm{xenon}\mathrm{in}\mathrm{excess}}{\mathrm{Xe}\left(\mathrm{g}\right)}+{\mathrm{F}}_2\left(\mathrm{g}\right)\stackrel{673\mathrm{K},1\mathrm{bar}}{\to }{\mathrm{XeF}}_2\left(\mathrm{s}\right)\\ \underset{(1:5\mathrm{ratio})}{\mathrm{Xe}\left(\mathrm{g}\right)}+{\mathrm{F}}_2\left(\mathrm{g}\right)\stackrel{873\mathrm{K},7\mathrm{bsr}}{\to }{\mathrm{XeF}}_4\left(\mathrm{s}\right)\\ \underset{(1:20\mathrm{ratio})}{\mathrm{Xe}\left(\mathrm{g}\right)}+{\mathrm{F}}_2\left(\mathrm{g}\right)\stackrel{573\mathrm{K},60-70\mathrm{bar}}{\to }{\mathrm{XeF}}_6\left(\mathrm{s}\right)\end{array}$

Q.34 With what neutral molecule is ClO⁻ isoelectronic? Is that molecule a Lewis base?

Ans.

ClO^– has (17 + 8 + 1) = 26 electrons and OF₂ has (8 + 2 x 9) = 26 electron system. ClO⁻ is isoelectronic to OF₂.

If we replace O^–(9 elecrtons) in ClO^– by F (9 electrons) atom, the resulting neutral molecule is ClF.

Since Cl in ClF can further donate electrons to two more F atoms to form ClF₃. Therefore, ClF acts as a Lewis base.

Q.35 How are XeO₃ and XeOF₄ prepared?

Ans.

(i) XeO₃ can be prepared in two ways as shown

6XeF₄ + 12H₂O → 4Xe + 2XeO₃ +24HF + 3O₂

XeF₆ + 3H₂O → XeO₃ + 6HF

(ii) XeOF₄ can be prepared using XeF₆.

XeF₆ + H­₂O → XeOF₄ +2HF

Q.36 Arrange the following in the order of property indicated for each set:
(i) F₂, Cl₂, Br₂, I₂ : increasing bond dissociation enthalpy.
(ii) HF, HCl, HBr, HI : increasing acid strength.
(iii) NH₃, PH₃, AsH₃, SbH₃, BiH₃ − increasing base strength.

Ans.

(i) Bond dissociation energy usually decreases on moving down a group as the atomic size increases. The bond dissociation energy is decreased as it requires less energy to break the bonds as we moving down the group (bond distance increases from F₂ to I₂). The bond dissociation energy of F₂ molecule is very less than Br₂ and Cl₂, it is due to the fact that F-atom is very small in size and hence the three lone pairs of electrons on each F atom repel the bond pairs holding the F-atom in the F₂ molecule.

Thus, the increasing order for bond dissociation energy among halogens is as follows:

I₂< F₂< Br₂< Cl₂

(ii) The relative acid strength of HF, HCl, HBr, and HI depends on the bond dissociation enthalpies. Since the bond dissociation enthalpies of H–X bond decreases from H–F to H–I as the size of the atom increases from F to I. Thus the acid strength increases in the following order.

HF < HCl < HBr < HI

(iii)Due to the presence of lone pair of electrons on the central atom in AH₃, all behave as Lewis bases. On moving from nitrogen to bismuth, the size of the atom increases while the electron density on the atom and as well electronegativity also decreases and hence the basic strength decreases as we move from NH₃ to BiH₃.

BiH₃< SbH₃< AsH₃< PH₃< NH₃

Q.37 Which one of the following does not exist?
(i)XeOF₄ (ii) NeF₂ (iii) XeF₂ (iv) XeF₆

Ans.

The sum of 1^st and 2^nd ionization enthalpies of Ne is much higher than those of Xe. Thus F₂ can oxidize Xe to Xe²⁺ but cannot oxidize Ne to Ne²⁺. So,NeF₂ does not exist but all the xenon fluorides (XeF₄ and XeF₆) and xenon oxyfluoride (XeOF₄) do exist.

Q.38 Give the formula and describe the structure of a noble gas species which is iso-structural with:
(i)ICl₄^– (ii) IBr₂^– (iii) BrO₃^–

Ans.

(i) ICl₄^– is iso-electronic with XeF₄ and a square planar geometry

(ii) XeF₂ is iso-electronic to IBr₂^– has a linear structure.

(iii)XeO₃ is iso-structural to BrO₃^– and has a pyramidal molecular structure.

Q.39 Why do noble gases have comparatively large atomic sizes?

Ans.

Noble gases are inert in nature; they don’t form molecules so the atomic radii of the noble gasses correspond to van der Waal’s radii. By definition, van der Waal’s radii are larger than covalent radii. That is why noble gases have comparatively large atomic sizes.

Q.40 List the uses of Neon and argon gases.

Ans.

Followings are the uses of neon gas:

(i) It is mainly used in discharge tubes and fluorescent lamps which are used in advertising purposes.
(ii) It is used in beacon lights.
(iii) It is filled in discharge tubes with characteristic colours.

Followings are the uses of argon gas:

(i) It is used to provide an inert temperature in a high metallurgical process. (ii) It is also used in laboratories to handle air-sensitive substances.
(iii)Argon along with nitrogen is used in gas-filled electric lamps. This is because Ar is more inert than N.