NCERT Solutions for Class 12 Chemistry The p-Block Elements [Old Book]

NCERT
SOLUTIONS
CLASS - 12th

aglase .co

Class : 12th
Subject : Chemistry
Chapter : 7
Chapter Name : The p-Block Elements

Q7.1 Why are pentahalides of P, As, Sb and Bi more covalent than their trihalides?

Answer. In pentahalides, the oxidation state is +5 and in trihalides, the oxidation state is +3. Since
the metal ion with a high charge has more polarizing power, pentahalides are more covalent than
trihalides.

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Q7.2 Why is BiH the strongest reducing agent amongst all the hydrides of Group 15 elements ?
3

Answer. As we move down a group, the atomic size increases and the stability of the hydrides of
group 15 elements decreases. Since the stability of hydrides decreases on moving from NH to 3

BiH , the reducing character of the hydrides increases on moving from NH to BiH .
3 3 3

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Q7.3 Why is N less reactive at room temperature?
2

Answer. The two N atoms in N are bonded to each other by very strong triple covalent bonds. The
2

bond dissociation energy of this bond is very high. As a result, N2 is less reactive at room
temperature.

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Q7.4 Mention the conditions required to maximise the yield of ammonia.

Answer. Ammonia iS prepared using the Haber's process. The yield of ammonia can be maximized
under the following conditions:
(i) High pressure 200 atm)
(ii) A temperature of -700 K
(iii) Use of a catalyst such as iron oxide mixed with small amounts of K O and Al O
2 2 3

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Q7.5 How does ammonia react with a solution Of Cu 2+
?

NH acts as a Lewis base. It donates its electron pair and forms a linkage with metal ion.
3
2+
2+
Cu (aq) ↔ [Cu(NH3 ) ]
4 (aq)
+ 4NH3(aq)
Blue Deep blue

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Q7.6 What is the covalence of nitrogen in N O ? 2 5

From the structure of N O , it is evident that the covalence of nitrogen is 4.
2 5

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Q7.7 (a) Bond angle in PH is higher than that in PH3. Why?
+

4

(b) What is formed when PH reacts with an acid?
3

Answer. In PH , P is sp is hybridized. Three orbitals are involved in bonding with three
3
3

hydrogen atoms and the fourth one contains a lone pair. As lone pair-bond pair repulsion is
stronger than bond pair-bond pair repulsion, the tetrahedral shape associated with Sp bonding is
3

changed to pyramidal. PH combines with a proton to form PH in which the lone pair is absent.
3
+

4

Due to the absence of lone pair in PH , there is no lone pair-bond pair repulsion. Hence, the bond
+

4

angle in PH is higher than the bond angle in PH .
+

4 3

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Q7.8 What happens when white phosphorus is heated with concentrated NaOH solution in an inert
atmosphere of CO ? 2

Answer. White phosphorous dissolves in boiling NaOH solution (in a CO atmosphere) to give
2

phosphine, PH . 3

PH3 + 3NaH2 PO2
P4 + 3NaOH + 3H2 O ⟶
Phosphine , Sodium hypophosphite

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Q7.9 What happens when PCl is heated?
5

Answer. All the bonds that are present in PCl are not similar. It has three equatorial and two
5

axial bonds. The equatorial bonds are stronger than the axial ones. Therefore, when PCl is 5

heated strongly, it decomposes to form PCl . 3

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Q7.10 Write a balanced equation for the hydrolytic reaction of PCl in heavy water.
5

PCl5 + D2 O ⟶ POCl3 + 2DCl2
Therefore the equation can be written as
POCl3 + 3D2 O ⟶ D3 PO4 + 3DCl

PCl5 + 4D2 O ⟶ D3 PO4 + 5DCl

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Q7.11 What is the basicity Of H PO ? 3 4

Since there are three OH groups present in H PO its basicity is three i.e., it is a tribasic acid.
3 4

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Q7.12 What happens when H PO is heated?
3 3

H3 PO3 , on heating, undergoes disproportionation reaction to form PH and H PO . The
3 3 4

oxidation numbers of P in H PO , PH , and H PO are +3, —3, and +5 respectively. As the
3 3 3 3 4

oxidation number of the same element is decreasing and increasing during a particular reaction,
the reaction is a disproportionation reaction.

4H3 PO3 3H3 PO4 + PH3
Δ
Orthophosphorousacid ⟶ Orthophosphoricacid Phosphine

(+3) (+5) (−3)

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Q7.13 List the important sources of sulphur.

Answer. Sulphur mainly exists in combined form in the earth's crust primarily as sulphates
[gypsum (CaSO ⋅ 2H O), Epsom salt (MgSO . 7H O), baryte (BaSO )] and sulphides [(galena
4 2 4 2 4

(PbS), zinc blends (ZnS), copper pyrites ( CuFeS )]. 2

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Q7.14 Write the order of thermal stability of the hydrides of Group 16 elements.

Answer. The thermal stability of hydrides decreases on moving down the group. This is due to a
decrease in the bond dissociation enthalpy (H—E) of hydrides on moving down the group .
Therefore,
H2 O ⏐
⏐
H2 S ⏐
⏐
⏐ Thermal stability
H2 Se ⏐
⏐ decreases
⏐
H2 Te ⏐
⏐
⏐
H 2 P0 ↓

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Q7.15 Why is H O a liquid and H S a gas ?
2 2

H O has oxygen as the central atom. Oxygen has smaller size and higher electronegativity as
2

compared to sulphur. Therefore, there is extensive hydrogen bonding in H O, which is absent in
2

H S. Molecules of H S are held together only by weak van der Waal's forces of attraction.
2 2

Hence, H O exists as a liquid while H S as a solid.
2 2

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Q7.16 Which of the following does not react with oxygen directly?
zn, Ti, Pt,

Answer. Pt is a noble metal and does not react very easily. All other elements, Zn, Ti, Fe, are quite
reactive. Hence, oxygen does not react with platinum (Pt) directly.

Page : 190 , Block Name : Intext Questions

Q7.17 Complete the reaction
(i)
C 2 H4 + O 2 →

(ii)
4Al + 3O2 →

C 2 H4 3O2 2CO2 2H2 O
(i) + ⟶ +
Ethene Oxygen Carbon dioxide Water

4Al + 3O2 ⟶ 2Al2 O3
(ii)
Aluminium Oxygen Alumina

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Q7.18 Why does O act as a powerful oxidising agent?
3

Answer. Ozone is not a very stable compound under normal conditions and decomposes readily on
heating to give a molecule of oxygen and nascent oxygen. Nascent oxygen, being a free radical, is
very reactive.
O3 Δ O2 [O]
⟶ +
Ozone Oxygen Nascent oxygen

Therefore, ozone acts as a powerful oxidising agent.

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Q7.19 How is O estimated quantitatively?
3

Answer. Quantitatively, ozone can be estimated with the help of potassium iodide. When ozone is
made to react with potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is
liberated. This liberated iodine can be titrated against a standard solution of sodium thiosulphate
using starch as an indicator. The reactions involved in the process are given below.
−
2I O3 −
+I2
+ H2 O + ⟶ 2OH + O2
Iodide Ozone Iodine

I2 + 2Na2 S2 O3 Na2 S4 O6

Sodium ⟶ Sodium + 2Nal

thiosulphate tetrathionate

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Q7.20 What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt?

Answer. SO: acts as a reducing agent when passed through an aqueous solution containing Fe(lll)
salt. It reduces Fe(lll) to Fe(ll) i.e., ferric ions to ferrous ions.

3+ 2+ 2− +
2Fe + SO2 + 2H2 O ⟶ 2Fe + SO + 4H
4

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Q7.21 Comment on the nature of two S—O bonds formed in SCh molecule. Are the two S—O bonds
in this molecule equal?

Answer. The electronic con guration of S is 1s 2s 2p 3s 3p . During the formation of SO , one
2 2 6 2 4
2

electron from 3p orbital goes to the 3d orbital and S undergoes spa hybridization. Two Of these
orbitals form sigma bonds with two oxygen atoms and the third contains a lone pair. p-orbital and
d-orbital contain an unpaired electron each. One of these electrons forms pn- pn bond with one
oxygen atom and the other forms pn- dn bond with the other molecule. This is the reason SO has 2

a bent structure. Also, it is a resonance hybrid of structures I and II.

30th S—O bonds are equal in length (143 pm) and have a multiple bond character.

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Q7.22 How is the presence Of SO detected?
2

SO is a colourless and pungent smelling gas.
2

It can be detected with the help of potassium permanganate solution. When SO is passed
2

through an acidi ed potassium permanganate solution, it decolonizes the solution As it reduce
− 2+
MnO ions to Mn
4
− 2− + 2+
5SO2 + 2MnO + 2H2 O ⟶ 5SO + 4H + 2Mn
4 4

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Q7.23 Mention three areas in which H SO plays an important role.
2 4

Answer. Sulphuric acid is an important industrial chemical and is used for a lot Of purposes. Some
important uses of sulphuric acid are given below.
(i) It is used in fertilizer industry. It is used to make various fertilizers such as ammonium sulphate
and calcium super phosphate.
(ii) It is used in the manufacture of pigments, paints, and detergents.
(iii) It is used in the manufacture of storage batteries.

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Q7.24 Write the conditions to maximize the yield of H SO by Contact process.
2 4

Answer. Manufacture of sulphuric acid by Contact process involves three steps.
1. Burning of ores to form SO 2

2. Conversion of SO to \mathrm { SO } _ { 3 } \) by the reaction of the former with O (Vac)' is
2 2

used in this process as a catalyst.)
3. Absorption Of SO in H SO to give oleum (H S O ) The key Step in this process is the
3 2 4 2 2 7

second Step. In this Step, two moles of gaseous reactants combine to give one mole of gaseous
product. Also, this reaction is exothermic. Thus, in accordance with Le Chatelier's principle, to
obtain the maximum amount of SO' gas, temperature should be low and pressure should be high.

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Q7.25 Why is K ai < Ka
i
for H2 SO4 in water ?

+ −
H2 SO4(aq) + H2 O(l) ⟶ H3 O + HSO ; Ka > 10
(aq) 4(aq) i

− + − −2
HSO + H2 O(l) ⟶ H3 O + SO ; Ko2 = 1.2 × 10
4(aq) (aq) 4(aq)

It can be noticed that K >> K This is because a neutral H SO has a much higher tendency
a1 a2 2 4

to lose a proton than the negatively charged HSO .Thus, the former is a much stronger acid than
−

4

the latter.

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Q7.26 Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and
hydration enthalpy, compare the oxidising power of F and Cl . 2 2

Answer. Fluorine iS a much stronger oxidizing agent than chlorine. The oxidizing power depends
on three factors.
1. Bond dissociation energy
2. Electron gain enthalpy
3. Hydration enthalpy
The electron gain enthalpy of Chlorine iS more negative than that Of uorine. However, the bond
dissociation energy Of uorine iS much lesser than that of Chlorine. Also, because of its small
Size, the hydration energy of uorine iS much higher than that of Chlorine.
Therefore, the latter two factors more than compensate for the less negative electron gain
enthalpy of uorine. Thus, uorine iS a much stronger oxidizing agent than Chlorine.

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Q7.27 Give two examples to show the anomalous behaviour of uorine.

Answer. Anomalous behaviour of uorine
(i) It forms only one oxoacid as compared to other halogens that form a number of oxoacids.
(ii) Ionisation enthalpy, electronegativity, and electrode potential of uorine are much higher than

expected.

Page : 202 , Block Name : Intext Questions

Q7.28 Sea is the greatest source of some halogens. Comment.

Answer. Sea water contains chlorides, bromides, and iodides of Na, K, Mg, and Ca. However, it
primarily contains NaCl. The deposits of dried up sea beds contain sodium chloride and carnallite,
KCl. MgCl . 6H O. Marine life also contains iodine in their systems. For example, sea weeds
2
2

contain upto 0.5% iodine as sodium iodide. Thus, sea is the greatest source of halogens.

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Q7.29 Give the reason for bleaching action of Cl .
2

Answer. When chlorine reacts with water, it produces nascent oxygen. This nascent oxygen then
combines with the coloured substances present in the organic matter to oxide them into colourless
substances.
Cl2 + H2 O ⟶ 2HCl + [O]

Coloured substances + [O] → Oxidized colourless substance

Page : 204 , Block Name : Intext Questions

Q7.30 Name two poisonous gases which can be prepared from chlorine gas.

Answer. Two poisonous gases that can be prepared from chlorine gas are
(i) Phosgene (COCl ) 2

(ii) Mustard gas (ClCH CH SCH CH Cl)
2 2 2 2

Page : 204 , Block Name : Intext Questions

Q7.31 Why is ICl more reactive than I ?2

Answer. ICl is more reactive than I because I—Cl bond in ICl is weaker than I—I bond in I .
2 2

Page : 207 , Block Name : Intext Questions

Q7.32 Why is helium used in diving apparatus?

Answer. Air contains a large amount of nitrogen and the solubility of gases in liquids increases
with increase in pressure. When sea divers dive deep into the sea, large amount of nitrogen
dissolves in their blood. When they come back to the surface, solubility of nitrogen decreases and
it separates from the blood and forms small air bubbles. This leads to a dangerous medical
condition called bends. Therefore, air in oxygen cylinders used for diving is diluted with helium

gas. This is done as He is sparingly less soluble in blood.

Page : 211 , Block Name : Intext Questions

Q7.33 Balance the following equation: XeF 6 + H2 O → XeO2 F2 + HF

Balanced equation

XeF 6 + 2H2 O → XeO2 F2 + 4HF

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Q7.34 Why has it been dif cult to study the chemistry of radon?

Answer. It is dif cult to study the chemistry' of radon because it is a radioactive substance having
a half-life of only 3.82 days. Also, compounds of radon such as RnF have not been isolated. They
2

have only been identi ed.

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Q7.1 Discuss the general characteristics of Group 15 elements with reference to their electronic
con guration, oxidation state, atomic size, ionisation enthalpy and electronegativity.

Answer. General trends in group 15 elements
(i) Electronic con guration: All the elements in group 15 have 5 valence electrons. Their general
electronic con guration is ns² np³.
(ii) Oxidation states: All these elements have 5 valence electrons and require three more electrons
to complete their octets. However, gaining electrons is very dif cult as the nucleus will have to
attract three more electrons. This can take place only with nitrogen as it is the smallest in 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 show —1 and -2 oxidation states.
All the elements present in this group show +3 and +5 oxidation states. However, the stability of
+5 oxidation state decreases down a group, whereas the stability of +3
oxidation state increases. This happens because of the inert pair effect.
(iii) Ionization energy and electronegativity First ionization decreases on moving down a group.
This is because of increasing atomic sizes. As we move down a group, electronegativity decreases,
owing to an increase in size.
(iv) Atomic size: On moving down a group, the atomic size increases. This increase in the atomic
size is attributed to an increase in the number of shells.

Page : 213 , Block Name : Exercise

Q7.2 Why does the reactivity of nitrogen differ from phosphorus?

Answer. Nitrogen is chemically less reactive. This is because of the high stability of its molecule,
N₂. In N₂, the two nitrogen atoms form a triple bond. This triple bond has very high bond strength,
which is very dif cult to break. It is because of nitrogen's small size that it is able to form Pit-pn
bonds with itself. This property is not exhibited by atoms such as phosphorus. Thus, phosphorus is
more reactive than nitrogen.

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Q7.3 Discuss the trends in chemical reactivity of group 15 elements.

Answer. General trends in chemical properties of group — 15
(i) Reactivity towards hydrogen:
The elements of group 15 react with hydrogen to form hydrides of type EH , where E z: N, P, As,
3

Sb, or Bi. The stability of hydrides decreases on moving down from NH to BiH .
3 3

(ii) Reactivity towards oxygen:
The elements of group 15 form two types of oxides: E O and E205, where E = N, P, As, Sb, or Bi.
2 3

The oxide with the element in the higher oxidation state is more acidic than the other. However,
the acidic character decreases on moving down a group.
(iii) Reactivity towards halogens:
The group 15 elements react with halogens to form two series of salts:EX and EX .
3 5

However, nitrogen does not form EX as it lacks the d-orbital. All trihalides (except NX ) are
5 3

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

Page : 213 , Block Name : Exercise

Q7.4 Why does NH3 form hydrogen bond but PH3 does not?

Answer. Nitrogen is highly electronegative as compared to phosphorus. This causes a greater
attraction of electrons towards nitrogen in NH than towards phosphorus in PH . Hence, the
3 3

extent of hydrogen bonding in PH is very less as compared to NH .
3 3

Page : 213 , Block Name : Exercise

Q7.5 How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions
involved.

Answer. An aqueous solution of ammonium chloride is treated with sodium nitrite.
NH4 Cl(aq) + NaNO2(aq) ⟶ N2(g) + 2H2 O(l) + NaCl(aq)

NO and HNO₃ are produced in small amounts. These are impurities that can be removed on
passing nitrogen gas through aqueous sulphuric acid, containing potassium dichromate.

Page : 213 , Block Name : Exercise

Q7.6 Ammonia is prepared on a large-scale by the Haber's process.

Answer. Ammonia is prepared on a large-scale by the Haber's process.
−1
N2(g] + 3H2(g) ⇌ 2NH3(g) ΔH = −46kJmol

The optimum conditions for manufacturing ammonia are:
(i) Pressure (around 200 x 105 Pa)
(ii) Temperature (4700 K)
(iii) Catalyst such as iron oxide with small amounts of Al O and K₂O
2 3

Page : 213 , Block Name : Exercise

Q7.7 Illustrate how copper metal can give different products on reaction with HNO₃.

Answer. Concentrated nitric acid is a strong oxidizing agent. It is used for oxidizing most metals.
The products of oxidation depend on the concentration of the acid, temperature, and also on the
material undergoing oxidation.
3Cu + 8HNO3(diute) ⟶ 3Cu(NO3 ) + 2NO + 4H2 O
2

Cu + 4HNO3(cons) ⟶ Cu(NO3 ) + 2NO + 2H2 O
2

Page : 213 , Block Name : Exercise

Q7.8 Give the resonating structures of NO and N O .
2 2 5

Page : 213 , Block Name : Exercise

Q7.9 The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be
explained on the basis of sp hybridisation in NH and only s–p bonding between hydrogen and
3
3

other elements of the group].

Answer. Hydride NH PH AsH SbH
3 3 3 3

H-M-H angle 107 92 91 90
∘ ∘ ∘ ∘

The above trend in the H—M-H bond angle can be explained on the basis of the electronegativity
of the central atom. 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. We know that electronegativity decreases on moving down a group.
Consequently, the repulsive interactions between the electron pairs decrease, thereby decreasing
the H—M—H bond angle.

Page : 213 , Block Name : Exercise

Q7.10 Why does R P = O exist but R N = O does not (R = alkyl group)?
3 3

Answer. N (unlike P) lacks the d-orbital. This restricts nitrogen to expand its coordination number
beyond four. Hence, R N = O does not exist.
3

Page : 213 , Block Name : Exercise

Q7.11 Explain why NH is basic while BiH is only feebly basic.
3 3

Answer. NH is distinctly basic while BiH is feebly basic.
3 3

Nitrogen has a 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.

Page : 213 , Block Name : Exercise

Q7.12 Nitrogen exists as diatomic molecule and phosphorus as P₄. Why?

Answer. Nitrogen owing to its small size has a tendency to form PTI—PTT multiple bonds with
itself. Nitrogen thus forms a very stable diatomic molecule, N₂. On moving down a group, the
tendency to form pπ − pπ bonds decreases (because of the large size of heavier elements).
Therefore, phosphorus (like other heavier metals) exists in the P₄ state.

Page : 213 , Block Name : Exercise

Q7.13 Write main differences between the properties of white phosphorus and red phosphorus.

Page : 213 , Block Name : Exercise

Q7.14 Why does nitrogen show catenation properties less than phosphorus?

Answer. Catenation is much more common in phosphorous compounds than in nitrogen
compounds. This is because of the relative weakness of the N-N single bond as compared to the P
—P single bond. Since nitrogen atom is smaller, there is greater repulsion of electron density of
two nitrogen atoms, thereby weakening the N-N single bond.

Page : 213 , Block Name : Exercise

Q7.15 Give the disproportionation reaction of H PO . 3 3

Answer. On heating, orthophosphorus acid (H PO ) disproportionates to give orthophosphoric
3 4

acid (H PO ) and phosphine (P H ). The oxidation states of P in various species involved in the
3 4 3

reaction are mentioned below.
4H3 PO3 ⟶ 3H3 PO4 + PH3

Page : 213 , Block Name : Exercise

Q7.16 Can PCl₅ act as an oxidising as well as a reducing agent? Justify.

Answer. PCl₅ can only act as an oxidizing agent. The highest oxidation state that P can show is +5.
In PCl₅ phosphorus is in its highest oxidation state (+5). However, it can decrease its oxidation
state and act as an oxidizing agent.

Page : 213 , Block Name : Exercise

Q7.17 Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms
of electronic con guration, oxidation state and hydride formation.

Answer. The elements of group 16 are collectively called chalcogens.
(i) Elements of group 16 have six valence electrons each. The general electronic con guration of
these elements is ns np , where n varies from 2 to 6.
2 4

(ii) Oxidation state:
As these elements have six valence electrons (ns np ), they should display an oxidation state of
2 4

—2. However, only oxygen predominantly shows the oxidation state of —2 owing to its high
electronegativity. It also exhibits the oxidation state of —1 (H O ), zero (O ), and +2 (OF ).
2 2 2 2

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 H E, where E = O, S, Se, Te, PO. Oxygen and sulphur
2

also form hydrides of type H E . These hydrides are quite volatile in nature.
2 2

Page : 213 , Block Name : Exercise

Q7.18 Why is dioxygen a gas but sulphur a solid?

Answer. Oxygen is smaller in size as compared to sulphur. Due to its smaller size, it can effectively
form pn—pn bonds and form O (O == O) molecule. Also, the intermolecular forces in oxygen are
2

weak van der Wall's, which cause it to exist as gas. On the other hand, sulphur does not form M 2

molecule but exists as a puckered structure held together by strong covalent bonds. Hence, it is a
solid.

Page : 213 , Block Name : Exercise

Q7.19 Knowing the electron gain enthalpy values for O → O and O → O - as —141 and 702 kJ
− 2−

mol-i respectively, how can you account for the formation of a large number of oxides having O 2−

species and not O ?
−

(Hint: Consider lattice energy factor in the formation of compounds).

Answer. Stability of an ionic compound depends on its lattice energy. More the lattice energy of a
compound, more stable it will be.
Lattice energy is directly proportional to the charge carried by an ion. When a metal combines
with oxygen, the lattice energy of the oxide involving O ion is much more than the oxide
2−

involving O ion. Hence, the oxide having O ions are more stable than oxides having O .
− 2− −

Hence, we can say that formation of is energetically more favourable than formation of O
−

Page : 213 , Block Name : Exercise

Q7.20 Which aerosols deplete ozone?

Answer. Freons or chloro uorocarbons (CFCs) are aerosols that 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.

Page : 213 , Block Name : Exercise

Q7.21 Describe the manufacture of H SO by contact process?
2 4

Answer. Sulphuric acid is manufactured by the contact process. It involves the following steps:
Step (i):
Sulphur or sulphide ores are burnt in air to form SO . 2

Step (ii):
By a reaction with oxygen, SO is converted into SO in the presence of V O as a catalyst.
2 3 2 5

v2 O5

2SO2(g) + O2(g) ⟶ 2SO3(g)

Step (iii):
SO produced is absorbed on H SO to give $H S O $ (oleum).
3 2 4 2 2 7

SO3 + H2 SO4 ⟶ H2 S2 O7

This oleum is then diluted to obtain H SO of the desired concentration. In practice, the plant is
2 4

operated at 2 bar (pressure) and 720 K (temperature). The sulphuric acid thus obtained is 96-98%
pure.

Page : 213 , Block Name : Exercise

Q7.22 How is SO an air pollutant?
2

Answer. Sulphur dioxide causes harm to the environment in many ways:
1. It combines with water vapour present in the atmosphere to form sulphuric acid. This causes
acid rain. Acid rain damages soil, plants, and buildings, especially those made of marble.
2. Even in very low concentrations, SO causes irritation in the respiratory tract. It causes throat
2

and eye irritation and can also affect the larynx to cause breathlessness.
3. It is extremely harmful to plants. Plants exposed to sulphur dioxide for a long time lose colour
from their leaves. This condition is known as chlorosis. This happens because the formation of
chlorophyll is affected by the presence of sulphur dioxide.

Page : 213 , Block Name : Exercise

Q7.23 Why are halogens strong oxidising agents?

Answer. The general electronic con guration of halogens is np , where n = 2-6. Thus, halogens
5

need only one more electron to complete their octet and to attain the stable noble gas
con guration. Also, halogens are highly electronegative with low dissociation energies and high
negative electron gain enthalpies. Therefore, they have a high tendency to gain an electron.
Hence, they act as strong oxidizing agents.

Page : 213 , Block Name : Exercise

Q7.24 Explain why uorine forms only one oxoacid, HOF.

Answer. Fluorine forms only one oxoacid i.e., HOF because of its high electronegativity and small
size.

Page : 213 , Block Name : Exercise

Q7.25 Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding
while chlorine does not.

Answer. Both chlorine and oxygen have almost the same electronegativity values, but chlorine
rarely forms hydrogen bonding. This is because in comparison to chlorine, oxygen has a smaller
size and as a result, a higher electron density per unit volume.

Page : 213 , Block Name : Exercise

Q7.26 Write two uses of ClO . −

2

Answer. Uses of ClO :
−

2

(i) It is used for purifying water.
(ii) It is used as a bleaching agent.

Page : 213 , Block Name : Exercise

Q7.27 Why are halogens coloured?

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

Page : 213 , Block Name : Exercise

Q7.28 Write the reactions of F and Cl with water.
2 2

HCl HOCl
(i) Cl 2 + H2 O ⟶ +
Hydrochloric acid Hypochlorous acid

(ii) 2F 2(g) + 2H2 O(l) ⟶ 4H
−

(aq)
+ 4F
−

(aq)
+ O2(g) + 4HF(aq)

Page : 213 , Block Name : Exercise

Q7.29 How can you prepare Cl from HCI and HCI from Cl Write reactions only.
2 2

Answer. (i) Cl can be prepared from HCI by Deacon's process.
2

CuCl2

4HCl + O2 ⟶ 2Cl2 + 2H2 O

(ii) HCI can be prepared from Cl on treating it with water.
2

HCl HoCl
Cl2 + H2 O ⟶ +
Hydrochloric acid Hypochlorous acid

Page : 213 , Block Name : Exercise

Q7.30 What inspired N. Bartlett for carrying out reaction between Xe and PtF ? 6

Answer. Neil Bartlett initially carried out a reaction between oxygen and PtF . This resulted in
6

the formation of a red compound,O [PtF ]
+ −

2 6

Later, he realized that the rst ionization energy of oxygen (1175 kJ/mol) and Xe (1170 kJ/mol) is
almost the same. Thus, he tried to prepare a compound with Xe and PtF . He Xe [PtF ] was
+ −
6 6

successful and a red-coloured compound, was formed.

Page : 213 , Block Name : Exercise

Q7.31 What are the oxidation states of phosphorus in the following:
(i) H3 PO3 (ii)PCl3 (iii)CaP2

(iv)Na3 PO4 (v)POF3 ?

Answer. Let the oxidation state of p be x
(i) H3 PO3

3 + x + 3(−2) = 0

3 + x − 6 = 0

x − 3 = 0

x = +3

(ii) PCl
3

x + 3(−1) = 0

x − 3 = 0

x = +3

(iv) Na3 PO4

3(+1) + x + 4(−2) = 0

3 + x − 8 = 0

x − 5 = 0

x = +5

(v) POF 3

x + (−2) + 3(−1) = 0

x − 5 = 0

x = +5

Page : 213 , Block Name : Exercise

Q7.32 Write balanced equations for the following:
(i) NaCl is heated with sulphuric acid in the presence of MnO . 2

(ii) Chlorine gas is passed into a solution of Nal in water.

(i) 4NaCl + MnO2 + 4H2 SO4 ⟶ MnCl2 + 4NaHSO4 + 2H2 O + Cl2

(ii) Cl2 + Nal ⟶ 2NaCl + I2

Page : 214 , Block Name : Exercise

Q7.33 How are xenon uorides XeF XeF 2, 4, and XeF6 obtained?

Answer. XeF XeF
2, 4, and XeF6 are obtained by a direct reaction between Xe and F 2. The
condition under which the reaction is carried out determines the product.
Xe(g)
673k . Ibar
+ F2(g) XeF2(s)
⟶
(Excess)

Xe(g)
873K.7bar
+ 2F2(g) XeF4(s)
⟶
(1 : 5 ratio )

Xe(g) + 3F2(z)
573K⋅60−706ar
XeF6(s)
⟶
(1 : 20 ratio )

Page : 214 , Block Name : Exercise

Q7.34 With what neutral molecule is Clo- isoelectronic? Is that molecule a Lewis base?

Answer. CIO¯ is isoelectronic to CIF. Also, both species contain 26 electrons in all as shown.
Total electrons ClO = 17 + 8 + 1 = 26
−

In CIF 17 +9 26
CIF acts like a Lewis base as it accepts electrons from F to form ClF . 3

Page : 214 , Block Name : Exercise

Q7.35 How are XeO 3 and XeOF4 prepared?

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

6XeF4 + 12H2 O ⟶ 4Xe + 2XeO3 + 24HF + 3O2

XeF6 + 3H2 O ⟶ XeO3 + 6HF

(ii) XeOF4 can be prepared using XeF6 .

XeF6 + H2 O ⟶ XeOF4 + 2HF

Page : 214 , Block Name : Exercise

Q7.36 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.

Answer. (i) Bond dissociation energy usually decreases on moving down a group as the atomic size
increases. However, the bond dissociation energy of F is lower than that of Cl and Br . This is
2 2 2

due to the small atomic size of uorine. Thus, the increasing order for bond dissociation energy
among halogens is as follows:
I2 < F2 < Br2 < Cl2

(ii) HF < HCI < H3r < HI

The bond dissociation energy of H-X molecules where X = F, Cl, ar, I, decreases with an increase in
the atomic size. Since H-I bond is the weakest, HI is the strongest acid.
(iii)
BiH3 ≤ SbH3 < AsH3 < PH3 < NH3

On moving from nitrogen to bismuth, the size of the atom increases while the electron
density on the atom decreases. Thus, the basic strength decreases.

Page : 214 , Block Name : Exercise

Q7.37 Which one of the following does not exist?
(i) XeOF4 (ii) NeF
2

(iii) XeF2 (iv) XeF6

NeF does not exist.
2

Page : 214 , Block Name : Exercise

Q7.38 Give the formula and describe the structure of a noble gas species which is isostructurz
with:
−
(i) ICl
4

−
(ii) IBr
2

−
(iii) BrO
3

Page : 214 , Block Name : Exercise

Q7.39 Why do noble gases have comparatively large atomic sizes?

Answer. Noble gases do not form molecules. In case of noble gases, the atomic radii corresponds to
van der Waals radii. On the other hand, the atomic radii of other elements correspond to their
covalent radii. By de nition, van der Waals radii are larger than covalent radii. It is for this reason
that noble gases are very large in size as compared to other atoms belonging to the same period.

Page : 214 , Block Name : Exercise

Q7.40 List the uses of Neon and argon gases.

Answer. Uses of neon gas:
(i) It is mixed with helium to protect electrical equipments from high voltage.
(ii) It is lled in discharge tubes with characteristic colours.
(iii) It is used in beacon lights.
Uses of Argon gas:
(i) Argon along with nitrogen is used in gas- lled electric lamps. This is because Ar is more inert
than N.
(ii) It is usually used to provide an inert temperature in a high metallurgical process.
(iii) It is also used in laboratories to handle air-sensitive substances.

Page : 214 , Block Name : Exercise