Some p-Block Elements - Solutions
CBSE Class–12 Subject Chemistry
NCERT Solutions
Chapter – 07
The p block elements
In-text question
1. Why are pentahalides more covalent than trihalides?
Ans. 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 in accodance with Fajans Rule..
2. Why is 
the strongest reducing agent amongst all the hydrides of Group 15 elements?
Ans. 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 
to , the reducing character of the hydrides increases on moving from to .
3. Why is 
less reactive at room temperature?
Ans. The two N atoms in are bonded to each other by very strong triple covalent bonds. The bond dissociation energy of this bond is very high. As a result, is less reactive at room temperature.
4. Mention the conditions required to maximize the yield of ammonia.
Ans. Ammonia is prepared using the Haber's process. The yield of ammonia can be maximized under the following conditions:
(i) High pressure
(ii) Low Tempeature
As Habers Process is exothermic in nature so low temperature ( appox. 700K) is required.
(iii) Use of a catalyst such as iron oxide mixed with small amounts of 
and 
5. How does ammonia react with a solution of 
?
Ans. acts as a Lewis base. It donates its electron pair and forms a linkage with metal ion.
6. What is the covalence of nitrogen in
?
Ans.
From the structure of , it is evident that the covalence of nitrogen is 4.
7. Bond angle in 
is higher than that in 
. Why?
Ans. In 
, P is sp3 hybridized. Three orbitals are involved in bonding with three 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 sp3 bonding is changed to pyramidal. combines with a proton to form 
in which the lone pair is absent. Due to the absence of lone pair in, there is no lone pair-bond pair repulsion. Hence, the bond angle in is higher ( 109.8o ) than the bond angle in 
( 92o )
8. What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of 
?
Ans. White phosphorous dissolves in boiling NaOH solution (in a atmosphere) to give phosphine, .
9. Write a balanced equation for the hydrolytic reaction of 
in heavy water.
Ans. 
Therefore, the net reaction can be written as
10. What happens when 
is heated?
Ans. All the bonds that are present in 
are not similar. It has three equatorial and two axial bonds. The equatorial bonds are stronger are shorter than the axial ones. Therefore, when is heated strongly, it decomposes to form 
.
11. What is the basicity of 
?
Ans. 
Since there are three OH groups present in , its basicity is three i.e., it is a tribasic acid.
Concept Insight: Basicity is the Number of hydrogen that are replaceable.
12. What happens when H3PO3 is heated?
Ans.H3PO3 on heating, undergoes disproportionation reaction to form and. The oxidation numbers of P in H3PO3,, and are +3, - 3, and +5 respectively. As the oxidation number of the same element is decreasing and increasing during a particular reaction, the reaction is a disproportionation reaction.
13. List the important sources of sulphur.
Ans. Sulphur mainly exists in combined form in the earth's crust primarily as sulphates [gypsum (
), Epsom salt 
, baryte 
] and sulphides [(galena (PbS), zinc blends (ZnS), copper pyrites 
].
14. Write the order of thermal stability of the hydrides of Group 16 elements.
Ans. The thermal stability of hydrides decreases on moving down the group. This is due to increase in bond length of E-H bond due to which there is decrease in the bond dissociation enthalpy (H-E) of hydrides on moving down the group.
Therefore,
15. Why is 
a liquid and 
a gas?
Ans. 
has oxygen as the central atom. Oxygen has smaller size and higher electronegativity as compared to sulphur. Therefore, there is extensive hydrogen bonding in , which is absent in 
. Molecules of are held together only by weak van der Waal's forces of attraction.
Hence, exists as a liquid while as a gas.
16. Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe
Ans. 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.
17. Complete the following reactions:
(i) 
(ii) 
Ans. (i) 
(ii) 
Therefore, ozone acts as a powerful oxidising agent.
18. Why does 
act as a powerful oxidizing agent?
Ans. 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 readical, is very reactive.
Therefore, ozone acts as a powerful oxidizing agent.
19. How is estimated quantitatively?
Ans. 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.
In-text question
20. What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt?
Ans. 
acts as a reducing agent when passed through an aqueous solution containing Fe(III) salt. It reduces Fe(III) to Fe(II) i.e., ferric ions to ferrous ions.
21. Comment on the nature of two S-O bonds formed in 
molecule. Are the two S-O bonds in this molecule equal?
Ans. The electronic configuration of S is 
.
During the formation of
, one electron from 3p orbital goes to the 3d orbital and S undergoes 
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
bond with one oxygen atom and the other forms 
bond with the other oxygen. This is the reason has a bent structure. Also, it is a resonance hybrid of structures I and II.
Both S-O bonds are equal in length (143 pm) and have a multiple bond character.
22. How is the presence of detected?
Ans. is a colourless and pungent smelling gas.
It can be detected with the help of potassium permanganate solution. When is passed through an acidified potassium permanganate solution, it decolorizes the solution as it reduces 
ions to 
ions.
23. Mention three areas in which 
plays an important role.
Ans. 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.
24. Write the conditions to maximize the yield of by Contact process.
Ans. Manufacture of sulphuric acid by Contact process involves three steps.
1. Burning of ores to form
2. Conversion of to 
by the reaction of the former with 
(
is used in this process as a catalyst.)
3. Absorption of in 
to give oleum 
The key step in this process is the 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 gas, temperature should be low and pressure should be high.
25. Why is 
for in water?
Ans. 
It can be noticed that 
This is because a neutral has a much higher tendency to lose a proton than the negatively charged 
. Thus, the former is a much stronger acid than the latter.
26. Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of 
and 
.
Ans. 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 fluorine. However, the bond dissociation energy of fluorine is much lesser than that of chlorine. Also, because of its small size, the hydration energy of fluorine is much higher than that of chlorine. Therefore, the latter two factors more than compensate for the less negative electron gain enthalpy of fluorine. Thus, fluorine is a much stronger oxidizing agent than chlorine.
27. Give two examples to show the anomalous behaviour of fluorine.
Ans. Anomalous behaviour of fluorine
(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 fluorine are much higher than expected.
28. Sea is the greatest source of some halogens. Comment.
Ans. 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, 
. Marine life also contains iodine in their systems. For example, sea weeds contain upto 0.5% iodine as sodium iodide. Thus, sea is the greatest source of halogens.
29. Give the reason for bleaching action of 
.
Ans. 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.
Coloured substances + [O]
’ Oxidized colourless substance
30. Name two poisonous gases which can be prepared from chlorine gas.
Ans. Two poisonous gases that can be prepared from chlorine gas are
(i) Phosgene 
(ii) Mustard gas 
31. Why is ICl more reactive than 
?
Ans. ICl is more reactive than 
because I-Cl bond in ICl is weaker than I-I bond in .
32. Why is helium used in diving apparatus?
Ans. 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.
33. Balance the following equation: 
Ans. Balanced equation: 
34. Why has it been difficult to study the chemistry of radon?
Ans. It is difficult 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 
have not been isolated. They have only been identified..
