[PDF] CBSE NCERT Solutions for Class 12 Chemistry Chapter 7

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Class- XII-CBSE-Chemistry The p- Block Elements

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Back of Chapter Questions

1.Why ar

e pentahalides more covalent than trihalides?

Solution:

The oxidation state of in pentahalides is more as compared to the oxidation state in trihalides. Due to the higher positive oxidation state of the central atom in pentahalide state, these atoms will have larger polarizing power than the halogen atom attached to them since the polarizing power is directly proportional to the charge. The central atom in pentahalide state will tend to polarize more the halide ion to which it is attached. In the case of trihalides, the central atom will polarize the halogen atom to a lesser extent as compared to a pentahalide state. More is the pol a r i z a t i on, larger will be the covalent character of the bond. Hence due to larger polarization of bond in pentahalide state as compared to trihalide state, the pentahalides are more covalent than trihalides.

2.Why is ଷ

the strongest reducing agent amongst all the hydrides of Group 15 elements?

Solution:

As we move down the group, the size of the element increases and, therefore, the bond length of the bond increases and its strength decreases(i.e. stability decreases). If the hydride is less stable, it is more reactive and possesses higher reducing strength. The reducing character of hydrides increases on moving from to ଷ . Hence, is the strongest reducing agent among hydrides of

Group-15

elements. 3.

Why is

less reactive at room temperature?

Solution:

Three electron pairs are shared between two nitrogen atoms to form a dinitrogen molecule. The two nitrogen atoms are joined by a strong triple bondؠ Since the nitrogen atom is very small in size, therefore the bond length is also quite small. Also, both the nitrogen complete their octet. So the bond dissociation energy is very less reactive at room temperature. 4.Me ntion the conditions required to maximise the yield of ammonia.

Solution:

For large scale production of ammonia Haber's process is used. The conditions required to maximize the yield of ammonia are as follows:

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1) In accordance with Le Chatelier's principle, low temperature favours the

reaction. It shifts the equil ibrium to the right because the reaction is exothermic. This gives a greater yield of ammonia. Therefore a temperature of about 700 K will be optimal for the preparation of ammonia.

2) High pressure on the reaction at equilibrium favours the shift of the

equilibrium to the right. It is because the forward reaction proceeds with a decrease in the number of moles. Therefore a pressure of about 200 atm will be optimal for a higher yield of ammonia.

3) A mixture of iron oxide with small amounts of K

O and Al

O should be used as a catalyst. 5.

How does ammonia react with a solution of Cu

Solution:

When ammonia reacts with a solution of Cu

, ammonia acts as a Lewis base due to the presence of lone pair of electrons on the nitrogen atom. Ammonia reacts with a s olution of Cu to from a deep blue coloured complex compound Cu(NH .The reaction can be written as follows: Cu 2+ (aq) +4NH 3 (aq)՜[Cu(NH 3 4 2+ (aq) 6. W h at is the covalence of nitrogen in N O

Solution:

Covalence is the number of electron pairs that an atom can share with other atoms.

The structure of N

O is shown below: From the above diagram, we see that the covalence of nitrogen is 4. 7. Bond angle in PH is higher than that in PH Why?

Solution:

Phosphorus(P) atom in both PH

and PH is sp hybridized. Due to the absence of lone pair-bond pair repulsion and the presence of four identical bond pair-bond pair interactions, PH has tetrahedral geometry with a bond angle of 109

28'. But PH

has three bond pairs and one lone pair around P. Due to greater lone pair-bond pair repulsion than bond pair-bond pair repulsion, the tetrahedral angle decreases from 109

28' to 93.6

. Therefore, PH is pyramidal in shape. Hence bond angle in PH is higher than that in PH

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

Solution:

When white phosphorus is heated with concentrated NaOH in an inert atmosphere of CO , the following reaction takes place: P +3NaOH+ 3H O՜ PH +3NaH PO Phosphine and sodium hypophosphite are formed as products. This reaction is an example of a disproportionation reaction in which the oxidation state of phosphorus decreases from 0 in P to -3 in PH , while it increases from 0 in P to +1 in NaH PO 9.

What happens when PCl

is heated?

Solution:

In PCl

, phosphorus undergoes sp d hybridization and has a trigonal bipyramidal structure. It has three equatorial PെCl bonds and two axial PെCl bonds which are different. The axial bonds are larger than equatorial bonds. PCl is thermally less stable than PCl .On heating, it sublimes but decomposes on stronger heating into phosphorus trichloride and chlorine. PCl

՜PCl

+ Cl 10.

Write a balanced equation for the reaction of PCl

with water.

Solution:

PCl partially hydrolyses to phosphorus oxychloride. PCl + H

O՜POCl

+2HCl and on complete hydrolysis, the POCl gets converted to phosphoric acid. POC l + 3H

O՜H

PO +3HCl 11.

What is the basicity of H

PO

Solution:

The basicity of a compound is defined as the number of acidic hydrogen atoms present in the compound. Acidic hydrogen is bonded to a strongly electronegative element like fluorine or oxygen atom.

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From the above figure, we see that one hydrogen atom is attached to each oxygen atom, which makes them acidic. Since the total number of acidic hydrogen is 3, the basicity of H PO is 3. 12.

What happens when H

PO is heated?

Solution:

Heating of H

PO is a disproportionate reaction. The products formed are PH and H PO . Phosphorus has oxidation states of +3,+5 and -3 in orthophosphorous acid H PO ), orthophosphoric acid (H PO ) and phosphine (PH ) respectively. Since 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.

Solution:

Sulphur mainly occurs in the earth's crust in the combined state primarily in the form of sulphates and sulphides.

Sulphates

are gypsum(CaSO .2H

O), Epsom salt MgSO

.7H

O, Baryte BaSO

Sulphides

are galena(PbS), zinc blende(ZnS), copper pyrites(CuFeS Traces of sulphur also occur as hydrogen sulphide in volcanoes. Organic materials such as eggs, proteins, garlic, onion, mustard, hair and wool also contain sulphur. 14. W r ite the order of thermal stability of the hydrides of Group 16 elements.

Solution:

On moving down t he gr oup, there is a decrease in the bond dissociation enthalpy (HെE) of hy drides with an increase in atomic size. (E = O,S,Se,Te,Po). Hence, the thermal stability of the hydrides decreases down the group.

The thermal stability is in the order H

O >H S > H Se> H Te> H Po. 15.

Why is H

O a liquid and H

S a gas?

Class- XII-CBSE-Chemistry The p- Block Elements

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Solution:

Weak Van der Waal's force is present in H

S molecules whereas intermolecular

hydrogen bonding is present in water (H

O). This is possible because the oxygen

atom has a smaller size and greater electronegativity than sulphur atom. Hence water is a liquid and H

S is a gas.

16. Which of the following does not react with oxygen directly?

Zn, Ti, Pt, Fe

Solution:

Pt is a noble metal (inert) and does not react with oxygen directly, whereas Zn, Ti, and Fe are active metals which react quickly with oxygen to form their respective oxides. 17.

Complete the following reactions:

(i) C H + O (ii) 4Al+ 3O

Solution:

(i) C 2 H

4୉୲୦ୟ୬ୣ

O 2

՜ 2CO

2 2H 2 O

Ethane on

reaction with oxygen gives carbon dioxide and water. (ii) 4Al + 3O 2

՜ 2Al

2 O

3alumina

Aluminium

combines with oxygen to form alumina. 18.

Why does O

act as a powerful oxidising agent?quotesdbs_dbs7.pdfusesText_13

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