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1Group 16 elements have lower ionisation enthalpy values compared to those of Group 15 in the corresponding

periods.

Ans.Group 15 elements have highly stable half filled sub shell configuration (ns

2np

3),whereas group 16 elements get half

filled sub shell configuration after losing one electron (ns2np

4).

2

Knowing the electron gain enthalpy values for O O

and O O2

as 141 and 702 kJ mol1

respectively, explain

these values and how can you account for the formation of a large number of oxides having O2

species and not O

?

Ans.

IInd

electron gain enthalpy of oxygen is positive because after gaining one electron , electron density increases which

causes electron-electron repulsion for next incoming electron.

Reason for the formation of a large number of oxides having O2

species and not O

is that after gaining 2 electrons

ox en ets noble as confi uration which is hi hl stable.

3 Oxygen has less negative electron gain enthalpy than sulphur.

Ans.

Due to small atomic size of oxygen there is high e-density so there is high electron-electron repulsion for incoming

electron.But due to bigger atomic size of sulpher there is low e- density so there is less electron-electron repulsionfor incoming electron. therefore oxygen has less negative electron gain enthalpy than sulphur.

4 Metallic character increases from oxygen to polonium.

Ans.Due to increase in atomic size down the group, ionisation enthalpy decreases so metallic character increases from

oxygen to polonium.

5 The stability of -2 oxidation state decreases down the group 16 .

Ans. It is due to decrease in electronegativity down the group.

6 The stability of + 6 oxidation state decreases down the group 16 and stability of + 4 oxidation state increase.

Ans. It is due to increased zeffdown the group (inert pair effect)

7 Oxygen limits its covalency to four but Sulphur can have this up to 6.

Ans. Oxygen has no vacant d-orbitals in its valence shell but sulphur has vacant d-orbitals.

8 Oxygen is a gas but Sulphur is a solid.

Ans.

Due to small atomic size of oxygen it is able to do p-pbonding and can form multiple bond , so it exists as

O=O which is very small molecule and so there are very weak dispersion forces work between O2

molecules therefore it is a gas.But due to bigger atomic size of sulpher, it is not able to do p-pbonding and can

not form multiple bond , so it exists as S8which is very large molecule and so there are very strong dispersion forces

work between S8molecules therefore it is a solid.

9 Inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.

Ans. Because oxygen has small atomic size but chlorine has bigger atomic size.

10 Sulphur forms SF6but Oxygen forms only OF2.

Ans.Oxygen has no vacant d-orbitals in its valence shell but sulphur has vacant d-orbitals. So sulphur can extend its

covalency up to 6.

11 Acidic character increases from H2O to H2Te.

Ans.It is due to decrease in bond enthalpy of M-H bond down the group (that is due to increase in atomic size of central

atom down the group). It causes easy release of H+ion

12 Reducing character increases from H2S to H2Te.

Ans.It is due to decrease in bond enthalpy of M-H bond down the group (that is due to increase in atomic size of central

atom down the group) It causes easy release of H atoms which cause reduction of other species.

13 H2S has lower boiling point than H2O.

Ans.In H2O there is H-bonding whereas in H2S there are dipole-dipole intractions. H-bonds are stronger intermolecular

forces than di ole-di ole intractions.14 H2O is liquid but H2S is gas.

Ans.In H2O there is H-bonding whereas in H2S there are dipole-dipole intractions. H-bonds are stronger intermolecular

forces than dipole-dipole intractions. So H2S has lower boiling point than H2O.

Reasoning based questions on Group 16 elements (The p-block elements)

By V.S.Verma PGT (Chemistry) K.V.No.1 Jaipur

8/9/2019 Group 16 Reasoning1 (1)

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15 H2S to H2Te boiling point increases.

Ans.It is due to increase in molecular size from H2S to H2Te, so dispersion forces among the molecules increase and it

causes increase in boiling point.

16 Reducing property of dioxide decreases from SO2to TeO2.

Ans.It is due to increased zeffdown the group (inert pair effect) due to which oxidation to +6 state become difficult as we

go down in the group.17 The stability of the halides decreases in the order F

> Cl

> Br

> I

.

Ans.As the size of halide ion increases, bond length of M-X bond increases and bond enthalpy decreases. So the stability

of the halides decreases in the order F

> Cl

> Br

> I

.

18 SF6is stable against hydrolysis but SF4and SeF6not.

Ans.

Because in SF6, sulphur is sterically protected from all sides and water molecule can not attack on it. But in SF4and

SeF6central atom is not fully sterically protected from all sides and water molecule can attack easily on them.

19 SF4is a gas, SeF4a liquid and TeF4a solid.

Ans.It is due to increase in molecular size from SF4to TeF4, so dispersion forces among the molecules increase and it

causes increase in boilin oint.

20 It is necessary to use a silent electrical discharge in preparation of O3.

Ans.While preparing O3silent electrical discharge( sparkless electral discharge) is carried out, as otherwise heat is

roduced which will convert O in to O .

21 Ozone is thermodynamically unstable with respect to oxygen.

Ans.

Ozone is thermodynamically unstable with respect to oxygen since its decomposition into oxygen results in the

liberation of heat (H is negative) and an increase in entropy (S is positive). These two effects reinforce each other,

resulting in large negative Gibbs energy change (G) for its conversion into oxygen.(G=H-TS)

22 O3acts as a powerful oxidising agent.

Ans.Ozone is thermodynamically unstable with respect to oxygen so it decomposes into O2and liberates nascent oxygen

atoms. O -> O + O

23 The two oxygen-oxygen bond lengths in the ozone molecule are identical (128 pm).Ans. It is due to resonance.

24 S2is paramagnetic like O2.

Ans. Because it has 2 unpaired electrons in its * molecular orbitals.

25 The two sulphur-oxygen bond lengths in the SO2molecule are identical.

Ans. It is due to resonance.

26 Ka2 Na2S2O3

By V.S.Verma PGT (Chemistry) K.V.No.1 Jaipur

8/9/2019 Group 16 Reasoning1 (1)

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8/9/2019 Group 16 Reasoning1 (1)

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