Q. 24 A5.0( 1 Vote )

Arrange the follo

Answer :

(i) The order of increasing bond dissociation enthalpy is given as:

I2 < F2 < Br2 < Cl2


Bond dissociation enthalpy should decrease as the bond distance increases from F2 to I2 due to the corresponding increase in the size of the atom as we move from F to I.

However, the F—F bond dissociation enthalpy is smaller than that of Cl—Cl bond and even smaller than that of Br—Br.

This is due to reason that the F atom is very small and hence the three lone pair of electrons on each F atom repel the bond air holding the F-atom in F2 molecule.

(ii) H2O, H2S, H2Se, H2Te (increasing acidic character)

The order of increasing acidic character of the given hydrides is:

H2O < H2S < H2Se < H2Te


The increase in acidic strength can be easily explained on the basis of their bond dissociation energy.

As the atomic size increase down the group, the bond length increase and hence the bond strength decreases.

Consequently, the cleavage of E—H bond (E= O, S, Se, Te) becomes easier.

As a result, the tendency to release hydrogen as proton increase down the group.

(b) As gas ‘A’ have a pungent odour and is highly soluble in water and its aqueous solution is weakly basic. As a weak base it precipitates the hydroxides of many metals from their salt solution. Gas ‘A’ finds application in detection of metal ions. It gives a deep blue colouration with copper ions.

Hence, Gas ‘A’ is ammonia (NH3)

(i) Gas ‘A’ (NH3) with copper ions

Cu2+ + 4NH3 [Cu(NH3)4]2+

(ii) Solution of gas ‘A’ (NH3) with ZnSO4 solution.

ZnSO4 + 2NH4OH Zn(OH)2 + (NH4)2SO4

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