Q. 24 A5.0( 1 Vote )
Arrange the follo
(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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