Question

The member of group 14 form tetrahalides of the type $ M{X_4} $. Which of the
following halides cannot be readily hydrolysed by water.
(a) $ C{X_4} $
(b) $ Si{X_4} $
(c) $ Ge{X_4} $
(d) $ Sn{X_4} $

Answer 1

Hint: We know that group 14 is the carbon family groups. They can form oxides, halides, hydrides and carbides. The group 14 can form tetrahalides of the type $ M{X_4} $ and few dihalides of the type $ M{X_2} $. All these tetrahalides are covalent (except $ {\text{PbB}}{{\text{r}}_4} $ and $ {\text{Pb}}{{\text{I}}_4} $ ) formed by $ s{p^3} $ hybridisation.

Complete Step by step answer:
Here in question it is asked which of the halides does not readily hydrolysed by water.
We can clearly say that carbon halides, $ C{X_4} $ cannot undergo hydrolysis due to non-availability of d-orbitals since carbon cannot extend its coordination number beyond four. But in case of $ Si{X_4} $ and the halides of the heavier metals can undergo hydrolysis due to availability of vacant $ d $ -orbitals, to which molecules can coordinate and hence their halides are hydrolysed by water.

Let us look into reaction of $ SiC{l_4} $, it hydrolyses to give silicic acid i.e., $ {H_4}S{i_4}\;{\text{or}}\;{H_2}Si{O_3}.{H_2}O $
 $ SiC{l_4} + 4{H_2}O \to Si{\left( {OH} \right)_4} + 4HCl $
Also $ Si{F_4} $ hydrolysis to give hydrofluorosilicic acid and silicic acid.

 $ 3Si{F_4} + 4{H_2}O \to Si{\left( {OH} \right)_4} + 2{H_2}Si{F_6} $

Hence the correct option is b.

Additional Information: All group elements except C and Si i.e., Ge, Sn and Pb also form dihalides $ M{X_2} $. The stability of these dihalides increases steadily as we move down the group from Ge to Pb
i.e., $ Pb{X_2} > > Sn{X_2} > > Ge{X_2} $.
It is due to the inert pair effect. In other words, we can say $ Ge{X_4} $ is more stable than $ Ge{X_2} $, Whereas $ Pb{X_2} $ is more stable than $ Pb{X_4} $. Dihalides are less volatile than corresponding tetrahalides. Dihalides of Ge and Sn act as reducing agents.

Note: In group 14 we have lead which is not mentioned in the option. Let us look into that condition. For lead $ PbB{r_4} $ and $ Pb{I_4} $ do not exist because $ P{b^{4 + }} $ ion cannot survive in presence of strong reductants $ B{r^ - } $ and $ {I^ - } $ and immediately get reduced to $ P{b^{2 + }} $ ion.