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

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.