Question

The structure and hybridisation of $Si(C{H}_3{)}_4$ is :

• A. octahedral, $s{p}^{3}d$
• B. tetrahedral, $s{p}^3$
• C. bent, $sp$
• D. trigonal, $s{p}^2$

Answer 1

The correct option is B tetrahedral, $s{p}^3$

The atomic number of silicon is $14$ $(2,8,4)$. It has four valence electrons with which it forms $4$ bond pairs with four $-C{H}_3$ groups.

So, it has hybridization of $s{p}^3$, with a tetrahedral geometry $\left({104.9}^{\circ }bondangle\right)$.

Hence, option (B) is correct.