Question

Which of the following compounds have same geometry ammong the following :
(i) $H_{2}O$ (ii) $AlC{l}_3$ (iii) $C{H}_4$ (iv) $N{F}_3$

• A. $\left(i\right),\left(ii\right)$
• B. $\left(i\right),\left(ii\right),\left(iv\right)$
• C. $\left(i\right),\left(iii\right),\left(iv\right)$
• D. $\left(iii\right),\left(iv\right)$

Answer 1

The correct option is C $\left(i\right),\left(iii\right),\left(iv\right)$

The compounds having same hybridisation of the central atom will have same geometry though their shape may be different.

(i) In $H_{2}O$, $O$ atoms has two lone pairs of electrons and two bonds with two hydrogen atoms. Thus it is $s{p}^3$ hybridised.
(ii) In $AlC{l}_3$ , $Al$ is bonded to three $H$ atom without any lone pair. Thus it is $s{p}^2$ hybridised.
(iii) In $C{H}_4$, $C$ is bonded to four $H$ atoms. Thus it $s{p}^3$ hybridised.
(iv) In $N{F}_3$, $N$ atom is bonded to three $F$ atoms and has one lone pair. Thus it is also $s{p}^3$ hybridised.

Thus $H_{2}O$, $C{H}_4$ and $N{F}_3$ have same geometry which is tetrahedral.