Question

Which of the following compounds have the same geometry among the following :
(i) $C_2{H}_4$
(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

Answer: (A), (D)

The correct options are
A $\left(i\right),\left(ii\right)$
D $\left(iii\right),\left(iv\right)$

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

(i) In $C_2{H}_4$, both the $C$ atoms have zero lone pairs of electrons, two bonds with two hydrogen atoms and 1 bond with another carbon. Thus it is $s{p}^2$ hybridised.

(ii) In $AlC{l}_3$, $Al$ is bonded to three $Cl$ atoms 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, $C{H}_4$ and $N{F}_3$ have the same geometry, which is tetrahedral.