Question

Which of the following is not tetrahedral in shape?

• A. $N{H}_4^{+}$
• B. $SiC{l}_4$
• C. $S{F}_4$
• D. $S{O}_4^{2-}$

Answer 1

The correct option is C $S{F}_4$

Analysing the options:

* One can use formula to find the hybridisation i.e., number of sigma bond + number of lone pairs on the central atom.

Option (A)
$N{H}_4^{+}$


Hybridisation of $N=s{p}^3\left(4\text{sigma bond}\right)$
Since the hybridisation of the central atom is $s{p}^3$, the structure is tetrahedral.

Option (B)
$SiC{l}_4$

Hybridisation of $Si=s{p}^3\left(4\text{sigma bond}\right)$
It has a tetrahedral shape due to $s{p}^3$ hybridisation.

Option (C).
$S{F}_4$


Hybridisation of S = Number of sigma bond + number of lone pairs $=4+1=5=s{p}^{3}d$

Option (D)
$S{O}_4^{2-}$


Hybridisation of S = Number of sigma bond + number of lone pairs $=4=s{p}^3$

$\therefore S{O}_4^{2-}$ has a tetrahedral shape.

So, option (C) is the correct answer.