Question

In which of the following molecules the central atom does not have $s{p}^3$ hybridization ?

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

Answer 1

Answer: (A)

When the number of hybrid orbitals, H is 4, the hybridization is $s{p}_3$. $H=\frac{1}{2}[V+M-C+A]$
where, V = number of valence electrons of central atom M = number of monovalent atoms C = total positive charge A = negative charge [a]$C{H}_4,$
$H=\frac{1}{2}[4+4-0+0]=4,$
thus $s{p}^3$
hybridization [b] For$S{F}_4,$
$H=\frac{1}{2}[6+4-0+0]=5,$
thus $s{p}^{3}d$ hybridization
[c] For $B{F}_4^{-},$
$H=\frac{1}{2}[3+4-0+1]=4,$
thus$s{p}^3$
hybridization [d] For $N{H}_4^{+},$
$H=\frac{1}{2}[5+4-1+0]=4,$
thus$s{p}^3$
hybridization Thus, only in $S{F}_4,$
the central atom does not have $s{p}^3$ hybridization.