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

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


In $S{F}_4$,
$\text{Total number of electrons around S}=6+4=10$
$\text{Number of bond pairs}=4$
$\text{Number of lone pairs}=1$
$\text{steric number}=5$
So hybridization of sulphur here is $s{p}^{3}d$
In $B{F}_4^{-}$
$\text{Total number of electrons around S}=3+4+1=8$
$\text{Number of bond pairs}=4$
$\text{Number of lone pairs}=0$
$\text{steric number}=4$
So hybridization of boron here is $s{p}^3$
In $N{H}_4^{+}$
$\text{Total number of electrons around N}=5+4-1=8$
$\text{Number of bond pairs}=4$
$\text{Number of lone pairs}=0$
$\text{Steric number}=4$
So hybridization of nitrogen here is $s{p}^3$
In $C{H}_4$
$\text{Total number of electrons around N}=4+4=8$
$\text{Number of bond pairs}=4$
$\text{Number of lone pairs}=0$
$\text{Steric number}=4$
So hybridization of carbon here is $s{p}^3$