Question

What hybridization is expected on the central atom of each of the following molecules ?
(i) $Be{H}_2$
(ii) $C{H}_{2}B{r}_2$
(iii) $P{F}_6^{-}$
(iv) $B{F}_3$

• A. $s{p}^2,sp,s{p}^3,s{p}^2$
• B. $sp,s{p}^3,s{p}^{3}d,s{p}^2$
• C. $sp,s{p}^3,s{p}^3{d}^2,s{p}^2$
• D. $s{p}^2,sp,s{p}^2,s{p}^3$

Answer 1

The correct option is C $sp,s{p}^3,s{p}^3{d}^2,s{p}^2$

$H=\frac{1}{2}[V+M-C+A]$
where,
H= Number of orbitals involved in hybridization.
V=Valence electrons of central atom.
M- Number of monovalent atoms linked to central atom.
C= Charge of cation.
A= Charge of anion.
Consider the hybridization of the options:-
A) $Be{H}_2$-
$H=\frac{1}{2}[2+2]=2$.
$\Rightarrow sp$ hybridized state.
B) $C{H}_{2}B{r}_2$
$H=\frac{1}{2}[4+4]=4$.
$\Rightarrow s{p}^3$ hybridized state.
C) $P{F}^{6-}$
$H=\frac{1}{2}[5+6+1]=6$.
$\Rightarrow s{p}^3{d}^2$ hybridized state.
D) $B{F}_3$
$H=\frac{1}{2}[3+3]=3$.
$\Rightarrow s{p}^2$ hybridized state.
Hence option C is the right answer.