Question

In $O{F}_2$, the number of bond pairs and lone pairs of electrons is, respectively :

• A. $2,6$
• B. $2,8$
• C. $2,10$
• D. $2,9$

Answer 1

The correct option is B $2,8$

$H=\frac{1}{2}[V+M-C+A]$
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.
Now consider
$O{F}_2$
$H=\frac{1}{2}[6+2]=4$.
$\Rightarrow s{p}^3$ hybridized state with 2 lone pairs on central atom.
Bent shape.
Hence it has 2 bond pairs and $2+3+3=8$ lone pairs.
Hence option B is the right answer.