Question

Find the number of molecule(s) among the following which is/are bent in shape but have different hybridization.

$Xe{F}_2,{B{F}_2}^{-},S{F}_2,H_{2}O,SnC{l}_2,I_3^{+},{I_3}^{-}$

Answer 1

Hybridization of the molecule is given by the formula,

$H=\frac{1}{2}[V+M-C+A]$

H = number of orbitals involved in hybridization

V = valence electrons of central atom

M = number of monovalent atoms linked to the central atom

C = charge on the cation

A = charge on the anion

Value of H and hybridization, $[H=2(sp),3(s{p}^2),4(s{p}^3),5(s{p}^{3}d\left)\right]$

Applying the above formula for given molecules,

$Xe{F}_2=\frac{1}{2}[8+2]=5$ $\left(s{p}^{3}d\right)\left[Linear\right]$

$B{F}_2^{-}=\frac{1}{2}[3+2+1]=3$ $\left(s{p}^2\right)\left[Bent\right]$

$S{F}_2=\frac{1}{2}[6+2]=4$ $\left(s{p}^3\right)\left[Bent\right]$

$H_{2}O=\frac{1}{2}[6+2]=4$ $\left(s{p}^3\right)\left[Bent\right]$

$SnC{l}_2=\frac{1}{2}[4+2]=3$$\left(s{p}^2\right)\left[Bent\right]$

$I_3^{+}=\frac{1}{2}[7+1+1-1]=4$ $\left(s{p}^3\right)\left[Bent\right]$

$I_3^{-}=\frac{1}{2}[7+1+1+1]=5$ $\left(s{p}^{3}d\right)\left[Linear\right]$

Therefore, five molecules have bent shape but have different hybridization.

Hence, the correct answer is $5$.