Question

Which of the following have ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridization?

• A. ${\mathrm{IF}}_7$
• B. ${{\mathrm{ICl}}_2}^{-}$
• C. ${{\mathrm{ICl}}_4}^{-}$
• D. ${{\mathrm{BrF}}_2}^{-}$

Answer 1

The correct option is C

${{\mathrm{ICl}}_4}^{-}$

Explanation for the correct option:

(C) ${{\mathrm{ICl}}_4}^{-}$

Step 1: Hybridization:

• “Hybridization is defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of a new set of orbitals of equivalent energies and shape.”

Step 2: Formula used for calculation:

• $\mathrm{Steric}\mathrm{number}=\frac{1}{2}\left[\mathrm{V}+\mathrm{M}-\mathrm{C}+\mathrm{A}\right]$
• Where V is the number of valence electrons, M is the monovalent electron, C is the positive charge and A represents the negative charge.

Step 3: Calculation for Hybridisation of ${{\mathrm{ICl}}_4}^{-}$:

• The total number of valence electrons of Idoine in ${{\mathrm{ICl}}_4}^{-}$is 7 and there are 4 monovalent atoms and a negative charge.
• $$\begin{gathered} \mathrm{Steric}\mathrm{number}=\frac{1}{2}\left[7+4+1\right] \\ \mathrm{Steric}\mathrm{number}=\frac{1}{2}\times 12=6 \end{gathered}$$
• The steric number is 6, thus the hybridization of ${{\mathrm{ICl}}_4}^{-}$ is ${\mathrm{sp}}^3{\mathrm{d}}^2$.

Explanation for incorrect options:

(A) ${\mathrm{IF}}_7$

Calculation for Hybridisation of ${\mathrm{IF}}_7$:

• The total number of valence electrons of iodine in ${\mathrm{IF}}_7$ is 7 and there are 7 monovalent atoms.
• $$\begin{gathered} \mathrm{Steric}\mathrm{number}=\frac{1}{2}\left[7+7\right] \\ \mathrm{Steric}\mathrm{number}=\frac{1}{2}\times 14=7 \end{gathered}$$
• The steric number is 7, thus the hybridization of ${\mathrm{IF}}_7$ is ${\mathrm{sp}}^3{\mathrm{d}}^3$.

(B) ${{\mathrm{ICl}}_2}^{-}$

Calculation for Hybridisation of ${{\mathrm{ICl}}_2}^{-}$:

• The total number of valence electrons of Iodine in ${{\mathrm{ICl}}_2}^{-}$ is 7 and there are 2 monovalent atoms with a negative charge.
• $$\begin{gathered} \mathrm{Steric}\mathrm{number}=\frac{1}{2}\left[7+2+1\right] \\ \mathrm{Steric}\mathrm{number}=\frac{1}{2}\times 10=5 \end{gathered}$$
• The steric number is 5, thus the hybridization of ${{\mathrm{ICl}}_2}^{-}$ is ${\mathrm{sp}}^3\mathrm{d}$.

(D) ${{\mathrm{BrF}}_2}^{-}$

Calculation for Hybridisation of ${{\mathrm{BrF}}_2}^{-}$:

• The total number of valence electrons of Iodine in ${{\mathrm{BrF}}_2}^{-}$ is 7 and there are 2 monovalent atoms with a negative charge.
• $$\begin{gathered} \mathrm{Steric}\mathrm{number}=\frac{1}{2}\left[7+2+1\right] \\ \mathrm{Steric}\mathrm{number}=\frac{1}{2}\times 10=5 \end{gathered}$$
• The steric number is 5, thus the hybridization of ${{\mathrm{BrF}}_2}^{-}$ is ${\mathrm{sp}}^3\mathrm{d}$.

Hence, the correct option is (C) i.e. ${{\mathrm{ICl}}_4}^{-}$has ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridization.