Question

What are the hybridizations of ${\text{ICl}}_2^{-}$ and ${\text{ClO}}_4^{-}$ respectively?
$$\begin{gathered} \text{A}\text{. s}{\text{p}}^3{\text{d}}^2\text{, s}{\text{p}}^3 \\ \text{B}\text{. s}{\text{p}}^2\text{, s}{\text{p}}^3\text{d} \\ \text{C}\text{. s}{\text{p}}^3\text{d, s}{\text{p}}^3 \end{gathered}$$
$\text{D}\text{.}$ None of these

Answer 1

Hint- Here, we will proceed by finding the number of valence electrons in the central atom, number of surrounding monovalent atoms, cationic charge and anionic charge for both the molecules given in the problem.

Complete answer:
Formula Used- H = ${\displaystyle \frac{1}{2}}$(V+M−C+A).
As we know that the number of orbitals involved in hybridization for any molecule is given by
H = ${\displaystyle \frac{1}{2}}$(V+M−C+A)
where H denotes the number of orbitals involved in hybridization, V denotes the number of valence electrons in the central atom, M denotes the number of surrounding monovalent atoms, C denotes the cationic charge and A denotes the anionic charge.
In this formula, only magnitudes of the cationic and anionic charges will be considered (the signs will be neglected).
If the value of H comes out to be 2 then the molecule is sp hybridised, 3 then it is $\text{s}{\text{p}}^2$ hybridised, 4 then it is $\text{s}{\text{p}}^3$ hybridized, 5 then it is $\text{s}{\text{p}}^3\text{d}$ hybridized, 6 then it is $\text{s}{\text{p}}^3{\text{d}}^2$ hybridised and so on.
For molecule ${\text{ICl}}_2^{-}$,
Here, the central atom is iodine (I) which has 7 valence electrons and this I atom is linked to 2 atoms of Cl (monovalent atom). The anionic charge on the molecule is clearly -1.
Here, V = 7, M = 2, C = 0 and A = 1
H = ${\displaystyle \frac{1}{2}}$(V+M−C+A) = ${\displaystyle \frac{1}{2}}$(7+2−0+1) = ${\displaystyle \frac{1}{2}}$(10) = 5
Therefore, the hybridization of ${\text{ICl}}_2^{-}$ is $\text{s}{\text{p}}^3\text{d}$.
For molecule ${\text{ClO}}_4^{-}$,
Here, the central atom is chlorine (Cl) which has 7 valence electrons and this Cl atom is linked to 4 atoms of O (divalent atom). The anionic charge on the molecule is clearly -1.
Here, V = 7, M = 0, C = 0 and A = 1
H = ${\displaystyle \frac{1}{2}}$(V+M−C+A) = ${\displaystyle \frac{1}{2}}$(7+0−0+1) = ${\displaystyle \frac{1}{2}}$(8) = 4
Therefore, the hybridization of ${\text{ClO}}_4^{-}$ is $\text{s}{\text{p}}^3$.

Hence, option C is correct.

Note -Orbital hybridization in chemistry is the concept of mixing atomic orbitals into new hybrid orbitals, suitable for pairing electrons to form chemical bonds in the theory of valence bonds. Atoms that only have one electron in the outermost shell, or those that require one electron to reach inert gas configuration, are considered monovalent atoms.