The hybridisation of metals involved in the following complexes, respectively are…

The hybridisation of metals involved in the following complexes, respectively are $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{CoF}_6\right]^{3-}$
  1. $s p^3 d^2, s p^3 d^2$
  2. $s p^3 d^2, d^2 s p^3$
  3. $d^2 s p^3, d^2 s p^3$
  4. $d^2 s p^3, s p^3 d^2$

Solution

In $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}, \mathrm{Mn}$ is in +3 oxidation state.
As $\mathrm{CN}^{-}$is a strong field ligand, it causes pairing of two $3 d$ electrons, thereby making two $d$-orbitals vacant, namely $d_{x^2-y^2}$ and $d_{z^2}$. These two $d$-orbitals and one $4 s$ and three $4 p$-orbitals hybridise and give $d^2 s p^3$-hybridisation. In $\left[\mathrm{Co}(\mathrm{F})_6\right]^{3-}, \mathrm{Co}$ is in +3 oxidation state.
As $\mathrm{F}^{-}$is a weak field ligand, so it cannot cause the pairing of unpaired $3 d$-electrons. Thus, one $4 s$, three $4 p$-orbitals and two $4 d$-orbitals hybridise and give $s p^3 d^2$-hybridisation.

Asked in: AP EAMCET 2022 (07 Jul Shift 2)

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