The hybridisation of metals involved in the f | vedclass

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(D) In $[Mn(CN)_6]^{3-}$,$Mn$ is in $+3$ oxidation state ($3d^4$ configuration). Since $CN^-$ is a strong field ligand,it causes the pairing of $3d$ e

Vedclass · Accepted Answer

$d^2sp^3, sp^3d^2$

Vedclass · Answer

(D) In $[Mn(CN)_6]^{3-}$,$Mn$ is in $+3$ oxidation state ($3d^4$ configuration). Since $CN^-$ is a strong field ligand,it causes the pairing of $3d$ electrons,making two $3d$-orbitals vacant. These two $3d$,one $4s$,and three $4p$-orbitals hybridise to give $d^2sp^3$ hybridisation. In $[CoF_6]^{3-}$,$Co$ is in $+3$ oxidation state ($3d^6$ configuration). Since $F^-$ is a weak field ligand,it cannot cause the pairing of unpaired $3d$ electrons. Thus,one $4s$,three $4p$,and two $4d$-orbitals hybridise to give $sp^3d^2$ hybridisation.

List-$I$	List-$II$
$A. \, Ni(CN)_5^{3-}$	$1. \, sp^3$
$B. \, CuCl_5^{3-}$	$2. \, dsp^2$
$C. \, AuCl_4^-$	$3. \, sp^3d_{z^2}$
$D. \, ClO_4^-$	$4. \, d_{x^2-y^2}sp^3$

The hybridisation of metals involved in the following complexes,$[Mn(CN)_6]^{3-}$ and $[CoF_6]^{3-}$,are respectively:

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Identify the coordination complexes in which the central metal ion has $d^4$ configuration.
$(A)$ $[FeO_4]^{2-}$
$(B)$ $[Mn(CN)_6]^{3-}$
$(C)$ $[Fe(CN)_6]^{3-}$
$(D)$ $[Cr(H_2O)_6]^{2+}$
$(E)$ $[NiF_6]^{2-}$
Choose the correct answer from the options given below:

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