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(B) $1$. The oxidation state of $Mn$ in $[Mn(CN)_6]^{3-}$ is calculated as: $x + 6(-1) = -3 \implies x = +3$.$2$. The electronic configuration of $Mn$

Vedclass · Accepted Answer

It is $d^2sp^3$ hybridised and octahedral.

Vedclass · Answer

(B) $1$. The oxidation state of $Mn$ in $[Mn(CN)_6]^{3-}$ is calculated as: $x + 6(-1) = -3 \implies x = +3$.$2$. The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$. Thus,$Mn^{3+}$ is $[Ar] 3d^4$.$3$. $CN^-$ is a Strong Field Ligand $(SFL)$,which causes pairing of electrons in the $3d$ orbitals.$4$. After pairing,two $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals are available for hybridization,resulting in $d^2sp^3$ hybridization.$5$. Since the coordination number is $6$,the geometry is octahedral.

Pick out the correct statement with respect to $[Mn(CN)_6]^{3-}.$

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