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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 $Mn$

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

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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 $Mn$ $(Z=25)$ 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.

List-$I$ $(Compounds)$	List-$II$ $(Hybridization)$
$(a)$ $[Ni(NH_3)_6]^{+2}$	$(i)$ $sp^3 d^2$
$(b)$ $[PtCl_4]^{-2}$	$(ii)$ $dsp^2$
$(c)$ $[Ni(CO)_4]$	$(iii)$ $sp^3$
$(d)$ $[Co(OX)_3]^{-3}$	$(iv)$ $d^2 sp^3$

List-$I$ (Complex)	List-$II$ (Hybridisation of central metal ion)
$A$. $[CoF_6]^{3-}$	$I$. $d^2sp^3$
$B$. $[NiCl_4]^{2-}$	$II$. $sp^3$
$C$. $[Co(NH_3)_6]^{3+}$	$III$. $sp^3d^2$
$D$. $[Ni(CN)_4]^{2-}$	$IV$. $dsp^2$

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

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Match List-$I$ with List-$II$.
List-$I$ (Complex) List-$II$ (Hybridisation of central metal ion)
$A$. $[CoF_6]^{3-}$ $I$. $d^2sp^3$
$B$. $[NiCl_4]^{2-}$ $II$. $sp^3$
$C$. $[Co(NH_3)_6]^{3+}$ $III$. $sp^3d^2$
$D$. $[Ni(CN)_4]^{2-}$ $IV$. $dsp^2$
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