Consider the following test for a group-IV ca | vedclass

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(D) The group-$IV$ cation forming a black precipitate with $H_2S$ is $Co^{2+}$.$Co^{2+} + H_2S \rightarrow CoS \downarrow$ (Black precipitate,$A$)$CoS

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(D) The group-$IV$ cation forming a black precipitate with $H_2S$ is $Co^{2+}$.$Co^{2+} + H_2S ightarrow CoS \downarrow$ (Black precipitate,$A$)$CoS + 	ext{aqua regia} ightarrow Co^{2+} + NOCl + S + H_2O$ ($B$ is $Co^{2+}$)$Co^{2+} + KNO_2 + CH_3COOH ightarrow K_3[Co(NO_2)_6] + NO + H_2O$ ($C$ is $K_3[Co(NO_2)_6]$)In $K_3[Co(NO_2)_6]$,the oxidation state of $Co$ is $+3$.Electronic configuration of $Co^{3+}$ is $3d^6$.Since $NO_2^-$ is a strong field ligand,it causes pairing of electrons,resulting in $d^2sp^3$ hybridization.Number of unpaired electrons $(n)$ = $0$.Magnetic moment = $\sqrt{n(n+2)} = \sqrt{0(0+2)} = 0 \ BM$.

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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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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