Consider that a d^{6} metal ion (M^{2+}) form | vedclass

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(B) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM.$ For $\mu = 4.90 \ BM,$ $n = 4$ unpaired electrons.For a $d^{6}$ ion,$4$ unpa

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tetrahedral and $-0.6 \Delta_{t}$

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(B) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM.$ For $\mu = 4.90 \ BM,$ $n = 4$ unpaired electrons.For a $d^{6}$ ion,$4$ unpaired electrons are possible in a high-spin tetrahedral complex $(e^{3}t_{2}^{3})$ or a high-spin octahedral complex $(t_{2g}^{4}e_{g}^{2})$.In a tetrahedral complex $[M(H_{2}O)_{4}]^{2+}$,the configuration is $e^{3}t_{2}^{3}$.$CFSE = (3 	imes -0.6 \Delta_{t}) + (3 	imes 0.4 \Delta_{t}) = -1.8 \Delta_{t} + 1.2 \Delta_{t} = -0.6 \Delta_{t}.$In an octahedral complex $[M(H_{2}O)_{6}]^{2+}$,the configuration is $t_{2g}^{4}e_{g}^{2}$.$CFSE = (4 	imes -0.4 \Delta_{0}) + (2 	imes 0.6 \Delta_{0}) = -1.6 \Delta_{0} + 1.2 \Delta_{0} = -0.4 \Delta_{0}.$Comparing with the given options,option $B$ matches the tetrahedral geometry and calculated $CFSE$.

Column $I$	Column $II$
$(A)$ $sp^3$	$(i)$ $[Co(NH_3)_6]^{3+}$
$(B)$ $dsp^2$	$(ii)$ $[Ni(CO)_4]$
$(C)$ $sp^3d^2$	$(iii)$ $[Pt(NH_3)_2Cl_2]$
$(D)$ $d^2sp^3$	$(iv)$ $[CoF_6]^{3-}$
	$(v)$ $[Fe(CO)_5]$

List-$I$	List-$II$
$A$. $[MnBr_4]^{2-}$	$I$. $d^2sp^3$ and diamagnetic
$B$. $[FeF_6]^{3-}$	$II$. $sp^3d^2$ and paramagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$	$III$. $sp^3$ and diamagnetic
$D$. $[Ni(CO)_4]$	$IV$. $sp^3$ and paramagnetic

Consider that a $d^{6}$ metal ion $(M^{2+})$ forms a complex with aqua ligands,and the spin-only magnetic moment of the complex is $4.90 \ BM.$ The geometry and the crystal field stabilization energy $(CFSE)$ of the complex are:

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