Match List-I with List-II: List-I (Compl | vedclass

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(A) The $CFSE$ is calculated using the formula: $CFSE = (-0.4 n_{t_{2g}} + 0.6 n_{e_g}) \Delta_0$.$A$. $[Cu(NH_3)_6]^{2+}$: $Cu^{2+}$ is $d^9$ $(t_{2g

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$A-I, B-IV, C-II, D-III$

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(A) The $CFSE$ is calculated using the formula: $CFSE = (-0.4 n_{t_{2g}} + 0.6 n_{e_g}) \Delta_0$.$A$. $[Cu(NH_3)_6]^{2+}$: $Cu^{2+}$ is $d^9$ $(t_{2g}^6 e_g^3)$. $CFSE = (-0.4 	imes 6 + 0.6 	imes 3) = -2.4 + 1.8 = -0.6 \Delta_0$. (Matches $I$)$B$. $[Ti(H_2O)_6]^{3+}$: $Ti^{3+}$ is $d^1$ $(t_{2g}^1 e_g^0)$. $CFSE = (-0.4 	imes 1 + 0.6 	imes 0) = -0.4 \Delta_0$. (Matches $IV$)$C$. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $d^5$ (strong field,$t_{2g}^5 e_g^0$). $CFSE = (-0.4 	imes 5 + 0.6 	imes 0) = -2.0 \Delta_0$. (Matches $II$)$D$. $[NiF_6]^{4-}$: $Ni^{2+}$ is $d^8$ $(t_{2g}^6 e_g^2)$. $CFSE = (-0.4 	imes 6 + 0.6 	imes 2) = -2.4 + 1.2 = -1.2 \Delta_0$. (Matches $III$)Therefore,the correct matching is $A-I, B-IV, C-II, D-III$.

List-$I$ (Complex)	List-$II$ ($CFSE$ in $\Delta_0$)
$A$. $[Cu(NH_3)_6]^{2+}$	$I$. $-0.6$
$B$. $[Ti(H_2O)_6]^{3+}$	$II$. $-2.0$
$C$. $[Fe(CN)_6]^{3-}$	$III$. $-1.2$
$D$. $[NiF_6]^{4-}$	$IV$. $-0.4$

Match List-$I$ with List-$II$:

List-$I$ (Complex) List-$II$ ($CFSE$ in $\Delta_0$)

$A$. $[Cu(NH_3)_6]^{2+}$ $I$. $-0.6$

$B$. $[Ti(H_2O)_6]^{3+}$ $II$. $-2.0$

$C$. $[Fe(CN)_6]^{3-}$ $III$. $-1.2$

$D$. $[NiF_6]^{4-}$ $IV$. $-0.4$

Choose the correct answer from the options given below:

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Match List-$I$ with List-$II$: List-$I$ (Complex) List-$II$ ($CFSE$ in $\Delta_0$) $A$. $[Cu(NH_3)_6]^{2+}$ $I$. $-0.6$ $B$. $[Ti(H_2O)_6]^{3+}$ $II$. $-2.0$ $C$. $[Fe(CN)_6]^{3-}$ $III$. $-1.2$ $D$. $[NiF_6]^{4-}$ $IV$. $-0.4$ Choose the correct answer from the options given below: