Match the LIST-I with LIST-II:

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(A) In a tetrahedral crystal field, the $d$-orbitals split into two sets: $e$ (lower energy) and $t_2$ (higher energy). The energy difference is $\Del

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

Vedclass · Answer

(A) In a tetrahedral crystal field, the $d$-orbitals split into two sets: $e$ (lower energy) and $t_2$ (higher energy). The energy difference is $\Delta_t$. The stabilization energy $(CFSE)$ is calculated as: $	ext{CFSE} = [n_e(-0.6) + n_{t_2}(+0.4)] \Delta_t$.
For $d^2$: $e^2 t_2^0 ightarrow 2(-0.6) = -1.2 \Delta_t$ $(A-III)$.
For $d^4$: $e^2 t_2^2 ightarrow 2(-0.6) + 2(0.4) = -1.2 + 0.8 = -0.4 \Delta_t$ $(B-IV)$.
For $d^6$: $e^3 t_2^3 ightarrow 3(-0.6) + 3(0.4) = -1.8 + 1.2 = -0.6 \Delta_t$ $(C-I)$.
For $d^8$: $e^4 t_2^4 ightarrow 4(-0.6) + 4(0.4) = -2.4 + 1.6 = -0.8 \Delta_t$ $(D-II)$.
Thus, the correct match is $A-III, B-IV, C-I, D-II$.

List-$I$ (Electronic configuration of tetrahedral metal ion)	List-$II$ (Crystal Field Stabilization Energy $(\Delta_t)$)
$A$. $d^2$	$I$. $-0.6$
$B$. $d^4$	$II$. $-0.8$
$C$. $d^6$	$III$. $-1.2$
$D$. $d^8$	$IV$. $-0.4$

$I$	$II$
$(a). [Cr(H_2O)_6]^{2+}$	$[Cr(H_2O)_6]^{3+}$
$(b). [Fe(H_2O)_6]^{3+}$	$[Fe(CN)_6]^{3-}$
$(c). [Fe(CN)_6]^{3-}$	$[Ru(CN)_6]^{3-}$
$(d). [NiF_6]^{4-}$	$[NiF_6]^{2-}$

Match the $LIST$-$I$ with $LIST$-$II$:

List-$I$ (Electronic configuration of tetrahedral metal ion) List-$II$ (Crystal Field Stabilization Energy $(\Delta_t)$)

$A$. $d^2$ $I$. $-0.6$

$B$. $d^4$ $II$. $-0.8$

$C$. $d^6$ $III$. $-1.2$

$D$. $d^8$ $IV$. $-0.4$

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Match the $LIST$-$I$ with $LIST$-$II$: List-$I$ (Electronic configuration of tetrahedral metal ion) List-$II$ (Crystal Field Stabilization Energy $(\Delta_t)$) $A$. $d^2$ $I$. $-0.6$ $B$. $d^4$ $II$. $-0.8$ $C$. $d^6$ $III$. $-1.2$ $D$. $d^8$ $IV$. $-0.4$