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

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(A) The Crystal Field Stabilization Energy $(CFSE)$ is calculated using the formula: $CFSE = [-0.4 	imes (n_{t2g}) + 0.6 	imes (n_{eg})] \Delta_0$.$

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

Vedclass · Answer

(A) The Crystal Field Stabilization Energy $(CFSE)$ is calculated using the formula: $CFSE = [-0.4 	imes (n_{t2g}) + 0.6 	imes (n_{eg})] \Delta_0$.$(A)$ $[Ti(H_2O)_6]^{2+}$: $Ti^{2+}$ is $3d^2$. $t_{2g}^2 e_g^0$. $CFSE = [-0.4 	imes 2 + 0.6 	imes 0] = -0.8 \Delta_0$ $(II)$.$(B)$ $[V(H_2O)_6]^{2+}$: $V^{2+}$ is $3d^3$. $t_{2g}^3 e_g^0$. $CFSE = [-0.4 	imes 3 + 0.6 	imes 0] = -1.2 \Delta_0$ $(I)$.$(C)$ $[Mn(H_2O)_6]^{3+}$: $Mn^{3+}$ is $3d^4$. $t_{2g}^3 e_g^1$. $CFSE = [-0.4 	imes 3 + 0.6 	imes 1] = -0.6 \Delta_0$ $(III)$.$(D)$ $[Fe(H_2O)_6]^{3+}$: $Fe^{3+}$ is $3d^5$. $t_{2g}^3 e_g^2$. $CFSE = [-0.4 	imes 3 + 0.6 	imes 2] = 0 \Delta_0$ $(IV)$.Thus,the correct match is $A-II, B-I, C-III, D-IV$.

List-$I$ (Complex)	List-$II$ ($CFSE$ in $\Delta_0$)
$A$. $[Ti(H_2O)_6]^{2+}$	$I$. $-1.2$
$B$. $[V(H_2O)_6]^{2+}$	$II$. $-0.8$
$C$. $[Mn(H_2O)_6]^{3+}$	$III$. $-0.6$
$D$. $[Fe(H_2O)_6]^{3+}$	$IV$. $0$

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

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

$A$. $[Ti(H_2O)_6]^{2+}$ $I$. $-1.2$

$B$. $[V(H_2O)_6]^{2+}$ $II$. $-0.8$

$C$. $[Mn(H_2O)_6]^{3+}$ $III$. $-0.6$

$D$. $[Fe(H_2O)_6]^{3+}$ $IV$. $0$

Similar Questions

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Which of the following are true about the energy of the given d-orbitals of a tetrahedral complex?
$A$. $d_{xy} = d_{xz} > d_{x^2-y^2}$
$B$. $d_{xy} = d_{yz} > d_{z^2}$
$C$. $d_{x^2-y^2} > d_{z^2} > d_{xz}$
$D$. $d_{x^2-y^2} = d_{z^2}$
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Match List-$I$ with List-$II$: List-$I$ (Complex) List-$II$ ($CFSE$ in $\Delta_0$) $A$. $[Ti(H_2O)_6]^{2+}$ $I$. $-1.2$ $B$. $[V(H_2O)_6]^{2+}$ $II$. $-0.8$ $C$. $[Mn(H_2O)_6]^{3+}$ $III$. $-0.6$ $D$. $[Fe(H_2O)_6]^{3+}$ $IV$. $0$