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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

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

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

Which one of the following is a strong field ligand?

The number of complexes from the following with no electrons in the $t_{2g}$ orbital is: $TiCl_4, [MnO_4]^{-}, [FeO_4]^{2-}, [FeCl_4]^{-}, [CoCl_4]^{2-}$

$[Fe(CN)_{6}]^{3-}$ is an inner orbital complex. Ignoring the pairing energy,the value of crystal field stabilization energy for this complex is $(-)\;\dots \;\Delta_{o}$.

The increasing order of the crystal field splitting power of some common ligands is

Carbon donor ligands are strong ligands and usually form low spin complexes. Among the complexes given below,select the high spin complex.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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