Among the following complexes,the one which s | vedclass

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

Question

(B) The $CFSE$ is calculated using the formula: $CFSE = [(-0.4 	imes n_{t_{2g}}) + (0.6 	imes n_{e_g})] \Delta_0$.For a complex to have zero $CFSE$,

Vedclass · Accepted Answer

$[Fe(H_2O)_6]^{3+}$

Vedclass · Answer

(B) The $CFSE$ is calculated using the formula: $CFSE = [(-0.4 	imes n_{t_{2g}}) + (0.6 	imes n_{e_g})] \Delta_0$.For a complex to have zero $CFSE$,the configuration must be such that the stabilization energy of electrons in $t_{2g}$ orbitals is exactly cancelled by the destabilization energy of electrons in $e_g$ orbitals.Let's analyze the electronic configurations of the metal ions in the given complexes (all are high-spin complexes with $H_2O$ as a weak field ligand):$A) [Mn(H_2O)_6]^{3+}: Mn^{3+} = 3d^4$. Configuration: $t_{2g}^3 e_g^1$. $CFSE = [(-0.4 	imes 3) + (0.6 	imes 1)] \Delta_0 = [-1.2 + 0.6] \Delta_0 = -0.6 \Delta_0$.$B) [Fe(H_2O)_6]^{3+}: Fe^{3+} = 3d^5$. Configuration: $t_{2g}^3 e_g^2$. $CFSE = [(-0.4 	imes 3) + (0.6 	imes 2)] \Delta_0 = [-1.2 + 1.2] \Delta_0 = 0$.$C) [Co(H_2O)_6]^{2+}: Co^{2+} = 3d^7$. Configuration: $t_{2g}^5 e_g^2$. $CFSE = [(-0.4 	imes 5) + (0.6 	imes 2)] \Delta_0 = [-2.0 + 1.2] \Delta_0 = -0.8 \Delta_0$.$D) [Co(H_2O)_6]^{3+}: Co^{3+} = 3d^6$. Configuration: $t_{2g}^4 e_g^2$. $CFSE = [(-0.4 	imes 4) + (0.6 	imes 2)] \Delta_0 = [-1.6 + 1.2] \Delta_0 = -0.4 \Delta_0$.Thus,$[Fe(H_2O)_6]^{3+}$ has zero $CFSE$.

Among the following complexes,the one which shows zero Crystal Field Stabilization Energy $(CFSE)$ is

Similar Questions

The ratio of the total energy of the $2^{\text{nd}}$ orbit electron for the hydrogen atom $1H^1$ to that of the helium ion $(He^{+})$ $2He^4$ is:

$A$ particle of mass $0.1 \,kg$ is subjected to a force which varies with distance as shown. If it starts its journey from rest at $x=0$,then its velocity at $x=12 \,m$ is .............. $m/s$.

Observe the following statements:
$A$. Three vectors are coplanar if one of them is expressible as a linear combination of the other two.
$R$. Any three coplanar vectors are linearly dependent.
Then,which of the following is true?

$A$ hyperbola passes through a focus of the ellipse $\frac{x^2}{169} + \frac{y^2}{25} = 1$. Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse. The product of their eccentricities is $1$. Then,the equation of the hyperbola is:

What type of life cycle in plants is indicated by the given figure?

Vedclass Test Series

Exam Paper Generator

Online Exam Module