The Crystal Field Stabilization Energy (CFSE) | vedclass

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

Question

(D) The magnetic moment $\mu = 3.87 \, BM$ corresponds to $n = 3$ unpaired electrons,as $\mu = \sqrt{n(n+2)} \, BM$. For an octahedral complex,the $CF

Vedclass · Accepted Answer

$Co^{2+}$

Vedclass · Answer

(D) The magnetic moment $\mu = 3.87 \, BM$ corresponds to $n = 3$ unpaired electrons,as $\mu = \sqrt{n(n+2)} \, BM$. For an octahedral complex,the $CFSE$ is given by the formula: $CFSE = (-0.4 	imes n_{t_{2g}} + 0.6 	imes n_{e_g}) \Delta_0$. For a $d^7$ ion in a weak field (aqua complex),the configuration is $t_{2g}^5 e_g^2$. $CFSE = (-0.4 	imes 5 + 0.6 	imes 2) \Delta_0 = (-2.0 + 1.2) \Delta_0 = -0.8 \Delta_0$. This matches the given $CFSE$ and the number of unpaired electrons $(3)$. Thus,the metal ion is $Co^{2+}$ ($d^7$ configuration).

List-$I$ (Complex)	List-$II$ (Electronic configuration of metal/ion)
$A. [Co(NH_3)_6]^{3+}$	$I. t_{2g}^5 e_g^0$
$B. [CoF_6]^{3-}$	$II. t_{2g}^6 e_g^0$
$C. [Ni(CO)_4]$	$III. t_{2g}^4 e_g^2$
$D. [Fe(CN)_6]^{3-}$	$IV. t^4 e^6$

The Crystal Field Stabilization Energy $(CFSE)$ and magnetic moment (spin-only) of an octahedral aqua complex of a metal ion $(M^{2+})$ are $-0.8\, \Delta_{0}$ and $3.87\, BM$,respectively. Identify $(M^{2+})$:

Similar Questions

The correct order of ligands arranged in increasing field strength.

Which of the following ion configurations has zero $CFSE$ in both strong and weak field ligands in an octahedral complex?

Which relationship is correct for the crystal field splitting energy between octahedral $(\Delta_0)$ and tetrahedral $(\Delta_t)$ complexes?

The correct order of energies of $d-$ orbitals of a metal ion in a square planar complex is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module