Question

Consider that a $d ^{6}$ metal ion $\left( M ^{2+}\right)$ forms a complex with aqua ligands, and the spin only magnetic moment of the complex is $4.90 BM$ The geometry and the crystal field stabilization energy of the complex is :

• A. tetrahedral and $-1.6 \Delta_{ t }+1 P$
• B. tetrahedral and $-0.6 \Delta_{ t }$
• C. octahedral and $-1.6 \Delta_{0}$
• D. octahcdral and $-2.4 \Delta_{0}+2 P$

Answer 1

Answer: (B)

If spin only magnetic moment of the complex is $4.90 \,BM$, it means number of unpaired electrons should be 4.

(A) In octahedral complex : $\underset{d^6}{[M(H_2O)_6]^{2+}}$



$ C.F.S.E. =\left(-0.4 \Delta_{0}\right) \times 4+\left(+0.6 \Delta_{0}\right) \times 2+0 \times P $

$=-0.4 \Delta_{0} $

(B) In tetrahedral complex : $\underset{d^6}{[M(H_2O)_4]^{2+}}$



C.F.S.E. $=\left(-0.6 \Delta_{t}\right) \times 3+\left(+0.4 \Delta_{t}\right) \times 3+0 \times P$

$=-0.6 \Delta_{ t } $