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 $4.90BM$ The geometry and the crystal field stabilization energy of the complex is :

• A. tetrahedral and $-1.6{\Delta }_t+1P$
• B. tetrahedral and $-0.6{\Delta }_t$
• C. octahedral and $-1.6{\Delta }_0$
• D. octahcdral and $-2.4{\Delta }_0+2P$

Answer 1

Answer: (B)

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

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



$C.F.S.E.=\left(-0.4{\Delta }_0\right)\times 4+\left(+0.6{\Delta }_0\right)\times 2+0\times P$

$=-0.4{\Delta }_0$

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



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

$=-0.6{\Delta }_t$