Consider that d 6 metal ion M 2 + 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 :

Question

Consider that d 6 metal ion M 2 + forms a complex with aqua ligands, and the spin only magnetic moment of the complex is 4 .90 &#8201; BM . The geometry and the crystal field stabilization energy of the complex is :

Aicharya · Accepted Answer

Since spin only magnetic moment is 4.90 BM so number of unpaired electrons must be 4. There are two possibilities:

(1) If the complex is octahedral, then it has to be high spin complex with configuration ${t_{2 g}}^{2, 1, 1} e_{g}^{1, 1},$ in that case $CFSE = 4 \times (- 0.4 Δ_{0}) + 2 \times 0.6 Δ_{0} = - 0.4 Δ_{0}$
(2) If the complex is tetrahedral then its electronic configuration will be $= e_{g} {}^{2, 1}t_{2 g}^{1, 1, 1}$ and $CFSE$ will $be = 3 X$ $(- 0.6 Δ_{t}) + 3 X (0.4 Δ_{t}) = - 0.6 Δ t$

Consider that d 6 metal ion M 2 + forms a complex with aqua ligands, and the spin only magnetic moment of…

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