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Consider that d6 metal ion (M2+) 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 :

(1) tetrahedral and –1.6$\Delta$t + 1P 

(2) octahedral and –2.4$\Delta$0 + 2P

(3) tetrahedral and –0.6$\Delta$t

(4) octahedral and –1.6$\Delta$0

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Coordination compounds

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Ans. (3) tetrahedral and –0.6$\Delta$t

Sol. Since spin only magnetic moment is 4.90 BM so number of unpaired electrons must be 4. so If the complex is octahedral, then it has to be high spin complex with configuration t2g2,1,1eg1,1, in that case

$CFSE=4\times(-0.4\Delta_0)+2\times0.6\Delta_0=-0.4\Delta_0$

If the complex is tetrahedral then its electronic configuration will be = eg2,1t2g1,1,1 and CFSE will be 

$=3\times(-0.6\Delta_t)+3\times(0.4\Delta_t)=-0.6\Delta_t$

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