Question

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

Comments

Coordination compounds

Answer 1

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 t_2g^2,1,1e_g^1,1, 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 = e_g^2,1t_2g^1,1,1 and CFSE will be 

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

Comments

Coordination chapter is an important chapter for NEET students
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why is the pairing energy not considered here?