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 BM . The geometry and the crystal field stabilization energy of the complex is :, octahedral and - 2 . 4 Δ 0 + 2 P, tetrahedral and − 0 .6 Δ t, octahedral and - 1 . 6 Δ 0, tetrahedral and - 1 . 6 Δ t + 1 P

MARKS App · 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 × - 0 . 4 Δ 0 + 2 × 0 . 6 Δ 0 = - 0 . 4 Δ 0 (2) If the complex is tetrahedral then its electronic configuration will be = e g t 2 g 2 , 1 1 , 1 , 1 and CFSE will be = 3 X - 0 . 6 Δ t + 3 X 0 . 4 Δ t = - 0 . 6 Δ t

Search any question & find its solution

Looking for more such questions to practice?