The species that has a spin-only magnetic moment of 5.9 BM, is: (T.H. tetrahedral)

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NiCN42−square planar

MnBr42−Tetrahedral

NiCl42−Tetrahedral

NiCO4Tetrahedral

answer is B.

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Detailed Solution

i) Ni(CN)42− No. of 'd' electrons =z−x−18 =28−2−18 =d8CN−→ strong ligand dx2−y2>dxy2>d−22>dyz2=dzx2 CN−→ strong ligand dx2−y2>dxy2>d−22>dyz2=dzx2No. of unpaired e-= 0Diamagnetic species ii) MnBr42− No. of electrons d=25−2−18=d5μ=n(n+2)B⋅M ii) MnBr42− No. of electrons d=25−2−18=d5μ=n(n+2)B⋅MBr−→weak ligand and obeys Hund's rule of multiplicity and contains 5 unpaired e−iii) NiCl42− No.of d e−=28−2−18=d8 No.of unpaired e−=2 iv) NiCO4 No. of d e− =28−0−18=d10 Dia magnetic species answer is MnBr42−Tetrahedral

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