The species that has a spin-only magnetic moment of 5.9 BM, is :
(Td = tetrahedral)
(Td = tetrahedral)
Show Hint
- [MnBr4]2– (Td)
- [NiCl4]2– (Td)
- Ni(CO)4 (Td)
- [Ni(CN)4]2– (square planar)
The Correct Option is A
Solution and Explanation
(1) [MnBr4]2-: Mn is +2, d5. Br- is a weak ligand and the complex is tetrahedral, so it is high spin with 5 unpaired electrons, \(\mu = 5.92\) BM.
(2) [NiCl4]2-: Ni2+ d8, tetrahedral, 2 unpaired electrons, \(\mu = 2.83\) BM.
(3) Ni(CO)4: Ni(0) d10, no unpaired electrons, \(\mu = 0\).
(4) [Ni(CN)4]2-: Ni2+ d8, square planar (strong field), no unpaired electrons.
Answer: (1) [MnBr4]2– (Td)
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