Total number of unpaired electrons present in the central metal atoms/ions of [Ni(CO)4], [NiCl4]2-, [PtCl2(NH3)2], [Ni(CN)4]2- and [Pt(CN)4]2- is .
Answer
2
Solution
[Ni(CO)4]: Ni(0), 3d84s2; CO (strong field) pairs the electrons 3d10, sp3, tetrahedral; unpaired electrons = 0
[NiCl4]2-: Ni2+, 3d8; Cl- is a weak-field ligand sp3, tetrahedral; unpaired electrons = 2
[PtCl2(NH3)2]: Pt2+, 5d8; 5d metal (large splitting) dsp2, square planar; unpaired electrons = 0
[Ni(CN)4]2-: Ni2+, 3d8; CN- strong field dsp2, square planar; unpaired electrons = 0
[Pt(CN)4]2-: Pt2+, 5d8 dsp2, square planar; unpaired electrons = 0
Total = 0 + 2 + 0 + 0 + 0 = 2
Answer: 2