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Question

The correct order of the calculated spin-only magnetic moments of complexes (A) to (D) is
(A)Ni(CO)4
(B)[Ni(H2O)6]Cl2
(C)Na2[Ni(CN)4]
(D)PdCl2(PPh3)2

A
(C)<(D)<(B)<(A)
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B
(A)(C)<(B)(D)
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C
(C)(D)<(B)<A)
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D
(A)(C)(D)<(B)
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Solution

The correct option is D (A)(C)(D)<(B)
(A)Ni(CO)4
Ni is in the zero oxidation state i.e., it has a configuration of 3d84s2. Since CO is a strong field ligand, it causes the pairing of unpaired 3d electrons. so number of unpaired electron is zero thus it is diamagnetic.
μ=0 B.M;diamagnetic.

(B) [Ni(H2O)6]Cl2
[Ni(H2O)6]2+
Ni2+=3d8 with weak field ligand.
Number of unpaired e=2
μ=n(n+2)
=2(2+2)=22 B.M.

(C) Na2[Ni(CN)4]
[Ni(CN)4]2
Ni2+=3d8 with strong field ligand CN. so it will paired up d electron.
Number of unpaired e=0
μ=0B.M
PdCl2(PPh3)2
Pd2+=4d8(diamagnetic)
μ=0B.M

(D)PdCl2(PPh3)2
Pd is in the +2 oxidation state i.e., it has a configuration of 4d85s0. Since (PPh3) is a strong field ligand, it causes the pairing of unpaired 3d electrons. so number of unpaired electron is zero thus it is diamagnetic.
μ=0 B.M;diamagnetic.

Hence, the correct order is ACD<B
Therefore, the correct option is D.

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