The correct order of the calculated spin-only magnetic moments of complexes (A) to (D) is : (A) Ni(CO)_{4} (B) \left [ Ni(H_{2}O)_{6} \right ]Cl_{2} (C) Na_{2}\left [ Ni(CN)_{4} \right ] (D) PdCl_{2}(PPh_{3})_{2}
Option: 1 (A)\approx (C)< (B)\approx (D)  
  
Option: 2 (C)\approx (D)< (B)< (A)

Option: 3 (C)< (D)< (B)< (A)

Option: 4 (A)\approx (C)\approx (D)< (B)
 

Answers (1)

As we have learnt,

[Ni(CO)_4] contains Ni(0) which has a d^{10} configuration having n=0 and \mu=0

[Ni(H_2O)_6]^{2+}contains Ni^{2+} which has a d^8 configuration. In a weak ligand field, it has n=2 and hence has \mu=\sqrt8 \ BM

[Ni(CN)]_4contains Ni^{2+}which has a d^8 configuration. In a strong ligand field, these electrons get paired up and hence n=0 and \mu =0

[Pd(PPh_3)_2Cl_2] contains Pd^{2+}which has a d^8 configuration. These electrons get paired in the presence of strong ligand like PPh_3 and hence, n=0 and correspondingly \mu=0.

Therefore, the correct order of spin only magnetic moment of the complexes is 

(B)>(A)=(C)=(D)

Therefore, Option(4) is correct.

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