Single Choice

The correct order of the calculated spin-only magnetic moments of complexes (A) to (D) is: (A) $$Ni{(CO)}_{4}$$ (B) $$[Ni{({H}_{2}O)}_{6}]{Cl}_{2}$$ (C) $${Na}_{2}[Ni{(CN)}_{4}]$$ (D) $$Pd{Cl}_{2}{(PP{h}_{3})}_{2}$$

A$$(C)< (D)< (B)< (A)$$
B$$(C)\approx (D)< (B)< (A)$$
C$$(A)\approx (C)< (B)\approx (D)$$
D$$(A)\approx (C)\approx (D)< (B)$$
Correct Answer

Solution

$$[Ni(CO)_4]$$ $$Ni=3d^84s^2$$ (strong field ligand) $$\mu=0\,B.M; \text{diamagnetic}.$$ $$[Ni(H_2O)_6]Cl_2$$ $$[Ni(H_2O)_6]^{2+}$$ $$Ni^{2+}=3d^8$$ (weak field ligand) $$\therefore$$ Number of unpaired $$e^-=2$$ $$\mu=\sqrt{n(n+2)}=\sqrt{2(2+2)}=2\sqrt{2}\,B.M$$ $$Na_2[Ni(CN)_4]$$ $$[Ni(CN)_4]^{2-}$$ $$Ni^{2+}=3d^8$$ (strong field ligand) Number of unpaired $$e^-=0$$ $$\therefore \mu=0\ B.M$$ $$PdCl_2(PPh_3)_2$$ $$Pd^{2+}=4d^8$$ (diamagnetic) $$\mu=0\,B.M$$ Hence, the correct order is $$A\approx C\approx D < B$$ Therefore, the correct option is $$D$$.

SIMILAR QUESTIONS

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