Question

The correct order of the calculated spin-only magnetic moments of complexes (A) to (D) is :
$(A )\, Ni(CO)_4$
$(B) \,[Ni(H_2O)_6]Cl_2$
$(C)\, Na_2[Ni(CN)_4]$
$(D)\, P dCl_2(PPh_3)_2$

• A. $\left(A\right)\approx\left(C\right) < \left(B\right) \approx \left(D\right)$
• B. $\left(C\right)\approx\left(D\right) < \left(B\right) \approx \left(A\right)$
• C. $\left(C\right)<\left(D\right) < \left(B\right) < \left(A\right)$
• D. $\left(A\right)\approx\left(C\right) \approx \left(D\right) < \left(B\right)$

Answer 1

Answer: (D)

$\left[ Ni ( CO )_{4}\right]$
$Ni =3 d ^{8} 4 s ^{2}$ (strong field ligand)
$\mu=0 B . M ;$ diamagnetic.
$\left[ Ni \left( H _{2} O \right)_{6}\right] Cl _{2}$
$\left[ Ni \left( H _{2} O \right)_{6}\right]^{2+}$
$Ni ^{2+}=3 d ^{8}$ (weak field ligand)
$\therefore $ Number of unpaired $e ^{-}=2$
$\mu=\sqrt{ n ( n +2)}=\sqrt{2(2+2)}=2 \sqrt{2} B . M$
$Na _{2}\left[ Ni ( CN )_{4}\right]$
$\left[ Ni ( CN )_{4}\right]^{2-}$
$Ni ^{2+}=3 d ^{8}$ (strong field ligand)
Number of unpaired $e ^{-}=0$
$\therefore \mu=0$ B.M
$P dCl _{2}\left( PPh _{3}\right)_{2}$
$Pd ^{2+}=4 d ^{8}$ (diamagnetic)
$\mu=0 B . M$
Hence, the correct order is $A \approx C \approx D < B$