Question

The correct order of the calculated spin-only magnetic moments of complexes (A) to (D) is :
$(A)Ni(CO{)}_4$
$(B)[Ni(H_{2}O{)}_6]C{l}_2$
$(C)N{a}_2[Ni(CN{)}_4]$
$(D)PdC{l}_2(PP{h}_3{)}_2$

• 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)

$\left[Ni(CO{)}_4\right]$
$Ni=3d^{8}4s^2$ (strong field ligand)
$\mu =0B.M;$ diamagnetic.
$\left[Ni{\left(H_{2}O\right)}_6\right]C{l}_2$
${\left[Ni{\left(H_{2}O\right)}_6\right]}^{2+}$
$N{i}^{2+}=3d^8$ (weak field ligand)
$\therefore$ Number of unpaired $e^{-}=2$
$\mu =\sqrt{n(n+2)}=\sqrt{2(2+2)}=2\sqrt{2}B.M$
$N{a}_2\left[Ni(CN{)}_4\right]$
${\left[Ni(CN{)}_4\right]}^{2-}$
$N{i}^{2+}=3d^8$ (strong field ligand)
Number of unpaired $e^{-}=0$
$\therefore \mu =0$ B.M
$PdC{l}_2{\left(PP{h}_3\right)}_2$
$P{d}^{2+}=4d^8$ (diamagnetic)
$\mu =0B.M$
Hence, the correct order is $A\approx C\approx D<B$