Question

The correct order of spin-only magnetic moments among the following is :
(Atomic number : Mn=25, Co=27, Ni=28, Zn=30)

• A. ${\left[ZnC{l}_4\right]}^{2-}>{\left[NiC{l}_4\right]}^{2-}>{\left[CoC{l}_4\right]}^{2-}>{\left[MnC{l}_4\right]}^2$
• B. ${\left[CoC{l}_4\right]}^{2^2}>{\left[MnC{l}_4\right]}^{2^{-}}>{\left[NiC{l}_4\right]}^{2^2}>{\left[ZnC{l}_4\right]}^{2-}$
• C. ${\left[NiC{l}_4\right]}^2>{\left[CoC{l}_4\right]}^{2-}>{\left[MnC{l}_4\right]}^2>{\left[ZnC{l}_4\right]}^{2-}$
• D. ${\left[MnC{l}_4\right]}^{2^{-}}>{\left[CoC{l}_4\right]}^{2-}>{\left[NiC{l}_4\right]}^{2-}>{\left[ZnC{l}_4\right]}^{2-}$

Answer 1

Answer: (D)

The spin only magnetic moment is given by $\mu =\sqrt{n(n+2)}$

where $n$ is the number of unpaired electrons and $\mu$ is the spin-only magnetic moment. The complex having higher number of unpaired electrons will have higher value of spin only magnetic moment,

In ${\left[ZnC{l}_4\right]}^{2-},Z{n}^{2+}$ has $d^{10}$ electronic configuration with 0 unpaired electrons.

In [NiCl ] 2 -, $N{i}^2$ thas $d^8$ electronic configuration with 2 unpaired electrons.

In ${\left[CoC{l}_4\right]}^2,C{o}^{2+}$ has $d^7$ electronic configuration with 3 unpaired electrons.

In $[MnC{l}_4{F}^2,M{n}^2$ 'has $d^5$ electronic configuration with 5 unpaired electrons. Hence, correct order is ${\left[MnC{l}_4\right]}^{2^{-}}>{\left[CoC{l}_4\right]}^{2-}>{\left[NiC{l}_4\right]}^{2-}>{\left[ZnC{l}_4\right]}^{2-}$