Question

The correct order of spin-only magnetic moments among the following is:

(Atomic number: $Mn=25,Co=27,Ni=28,Zn=30)$.

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

Answer 1

The correct option is D $[MnC{l}_4{]}^{2-}>[CoC{l}_4{]}^{2-}>[NiC{l}_4{]}^{2-}>[ZnC{l}_4{]}^{2-}$

The complex having higher number of unpaired electrons will have higher value of spin-only magnetic moment.

The correct order of spin-only magnetic moments is $[MnC{l}_4{]}^{2-}>[CoC{l}_4{]}^{2-}>[NiC{l}_4{]}^{2-}>[ZnC{l}_4{]}^{2-}$.

In these complexes, the central metal ion is in $+2$ oxidation state.

$Z{n}^{2+}$ has $3d^{10}$ outer electronic configuration with $0$ unpaired electrons.

$N{i}^{2+}$ has $3d^8$ outer electronic configuration with $2$ unpaired electrons.

$C{o}^{2+}$ has $3d^7$ outer electronic configuration with $3$ unpaired electrons.

$M{n}^{2+}$ has $3d^5$ outer electronic configuration with $5$ unpaired electrons.

Hence, the correct option is D.