Question

The correct order of magnetic moments (spin only values in Bohr magneton) among the following is:

• A. $[MnC{l}_4{]}^{2-}>[CoC{l}_4{]}^{2-}>\left[Fe\right(CN{)}_6{]}^{4-}$
• B. $[MnC{l}_4{]}^{2-}>[Fe(CN{)}_6{]}^{4-}>[CoC{l}_4{]}^{2-}$
• C. $\left[Fe\right(CN{)}_6{]}^{4-}>[MnC{l}_4{]}^{2-}>[CoC{l}_4{]}^{2-}$
• D. $\left[Fe\right(CN{)}_6{]}^{4-}>[CoC{l}_4{]}^{2-}>[MnC{l}_4{]}^{2-}$

Answer 1

The correct option is A $[MnC{l}_4{]}^{2-}>[CoC{l}_4{]}^{2-}>\left[Fe\right(CN{)}_6{]}^{4-}$

In $[MnC{l}_4{]}^{2-}$, Mn is in +2 state which is a $d^5$ configuration. It forms a high spin complex with 5 unpaired electrons.
In $[CoC{l}_4{]}^{2-}$, Co is in +2 state which is a $d^7$ configuration. It forms a high spin complex with 3 unpaired electrons.
In $\left[Fe\right(CN{)}_6{]}^{4-}$, Fe is in +2 state which is a $d^6$ configuration. It forms a low spin complex with zero unpaired electron.
So, the order of spin only magnetic moment is $[MnC{l}_4{]}^{2-}>[CoC{l}_4{]}^{2-}>\left[Fe\right(CN{)}_6{]}^{4-}$.

Hence option A is correct.