Question

The correct order of magnetic moments (spin only values in B.M.) among is :

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

Answer 1

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

Option (C) is correct.

$\left(A\right)$ : $[MnC{l}_4{]}^{2-}$ has 5 unpaired electrons and $C{l}^{-}$ is a weak ligands,so electrons does not paired up

$\mu =\sqrt{n(n+2)}$

$=\sqrt{5(5+2)}$

$=\sqrt{35}$

$\left(B\right)$ : $[CoC{l}_4{]}^{2-}$ has 3 unpaired electrons and $C{l}^{-}$ is a weak ligands,so electrons does not paired up

$\mu =\sqrt{n(n+2)}$

$=\sqrt{3(3+2)}$

$=\sqrt{15}$

$\left(C\right)$ : $\left[Fe\right(CN{)}_6{]}^{4-}$ has 0 unpaired electrons and $C{N}^{-}$ is a strong ligands,so electrons gets paired up

$\mu =\sqrt{n(n+2)}$

$=\sqrt{0(0+2)}$

$=0$