Question

The correct order of magnetic moment (spin only values in B.M.) among is: (Atomic no.s : $Mn=25;Fe=26;Co=27$)

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

$\left[Fe\right(CN{)}_6{]}^{4-}\to F{e}^{2+}\to \left[Ar\right]3d^6$
(Due to strong field ligand of $\left(C{N}^{-}\right)$ all unpaired electrons get paired.)
$[MnC{l}_4{]}^{2-}\to M{n}^{2+}\to [Ar]3d^5$
(due to weak field ligand of $\left(C{l}^{-}\right)$ no change in unpaired electrons
$[CoC{l}_4{]}^{2-}\to C{o}^{2+}\to [Ar]3d^7$
(Due to weak field ligand of $\left(C{l}^{-}\right)$ no change in unpaired electrons)
Magnetic moment $=\sqrt{n(n+2)}B.M.$
Where, n = number of unpaired electrons.
In $[MnC{l}_4{]}^{2-}$, $Mn$ contains maximum no. of electrons
So, order of magnetic moment $[MnC{l}_4{]}^{2-}>[CoC{l}_4{]}^{2-}>\left[Fe\right(CN{)}_6{]}^{4-}$