Question

Arrange $\left[Fe\right(CN{)}_6{]}^{4-}$, $\left[Fe\right(CN{)}_6{]}^{3-}$, $\left[Ni\right(CN{)}_4{]}^{2-}$ and $\left[Ni\right(H_{2}O{)}_4{]}^{2+}$ in increasing order of magnetic moment.

• A. $\left[Ni\right(H_{2}O{)}_4{]}^{2+}$ < $\left[Fe\right(CN{)}_6{]}^{4-}$ = $\left[Ni\right(CN{)}_4{]}^{2-}$ < $\left[Fe\right(CN{)}_6{]}^{3-}$
• B. $\left[Fe\right(CN{)}_6{]}^{4-}$ < $\left[Fe\right(CN{)}_6{]}^{3-}$ = $\left[Ni\right(CN{)}_4{]}^{2-}$ < $\left[Ni\right(H_{2}O{)}_4{]}^{2+}$
• C. $\left[Fe\right(CN{)}_6{]}^{4-}$ = $\left[Ni\right(CN{)}_4{]}^{2-}$ < $\left[Fe\right(CN{)}_6{]}^{3-}$ < $\left[Ni\right(H_{2}O{)}_4{]}^{2+}$
• D. $\left[Fe\right(CN{)}_6{]}^{4-}$ < $\left[Fe\right(CN{)}_6{]}^{3-}$ < $\left[Ni\right(CN{)}_4{]}^{2-}$ = $\left[Ni\right(H_{2}O{)}_4{]}^{2+}$

Answer 1

The correct option is C $\left[Fe\right(CN{)}_6{]}^{4-}$ = $\left[Ni\right(CN{)}_4{]}^{2-}$ < $\left[Fe\right(CN{)}_6{]}^{3-}$ < $\left[Ni\right(H_{2}O{)}_4{]}^{2+}$

In ${\left[Ni{\left(H_{2}O\right)}_4\right]}^{2+}$, there are unpaired electrons due to the unpairing of weak ligand $H_{2}O$. These electrons make d-d transition, hence, paramagnetic.
In ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$, Ni has +2 configuration and no unpaired electrons due to the pairing of electrons by strong ligand $\left(C{N}^{-}\right)$. Even after pairing of electrons, d-d transition can occur because of vacant d-orbitals.

Similarly in ${\left[Fe{\left(CN\right)}_6\right]}^{4-}[\mu =\sqrt{n(n+2)};n=3]$ ${\left[Fe{\left(CN\right)}_6\right]}^{3-}$ with ${Fe}^{+3}$ has $3d^7$ outer configuration thus, has more magnetic moment than ${\left[Fe{\left(CN\right)}_6\right]}^{4-}$