JEE Main 2023 — Coordination Compounds Question with Solution

The spin-only magnetic moment of a complex ion is calculated using the formula $\mu =\sqrt{n(n+2)}$, where $n$ is the number of unpaired electrons in the complex ion. In the case of transition metal complexes, the number of unpaired electrons and hence the magnetic moment can be influenced by the field strength of the ligands (the Ligand Field Theory).

Here, CN${}^{-}$ is a strong-field ligand (or high-spin ligand) that can cause pairing of electrons, while F${}^{-}$ and Br${}^{-}$ are considered weak-field ligands (or low-spin ligands) that do not promote electron pairing.

1. ${\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{3-}$: In this complex, Fe is in +3 oxidation state and has 5 d-electrons. Due to the strong field ligand, CN${}^{-}$, all the 5 electrons are paired, and there are 0 unpaired electrons. The spin-only magnetic moment is $\mu =\sqrt{0(0+2)}=0$.




2. ${\left[{\mathrm{CoF}}_6\right]}^{3-}$: Here, Co is in +3 oxidation state and has 6 d-electrons. As F${}^{-}$ is a weak field ligand, no electron pairing occurs, and hence there are 4 unpaired electrons. The spin-only magnetic moment is $\mu =\sqrt{4(4+2)}=\sqrt{24}$.




3. ${\left[{\mathrm{MnBr}}_4\right]}^{2-}$: Here, Mn is in +2 oxidation state and has 5 d-electrons. Since Br${}^{-}$ is a weak field ligand, no electron pairing occurs, and hence there are 5 unpaired electrons. The spin-only magnetic moment is $\mu =\sqrt{5(5+2)}=\sqrt{35}$.




4. ${\left[\mathrm{Mn}(\mathrm{CN}{)}_6\right]}^{3-}$: In this complex, Mn is in +3 oxidation state and has 4 d-electrons. Due to the strong field ligand, CN${}^{-}$, all the 4 electrons are paired, and there are 0 unpaired electrons. The spin-only magnetic moment is $\mu =\sqrt{0(0+2)}=0$.

So the correct order of spin-only magnetic moments is:

$${\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{3-}<{\left[\mathrm{Mn}(\mathrm{CN}{)}_6\right]}^{3-}<{\left[{\mathrm{CoF}}_6\right]}^{3-}<{\left[{\mathrm{MnBr}}_4\right]}^{2-}$$