JEE Chemistry 2025 Question with Solution

Given: The complexes are $K_3[Fe(OH)_6]$ and $K_4[Fe(OH)_6]$.

Find: The spin-only magnetic moments of both complexes.

First determine the oxidation state of iron in each complex.

For $K_3[Fe(OH)_6]$, the complex ion has charge $-3$ because it is balanced by $3K^+$ ions.

$$x + 6(-1) = -3$$

So, $x = +3$. Therefore iron is $Fe^{3+}$.

For $K_4[Fe(OH)_6]$, the complex ion has charge $-4$ because it is balanced by $4K^+$ ions.

$$x + 6(-1) = -4$$

So, $x = +2$. Therefore iron is $Fe^{2+}$.

The ground-state electronic configuration of iron is $[Ar] \, 3d^6 4s^2$.

• For $Fe^{3+}$: remove three electrons to get $[Ar] \, 3d^5$, which has $5$ unpaired electrons.
• For $Fe^{2+}$: remove two electrons to get $[Ar] \, 3d^6$, which has $4$ unpaired electrons.

Use the spin-only magnetic moment formula:

$$\mu = \sqrt{n(n+2)} \, \text{B.M.}$$

For $K_3[Fe(OH)_6]$:

$$\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92 \, \text{B.M.}$$

For $K_4[Fe(OH)_6]$:

$$\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90 \, \text{B.M.}$$

Therefore, the magnetic moments are $5.92 \, \text{B.M.}$ and $4.90 \, \text{B.M.}$, so the correct option is B.

Given: Spin-only magnetic moment depends on the number of unpaired electrons.

Find: The magnetic moments of $K_3[Fe(OH)_6]$ and $K_4[Fe(OH)_6]$.

The required relation is:

$$\mu = \sqrt{n(n+2)} \, \text{B.M.}$$

where $n$ is the number of unpaired electrons.

For $K_3[Fe(OH)_6]$:

• Each $OH^-$ ligand contributes charge $-1$.
• Total ligand charge is $6(-1) = -6$.
• The complex ion charge is $-3$.

$$x - 6 = -3$$ $$x = +3$$

So iron is $Fe^{3+}$.

For $K_4[Fe(OH)_6]$:

• The complex ion charge is $-4$.

$$x - 6 = -4$$ $$x = +2$$

So iron is $Fe^{2+}$.

Now write the electronic configurations:

• $Fe^{3+} : [Ar] \, 3d^5$
• $Fe^{2+} : [Ar] \, 3d^6$

Thus, the numbers of unpaired electrons are:

• $Fe^{3+} \rightarrow n = 5$
• $Fe^{2+} \rightarrow n = 4$

Substitute into the formula:

$$\mu(Fe^{3+}) = \sqrt{5(7)} = \sqrt{35} = 5.92 \, \text{B.M.}$$ $$\mu(Fe^{2+}) = \sqrt{4(6)} = \sqrt{24} = 4.90 \, \text{B.M.}$$

Hence, $K_3[Fe(OH)_6]$ has magnetic moment $5.92 \, \text{B.M.}$ and $K_4[Fe(OH)_6]$ has magnetic moment $4.90 \, \text{B.M.}$. The correct option is B.