JEE Chemistry 2026 Question with Solution

Given: Two statements about CFSE and spin-only magnetic moment are to be checked.

Find: Which option correctly identifies the truth values of Statement I and Statement II.

For octahedral complexes,

$$\text{CFSE} = \left[-0.4\,n(t_{2g}) + 0.6\,n(e_g)\right]\Delta_o$$

and spin-only magnetic moment depends on the number of unpaired electrons $n$:

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

Statement I:

In $[Cr(H_2O)_6]^{2+}$, chromium is $Cr^{2+}$, so it is a $d^4$ ion. Since $H_2O$ is a weak field ligand, the complex is high spin.

$$t_{2g}^3 e_g^1$$

Therefore,

$$\text{CFSE} = \left[3(-0.4) + 1(0.6)\right]\Delta_o = -0.6\Delta_o$$

In $[Mn(H_2O)_6]^{2+}$, manganese is $Mn^{2+}$, so it is a $d^5$ ion. With weak field ligand $H_2O$, it is also high spin.

$$t_{2g}^3 e_g^2$$

Therefore,

$$\text{CFSE} = \left[3(-0.4) + 2(0.6)\right]\Delta_o = 0$$

The magnitude of CFSE for $[Cr(H_2O)_6]^{2+}$ is greater than that for $[Mn(H_2O)_6]^{2+}$. So Statement I is true.

Statement II:

Potassium ferricyanide is $K_3[Fe(CN)_6]$. Here iron is in the $+3$ oxidation state, so it is $d^5$. Since $CN^-$ is a strong field ligand, the complex is low spin.

$$t_{2g}^5 e_g^0$$

Number of unpaired electrons is $n=1$.

Sodium ferrocyanide is $Na_4[Fe(CN)_6]$. Here iron is in the $+2$ oxidation state, so it is $d^6$. With strong field ligand $CN^-$, the complex is low spin.

$$t_{2g}^6 e_g^0$$

Number of unpaired electrons is $n=0$.

Since ferricyanide has more unpaired electrons than ferrocyanide, its spin-only magnetic moment is greater. So Statement II is true.

Therefore, both Statement I and Statement II are true. The correct option is C.

Given: Weak field ligand $H_2O$ and strong field ligand $CN^-$ are involved.

Find: Compare CFSE values and magnetic moments.

A weak field ligand gives a high spin complex, while a strong field ligand gives a low spin complex.

1. For $[Cr(H_2O)_6]^{2+}$:

• $Cr^{2+} \to d^4$
• High spin octahedral arrangement: $t_{2g}^3 e_g^1$
• CFSE:

$$[3(-0.4) + 1(0.6)]\Delta_o = -0.6\Delta_o$$

2. For $[Mn(H_2O)_6]^{2+}$:

• $Mn^{2+} \to d^5$
• High spin octahedral arrangement: $t_{2g}^3 e_g^2$
• CFSE:

$$[3(-0.4) + 2(0.6)]\Delta_o = 0$$

So, Statement I is true.

3. For $K_3[Fe(CN)_6]$:

• Iron oxidation state $= +3$
• $Fe^{3+} \to d^5$
• Strong field, low spin: $t_{2g}^5 e_g^0$
• Unpaired electrons $=1$

4. For $Na_4[Fe(CN)_6]$:

• Iron oxidation state $= +2$
• $Fe^{2+} \to d^6$
• Strong field, low spin: $t_{2g}^6 e_g^0$
• Unpaired electrons $=0$

Hence ferricyanide has greater magnetic moment than ferrocyanide. Statement II is also true.

Therefore, the correct option is C.