JEE Main 2025 — Coordination Compounds Question with Solution

Identify the oxidation state and d-electron count for each complex:

For $${\left[\mathrm{Co}(\mathrm{en}{)}_3\right]}^{3+}$$:

Ethylenediamine (en) is neutral.

Oxidation state of cobalt is +3.

Cobalt (atomic number 27) in the +3 state gives a $$d^6$$ configuration.

With strong/moderate field ligands like en, the complex is low spin, leading to a configuration of $$t_{2g}^6{e}_g^0$$.

For $${\left[{\mathrm{CoF}}_6\right]}^{3-}$$:

Each fluoride ion is $${\mathrm{F}}^{-}$$ so, with six of them, Co must be in the +3 state again.

However, since fluoride is a weak field ligand, this complex adopts a high spin configuration for $$d^6$$, resulting in $$t_{2g}^4{e}_g^2$$.

For $${\left[\mathrm{Mn}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$$:

Water is neutral.

Manganese in the +2 state gives a $$d^5$$ configuration.

With weak field water, the $$d^5$$ configuration is high spin: $$t_{2g}^3{e}_g^2$$.

For $${\left[\mathrm{Zn}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$$:

Zinc in the +2 state has a $$d^{10}$$ configuration.

In an octahedral field, all 10 electrons fill the orbitals as $$t_{2g}^6{e}_g^4$$.

Calculate the crystal field stabilization energy (CFSE) for each case:

The CFSE in an octahedral field can be estimated as:

$$\mathrm{CFSE}=(n_{t_{2g}}\times -0.4{\Delta }_o)+(n_{e_g}\times 0.6{\Delta }_o)$$

For $$t_{2g}^6{e}_g^0$$ (low-spin $$d^6$$ in $${\left[\mathrm{Co}(\mathrm{en}{)}_3\right]}^{3+}$$):

$$\mathrm{CFSE}=6(-0.4{\Delta }_o)+0(0.6{\Delta }_o)=-2.4{\Delta }_o$$

For $$t_{2g}^4{e}_g^2$$ (high-spin $$d^6$$ in $${\left[{\mathrm{CoF}}_6\right]}^{3-}$$):

$$\mathrm{CFSE}=4(-0.4{\Delta }_o)+2(0.6{\Delta }_o)=-1.6{\Delta }_o+1.2{\Delta }_o=-0.4{\Delta }_o$$

For $$t_{2g}^3{e}_g^2$$ (high-spin $$d^5$$ in $${\left[\mathrm{Mn}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$$):

$$\mathrm{CFSE}=3(-0.4{\Delta }_o)+2(0.6{\Delta }_o)=-1.2{\Delta }_o+1.2{\Delta }_o=0$$

For $$t_{2g}^6{e}_g^4$$ (for $$d^{10}$$ in $${\left[\mathrm{Zn}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$$):

$$\mathrm{CFSE}=6(-0.4{\Delta }_o)+4(0.6{\Delta }_o)=-2.4{\Delta }_o+2.4{\Delta }_o=0$$

Compare the CFSE values:

$$t_{2g}^6{e}_g^0$$ gives a CFSE of $$-2.4{\Delta }_o$$ (most negative, hence highest stabilization).

The others result in less stabilization (or net zero).

Conclusion:

The complex with the highest CFSE is the one with the configuration $$t_{2g}^6{e}_g^0$$, which corresponds to $${\left[\mathrm{Co}(\mathrm{en}{)}_3\right]}^{3+}$$.

Thus, the answer is:

Option A: $$t_{2g}^6{e}_g^0$$.