JEE Chemistry 2025 Question with Solution

Given: The metal M is chosen from Mn, Cr, Co, and Fe and has the highest standard electrode potential for the $M^{3+}/M^{2+}$ couple. The cyanide complex discussed in the solution is $[\text{M(CN)}_6]^{4-}$.

Find: The number of electrons in the $e_g$ orbitals of the complex.

From the standard reduction potentials listed in the solution:

• $E^\circ(\text{Mn}^{3+}/\text{Mn}^{2+}) = +1.51 \, \text{V}$
• $E^\circ(\text{Cr}^{3+}/\text{Cr}^{2+}) = -0.41 \, \text{V}$
• $E^\circ(\text{Co}^{3+}/\text{Co}^{2+}) = +1.82 \, \text{V}$
• $E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77 \, \text{V}$

So the metal is Co.

Now use the complex given in the solution, $[\text{Co(CN)}_6]^{4-}$. Let the oxidation state of cobalt be $x$.

Therefore the metal ion is $\text{Co}^{2+}$.

Neutral cobalt has configuration $[\text{Ar}]\,3d^7 4s^2$. Hence,

So this is a $d^7$ system.

Since $\text{CN}^-$ is a strong-field ligand, the octahedral complex is low spin. Therefore the electron distribution is

So the number of electrons in the $e_g$ orbitals is 1.

The solution concludes 1, but this value is not present in the given options 6, 5, 4, 3. The provided answer key marks option (1), which corresponds to A = 6. Hence there is a discrepancy between the worked solution and the listed options/answer key. Following the source answer key mapping, the recorded correct option is A.