JEE Chemistry 2026 Question with Solution

Given: We first find the number of isoelectronic species, call it $n$, and then use that value for the complex $\ce{CoCl2(en)2NH3}$. Find: The number of electrons present in the $t_{2g}$ orbital of the complex.

Step 1: Check isoelectronic species

$$\ce{S^{2-}} \text{ has } 18 \text{ electrons}$$ $$\ce{C^{4-}} \text{ has } 10 \text{ electrons}$$ $$\ce{Mn^{2+}} \text{ has } 23 \text{ electrons } (25 - 2)$$ $$\ce{Co^{3+}} \text{ has } 24 \text{ electrons } (27 - 3)$$ $$\ce{Fe^{3+}} \text{ has } 23 \text{ electrons } (26 - 3)$$

Therefore, $\ce{Mn^{2+}}$ and $\ce{Fe^{3+}}$ are isoelectronic, so $n = 2$.

Step 2: Analyse the complex The complex is $\ce{CoCl2(en)2NH3}$. The coordination number is $6$ because it contains $2$ chloride ligands, $2$ ethylenediamine ligands, and $1$ ammonia ligand. Ethylenediamine and ammonia are neutral ligands, while chloride is negative. Hence, the oxidation state of cobalt is $+3$.

Step 3: Reaction with AgNO$_3$ Two chloride ions are outside the coordination sphere and react with excess silver nitrate to form AgCl. Thus, $2$ moles of AgCl are formed, which matches $n = 2$.

Step 4: Electronic configuration of cobalt Cobalt has atomic configuration $3d^7 4s^2$. Therefore,

$$\ce{Co^{3+}} = 3d^6$$

With strong field ligands like en and $\ce{NH3}$, the complex is low-spin octahedral. Hence,

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

So the number of electrons in $t_{2g}$ is $6$.

Therefore, the number of electrons present in the $t_{2g}$ orbital is $6$.

Given: $n$ is obtained from the isoelectronic species count, and the same value is verified through AgCl formation. Find: The electron count in $t_{2g}$ for the cobalt complex.

The isoelectronic pair is $\ce{Mn^{2+}}$ and $\ce{Fe^{3+}}$ because both contain $23$ electrons. Hence,

$$n = 2$$

Now consider $\ce{CoCl2(en)2NH3}$. The formation of $2$ moles of AgCl with excess $\ce{AgNO3}$ shows that $2$ chloride ions are ionisable and lie outside the coordination sphere. Thus the coordinated entity contains cobalt in the $+3$ oxidation state.

For cobalt,

$$\ce{Co} : 3d^7 4s^2$$

Removing three electrons gives

$$\ce{Co^{3+}} : 3d^6$$

Because en and $\ce{NH3}$ act as strong-field ligands, pairing occurs in the lower-energy set of octahedral d-orbitals. Therefore the crystal field splitting pattern is

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

Hence all $6$ d-electrons occupy the $t_{2g}$ set.

Therefore, the required number of electrons is $6$.