JEE Main 2024 — Coordination Compounds Question with Solution

To determine the number of complexes with an even number of electrons in the $${\mathrm{t}}_{2g}$$ orbitals, we first need to determine the electronic configuration of the metal ions in each complex and then find the distribution of electrons among the $${\mathrm{t}}_{2g}$$ and $${\mathrm{e}}_{\mathrm{g}}$$ orbitals.

Let's consider each complex one by one:

1. $${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$$

The oxidation state of Fe is +2. The electronic configuration of $${\mathrm{Fe}}^{2+}$$ is $$\left[\mathrm{Ar}\right]3d^6$$.

In an octahedral field, the splitting pattern for the 3d orbitals is such that: $$t_{2g}$$ (lower energy) and $$e_g$$ (higher energy).

So, the electronic configuration in octahedral field will be: $$t_{2g}^4{e}_g^2$$. Therefore, there are 4 electrons in the $$t_{2g}$$ orbitals (even).

2. $${\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$$

The oxidation state of Co is +2. The electronic configuration of $${\mathrm{Co}}^{2+}$$ is $$\left[\mathrm{Ar}\right]3d^7$$.

In an octahedral field, the splitting pattern for the 3d orbitals will be: $$t_{2g}^5{e}_g^2$$. Therefore, there are 5 electrons in the $$t_{2g}$$ orbitals (odd).

3. $${\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}$$

The oxidation state of Co is +3. The electronic configuration of $${\mathrm{Co}}^{3+}$$ is $$\left[\mathrm{Ar}\right]3d^6$$.

In an octahedral field, the splitting pattern for the 3d orbitals will be: $$t_{2g}^4{e}_g^2$$. Therefore, there are 4 electrons in the $$t_{2g}$$ orbitals (even).

4. $${\left[\mathrm{Cu}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$$

The oxidation state of Cu is +2. The electronic configuration of $${\mathrm{Cu}}^{2+}$$ is $$\left[\mathrm{Ar}\right]3d^9$$.

In an octahedral field, the splitting pattern for the 3d orbitals will be: $$t_{2g}^6{e}_g^3$$. Therefore, there are 6 electrons in the $$t_{2g}$$ orbitals (even).

5. $${\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$$

The oxidation state of Cr is +2. The electronic configuration of $${\mathrm{Cr}}^{2+}$$ is $$\left[\mathrm{Ar}\right]3d^4$$.

In an octahedral field, the splitting pattern for the 3d orbitals will be: $$t_{2g}^3{e}_g^1$$. Therefore, there are 3 electrons in the $$t_{2g}$$ orbitals (odd).

So, the complexes with an even number of electrons in the $${\mathrm{t}}_{2g}$$ orbitals are:

1. $${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$$ (4 electrons)

1. $${\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}$$ (4 electrons)
   


2. $${\left[\mathrm{Cu}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$$ (6 electrons)

Hence, there are 3 complexes with an even number of electrons in the $${\mathrm{t}}_{2g}$$ orbitals. Therefore, the correct answer is option A.