Number of Complexes with even number of electrons in $\mathrm{t_{2 g}}$ orbitals is -

$$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$$

<p>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 $\mathrm{e_{g}}$ orbitals.</p> <p>Let's consider each complex one by one:</p> <p><strong>1. $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$</strong></p> <p>The oxidation state of Fe is +2. The electronic configuration of $\mathrm{Fe}^{2+}$ is $\left[\mathrm{Ar}\right]3d^6$.</p> <p>In an octahedral field, the splitting pattern for the 3d orbitals is such that: $t_{2g}$ (lower energy) and $e_g$ (higher energy).</p> <p>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).</p> <p><strong>2. $\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$</strong></p> <p>The oxidation state of Co is +2. The electronic configuration of $\mathrm{Co}^{2+}$ is $\left[\mathrm{Ar}\right]3d^7$.</p> <p>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).</p> <p><strong>3. $\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$</strong></p> <p>The oxidation state of Co is +3. The electronic configuration of $\mathrm{Co}^{3+}$ is $\left[\mathrm{Ar}\right]3d^6$.</p> <p>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).</p> <p><strong>4. $\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$</strong></p> <p>The oxidation state of Cu is +2. The electronic configuration of $\mathrm{Cu}^{2+}$ is $\left[\mathrm{Ar}\right]3d^9$.</p> <p>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).</p> <p><strong>5. $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$</strong></p> <p>The oxidation state of Cr is +2. The electronic configuration of $\mathrm{Cr}^{2+}$ is $\left[\mathrm{Ar}\right]3d^4$.</p> <p>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).</p> <p>So, the complexes with an even number of electrons in the $\mathrm{t_{2g}}$ orbitals are:</p> <p>1. $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ (4 electrons) <br> <ol> <li>$\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ (4 electrons) <br></li><br> <li>$\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ (6 electrons) </p></li> </ol> <p>Hence, there are 3 complexes with an even number of electrons in the $\mathrm{t_{2g}}$ orbitals. Therefore, the correct answer is option A.</p>