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.

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.