1. Find the oxidation state and $d$-electron count of each metal ion

Since $\mathrm{H_2O}$ is a neutral ligand, the oxidation state is just the charge on the complex.

• $\left[\mathrm{Fe(H_2O)_6}\right]^{2+}$: $\mathrm{Fe^{2+}} \Rightarrow d^6$
• $\left[\mathrm{Co(H_2O)_6}\right]^{2+}$: $\mathrm{Co^{2+}} \Rightarrow d^7$
• $\left[\mathrm{Co(H_2O)_6}\right]^{3+}$: $\mathrm{Co^{3+}} \Rightarrow d^6$
• $\left[\mathrm{Cu(H_2O)_6}\right]^{2+}$: $\mathrm{Cu^{2+}} \Rightarrow d^9$
• $\left[\mathrm{Cr(H_2O)_6}\right]^{2+}$: $\mathrm{Cr^{2+}} \Rightarrow d^4$

2. Decide spin state in octahedral field with $\mathrm{H_2O}$ ligand

$\mathrm{H_2O}$ is a weak field ligand, so these are generally high-spin octahedral complexes.

Thus, electron distribution in $t_{2g}$ and $e_g$ is:

• $d^4$: $t_{2g}^3 e_g^1$
• $d^6$ (high spin): $t_{2g}^4 e_g^2$
• $d^7$ (high spin): $t_{2g}^5 e_g^2$
• $d^9$: $t_{2g}^6 e_g^3$

3. Count electrons in $t_{2g}$ for each complex

• $\left[\mathrm{Fe(H_2O)_6}\right]^{2+}$: $d^6 \Rightarrow t_{2g}^4$ → even

• $\left[\mathrm{Co(H_2O)_6}\right]^{2+}$: $d^7 \Rightarrow t_{2g}^5$ → odd

• $\left[\mathrm{Co(H_2O)_6}\right]^{3+}$: $d^6$

  Here we should check carefully: $\mathrm{Co^{3+}}$ with octahedral coordination often has a relatively large $\Delta_o$, and even with $\mathrm{H_2O}$ it is typically low spin.

  So for $\mathrm{Co^{3+}}$: $d^6 \Rightarrow t_{2g}^6 e_g^0$ Hence $t_{2g}$ has 6 electrons, which is even.

• $\left[\mathrm{Cu(H_2O)_6}\right]^{2+}$: $d^9 \Rightarrow t_{2g}^6$ → even

• $\left[\mathrm{Cr(H_2O)_6}\right]^{2+}$: $d^4 \Rightarrow t_{2g}^3$ → odd

4. Total number with even number of electrons in $t_{2g}$

The complexes with even number of $t_{2g}$ electrons are:

• $\left[\mathrm{Fe(H_2O)_6}\right]^{2+}$
• $\left[\mathrm{Co(H_2O)_6}\right]^{3+}$
• $\left[\mathrm{Cu(H_2O)_6}\right]^{2+}$

So the total is: $3$

5. Option check

• A: $3$ ✅
• B: $2$ ❌
• C: $1$ ❌
• D: $5$ ❌

Therefore, the correct answer is A.