1. The complex described is the homoleptic octahedral aqua complex of cobalt(II): $[\mathrm{Co}(\mathrm{H_2O})_6]^{2+}$

2. Determine the electronic configuration of $\mathrm{Co}^{2+}$.

   Cobalt has atomic number $27$: $\mathrm{Co}: [\mathrm{Ar}]\,3d^7 4s^2$

   For $\mathrm{Co}^{2+}$, remove two electrons from $4s$ first: $\mathrm{Co}^{2+}: [\mathrm{Ar}]\,3d^7$

3. In an octahedral field, the $d$-orbitals split into: $t_{2g} \text{ (lower energy)} \quad \text{and} \quad e_g \text{ (higher energy)}$

4. Since $\mathrm{H_2O}$ is a weak field ligand, the complex is high spin.

   So for $d^7$ high-spin octahedral configuration: $t_{2g}^5 e_g^2$

5. Now place 5 electrons in the three $t_{2g}$ orbitals according to Hund's rule:

   First three electrons occupy singly: $\uparrow\quad \uparrow\quad \uparrow$

   The next two pair up in two of them: $\uparrow\downarrow \quad \uparrow\downarrow \quad \uparrow$

6. Therefore, in the $t_{2g}$ set, there is exactly 1 unpaired electron.

Hence, the required number of unpaired electrons in the $t_{2g}$ set is: $\boxed{1}$