Considering $H_2O$ as a weak field ligand, the number of unpaired electrons in $[Mn(H_2O)_6]^{2+}$ will be: (At. no. of Mn = 25)

In the complex $[Mn(H_2O)_6]^{2+}$, the central manganese atom is present as $Mn^{2+}$ ion. The atomic number of Mn is 25, so its ground state electronic configuration is $[Ar] 3d^5 4s^2$. Therefore, the electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$. The given ligand, $H_2O$, is a weak field ligand, which means the crystal field splitting energy ($\Delta_o$) is less than the pairing energy ($P$). In an octahedral field, the five d-electrons will occupy the $t_{2g}$ and $e_g$ orbitals singly before any pairing occurs, leading to a high-spin complex with the configuration $t_{2g}^3 e_g^2$. Thus, there are 5 unpaired electrons.