The 'd' electronic configuration of $\left[\mathrm{Ru}(\mathrm{en})_3\right] \mathrm{Cl}_2$ and $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_2$ respectively are _____ and ____ . [Given: Atomic number of $\mathrm{Ru}=44$ and $\mathrm{Fe}=26$ ]

For $\left[\mathrm{Ru}(\mathrm{en})_3\right] \mathrm{Cl}_2$, the oxidation state of $\mathrm{Ru}$ is $+2$. The atomic number of $\mathrm{Ru}$ is $44$, so its electronic configuration is $[\mathrm{Kr}] 4d^7 5s^1$. For $\mathrm{Ru}^{2+}$, the configuration is $4d^6$. Since $\mathrm{Ru}$ is a $4d$ transition metal, it forms low-spin complexes regardless of the ligand field strength. Thus, the $d^6$ configuration in an octahedral field results in $t_{2g}^6 e_g^0$.

For $\left[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6\right] \mathrm{Cl}_2$, the oxidation state of $\mathrm{Fe}$ is $+2$. The atomic number of $\mathrm{Fe}$ is $26$, so its electronic configuration is $[\mathrm{Ar}] 3d^6 4s^2$. For $\mathrm{Fe}^{2+}$, the configuration is $3d^6$. Since $\mathrm{H}_2\mathrm{O}$ is a weak field ligand, the complex is high-spin. The $d^6$ configuration in a high-spin octahedral field is $t_{2g}^4 e_g^2$.

Therefore, the configurations are $t_{2g}^6 e_g^0$ and $t_{2g}^4 e_g^2$.

Answer: $t_{2 g}{ }^6 e_g^0 ; t_{2 g}{ }^4 e_g^2$