The $\Delta_0$ of a coordination complex of a metal ion $\left(3 d^1\right)$ is $1000 \mathrm{~kJ}…
The $\Delta_0$ of a coordination complex of a metal ion $\left(3 d^1\right)$ is $1000 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If the energy of $t_{2 g}$ orbitals is $-400 \mathrm{~kJ} \mathrm{~mol}^{-1}$, the energy (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) of $e_g$ orbitals is
-600
600
1000
400
Solution
Given, $\Delta_0$ for $\left(3 d^1\right)$ :
$
\Delta_0=1000 \mathrm{~kJ} \mathrm{~mol}^{-1}
$
Energy of $\left(t_{2 g}\right)$, i.e. $E\left(t_{2 g}\right)$.
$
\begin{aligned}
E\left(t_{2 g}\right) & =-400 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
\Delta_0 & =E_{\left(e_g\right)}-E\left(t_{2 g}\right)
\end{aligned}
$
(Energy difference) (Higher energy) (Lower energy)
where, $E_{\left(e_g\right)}$ is the energy of orbitals. Therefore,
or,
$
\begin{aligned}
& 1000=E_{\left(e_g\right)}-(-400) \\
& 1000=E_{\left(e_g\right)}+400 \\
& E_{\left(e_g\right)}=1000-400 \\
& E_{\left(e_g\right)}=600 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$
Hence, option (b) is the correct answer