In the circuit shown (figure), what is the change of total electrical energy stored in the capacitors when the key is pressed?

Initially, all three capacitors are in parallel
${\mathrm{E}}_{\mathrm{i}}=\frac{1}{2}\times 3{\mathrm{CV}}^2=\frac{3}{2}{\mathrm{CV}}^2$
When key is closed, two capacitors are in series.
$\begin{array}{l}{\mathrm{E}}_{\mathrm{f}}=\frac{1}{2}\mathrm{C}{\left(\frac{\mathrm{V}}{2}\right)}^2\times 2=\frac{{\mathrm{CV}}^2}{4}\\ \mathrm{\Delta E}={\mathrm{E}}_{\mathrm{i}}-{\mathrm{E}}_{\mathrm{f}}=\frac{5}{4}{\mathrm{CV}}^2\end{array}$