($i$)

$C^{\prime}=\frac{\varepsilon_0 A}{d^{\prime}}, C=\frac{\varepsilon_0 A}{d}$

$d^{\prime}>d$

$ C^{\prime}>C$

Hence, final capacitance greater than initial capacitance,

($ii$)

$U^{\prime}=\frac{1}{2} C^{\prime} V^2$

$U=\frac{1}{2} C V^2$

$U^{\prime}>U$

Hence final energy is greater than initial energy

($iii$)

$\frac{Q^{\prime}}{V^{\prime}}=C^{\prime} \text { and } \frac{Q}{V}=C$

$\frac{Q^{\prime}}{V^{\prime}} \neq \frac{Q}{V}$

($iv$) Product of charge and voltage

$X^{\prime}=Q^{\prime} V=C^{\prime} V^2$

$X=Q V=C V^2$

$X^{\prime}>X$