A

Step 1: When a dielectric slab is inserted into capacitor 1, the capacitance of capacitor 1 increases. In a series circuit, the voltage across each capacitor is proportional to its inverse capacitance. Thus, capacitor 2 would see an increase in charge as a result of this change in capacitance in capacitor 1.
Therefore, the effect on capacitor 2 is that the charge on capacitor increases.

Step 2: For change (ii), if the separation between the plates of capacitor 1 is increased, the capacitance decreases, which leads to a decrease in charge on capacitor 2. Hence, the corresponding effect is that the charge on capacitor 2 decreases.

Step 3: For change (iii), inserting a metal plate connecting both plates of capacitor 1 effectively creates a short circuit for capacitor 1, causing the charge on capacitor 2 to drop significantly or be eliminated altogether. This change would correspond to the energy stored in capacitor 2 decreasing.

Step 4: For change (iv), if the separation between the plates of capacitor 1 is decreased, the capacitance increases leading to an initial increase in charge on capacitor 2. Thus, the electric field between the plates of capacitor 2 would increase as a result of this change.
Therefore, the correct matches are:
(i)-[A], (ii)-[B], (iii)-[C], (iv)-[D].

Therefore, the final matched answer is:
(i)-[A], (ii)-[B], (iii)-[C], (iv)-[D].