The formula of the following compound: Mercury (II) chloride is :

We are aware of the procedures for creating a compound's formula, which include writing the symbols of the involved radicals side by side, with their valencies at the top, the positive radical to the left and the negative radical to the right, and their valencies at the top. Second, to get a straightforward ratio of the valencies of combining radicals, we must eliminate the common component, if there is any. Finally, we cross the valencies, or move them across to the lower right of the radical.

In the given compound, the oxidation state of Hg is II.

We know that the valency of Cl is 1.

Mercury is in +2 state and Chlorine is in −1 state.

${\mathrm{Hg}}^{+}+2{\mathrm{Cl}}^{-}\to {\mathrm{HgCl}}_2$

Thus the molecular formula of the compound will be Hg(II)Cl₂

$HgC{l}_2$

​Hence, option B is correct.