Answer
$\therefore \text{Correct name of the compound $Hg_2CrO_4$ is } \textbf{Mercury(I) chromate}$
Work Step by Step
Charge on Mercury (i.e Hg) is $\textbf{+1}$
Charge on chromate (i.e $CrO_4$ ) is$\textbf{-2}$
$\therefore \text{Correct name of the compound $Hg_2CrO_4$ is } \textbf{Mercury(I) chromate}$
