To determine which compound is not soluble in sodium hydrogen carbonate (NaHCO$_3$), we need to consider the acidity of each compound.$\\$Sodium hydrogen carbonate is a weak base, and it will react with acids that are stronger than itself.$\\$The reaction is typically:$\\$Acid$+$NaHCO$_3 \rightarrow$Salt$+$CO$_2 +$H$_2$O$\\$Let's analyze each option:$\\$• Option 1: 2,4,6-Trinitrophenol (Picric acid)$\\$This is a very strong acid due to the presence of electron-withdrawing nitro groups.$\\$It will react with NaHCO$_3$ and is soluble.$\\$• Option 2: Benzoic acid$\\$Benzoic acid is a moderately strong acid and will react with NaHCO$_3$ to form sodium benzoate,$\\$which is soluble.$\\$• Option 3: o-Nitrophenol$\\$o-Nitrophenol is a weak acid. The presence of a single nitro group does not make it acidic enough$\\$to react with NaHCO$_3$. Therefore, it will not be soluble.$\\$• Option 4: Benzenesulphonic acid$\\$This is a strong acid due to the presence of the sulfonic group, which is highly electron-withdrawing.$\\$It will react with NaHCO$_3$ and is soluble.$\\$Therefore, the compound that will not be soluble in sodium hydrogen carbonate is o-nitrophenol.$\\$This corresponds to Option 3.