NEET Practice Questions (MCQs) with Answers & Solutions

In the test for sulphate ion ($SO_4^{2-}$), the white precipitate formed, which is barium sulphate ($BaSO_4$), is insoluble in concentrated acids like HCl, $H_2SO_4$, and $HNO_3$. Therefore, the correct answer is 'None of these'.

Lead(II) nitrate ($Pb(NO_3)_2$) is used to remove sulphate ions ($SO_4^{2-}$) and chloride ions ($Cl^-$) from a solution. This is because $Pb^{2+}$ ions react with $SO_4^{2-}$ and $Cl^-$ ions to form insoluble lead(II) sulphate ($PbSO_4$) and lead(II) chloride ($PbCl_2$) respectively, which precipitate out of the solution.

When potassium iodide (KI) is heated with concentrated $H_2SO_4$, iodine ($I_2$) is formed. The reaction is as follows:

$$ 2KI + 2H_2SO_4 ightarrow I_2 + SO_2 + 2H_2O + K_2SO_4 $$

The iodine can be observed as a violet vapor or solid.

The chromyl chloride test is specifically used to confirm the presence of chloride ions (Cl⁻) in a sample. When a chloride salt is heated with potassium dichromate and concentrated sulfuric acid, a red vapor of chromyl chloride (CrO2Cl2) is produced, indicating the presence of chloride ions.

When ammonia (NH3) reacts with excess chlorine (Cl2), nitrogen trichloride (NCl3) and hydrochloric acid (HCl) are formed. The reaction is: $$ 8 NH_3 + 3 Cl_2 ightarrow N_2 + 6 NH_4Cl + NCl_3 $$ However, in the presence of excess chlorine, $$ NH_3 + 3 Cl_2 ightarrow NCl_3 + 3 HCl $$

The anion $S_2^-$ (sulfide) can decolourise acidified $KMnO_4$ solution as it acts as a reducing agent. The reaction involves the reduction of $KMnO_4$ (potassium permanganate) to $Mn^{2+}$ while the sulfide ion is oxidised to sulfur. The purple colour of $KMnO_4$ fades away as it is reduced.

The gas NO (nitric oxide) is absorbed by ferrous sulphate solution ($FeSO_4$) giving a blackish brown colour due to the formation of a complex compound, nitroso ferrous sulphate ($FeSO_4.NO$). This reaction is characteristic of NO gas.

The anion $NO_3^-$ (nitrate) is not easily removed from aqueous solutions by precipitation because it forms very soluble salts with most cations. Unlike anions such as $Cl^-$, $CO_3^{2-}$, and $SO_4^{2-}$ which can form insoluble precipitates with certain cations, $NO_3^-$ remains in solution.