NEET Practice Questions (MCQs) with Answers & Solutions

When a salt containing iodide ions ($I^-$) is treated with concentrated sulfuric acid ($H_2SO_4$), violet vapors of iodine ($I_2$) are produced. This reaction is specific to iodides and is a common test to identify the presence of iodide ions in a salt.

When water is added to a salt solution containing chloroform, and the chloroform layer turns violet, it indicates the presence of iodide ions ($I^-$). The iodide ions react with chloroform in the presence of water to form iodine, which gives the violet color.

When phosphate radicals react with ammonium molybdate, they form a canary yellow precipitate of ammonium phosphomolybdate. This is a characteristic color used to confirm the presence of phosphate ions.

Silver chloride (AgCl) is soluble in ammonium hydroxide (NH4OH). The solubility is due to the formation of the complex ion [Ag(NH3)2]+ when AgCl reacts with NH4OH. The reaction is as follows: $$ AgCl + 2NH_4OH → [Ag(NH_3)_2]Cl + 2H_2O $$ This complex ion keeps the silver ion in solution, making AgCl soluble in NH4OH.

Nitrates of all metals are soluble in water. This is a general property of nitrates, making them easily dissociate into their respective ions when dissolved in water. This is due to the strong ionic nature of nitrate salts and the ability of water to stabilize the ions formed.

The brown ring test for nitrates is due to the formation of a complex between ferrous sulfate and nitric oxide. When a nitrate is mixed with iron(II) sulfate and concentrated sulfuric acid is added, a brown ring forms at the interface. The brown color is due to the formation of the complex ion $[FeSO_4NO]$. The reaction involved is: $$ Fe^{2+} + NO + SO_4^{2-} → [FeSO_4NO] $$ This complex ion gives the characteristic brown color observed in the nitrate confirmation test.

When an aqueous solution of a salt containing chloride ions ($Cl^-$) is treated with silver nitrate ($AgNO_3$), a white precipitate of silver chloride ($AgCl$) is formed. This white precipitate dissolves in ammonium hydroxide ($NH_4OH$) forming a complex ion. The reaction is as follows:

$Cl^- + AgNO_3 → AgCl (white ppt.) + NO_3^-$

$AgCl + 2NH_4OH → [Ag(NH_3)_2]^+ + Cl^- + 2H_2O$

When carbon dioxide ($CO_2$) is passed into lime water (a solution of calcium hydroxide, $Ca(OH)_2$), it forms calcium carbonate ($CaCO_3$), which is insoluble in water and makes the solution appear milky. The chemical reaction is:

$Ca(OH)_2 + CO_2 → CaCO_3 (milky) + H_2O$

When an excess of $CO_2$ is passed, the calcium carbonate reacts with the $CO_2$ and water to form calcium bicarbonate ($Ca(HCO_3)_2$), which is soluble in water, hence the milkiness disappears. The reaction is:

$CaCO_3 + CO_2 + H_2O → Ca(HCO_3)_2$

When a mixture containing bromide ions ($Br^-$) is heated with concentrated sulfuric acid ($H_2SO_4$) and manganese dioxide ($MnO_2$), brown fumes of bromine ($Br_2$) are formed. The reaction is:

$2KBr + 2H_2SO_4 + MnO_2 → Br_2 (brown fumes) + MnSO_4 + K_2SO_4 + 2H_2O$

This indicates that the bromide ion ($Br^-$) is responsible for the formation of brown fumes.