Q.Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by the lead acetate test?

The key concept here is metallurgical extraction sequences. Each metal has a specific set of chemical reactions used to purify it from its ore. Copper extraction from chalcopyrite or copper glance involves roasting, smelting, and converting — but never uses calcium oxide. The reactions in (A), (B), and (C) are all directly part of the copper extraction process, while (D) is a slag-forming reaction from the iron extraction process.

Let’s go through each option step by step.

1. Option (A):

   $2\text{Cu}_2\text{S} + 3\text{O}_2 \rightarrow 2\text{Cu}_2\text{O} + 2\text{SO}_2$

   This is the roasting of copper(I) sulfide (from ores like chalcocite). It converts sulfide to oxide and removes sulfur as SO₂. This is a core step in copper extraction.

2. Option (B):

   $2\text{FeS} + 3\text{O}_2 \rightarrow 2\text{FeO} + 2\text{SO}_2$

   Copper ores often contain iron sulfides (e.g., chalcopyrite CuFeS₂). During roasting, iron sulfide is oxidized to iron(II) oxide. This is essential to later remove iron as slag. So this reaction is part of copper extraction.

3. Option (C):

   $\text{FeO} + \text{SiO}_2 \rightarrow \text{FeSiO}_3$

   In copper smelting, silica (SiO₂) is added as a flux to combine with FeO, forming iron silicate slag (FeSiO₃). This removes iron impurities from the molten copper. So this reaction is part of copper extraction.

4. Option (D):

   $\text{SiO}_2 + \text{CaO} \rightarrow \text{CaSiO}_3$ …

The question asks which compound, when reacted with HCl, produces the most number of different oxides (as gases). We need to consider the products of each reaction, focusing on the gaseous oxides formed.

Concept & Intuition:

When a salt of a weak acid reacts with a strong acid like HCl, the weak acid is displaced. If the weak acid is unstable, it decomposes into an oxide (often a gas) and water. The "number of oxides" here means the distinct chemical species that are oxides (e.g., CO₂, SO₂, NO₂). Some reactions may yield more than one oxide if the salt contains an element that can exist in multiple oxidation states in its gaseous products.

Step-by-step reasoning:

1. Reaction of Na₂CO₃ with HCl Sodium carbonate reacts with HCl to give carbonic acid, which immediately decomposes:

Only one oxide is produced: CO₂.

2. Reaction of NaNO₂ with HCl Sodium nitrite reacts with HCl to form nitrous acid (HNO₂), which is unstable and decomposes in multiple ways:
   • Primary decomposition:

 This yields **two oxides**: NO (nitric oxide) and NO₂ (nitrogen dioxide).  

• Additionally, in the presence of excess HCl, some NO₂ can further react, but the key point is that at least two distinct nitrogen oxides are formed directly from the decomposition. Thus, NaNO₂ gives two different oxides.

3. Reaction of Na₂SO₃ with HCl Sodium sulfite reacts with HCl to give sulfurous acid, which decomposes: $\text{Na}_2\text{SO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{SO}_2 \uparrow$ …

The question tests your knowledge of the everyday uses of three important sodium compounds. Each of these uses is a direct consequence of the compound’s chemical properties — not something to memorise blindly, but something to connect to the chemistry you already know.

Let’s examine each statement one by one.

1. NaHCO₃ – used in fire extinguisher Sodium bicarbonate decomposes on heating to give CO₂ gas:

The CO₂ produced is heavier than air and smothers the fire by cutting off the oxygen supply. This is the principle behind soda-acid fire extinguishers. Statement I is correct.

2. Na₂CO₃ – water softening Hard water contains Ca²⁺ and Mg²⁺ ions. Sodium carbonate (washing soda) precipitates these as insoluble carbonates:

This removes the hardness permanently. Statement II is correct.

3. NaOH – purification of bauxite …

Concept and Intuition

Thermal decomposition reactions often produce gases, and the identity of those gases depends on the stability of the anion and the metal’s reactivity. For dinitrogen (N₂) to be released, the compound must contain nitrogen in a form that can pair up into N≡N bonds upon heating. Nitrates (NO₃⁻) tend to decompose to oxygen and nitrogen dioxide (or nitrites), not N₂. Azides (N₃⁻) are famous for yielding N₂ because the azide ion is inherently unstable. Ammonium dichromate is a special case: the ammonium ion (NH₄⁺) acts as a reducing agent, and the dichromate (Cr₂O₇²⁻) oxidizes it, producing N₂, water, and chromium(III) oxide. So we check each compound’s decomposition pathway.

Step-by-step reasoning

1. Sodium nitrate (NaNO₃)
   • Sodium is a highly reactive metal (Group 1). On heating, sodium nitrate decomposes to sodium nitrite and oxygen:

• No nitrogen-containing gas is produced; the nitrogen remains in the nitrite ion. Thus, no dinitrogen is liberated.

2. Ammonium dichromate ((NH₄)₂Cr₂O₇)
   • This is a classic “volcano” reaction. The ammonium ion (oxidation state –3) is a strong reducing agent, and dichromate (Cr⁺⁶) is a strong oxidizer. Upon heating, they undergo an internal redox reaction:

• The nitrogen in NH₄⁺ is oxidized from –3 to 0 (N₂), while chromium is reduced from +6 to +3. Dinitrogen gas is clearly liberated.

3. Barium azide (Ba(N₃)₂) …

The key to this question lies in understanding how the alkali metal nitrates behave differently on heating. Lithium, being the smallest alkali metal, has a significantly different chemistry from its heavier cousins — it behaves more like magnesium in many ways. Sodium, potassium, rubidium, and caesium follow a different decomposition pattern.

When you heat a nitrate, the central nitrogen atom is already in its +5 oxidation state. The question is: does the metal cation force the nitrate to break down all the way to the oxide, or can it stop at the nitrite stage? That depends on the polarising power of the cation.

Lithium ion is tiny and highly polarising — it pulls electron density so strongly that the nitrate ion cannot hold together. It decomposes completely, giving lithium oxide, nitrogen dioxide, and oxygen. Sodium ion is larger and less polarising; its nitrate merely loses one oxygen atom to form the nitrite, releasing only oxygen gas.

Let’s work through each case.

1. Lithium nitrate decomposition Lithium nitrate ($\text{LiNO}_3$) on strong heating gives:

Count the products: lithium oxide ($\text{Li}_2\text{O}$), nitrogen dioxide ($\text{NO}_2$), and oxygen ($\text{O}_2$) — that’s three distinct products.

2. Sodium nitrate decomposition Sodium nitrate ($\text{NaNO}_3$) on heating gives:

The products are sodium nitrite ($\text{NaNO}_2$) and oxygen ($\text{O}_2$) — that’s two distinct products. …

The question describes the classic brown ring test for nitrate ions — a staple in qualitative inorganic analysis. When dilute ferrous sulphate is added to a solution containing nitrate, and then concentrated sulphuric acid is carefully poured down the side of the test tube, a brown ring appears at the junction of the two liquids. That ring is not a simple oxide of nitrogen floating around; it is a coordination complex. But the oxide that forms first, right there in the solution, is what we need to identify.

The key insight: the nitrate ion ($\ce{NO3-}$) is a poor oxidising agent in dilute acid, but concentrated sulphuric acid provides a strongly acidic medium that reduces nitrate to nitric oxide (NO). The Fe(II) ions reduce the nitrate, and in the process get oxidised to Fe(III) — but the NO that is produced immediately binds to excess Fe(II) to form a stable, coloured complex.

Let’s walk through the chemistry step by step.

1. The role of concentrated sulphuric acid.

   Conc. $\ce{H2SO4}$ is added carefully so it forms a separate layer below the aqueous mixture. It does two things: it provides a high concentration of $\ce{H+}$ ions, and it dehydrates the medium, making the reduction of nitrate possible. Without this strong acid, the reaction is too slow to be observed.

2. Reduction of nitrate to nitric oxide.

   In the strongly acidic environment, the nitrate ion is reduced by ferrous ions:

The product is nitric oxide (NO), a neutral gas. This is the oxide that forms "in situ" — right there in the test tube, not added from outside.

3. Formation of the brown ring complex. The freshly formed NO is not free for long. It coordinates with excess $\ce{Fe^{2+}}$ ions (which are present in the dilute FeSO$_4$ solution) to form the brown-coloured nitrosyl complex: $\ce{[Fe(H2O)6]^{2+} + NO -> [Fe(H2O)5(NO)]^{2+} + H2O}$ …

Concept & Intuition

Gold is a “noble” metal — it resists oxidation by either HCl or HNO₃ alone. But when you mix them in a 3:1 ratio (aqua regia), you get a powerful one-two punch:

• HNO₃ oxidizes gold to Au³⁺.
• HCl provides Cl⁻ ions that immediately complex with Au³⁺ to form the very stable $\mathrm{AuCl_4^-}$ ion, pulling the equilibrium forward. Meanwhile, the HNO₃ is reduced. In concentrated acid, the typical reduction product is NO (not NO₂, which forms in dilute conditions). The reaction also produces water and sometimes traces of nitrosyl chloride (NOCl), but the net ionic equation gives NO as the main gaseous product.

Let’s walk through the reasoning step by step.

1. Identify the oxidizing and complexing agents

   • HNO₃ is the oxidizer: it takes electrons from Au.
   • HCl supplies Cl⁻, which binds Au³⁺ tightly as $\mathrm{AuCl_4^-}$ (stability constant ~10²⁵). This lowers the effective concentration of Au³⁺, making the oxidation thermodynamically favorable.

2. Write the half-reactions

   • Oxidation: $\mathrm{Au(s) \rightarrow Au^{3+} + 3e^-}$
   • Reduction of nitrate in concentrated acid: $\mathrm{NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O}$ (This is the standard reduction in strong acid; NO₂ forms only when HNO₃ is dilute or heated.)

3. Combine and add chloride

   • Balance electrons: multiply the gold oxidation by 1 and the nitrate reduction by 1 (both involve 3 electrons).
   • Net ionic (before complexation): $\mathrm{Au + NO_3^- + 4H^+ \rightarrow Au^{3+} + NO + 2H_2O}$
   • Now add Cl⁻: Au³⁺ immediately forms $\mathrm{AuCl_4^-}$: $\mathrm{Au^{3+} + 4Cl^- \rightarrow AuCl_4^-}$
   • Overall: $\mathrm{Au + NO_3^- + 4H^+ + 4Cl^- \rightarrow AuCl_4^- + NO + 2H_2O}$

4. Check the options …