Q.Identify A to E and also explain the reactions involved.
(The Exemplar gives the following reaction scheme:)
$CuCO_3$ gives $CuO$ on one branch and gas (D) on the other.
Left branch: $CuO$, on heating with $CuS$, gives (A); (A) with $HNO_3$ (conc.) gives (B); (B) with $NH_3$ (aq.) gives (C), a blue solution.
Right branch: (D) with $Ca(OH)_2$ gives (E), which appears milky; (E) with $CO_2$ gives $Ca(HCO_3)_2$, a clear solution.

Concept: Inorganic Reaction Pathways (Copper and Carbonate Chemistry)

Step 1 – Thermal decomposition of $CuCO_3$

$CuCO_3 \xrightarrow{\Delta} CuO + CO_2 \uparrow$

So D is $CO_2$.

Step 2 – Left branch (copper series)

$CuO + CuS \xrightarrow{\Delta} 2Cu + SO_2$

Thus A is copper metal ($Cu$).

$Cu + 4HNO_3(\text{conc.}) \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O$

So B is $Cu(NO_3)_2$.

$Cu(NO_3)_2 + 4NH_3(aq.) \rightarrow [Cu(NH_3)_4](NO_3)_2$ (deep blue solution)

Hence C is tetraamminecopper(II) nitrate.

Step 3 – Right branch (carbonate series) …

The key to this problem is recognising that copper carbonate decomposes on heating into two products — a solid and a gas — and that each branch then undergoes a series of characteristic reactions. Let’s walk through each step with the chemistry clearly explained.

1. Thermal decomposition of $CuCO_3$ When copper(II) carbonate is heated, it breaks down into copper(II) oxide and carbon dioxide gas:

This is a common decomposition for metal carbonates. The solid residue is black $CuO$, and the gas evolved is colourless $CO_2$. So the gas (D) is carbon dioxide.

2. Left branch: from $CuO$ to (A) $CuO$ is heated with copper(I) sulfide ($CuS$). This is a redox reaction where copper(II) oxide oxidises the sulfide, and both copper centres are reduced to elemental copper:

The product (A) is copper metal, $Cu$. (The $SO_2$ gas is not part of the scheme here.)

3. Reaction of (A) with conc. $HNO_3$ to give (B) Copper metal reacts vigorously with concentrated nitric acid. The acid acts as both an oxidiser and a source of nitrate ions. The reaction produces copper(II) nitrate, nitrogen dioxide (brown fumes), and water:

So (B) is copper(II) nitrate, $Cu(NO_3)_2$.

4. Reaction of (B) with aqueous $NH_3$ to give (C) When aqueous ammonia is added to a solution of copper(II) nitrate, a pale blue precipitate of copper(II) hydroxide forms first:

On adding excess ammonia, the precipitate dissolves to form the deep blue tetraamminecopper(II) complex:

The overall solution contains $[Cu(NH_3)_4](NO_3)_2$, which is the intense blue solution (C). This is a classic test for copper(II) ions.

5. Right branch: gas (D) with $Ca(OH)_2$ to give (E) …

Method: Reaction Pathway Analysis (Step-by-Step Deduction)

This method involves identifying each compound by working forward from the given starting material, using known chemical reactions and properties.

Steps

1. Start with the decomposition of $CuCO_3$

   • $CuCO_3 \xrightarrow{\Delta} CuO + CO_2 \uparrow$
   • So, gas (D) is $CO_2$.

2. Left branch — from $CuO$ to (A)

   • $CuO$ is heated with $CuS$:

 But here, the product (A) is likely **$Cu$ (copper metal)**.  
 *(Alternative: $CuO + CuS \rightarrow Cu_2S + SO_2$ — but the common Exemplar reaction gives copper metal.)*

3. (A) with conc. $HNO_3$ gives (B)

• $Cu + 4HNO_3 (conc.) \rightarrow Cu(NO_3)_2 + 2NO_2 \uparrow + 2H_2O$
• So, (B) is $Cu(NO_3)_2$.

4. (B) with aqueous $NH_3$ gives (C), a blue solution

   • $Cu(NO_3)_2 + 4NH_3 (aq.) \rightarrow [Cu(NH_3)_4](NO_3)_2$
   • This is the deep blue tetraamminecopper(II) complex.
   • So, (C) is $[Cu(NH_3)_4](NO_3)_2$.

5. Right branch — (D) $CO_2$ with $Ca(OH)_2$ gives (E)

   • $CO_2 + Ca(OH)_2 \rightarrow CaCO_3 \downarrow + H_2O$
   • The milky appearance is due to $CaCO_3$ precipitate.
   • So, (E) is $CaCO_3$.

6. (E) with excess $CO_2$ gives clear solution

   • $CaCO_3 + CO_2 + H_2O \rightarrow Ca(HCO_3)_2$ (soluble, clear)
   • This confirms (E) as $CaCO_3$.

Final Identification

--- …

🧠 Common Mistake #1: Misidentifying Gas (D) from $CuCO_3$ Decomposition

The Mistake:

Students often think $CuCO_3$ decomposes to give $CO$ or a mixture of $CO$ and $CO_2$.

Why it’s wrong:

Copper(II) carbonate decomposes cleanly:

• Gas (D) is $CO_2$, not $CO$.
• $CO$ forms only under reducing conditions (e.g., with carbon), not here.

How to avoid:

Memorise the thermal decomposition pattern for carbonates:

• Most metal carbonates (except Group 1) give metal oxide + $CO_2$.

🧠 Common Mistake #2: Wrong Product (A) from $CuO + CuS$ Reaction

The Mistake:

Students write $CuSO_4$ or $Cu_2O$ as product (A).

Why it’s wrong:

Heating $CuO$ with $CuS$ is a redox reaction where sulfur is oxidised and copper is reduced:

• (A) is copper metal ($Cu$).
• $CuS$ acts as a reducing agent (S²⁻ → S⁴⁺ in $SO_2$).

How to avoid:

Check oxidation states:

• In $CuO$: Cu = +2
• In $CuS$: Cu = +2, S = –2
• Product $Cu$: Cu = 0 → reduction
• $SO_2$: S = +4 → oxidation

If both reactants contain Cu²⁺, the only way to get a new copper compound is by reduction.

🧠 Common Mistake #3: Forgetting $NO_2$ Evolution in Step (A) → (B)

The Mistake:

Students write $Cu + HNO_3$ (conc.) → $Cu(NO_3)_2 + H_2$ or $Cu(NO_3)_2 + NO$.

Why it’s wrong:

Concentrated $HNO_3$ is a strong oxidising agent. With copper:

• (B) is $Cu(NO_3)_2$ (not $CuO$ or $CuOH$).
• Gas evolved is brown $NO_2$, not $H_2$ or $NO$.

How to avoid:

Remember the metal + acid rule changes for oxidising acids:

• Dilute $HNO_3$ → $NO$
• Conc. $HNO_3$ → $NO_2$
• Never $H_2$ with $HNO_3$ (it gets oxidised).

🧠 Common Mistake #4: Wrong Formula for the Blue Solution (C)

The Mistake:

Students write $Cu(OH)_2$ as the product when $NH_3$ (aq.) is added to $Cu(NO_3)_2$.

Why it’s wrong:

Initially, $Cu(OH)_2$ (pale blue precipitate) forms, but excess $NH_3$ dissolves it:

• (C) is the deep blue tetraamminecopper(II) complex, $[Cu(NH_3)_4](NO_3)_2$.

How to avoid:

Always check if the reagent is in excess. Aqueous $NH_3$ first gives a precipitate, then a soluble complex — the question says “blue solution”, so it’s the complex.

🧠 Common Mistake #5: Confusing (E) with $CaCO_3$ Only

The Mistake:

Students write (E) = $CaCO_3$ and stop there.

Why it’s incomplete:

$CO_2$ with $Ca(OH)_2$ gives milky $CaCO_3$: …