A coloured solution known to contain two metal ions, was treated with excedss cold sodium hydroxide solution. When filtered a whitish solid, slowly changing to brown, was retained on the filter paper and a colourless solution collected as the filtrate. dropwise addition of hydrochloric acid to the filtrate produced a white ppt. which dissolved in excess acid. treatment of the residue on filter paper with a solution of strong oxidier produced a reddishh-violet solution.. indicate any pairs of ions:

Options

A. `Zn^(2+) and Mn^(2+)` ions
B. `Mg^(2+) and Zn^(2+)` ions
C. `Mn^(2+) and Mg^(2+)` ions
D. `Fe^(2+) and Zn^(2+)` ions

Answer

To solve the problem step by step, we will analyze the reactions and identify the metal ions involved based on the observations given.

### Step 1: Treatment with Sodium Hydroxide
When the colored solution containing two metal ions is treated with excess cold sodium hydroxide (NaOH), a whitish solid is formed that slowly changes to brown.

**Analysis:**
- The whitish solid indicates the formation of a hydroxide precipitate.
- Manganese ions (Mn²⁺) react with NaOH to form manganese hydroxide (Mn(OH)₂), which is a white precipitate.
- Over time, Mn(OH)₂ can oxidize in the presence of air to form manganese dioxide (MnO₂), which is brown.

**Hint:** Look for metal ions that form white precipitates with NaOH and can oxidize to a brown form.

### Step 2: Filtration
The solution is then filtered, retaining the whitish solid (Mn(OH)₂) on the filter paper and collecting a colorless solution as the filtrate.

**Analysis:**
- The colorless filtrate contains the remaining metal ions that did not precipitate.
- In this case, zinc ions (Zn²⁺) remain in solution as they do not form a precipitate with NaOH in excess.

**Hint:** Identify which metal ions remain in solution after filtration.

### Step 3: Addition of Hydrochloric Acid
Next, dropwise addition of hydrochloric acid (HCl) to the filtrate produces a white precipitate that dissolves in excess acid.

**Analysis:**
- The white precipitate formed upon addition of HCl is likely zinc chloride (ZnCl₂) or zinc hydroxide (Zn(OH)₂) which is soluble in excess acid.
- Zinc hydroxide precipitates as a white solid and dissolves in excess HCl to form soluble zinc ions.

**Hint:** Consider the behavior of zinc compounds in acidic solutions.

### Step 4: Treatment of Residue with Strong Oxidizer
The residue on the filter paper (Mn(OH)₂) is treated with a strong oxidizer, producing a reddish-violet solution.

**Analysis:**
- The strong oxidizer likely oxidizes Mn(OH)₂ to permanganate ions (MnO₄⁻), which are known to produce a reddish-violet color in solution.

**Hint:** Think about the oxidation states of manganese and the color changes associated with them.

### Conclusion
Based on the observations and analysis, the metal ions present in the colored solution are:

- **Manganese (Mn²⁺)**: Forms a white precipitate (Mn(OH)₂) that turns brown (MnO₂) upon oxidation.
- **Zinc (Zn²⁺)**: Remains in solution and forms a white precipitate (Zn(OH)₂) that dissolves in excess acid.

### Final Answer
The pairs of ions indicated are **Zinc (Zn²⁺) and Manganese (Mn²⁺)**.