`Br_(2)` water in `NaOH` soin .Oxidises `Mn(OH)_(2)` to a…. Ppt due to the formation of `MnO(2)`

Options

A. Black
B. violet
C. Blue
D. white

Answer

To solve the given question step by step, we will analyze the reaction of bromine water in sodium hydroxide (NaOH) with manganese hydroxide (Mn(OH)₂) and identify the resulting precipitate.

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Bromine water (Br₂ in water)
- Sodium hydroxide (NaOH)
- Manganese hydroxide (Mn(OH)₂)

### Step 2: Understand the Reaction
When bromine water is added to a solution containing manganese hydroxide in sodium hydroxide, an oxidation reaction occurs. Bromine acts as an oxidizing agent.

### Step 3: Write the Reaction Equation
The reaction can be represented as:
\[ \text{Mn(OH)}_2 + \text{Br}_2 + 2 \text{NaOH} \rightarrow \text{MnO}_2 + 2 \text{NaBr} + 2 \text{H}_2\text{O} \]

### Step 4: Analyze the Products
In this reaction:
- Manganese hydroxide (Mn(OH)₂) is oxidized to manganese dioxide (MnO₂).
- Manganese dioxide (MnO₂) is known to form a precipitate.

### Step 5: Identify the Color of the Precipitate
Manganese dioxide (MnO₂) typically forms a black precipitate. Therefore, the precipitate formed in this reaction is black.

### Step 6: Conclusion
The question asks for the color of the precipitate formed due to the oxidation of Mn(OH)₂ by bromine water in NaOH solution. The answer is that it forms a black precipitate.

### Final Answer
The precipitate formed is **Black**.

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