In IV th group the ppt of `Mn(OH)_(2)` in excess of `NaOH` , turns brown or blue in air due to the formation of

Options

A. `MnO_(2).xH_(2)O`
B. `MnO_(2)`
C. `MnO_(2).H_(2)O`
D. All

Answer

To solve the question regarding the behavior of manganese hydroxide in excess sodium hydroxide, we can break it down into the following steps:

### Step 1: Identify the Compound
In the fourth group of qualitative analysis, we deal with manganese ions (Mn²⁺). When manganese ions react with sodium hydroxide (NaOH), manganese hydroxide (Mn(OH)₂) precipitates out.

### Step 2: Understand the Reaction with Air
When manganese hydroxide (Mn(OH)₂) is exposed to air, it undergoes oxidation. The oxygen present in the air reacts with manganese hydroxide.

### Step 3: Formation of Manganese Oxides
The oxidation of manganese hydroxide leads to the formation of manganese oxides. Specifically, manganese dioxide (MnO₂) is formed as a result of this reaction. The manganese hydroxide can also form hydrated manganese oxides, which can be represented as MnO₂·xH₂O, where x indicates the number of water molecules associated with the oxide.

### Step 4: Color Change Observation
As manganese hydroxide is oxidized to manganese dioxide in the presence of air, the color of the precipitate changes. Manganese hydroxide is typically a pale pink color, but upon oxidation, it can turn brown or blue depending on the specific conditions and the presence of water.

### Conclusion
Thus, the answer to the question is that the precipitate of Mn(OH)₂ in excess NaOH turns brown or blue in air due to the formation of manganese oxide (MnO₂).

### Final Answer
The precipitate of Mn(OH)₂ in excess NaOH turns brown or blue in air due to the formation of manganese oxide (MnO₂).

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