In `KMnO_(2)` on heating to red hot gives

Options

A. `+2`
B. `+4`
C. `+6`
D. `+7`

Answer

To solve the question regarding the decomposition of potassium permanganate (KMnO₂) upon heating, we will follow these steps:

### Step 1: Write the decomposition reaction
When potassium permanganate (KMnO₂) is heated to red hot, it decomposes. The balanced chemical equation for this reaction is:

\[ 2 \, KMnO_4 \, \xrightarrow{\text{heat}} \, K_2MnO_4 + O_2 \]

### Step 2: Identify the products
From the balanced equation, we can see that the products formed are potassium manganate (K₂MnO₄) and oxygen gas (O₂).

### Step 3: Determine the oxidation state of manganese in K₂MnO₄
To find the oxidation state of manganese (Mn) in potassium manganate (K₂MnO₄), we can use the following formula:

Let the oxidation state of Mn be \( x \).

The oxidation state of potassium (K) is +1, and there are 2 potassium atoms, contributing +2 to the overall charge. The oxidation state of oxygen (O) is -2, and there are 4 oxygen atoms, contributing -8 to the overall charge.

Setting up the equation for the overall charge, we have:

\[ 2(+1) + x + 4(-2) = 0 \]

### Step 4: Solve for x
Now we can simplify the equation:

\[ 2 + x - 8 = 0 \]

This simplifies to:

\[ x - 6 = 0 \]

Thus, we find:

\[ x = +6 \]

### Step 5: Conclusion
The oxidation state of manganese in K₂MnO₄ is +6. Therefore, the answer to the question is that upon heating KMnO₂, it gives K₂MnO₄ with manganese having an oxidation state of +6.