The oxidation state of iodine in compound which is obtained by heating `HIO_3` at `170^@C` is +n.
Give value of n

Explanation

To determine the oxidation state of iodine in the compound formed by heating HIO3 at 170°C, we follow these steps:

### Step 1: Identify the compound formed
When HIO3 (iodic acid) is heated at 170°C, it decomposes to form I2O5 (iodine pentoxide).

### Step 2: Write the formula for the compound
The formula for the compound is I2O5.

### Step 3: Set up the oxidation state equation
Let the oxidation state of iodine (I) in I2O5 be represented by \( x \). Since there are 2 iodine atoms in the compound, the total contribution from iodine will be \( 2x \).

### Step 4: Consider the oxidation state of oxygen
Oxygen typically has an oxidation state of -2. In I2O5, there are 5 oxygen atoms, contributing a total of \( 5 \times (-2) = -10 \).

### Step 5: Set up the equation based on the overall charge
The overall charge of the compound I2O5 is neutral (0), so we can set up the equation:
\[ 2x + (-10) = 0 \]

### Step 6: Solve for \( x \)
Rearranging the equation gives:
\[ 2x - 10 = 0 \]
\[ 2x = 10 \]
\[ x = 5 \]

### Conclusion
The oxidation state of iodine in I2O5 is +5. Therefore, the value of \( n \) is 5.

### Final Answer
The value of \( n \) is 5.

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