For `1.34xx10^(-3)` moles of `KBrO_3` to reduce into bromide `4.02 xx 10^(-3)` mole of `X^(n+)` ion is needed. New oxidation state of X is:

Options

A. `n+2`
B. `n-2`
C. 2
D. `-2`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the oxidation state of bromine in KBrO₃.
In KBrO₃, the oxidation state of bromine (Br) can be calculated as follows:
- The oxidation state of oxygen (O) is -2.
- There are three oxygen atoms, contributing a total of -6.
- Let the oxidation state of bromine be \( x \).

The equation can be set up as:
\[ x + 3(-2) = -1 \]
\[ x - 6 = -1 \]
\[ x = +5 \]

### Step 2: Calculate the change in oxidation state for bromine.
The bromine in KBrO₃ is reduced to bromide (Br⁻), which has an oxidation state of -1. The change in oxidation state is:
\[ \text{Change in oxidation state} = \text{Final state} - \text{Initial state} = -1 - (+5) = -6 \]

### Step 3: Determine the n-factor for KBrO₃.
The n-factor for KBrO₃ (or BrO₃⁻) is equal to the change in oxidation state, which we calculated as 6.

### Step 4: Set up the relationship between moles and equivalents.
We know that the equivalents of KBrO₃ will equal the equivalents of Xⁿ⁺. The formula for equivalents is:
\[ \text{Equivalents} = \text{Number of moles} \times \text{n-factor} \]

For KBrO₃:
\[ \text{Equivalents of KBrO₃} = 1.34 \times 10^{-3} \text{ moles} \times 6 \]

For Xⁿ⁺:
\[ \text{Equivalents of Xⁿ⁺} = 4.02 \times 10^{-3} \text{ moles} \times (n_{\text{final}} - n) \]

### Step 5: Set the equivalents equal to each other.
Now we can set the two equations equal to each other:
\[ 1.34 \times 10^{-3} \times 6 = 4.02 \times 10^{-3} \times (n_{\text{final}} - n) \]

### Step 6: Solve for \( n_{\text{final}} - n \).
Calculating the left side:
\[ 1.34 \times 6 = 8.04 \]
So,
\[ 8.04 \times 10^{-3} = 4.02 \times 10^{-3} \times (n_{\text{final}} - n) \]

Dividing both sides by \( 4.02 \times 10^{-3} \):
\[ \frac{8.04 \times 10^{-3}}{4.02 \times 10^{-3}} = n_{\text{final}} - n \]
\[ 2 = n_{\text{final}} - n \]

### Step 7: Find the new oxidation state of X.
This means:
\[ n_{\text{final}} = n + 2 \]

### Conclusion:
The new oxidation state of X is \( n + 2 \).