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 need to determine the new oxidation state of the ion \(X^{n+}\) based on the reduction of \(KBrO_3\) to bromide. Here’s a step-by-step solution:

### Step 1: Determine the oxidation state of bromine in \(KBrO_3\)

In \(KBrO_3\), the oxidation state of bromine (Br) can be calculated as follows:

- Let the oxidation state of Br be \(x\).
- The oxidation state of oxygen (O) is \(-2\).
- There are three oxygen atoms, so their total contribution is \(3 \times (-2) = -6\).
- The overall charge of the \(BrO_3^-\) ion is \(-1\).

Setting up the equation:
\[ x + (-6) = -1 \]
\[ x - 6 = -1 \]
\[ x = +5 \]

Thus, the oxidation state of bromine in \(KBrO_3\) is +5.

### Step 2: Determine the change in oxidation state of bromine

When \(KBrO_3\) is reduced to bromide (\(Br^{-}\)), the oxidation state of bromine changes from +5 to -1.

The change in oxidation state is:
\[ \Delta \text{Oxidation State} = +5 - (-1) = +5 + 1 = +6 \]

### Step 3: Calculate the n-factor for bromine

The n-factor for a substance in a redox reaction is defined as the total number of electrons gained or lost per formula unit. Since bromine goes from +5 to -1, it gains 6 electrons.

Thus, the n-factor of \(BrO_3^{-}\) is 6.

### Step 4: Use the given moles to find the equivalent of \(KBrO_3\)

We have:
- Moles of \(KBrO_3 = 1.34 \times 10^{-3}\)
- n-factor of \(KBrO_3 = 6\)

The total equivalents of \(KBrO_3\) can be calculated as:
\[ \text{Equivalents of } KBrO_3 = \text{Moles} \times \text{n-factor} = 1.34 \times 10^{-3} \times 6 = 8.04 \times 10^{-3} \text{ equivalents} \]

### Step 5: Determine the equivalents of \(X^{n+}\)

Let the n-factor of \(X^{n+}\) be \(n_f\). We know that:
\[ \text{Equivalents of } X^{n+} = \text{Moles of } X^{n+} \times n_f \]
Given that moles of \(X^{n+} = 4.02 \times 10^{-3}\), we can write:
\[ \text{Equivalents of } X^{n+} = 4.02 \times 10^{-3} \times n_f \]

### Step 6: Set the equivalents equal to each other

Since the equivalents of \(KBrO_3\) and \(X^{n+}\) must be equal:
\[ 8.04 \times 10^{-3} = 4.02 \times 10^{-3} \times n_f \]

### Step 7: Solve for \(n_f\)

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

### Step 8: Determine the new oxidation state of \(X\)

Since the n-factor \(n_f\) represents the change in oxidation state, and we know that \(X^{n+}\) has gained electrons, we can find the new oxidation state of \(X\):
\[ \text{New oxidation state of } X = n - n_f = n - 2 \]
Assuming the original oxidation state of \(X\) was \(n\), the new oxidation state is \(n - 2\).

### Final Answer

The new oxidation state of \(X\) is \(n - 2\).

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