Oxidation number of Xe in `XeF_(5)^(-)` is :

Options

A. `+1`
B. `+2`
C. `+3`
D. `+4`

Answer

To find the oxidation number of xenon (Xe) in the ion \(XeF_5^-\), we can follow these steps:

### Step 1: Assign the oxidation number to xenon
Let the oxidation number of xenon be \(x\).

### Step 2: Determine the oxidation number of fluorine
The oxidation number of fluorine (F) is always \(-1\).

### Step 3: Set up the equation based on the overall charge
The ion \(XeF_5^-\) has an overall charge of \(-1\). Since there are 5 fluorine atoms, we can express the total contribution of fluorine to the oxidation state as:
\[ 5 \times (-1) = -5 \]

### Step 4: Write the equation for the total oxidation state
Now, we can write the equation based on the oxidation states:
\[ x + (-5) = -1 \]

### Step 5: Solve for \(x\)
Rearranging the equation gives:
\[ x - 5 = -1 \]
Adding 5 to both sides:
\[ x = -1 + 5 \]
Thus,
\[ x = +4 \]

### Conclusion
The oxidation number of xenon in \(XeF_5^-\) is \(+4\).

### Final Answer
The oxidation number of Xe in \(XeF_5^-\) is \(+4\).

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