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Oxidation number of Xe in XeF(5)^(-) is ...

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

A

`+1`

B

`+2`

C

`+3`

D

`+4`

Text Solution

AI Generated Solution

The correct Answer is:
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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