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Noble gases have completely filled valan...

Noble gases have completely filled valance shall i.e. `ns^(2)np^(6)` exceps He. Noble gases are monoatomic under normal conditions .Law boiling point of the ligher noble gases are due to weak vander walls forces between the atoms and absence of any interatomic interactions `Xe` reacts with `F_(2)` so give a source of flouoride mainly `XeF_(2),XeF_(4),XeF_(4),XeF_(3)` on complete hydrolyses gives `XeF_(3)`,
Oxidation state of `Xe` in `XeF_(2)` is

A

`+2`

B

`+4`

C

`+6`

D

`+8`

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To determine the oxidation state of xenon (Xe) in the compound xenon difluoride (XeF₂), we can follow these steps: ### Step 1: Write the formula and assign variables We have the compound XeF₂. Let's assign the oxidation state of xenon as \( X \). ### Step 2: Identify the oxidation state of fluorine Fluorine (F) is known to have an oxidation state of -1 in compounds. Since there are two fluorine atoms in XeF₂, their total contribution to the oxidation state will be: \[ 2 \times (-1) = -2 \] ### Step 3: Set up the equation The overall charge of the molecule XeF₂ is neutral (0). Therefore, we can set up the equation based on the oxidation states: \[ X + (-2) = 0 \] ### Step 4: Solve for \( X \) Now, we can solve for \( X \): \[ X - 2 = 0 \\ X = +2 \] ### Conclusion The oxidation state of xenon in XeF₂ is +2. ### Final Answer The answer is \( +2 \). ---
To determine the oxidation state of xenon (Xe) in the compound xenon difluoride (XeF₂), we can follow these steps: ### Step 1: Write the formula and assign variables We have the compound XeF₂. Let's assign the oxidation state of xenon as \( X \). ### Step 2: Identify the oxidation state of fluorine Fluorine (F) is known to have an oxidation state of -1 in compounds. Since there are two fluorine atoms in XeF₂, their total contribution to the oxidation state will be: \[ ...
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