Draw the structure of the following :
(i) `XeF_(2)`
(ii) `BrF_(3)`

Explanation

To draw the structures of `XeF₂` (xenon difluoride) and `BrF₃` (bromine trifluoride), we will follow these steps:

### Step 1: Determine the Central Atom and Valence Electrons
- For `XeF₂`, the central atom is xenon (Xe), which has 8 valence electrons. Each fluorine (F) atom has 7 valence electrons. Therefore, the total number of valence electrons in `XeF₂` is:
\[ 8 \, (\text{Xe}) + 2 \times 7 \, (\text{F}) = 22 \, \text{valence electrons} \]

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