How many of the following compound(s) cannot be prepared:
1. $X e F_{2}$
2. $X e F_{3}$
3. $X e F_{5}$
4. $X e F_{6}$

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Solution

Xenon forms three binary fluorides, $X e F_{2}, X e F_{4}$ and $X e F_{6}$ by the direct reaction of elements under appropriate experimental conditions. Xe has a complete filled 5p configuration. As a result, when it undergoes bonding with an odd number (3 or 5) of F atoms it leaves behind one unpaired electron. This causes the molecule to become unstable. As a result $X e F_{3}$ , $X e F_{5}$ are not possible.
$X e (g) + F_{2} (g) 673 K, 1 b a r - ----- \to X e F_{2} (s)$
(xenon in excess)
$X e (g) + 2 F_{2} (g) 873 K, 7 b a r - ----- \to X e F_{4} (s)$
(1:5 ratio)
$X e (g) + 3 F_{2} (g) 573 K, 60 - 70 b a r - -------- \to X e F_{6} (s)$
(1:20 ratio)

Only even number of fluorides can combine with Xe. Therefore, $X e F_{3}$ and $X e F_{5}$ cannot be formed.

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