Which of the following fluorides of xenon is impossible?

X e F_{2}

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X e F_{3}

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X e F_{4}

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X e F_{6}

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Solution

The correct option is C $X e F_{3}$
Xenon is an inert gas. Its electronic configuration is $[K r] 4 d^{10} 5 s^{2} 5 p^{6}$ .
All orbitals that are filled have paired electrons.

When one, two or three electrons are promoted from 5p (filled) to 5d (empty) orbitals, two, four and six half-filled orbitals are formed.

Xenon can combine with even number of F atoms to form $X e F_{2}$ , $X e F_{4}$ or $X e F_{6}$ . It cannot combine with odd number of F atoms.

Thus, formation of $X e F_{3}$ and $X e F_{5}$ is not possible.

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