Which of the possible following florides of xenon is impossible ?

Options

A. `XeF_(6)`
B. `XeF_(4)`
C. `XeF_(3)`
D. `XeF_(2)`

Answer

C

Explanation

In excited state, the promotion of 1, 2 or 3 electrons from 5p to 5d orbital will result in 2,4 or 6 unpaired electrons. Now these half–filled p and d – orbitals can form `sigma`–bonds with 2,4 or 6 numbers of fluorine atoms. Thus the compound `XeF_(3)` is not possible.