A xenon compound ‘A’ upon partial hydrolysis gives $X e O_{2} F_{2}$ . The number of lone pairs of electrons present in compound A is ______. (Round off to the Nearest Integer).

19.0

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19.00

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Solution

Answer: 19

Partial Hydrolysis:

$X e F_{6} + H_{2} O \to X e O F_{4} + 2 H F$

$X e F_{6} + 2 H_{2} O \to X e O_{2} F_{2} + 4 H F$

$Complete hydrolysis$

$X e F_{6} + 3 H_{2} O \to X e O_{3} + 6 H F$

Thus, compound A is $X e F_{6}$

Each $F$ has 3 lone pair of electrons. So six $F$ atom give 18 lone pair of electrons.
The central $X e$ atom has one lone pair of electron. So the total number of lone pairs of electrons in compound (A) is 19.

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