A xenon compound A upon partial hydrolysis gives XeO$$_2$$F$$_2$$. The number of lone pair of electrons present in compound A is ___ (Round off to the Nearest integer)

The compound A undergoes partial hydrolysis to give $$\text{XeO}_2\text{F}_2$$. The xenon compound that gives $$\text{XeO}_2\text{F}_2$$ on partial hydrolysis is $$\text{XeF}_6$$, since $$\text{XeF}_6 + 2\text{H}_2\text{O} \to \text{XeO}_2\text{F}_2 + 4\text{HF}$$. So compound A is $$\text{XeF}_6$$.

Now we count the total number of lone pairs in $$\text{XeF}_6$$. Each fluorine atom has 7 valence electrons; one is used in bonding to Xe, leaving 6 non-bonding electrons, i.e., 3 lone pairs per F atom. With 6 fluorine atoms, that gives $$6 \times 3 = 18$$ lone pairs from fluorine.

For the central xenon atom in $$\text{XeF}_6$$: Xe has 8 valence electrons, 6 are used in bonding with the six F atoms, leaving 2 non-bonding electrons, which is 1 lone pair on Xe.

The total number of lone pairs in $$\text{XeF}_6$$ is $$18 + 1 = 19$$.