In XeO$$_3$$ F$$_2$$, the number of bond pair(s), $$\pi$$-bond(s) and lone pair(s) on Xe atom respectively are:

We begin with the central atom xenon. Xenon belongs to group 18, so it has $$8$$ valence electrons.

The molecule is $$\mathrm{XeO_3F_2}$$, that is, xenon is surrounded by three oxygen atoms and two fluorine atoms – five atoms in all. Therefore xenon will furnish five $$\sigma$$-bonds, one to each surrounding atom.

First we count the total number of valence electrons present in the molecule.

$$\begin{aligned} \text{Electrons from Xe}&=&8\\ \text{Electrons from }3\,\text{O}&=&3\times6=18\\ \text{Electrons from }2\,\text{F}&=&2\times7=14\\[2pt] \text{Total valence electrons}&=&8+18+14=40 \end{aligned}$$

Now we draw an initial skeleton with five single bonds ($$5$$ bond pairs). Forming a single bond uses one pair of electrons, so

$$\text{Electrons consumed in }5\;\sigma\text{-bonds}=5\times2=10$$

$$\text{Electrons still to be placed}=40-10=30$$

Next we complete the octets of the surrounding atoms:

• Each fluorine already has two electrons in the Xe–F bond and therefore needs six more (three lone pairs). Thus $$2\times6=12$$ electrons are needed for both fluorine atoms.

• Each oxygen, if left with only a single bond, would need six more electrons to fill its octet. Hence $$3\times6=18$$ electrons are needed for the three oxygens.

The total demanded by the outer atoms is

$$12+18=30$$

and exactly the same $$30$$ electrons are still unassigned, so every outer atom’s octet can indeed be completed. At this stage xenon has

$$5$$ bond pairs and $$0$$ lone pairs, carrying a positive formal charge, while each oxygen carries a negative formal charge. To remove these opposite charges we invoke the common octet-expansion rule for elements of the 3rd period and beyond: we convert lone-pair electrons on oxygen into $$\pi$$-bonds with xenon.

Moving one lone pair from each oxygen toward xenon produces three Xe=O double bonds. Each new double bond keeps its original $$\sigma$$-bond and introduces one $$\pi$$-bond. The formal-charge accounting now becomes

$$ \begin{aligned} \text{Formal charge on O}&=&6-(4+2)=0,\\ \text{Formal charge on Xe}&=&8-(0+8)=0, \end{aligned} $$

so the structure is completely neutral.

Finally we list the required counts for the xenon atom:

• Bond pairs (i.e. $$\sigma$$-bonds): Xe is attached to each of five terminal atoms, so $$5$$ bond pairs are present.
• $$\pi$$-bonds: every Xe=O double bond adds one $$\pi$$-bond, and there are three such doubles, so $$3$$ $$\pi$$-bonds occur.
• Lone pairs on Xe: all eight of xenon’s valence electrons participate in bonding (five $$\sigma$$ pairs and three $$\pi$$ pairs), leaving $$0$$ lone pairs.

Thus the ordered set (bond pair, $$\pi$$-bond, lone pair) on xenon is $$\bigl(5,\;3,\;0\bigr)$$.

Hence, the correct answer is Option A.