Match List - I with List - II:
List-I (Species)   List-II (Number of lone pairs of electrons on the central atom)
(a) XeF$$_2$$       (i) 0
(b) XeO$$_2$$F$$_2$$   (ii) 1
(c) XeO$$_3$$F$$_2$$   (iii) 2
(d) XeF$$_4$$       (iv) 3
Choose the most appropriate answer from the options given below:

For every xenon compound we first count the eight valence electrons that xenon already possesses. Whenever xenon forms a bond it in effect “uses up’’ one of these electrons. Hence, after we know xenon’s oxidation state (the number of electrons it has formally donated to the surrounding atoms) we can find the electrons that remain as non-bonding or lone-pair electrons.

The relation we shall use is stated first: $$\text{Number of electrons left on Xe} = 8 - (\text{oxidation number of Xe}).$$ Because two electrons make one lone pair, the number of lone pairs is then $$\text{lone pairs} = \dfrac{8-\text{oxidation number}}{2}.$$

(a) XeF$$_2$$

Let the oxidation number of xenon be $$x.$$ For the neutral molecule we write $$x + 2(-1) = 0 \;\;\Longrightarrow\;\; x = +2.$$ Now, using the formula, the electrons that remain on xenon are $$8-2 = 6\text{ electrons}.$$ Dividing by two, the lone pairs are $$\dfrac{6}{2}=3.$$ So XeF$$_2$$ possesses three lone pairs.

(b) XeO$$_2$$F$$_2$$

The oxidation-number equation is $$x + 2(-2) + 2(-1) = 0 \;\;\Longrightarrow\;\; x = +6.$$ Electrons left on xenon: $$8-6 = 2\text{ electrons}.$$ Therefore the lone pairs are $$\dfrac{2}{2}=1.$$ XeO$$_2$$F$$_2$$ has one lone pair.

(c) XeO$$_3$$F$$_2$$

Writing the oxidation-number balance, $$x + 3(-2) + 2(-1) = 0 \;\;\Longrightarrow\;\; x = +8.$$ Electrons still on xenon: $$8-8 = 0.$$ Hence the lone pairs are $$\dfrac{0}{2}=0.$$ XeO$$_3$$F$$_2$$ carries no lone pair at all.

(d) XeF$$_4$$

The oxidation-number calculation is $$x + 4(-1) = 0 \;\;\Longrightarrow\;\; x = +4.$$ Electrons left on xenon: $$8-4 = 4\text{ electrons}.$$ Thus the lone pairs are $$\dfrac{4}{2}=2.$$ XeF$$_4$$ therefore contains two lone pairs.

Collecting all the results, we have $$\begin{aligned} \text{XeF}_2 &\longrightarrow 3 \text{ lone pairs} \;(iv)\\ \text{XeO}_2\text{F}_2 &\longrightarrow 1 \text{ lone pair} \;(ii)\\ \text{XeO}_3\text{F}_2 &\longrightarrow 0 \text{ lone pairs} \;(i)\\ \text{XeF}_4 &\longrightarrow 2 \text{ lone pairs} \;(iii) \end{aligned}$$

Matching List-I with List-II we obtain

(a) XeF$$_2$$ - (iv), (b) XeO$$_2$$F$$_2$$ - (ii), (c) XeO$$_3$$F$$_2$$ - (i), (d) XeF$$_4$$ - (iii).

Hence, the correct answer is Option D.