From the given covalent compounds (A to F) identify the pair of molecules which have: (i) Two lone pairs of electrons on the central atom. (ii) One lone pair of electrons on the central atom. A. $\mathrm{SO}_2$ B. $\mathrm{ClF}_3$ C. $\mathrm{BF}_3$ D. $\mathrm{BrF}_5$ E. $\mathrm{XeF}_4$ F. $\mathrm{SF}_6$

The number of lone pairs on the central atom is determined by the formula: Lone pairs = $\dfrac{1}{2}(V - N - C)$, where $V$ is the number of valence electrons of the central atom, $N$ is the number of monovalent atoms bonded to it, and $C$ is the charge on the molecule.

A. $\mathrm{SO}_2$: $V=6$, $N=0$ (oxygen is divalent). Lone pairs = $\dfrac{1}{2}(6 - 0) = 3$ (considering bonding electrons). More accurately, $\mathrm{S}$ has 6 valence electrons, 2 are used for double bonds with each O, leaving 1 lone pair.

B. $\mathrm{ClF}_3$: $V=7$, $N=3$. Lone pairs = $\dfrac{1}{2}(7 - 3) = 2$.

C. $\mathrm{BF}_3$: $V=3$, $N=3$. Lone pairs = $\dfrac{1}{2}(3 - 3) = 0$.

D. $\mathrm{BrF}_5$: $V=7$, $N=5$. Lone pairs = $\dfrac{1}{2}(7 - 5) = 1$.

E. $\mathrm{XeF}_4$: $V=8$, $N=4$. Lone pairs = $\dfrac{1}{2}(8 - 4) = 2$.

F. $\mathrm{SF}_6$: $V=6$, $N=6$. Lone pairs = $\dfrac{1}{2}(6 - 6) = 0$.

Comparing the results:
(i) Two lone pairs: B ($\mathrm{ClF}_3$) and E ($\mathrm{XeF}_4$).
(ii) One lone pair: A ($\mathrm{SO}_2$) and D ($\mathrm{BrF}_5$).

The pair matching (i) is B & E and (ii) is A & D.

Answer: $(i) B \& E \quad (ii) \mathrm{A} \& \mathrm{D}$