Compounds with Two Lone Pairs on Central Atom

To determine the number of lone pairs of electrons on the central atom for each compound, we can use VSEPR theory. The formula to calculate the number of lone pairs ($L$) on the central atom is given by: $L = \frac{V - B}{2}$ where $V$ is the number of valence electrons of the central atom and $B$ is the number of single covalent bonds formed with surrounding monovalent atoms (in this case, fluorine atoms).

1. For $\text{BrF}_5$:

   • Central atom: Bromine ($\text{Br}$), which belongs to Group 17 and has $V = 7$ valence electrons.
   • Number of bonded fluorine atoms, $B = 5$.
   • Number of lone pairs: $L = \frac{7 - 5}{2} = 1$
   • Thus, $\text{BrF}_5$ has 1 lone pair on the central atom.

2. For $\text{ClF}_3$:

   • Central atom: Chlorine ($\text{Cl}$), which belongs to Group 17 and has $V = 7$ valence electrons.
   • Number of bonded fluorine atoms, $B = 3$.
   • Number of lone pairs: $L = \frac{7 - 3}{2} = 2$
   • Thus, $\text{ClF}_3$ has 2 lone pairs on the central atom.

3. For $\text{XeF}_4$:

   • Central atom: Xenon ($\text{Xe}$), which belongs to Group 18 and has $V = 8$ valence electrons.
   • Number of bonded fluorine atoms, $B = 4$.
   • Number of lone pairs: $L = \frac{8 - 4}{2} = 2$
   • Thus, $\text{XeF}_4$ has 2 lone pairs on the central atom.

4. For $\text{SF}_4$:

   • Central atom: Sulfur ($\text{S}$), which belongs to Group 16 and has $V = 6$ valence electrons.
   • Number of bonded fluorine atoms, $B = 4$.
   • Number of lone pairs: $L = \frac{6 - 4}{2} = 1$
   • Thus, $\text{SF}_4$ has 1 lone pair on the central atom.

Therefore, the compounds possessing two lone pairs of electrons on the central atom are $\text{ClF}_3$ and $\text{XeF}_4$.

Correct Options: (B) and (C)