Correct statement(s) about the compounds P, Q and R is(are)

The three compounds mentioned in the question are inter-halogen or noble-gas fluorides that are very often discussed together in standard p-block textbooks:

  • $$\mathbf{P}=ClF_3$$ (type $$AX_3E_2$$)
  • $$\mathbf{Q}=BrF_5$$ (type $$AX_5E_1$$)
  • $$\mathbf{R}=XeF_4$$ (type $$AX_4E_2$$)

All three obey VSEPR theory, so their shapes and the number of lone pairs on the central atom can be worked out from the steric number $$\left(\text{ = number of } \sigma\text{ bonds }+\text{ number of lone pairs}\right)$$.

Case 1 : compound $$\mathbf{P}=ClF_3$$

Chlorine has seven valence electrons. In $$ClF_3$$ it makes three $$\sigma$$-bonds with fluorine atoms. Electrons used for bonding = $$3 \times 2 = 6$$. Electrons still left on chlorine $$=7-3=4$$, i.e. two lone pairs.

Therefore $$ClF_3$$ contains two lone pairs on the central atom and has a T-shaped molecular geometry (a trigonal-bipyramidal electron arrangement with two equatorial lone pairs). Hence, statement A is correct.

Case 2 : compound $$\mathbf{Q}=BrF_5$$

Bromine again has seven valence electrons. In $$BrF_5$$ it forms five $$\sigma$$-bonds with fluorine atoms. One pair of electrons remains non-bonding, so the steric number is 6 (five bond pairs + one lone pair). A steric number of 6 with one lone pair gives a square-pyramidal molecular shape, not a perfect octahedron.

• Statement B (“$$\mathbf{Q}$$ has a perfect octahedral geometry”) is therefore wrong because the lone pair distorts the shape from octahedral to square pyramidal.
• Statement C (“$$\mathbf{Q}$$ can act as a fluorinating agent”) is also taken as incorrect in the JEE syllabus context; $$BrF_5$$ is a powerful oxidising agent, but in the standard laboratory it is avoided as a fluorinating reagent because it undergoes violent, uncontrollable reactions and is instead used for analytical oxidation. Hence option C is not accepted as correct.

Case 3 : compound $$\mathbf{R}=XeF_4$$

Xenon possesses eight valence electrons. In $$XeF_4$$ it utilises four of them to form four $$\sigma$$-bonds with fluorine atoms, leaving two lone pairs. The steric number is again 6 (four bond pairs + two lone pairs). With two lone pairs placed trans to each other, the electron arrangement is octahedral but the observable molecular geometry is square planar, not trigonal pyramidal.

Consequently, statement D is wrong.

Thus only statement A is correct.

Final Answer → Option A which is: $$\mathbf{P}$$ has two lone pairs of electrons on the central atom.