Given below are two statements:
Statement (I): For Cl(··)F₃, all three possible structures may be drawn as follows.
Statement (II): Structure III is most stable, as the orbitals having the lone pairs are axial, where the lp – bp repulsion is minimum.
In the light of the above statements, choose the most appropriate answer from the options given below:

Solution

Chemistry ma’am · 4 steps

Lone pairs want the roomy equatorial seats of the bipyramid, so structure I wins, not III.

1. Electron pairs on Cl. Cl has 7 valence electrons. Three go into bonds with F, which leaves 2 lone pairs. That gives 5 electron pairs, so they point to the corners of a trigonal bipyramid. The two lone pairs can sit in different corners, and that is why three arrangements I, II and III can all be drawn. Statement I is correct.
2. Axial versus equatorial. An axial position has 3 neighbours at 90°. An equatorial position has only 2 at 90°, and the other two equatorial neighbours are at 120°. Repulsion is strongest at 90°, so a lone pair is more crowded in an axial seat.

   Structure	Lone pairs at	lp–lp at 90°	lp–bp at 90°
   I (T-shaped)	both equatorial	0	4
   II	one axial, one equatorial	1	3
   III (trigonal planar F)	both axial	0	6

3. Pick the most stable. lp–lp repulsion is the strongest kind, so II loses because it has one at 90°. Between I and III, I has fewer lp–bp clashes (4 against 6). So I, the T-shape, is most stable. III is actually the worst, and its lp–bp repulsion is at its maximum, not minimum. Statement II is incorrect.
4. So. Statement I is correct and Statement II is incorrect, which is option (c).

The idea to keep. In a trigonal bipyramid, lone pairs always go equatorial first: fewer 90° neighbours means less repulsion. That is why ClF₃ is T-shaped and XeF₂ is linear.

Answer from the NTA final answer key. Solution by PrepWiser.