Chemistry · Chemical Bonding · NEET
Only the 90 degree repulsions really matter (120 and 180 degree ones are weak because the pairs are far apart). Count the 90 degree neighbours: an AXIAL position has 3 neighbours at 90 degrees, while an EQUATORIAL position has only 2 neighbours at 90 degrees. Lone pairs push harder than bond pairs (lp-bp > bp-bp). So putting the lone pair equatorial gives fewer strong lp-bp repulsions, lower energy, and more stability. NCERT states this directly for SF4: axial lone pair = three lp-bp repulsions at 90 degrees, equatorial lone pair = only two, so equatorial wins.
Repulsion between electron pairs falls off very fast with angle. At 90 degrees the pairs are close, so the push is strong. At 120 degrees they are further apart and the push is much weaker; at 180 degrees it is almost nothing. So when we compare axial versus equatorial, we only count how many strong 90 degree lp-bp repulsions each choice makes. The choice with fewer 90 degree lp-bp repulsions wins. This is why we ignore the 120 degree pairs in the count.
In a trigonal bipyramid there are 2 axial spots (top and bottom) and 3 equatorial spots (the middle belt). An AXIAL spot points at all 3 equatorial spots at 90 degrees = 3 neighbours at 90 degrees. An EQUATORIAL spot points at the 2 axial spots at 90 degrees, and at the other 2 equatorial spots at 120 degrees = only 2 neighbours at 90 degrees. Fewer 90 degree neighbours is why the lone pair chooses equatorial.
All three are built on a trigonal bipyramid (sp3d, 5 electron pairs) and all lone pairs sit in equatorial positions. SF4 = AB4E: 1 lone pair equatorial, giving a see-saw shape. ClF3 = AB3E2: 2 lone pairs equatorial, giving a T-shape. XeF2 = AB2E3: 3 lone pairs all equatorial, giving a linear shape (the 2 fluorines end up axial). Notice: as lone pairs fill the equatorial belt, the F atoms get pushed to the axial line.
SF4 has 4 bond pairs plus 1 lone pair (AB4E). The lone pair takes one equatorial spot, so only 4 F atoms remain: 2 axial and 2 equatorial. The lone pair squeezes the nearby bonds, so the shape is a distorted tetrahedron, also called a see-saw (or folded square). It is NOT a perfect tetrahedron like CH4, because CH4 has no lone pair. A common NEET trap is to say SeF4 and CH4 have the same shape - they do not.
The equatorial preference rule is about LONE pairs. In PCl5 (AB5, no lone pair) all 5 positions are bond pairs, so the shape is a regular trigonal bipyramid. But a related fact still shows up: the axial P-Cl bonds are longer than the equatorial ones, because each axial bond feels 3 repulsions at 90 degrees while each equatorial bond feels only 2. Same 90 degree counting idea, applied to bonds instead of lone pairs.
Identify the correct statement about ClF3 from the following options:
Amongst the following, which one will have maximum 'lone pair-lone pair' electron repulsion?
In the structure of ClF3, the number of lone pairs of electrons on the central atom Cl is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. For any molecule based on 5 electron pairs (sp3d), every lone pair goes to an equatorial position. This holds for SF4 (1 lp), ClF3 (2 lp) and XeF2 (3 lp). It gives the fewest strong 90 degree lone pair repulsions and the most stable shape.
1 equatorial lone pair gives see-saw (SF4). 2 equatorial lone pairs give T-shape (ClF3). 3 equatorial lone pairs give linear (XeF2). Learn these three as a set for NEET.
Electron-pair repulsion drops steeply as the angle grows. At 90 degrees the pairs are close and push hard; at 120 degrees they are further apart and push much less. So only the number of 90 degree neighbours decides axial versus equatorial.
Yes, it is a very common exam idea. Once you know a molecule is trigonal bipyramidal, place lone pairs equatorial first, then read off the shape of the remaining atoms. This directly gives see-saw, T-shape or linear, which NEET asks about almost every year.
