Q. 13 According to VSEPR theory, the species that has the smallest F-X-F angle (where 𝐗 = central atom)is
(A) BF3
(B) PF3
(C) BF4−
(D) IF4−

The correct answer is (D) IF4−, which has the smallest F–X–F bond angle among the given species according to VSEPR theory. The presence of lone pairs on the central atom in IF4− causes strong lone pair–bond pair repulsions, compressing the F–I–F angles more than in the other options.​​

Question and correct answer

According to VSEPR theory, the species that has the smallest F–X–F angle (where X is the central atom) is IF4−, option (D).​

• IF4− has:

  • Central atom I with 7 valence electrons plus 1 extra (negative charge) and 4 I–F bonds.

  • Total 8 electrons (4 pairs) used in bonding and 2 electrons (1 lone pair) left, so there are 5 electron pairs (4 bond pairs + 1 lone pair).

  • Electron-pair geometry: trigonal bipyramidal; molecular shape: seesaw or distorted square planar description depending on representation, but in all realistic VSEPR treatments, one lone pair on a 5‑pair arrangement causes significant bond angle compression.​

• Lone pair–bond pair repulsions are stronger than bond pair–bond pair repulsions, so the F–I–F angles become smaller than typical tetrahedral or trigonal planar angles.​

Because IF4− has both a high coordination (four F atoms) and at least one lone pair on a relatively large central atom, the F–I–F angle is most strongly compressed, giving the smallest F–X–F angle among the four species.​

Option (A) BF3: geometry and F–B–F angle

• BF3 is a classic trigonal planar molecule.​

• Boron has 3 valence electrons and forms three B–F bonds with no lone pair on boron.​

• Electron pair arrangement: 3 bond pairs, 0 lone pairs.

• Predicted bond angle from VSEPR: F–B–F angles are 120° in a perfect trigonal planar geometry, so there is no compression by lone pairs.​

Because 120° is relatively large and symmetric with only bond pairs, BF3 does not have the smallest F–X–F angle; instead, its angle is larger than in species where lone pairs compress the angle, such as PF3 or IF4−.​

Option (B) PF3: geometry and F–P–F angle

• PF3 has a central phosphorus atom with 5 valence electrons forming three P–F bonds and holding one lone pair.

• This gives 3 bond pairs + 1 lone pair, so the electron pair geometry is tetrahedral, but the molecular shape is trigonal pyramidal, analogous to NH3.​

• In a perfect tetrahedral environment, bond angles are 109.5°, but the presence of a lone pair increases repulsion and pushes the bonding pairs closer together, so the F–P–F angle becomes less than 109.5°.​

Because there is one lone pair, PF3 does have a compressed bond angle compared with BF3 (120°), but the compression is not as extreme as in a 5‑pair system with a lone pair such as IF4−, so PF3 does not show the smallest F–X–F angle in this list.​

Option (C) BF4−: geometry and F–B–F angle

• In BF4−, boron has 3 valence electrons plus one more bonding interaction, forming four B–F bonds, and the extra negative charge corresponds to an extra electron in the bonding system; boron effectively has 4 bond pairs and no lone pairs.

• Electron pair geometry: tetrahedral with 4 bond pairs, 0 lone pairs.​

• Ideal F–B–F bond angle: 109.5° in a regular tetrahedron, because no lone pairs are present to distort the geometry.​

Since 109.5° is still larger than the angle compressed by lone pairs in IF4− and there is no lone pair to cause extra compression, BF4− cannot have the smallest F–X–F angle among these species. Its angle lies between BF3 (120°) and lone‑pair–containing, more distorted systems like IF4−.​

Option (D) IF4−: geometry, lone pairs and smallest F–I–F angle

• For IF4−:

  • Iodine (group 17) has 7 valence electrons, plus one extra due to the negative charge, making 8 valence electrons.

  • It forms four I–F bonds, using 4 electron pairs; 4 pairs used for bonding leave 0 or 2 electrons depending on the specific distribution, so at least one lone pair exists on iodine in the common VSEPR model for such polyfluorides.

• With four bond pairs and at least one lone pair on a central atom in a higher coordination environment (5 total electron domains), the molecular shape is described as a distorted seesaw/square‑planar‑like arrangement and the F–I–F angles are significantly compressed.​

• Lone pair–bond pair repulsions in this 5‑domain system are stronger and act over more directions than in a simple tetrahedral system, causing F–I–F angles to drop below the typical 109.5° tetrahedral value and even lower than in PF3.​

Therefore, IF4− experiences the greatest angular compression and is correctly identified as the species with the smallest F–X–F angle in this question, making option (D) the correct answer.​