Sulfur tetrafluoride adopts a see-saw geometry with two axial F atoms with a angle of about 180º and two equatorial F atoms at about 90º from the axial fluorines. Which statement most accurately describes the axial and equatorial bonds?
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Explanation
draftIn , sulfur is best described as hypervalent: the axial S–F bonds are not simple 2‑center‑2‑electron bonds like the equatorial ones. Instead, the two axial fluorines share a single p‑orbital on sulfur in a 3‑center‑4‑electron (3c‑4e) bonding scheme, similar to the axial bonding in or .
- Each axial S–F bond gets, on average, only about half the orbital overlap that a normal 2c‑2e bond would have.
- Weaker overlap per bond means a longer, weaker bond.
- Experimentally, the axial S–F bonds (~1.65 Å) are indeed longer than the equatorial S–F bonds (~1.55 Å), consistent with this bonding picture rather than simple electron‑pair repulsion arguments.
Thus the axial bonds are longer because of the shared, delocalized 3‑center bonding, not because of extra steric repulsion.
Correct: the axial bonds are longer because the two axial F atoms share bonding through a single p‑orbital (3‑center‑4‑electron bonding), weakening and lengthening each axial S–F bond.
This invokes simple VSEPR repulsion, but axial positions in a trigonal‑bipyramidal-like arrangement actually experience less steric crowding than equatorial ones (90° neighbors), not more, so this reasoning is backwards.
This confuses bond length with bond angle magnitude; the equatorial F–S–F angle is not the smallest (the axial–equatorial angle, ~90°, is), and bond length differences here arise from bonding type, not angle size.
This assumes lone-pair repulsion dominates bond length, but the lone pair (in an equatorial-like position) primarily distorts angles, not the fundamental bond order/overlap difference that actually causes the axial bonds to be longer.
USNCO 2018 National Part I exam, question 54. © American Chemical Society; reproduced for non-commercial practice.