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Ammonia, phosphine, water and hydrogen sulphide all have four electron domains with at least one lone pair, so VSEPR predicts angles a little under 109.5° for all four. The measured angles say otherwise: phosphine and hydrogen sulphide sit close to 90°. Lone-pair counting cannot explain a drop of about 13° down a group. This post gives the two rules that do, one calculation that makes them concrete, and the exception examiners like to test.
Where is this in the syllabus?
The NCERT Class 11 chapter Chemical Bonding and Molecular Structure (2026-27 reprint) teaches VSEPR and asks why water's angle is smaller than ammonia's. The p-block chapter that used to discuss phosphine and hydrogen sulphide is no longer in the Class 12 NCERT Chemistry Part I, whose contents now run from Solutions to Coordination Compounds.
The entrance exams still cover it. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list VSEPR and a p-block unit on trends "down the groups". The JEE (Advanced) 2026 brochure lists VSEPR shapes and, under group 15, phosphine. So group comparisons remain fair game even though the textbook chapter has gone. Check the current year's bulletin.
What do the measured angles show?
Experimental geometries from the NIST Computational Chemistry Comparison and Benchmark Database:
Two patterns stand out. Angles fall down a group, sharply from period 2 to period 3 and only slightly after that. And replacing H by F lowers the angle at nitrogen but raises it at phosphorus.
Rule 1: why does a bigger central atom give a smaller angle?
Nitrogen is smaller and more electronegative than phosphorus. In NH₃ the bonding pairs are pulled in close to nitrogen, and so close to one another. Bond pair–bond pair repulsion is strong and holds the angle open, near the tetrahedral value. In PH₃ the bonds are longer (1.421 Å against 1.012 Å) and the bonding pairs sit further from the central atom, so they repel each other less and the lone pair pushes them together to about 93°. Oxygen and sulphur follow the same pattern.
Some textbooks describe the same result by saying phosphorus and sulphur bond with orbitals of mostly p character, which lie at 90° to each other. Our post on hybridisation and orbital shapes covers that language.
Worked example: how far apart are the hydrogens?
The distance between two hydrogens bonded to the same atom is d = 2r sin(θ/2), where r is the bond length and θ the angle.
NH₃: 2 × 1.012 × sin 53.34° = 2 × 1.012 × 0.802 ≈ 1.62 Å
PH₃: 2 × 1.421 × sin 46.65° = 2 × 1.421 × 0.727 ≈ 2.07 Å
H₂O: 2 × 0.958 × sin 52.24° ≈ 1.51 Å
H₂S: 2 × 1.336 × sin 46.06° ≈ 1.92 Å
Even with a smaller angle, the hydrogens in PH₃ are about 0.44 Å further apart than in NH₃. The bonds have room to spare, so little resists the lone pair. That is Rule 1 in numbers.
Rule 2: what do electronegative outer atoms do?
In NF₃ the fluorine atoms pull the bonding pairs away from nitrogen towards themselves. The pairs then repel each other less, and the angle closes: 102.37° against 106.67° in NH₃.
Now try the rule on phosphorus. It predicts PF₃ below PH₃, but the data say 97.7° against 93.3°. Explanations for this vary between textbooks, and they often invoke extra bonding between phosphorus and fluorine. For exams, learn PF₃ versus PH₃ as the known exception and quote the numbers.
Which mistakes cost marks?
Ranking by lone pairs alone. NH₃ and PH₃ both have one lone pair; the count cannot separate them.
Calling PH₃ "less pyramidal". A smaller angle means a sharper pyramid.
Applying Rule 2 to PF₃ without checking.
Mixing up angle and polarity trends. NF₃ has both a smaller angle and a smaller dipole than NH₃, but the dipole difference comes from the lone pair's direction. NCERT gives 1.47 D for NH₃ and 0.23 D for NF₃.
Memorising 107° and 104.5° as exact. Use the values a question gives you.
For the four-step VSEPR method itself, see our guide to predicting molecular shapes. If you want to turn these dipoles into numbers, our post on percentage ionic character shows how.
A practice routine
Rank these sets before looking at the data: (NH₃, PH₃, AsH₃), (H₂O, H₂S, H₂Se), (NH₃, NF₃), (PH₃, PF₃).
Compute the H to H distance in AsH₃ (bond 1.511 Å, angle 92.08°). You should get about 2.18 Å.
Write two sentences for each ranking: which rule applies and why.
Explain PF₃ versus PH₃ aloud as an exception, with both numbers.
If electron domains themselves still feel shaky, the Fermi Concept Library has every concept in chemistry, each with its own lesson and its place in the map.
Today, write the eight angles above on a card in two columns, central atom down and outer atom across, and explain each step with one rule.
