Effect of lone pairs on geometry (e.g. H₂O bent, NH₃ pyramidal): Question bank

Definition Shorthand you'll see on every line below

An electron domain = one region of electron density around the central atom (a bond of any order, or a lone pair).
Valence electrons = the electrons in the outermost shell of an atom, the ones available to form bonds or sit as lone pairs. Carbon has 4, nitrogen 5, oxygen 6, fluorine 7 — you read this off the atom's group in the periodic table.
SN = steric number = the total count of domains = (atoms bonded) + (lone pairs on the central atom). It fixes the electron geometry. "SN 4" just means four domains.
BP = bonding pair (a shared pair sitting in a bond); LP = lone pair (a pair held by the central atom alone). So "LP–BP > BP–BP" is comparing how hard those pairs repel.
AB₂, AB₃ … is a naming convention: A is the central atom, B is each attached atom, and the subscript is how many B's there are. So H₂O is an "AB₂" molecule (one central O, two H's) — note this counts atoms only, never lone pairs. When a lone pair matters we write it as E: H₂O is "AB₂E₂".

Lone pairs repel harder than bonds because they are held by only one nucleus, so their cloud bulges closer to the central atom and elbows more space.

Figure — Effect of lone pairs on geometry (e.g. H₂O bent, NH₃ pyramidal)

True or false — justify

TF: H₂O and CO₂ are both AB₂ molecules, so both must be linear.

False — "AB₂" only counts atoms, not domains; CO₂'s carbon has SN 2 (no lone pairs) so it is linear, while water's oxygen has SN 4 (two lone pairs) so it is bent at 104.5°.

TF: A double bond, having more electrons, counts as two electron domains.

False — all bonds to the same neighbour point in one direction, so a double bond is one domain; the extra electron density only makes that single domain repel slightly harder.

TF: NH₃ has three bonds, so like BF₃ it should be trigonal planar.

False — BF₃ boron has SN 3 (no lone pair, truly planar), but NH₃ nitrogen has SN 4 with one lone pair that pushes the three N–H bonds down into a pyramid at 107°.

TF: Removing water's two lone pairs would leave the same 104.5° bend.

False — the lone pairs are exactly what squeezes the angle below tetrahedral; with 0 lone pairs the two bonds would relax toward the symmetric arrangement, not stay bent.

TF: NH₄⁺ has a wider bond angle than NH₃.

True — NH₄⁺ has 0 lone pairs (N donated its pair to H⁺), so all four domains are equivalent bonds and relax to the perfect tetrahedral 109.5°, wider than NH₃'s 107°.

TF: The electron geometry of NH₃, H₂O and CH₄ is the same.

True — all three have SN 4, so all have tetrahedral electron geometry; they differ only in molecular shape because lone pairs are invisible in the reported shape.

TF: Lone pairs always sit at an axial site in a trigonal bipyramid.

False — lone pairs prefer equatorial sites because equatorial has only two close 90° neighbours instead of three, minimising strong 90° repulsion (see Trigonal bipyramidal geometry).

TF: A molecule with lone pairs on the central atom is always polar.

False — lone pairs break symmetry in H₂O and NH₃ (polar), but in symmetric cases like XeF₄ the two lone pairs sit opposite each other and their effects, together with the bond dipoles, can cancel (see Dipole moment).

Spot the error

Find the flaw: "O in water has 6 valence electrons and 2 bonds, so it has 6 − 2 = 4 lone pairs."

The error is skipping the halving — 4 leftover electrons pair up into (6 − 2)/2 = 2 lone pairs, since each lone pair holds two electrons.

Find the flaw: "H₂O's angle is 90° because oxygen uses two pure p orbitals for bonding."

The pure-p picture predicts 90°, but oxygen is sp³ hybridised (see Hybridization); the real angle is 104.5°, close to tetrahedral, not 90°.

Find the flaw: "NH₃ is 107° and PH₃ is also 107° because both have one lone pair."

One lone pair alone does not fix the exact angle; PH₃ is near 93° because the larger P atom and different orbital sizes weaken bond-pair repulsion, letting the lone pair squeeze bonds much further (see Bond angle and electronegativity).

Find the flaw: "SF₄ has SN 5 so it is trigonal bipyramidal in shape."

SN 5 gives the electron geometry as trigonal bipyramidal, but with one lone pair the molecular shape is seesaw — you must delete the lone-pair position before naming the shape.

Find the flaw: "Each lone pair drops the bond angle by exactly 2.5°, always."

The ~2.5° figure is an approximate trend for the C→N→O series only; actual drops depend on atom size and electronegativity, so it is a rule of thumb, not a law.

Find the flaw: "H₃O⁺ has the same shape as H₂O because both are oxygen-centred."

H₃O⁺ has three bonds and one lone pair (SN 4) making it pyramidal like NH₃, whereas H₂O has two bonds and two lone pairs, so it is bent — the domain counts differ.

Why questions

Why does a lone pair repel more strongly than a bonding pair?

A lone pair is attracted by only one nucleus, so its cloud is not pulled thin between two atoms and instead bulges out closer to the central atom, occupying more angular space (Figure 1).

Why do we count all domains for electron geometry but ignore lone pairs when naming the shape?

Domains set the geometry that minimises repulsion, but we can only "see" atoms, so the reported molecular shape describes atom positions and omits the invisible lone-pair positions (Figure 2).

Why is the repulsion order LP–LP > LP–BP > BP–BP?

Both electrons of a lone pair (LP) sit near one atom (fattest cloud), a bond pair (BP) is split between two atoms (thinnest), so two lone-pair clouds elbow each other hardest and two bond pairs least.

Why does the lone pair in SF₄ choosing equatorial reduce strain?

An equatorial site faces only two neighbours at the strongly repulsive 90° angle, whereas an axial site faces three, so equatorial minimises the count of hard 90° lone-pair repulsions (Figure 3).

Why does adding a second lone pair (NH₃ → H₂O) shrink the angle further?

The second lone pair adds another fat cloud and an extra LP–LP repulsion, together shoving the remaining bonds even closer than a single lone pair does (107° → 104.5°).

Why is CO₂ linear despite carbon forming double bonds?

Each C=O double bond is one domain and carbon has no lone pairs (SN 2), so the two domains point in exactly opposite directions to maximise separation, giving 180°.

Edge cases

Edge: what is the shape when SN = 4 but there are 0 lone pairs?

Tetrahedral both in electron geometry and molecular shape, with the symmetric ideal angle 109.5° (e.g. CH₄, NH₄⁺).

Edge: what happens to the angle prediction if the central atom has no lone pairs at all?

The bonds relax to the ideal symmetric angle for that SN (109.5° for SN 4, 120° for SN 3, 180° for SN 2), since there is no extra lone-pair repulsion to shrink them.

Edge: what are the ideal angles for a true SN 5 with zero lone pairs (e.g. PF₅)?

A perfect trigonal bipyramid has two distinct angles — axial-to-equatorial bonds are 90° and equatorial-to-equatorial bonds are 120° — because the two axial and three equatorial sites are geometrically different.

Edge: what are the ideal angles for a true SN 6 with zero lone pairs (e.g. SF₆)?

A perfect octahedron has every adjacent bond at 90° (and opposite bonds at 180°); all six sites are equivalent, so there is only one bond angle to report.

Edge: what shape is SN 3 with one lone pair (AB₂E, e.g. SO₂ or NO₂⁻)?

The three domains start as a trigonal plane, but deleting one lone-pair position leaves a bent molecule with an angle a little under 120° (about 118° for SO₂), since the lone pair squeezes the two bonds together.

Edge: how do you count lone pairs on a charged central atom like N in NH₄⁺?

Adjust valence electrons for charge first — N normally has 5, but the +1 charge removes one to give 4, all used in four bonds, leaving 0 lone pairs and a perfect tetrahedron.

Edge: does a triple bond count differently from a single bond in acetylene (HC≡CH)?

No — each carbon's triple bond plus one C–H bond are two domains (SN 2, no lone pairs), so acetylene is linear at 180°; the triple bond, like any bond order, still points in just one direction and counts as one domain.

Edge: a central atom has SN 5 with two lone pairs (e.g. ClF₃) — where do both go?

Both lone pairs take equatorial sites to avoid 90° neighbours, leaving the shape T-shaped (see Steric number and molecular shape).

Edge: XeF₄ has SN 6 with two lone pairs — why is it square planar and nonpolar?

The two lone pairs sit on opposite axial sites (180° apart) to minimise repulsion, leaving four fluorines in a plane whose bond dipoles cancel, so the molecule is symmetric and nonpolar.

Edge: what shape results from SN 2 with one lone pair?

The two domains (one bond, one lone pair) still point 180° apart, so the electron geometry is linear; but with only one attached atom there is no bond angle to report — you have a diatomic fragment, so VSEPR simply calls the atom arrangement "linear" by default.

Recall The one-sentence summary of this whole bank

Always convert the molecule into an honest domain count: valence electrons of the central atom (adjusted for charge) minus the electrons used in bonds, then divide the leftover by two to get lone pairs; add lone pairs to bonded atoms for SN, fix the electron geometry, delete lone-pair positions to name the shape, and remember lone pairs still squeeze the angle harder than bonds do.

Connections

VSEPR Theory — every trap here is a mis-application of VSEPR's domain counting.
Steric number and molecular shape — the counting recipe these traps test.
Hybridization — resolves the "90° vs 104.5°" pure-p trap.
Bond angle and electronegativity — explains PH₃ vs NH₃ anomalies.
Dipole moment — settles the "lone pairs ⇒ always polar" trap.
Trigonal bipyramidal geometry — the equatorial-lone-pair edge cases (SF₄, ClF₃, XeF₄).