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

This page is a drill hall. The parent note (Effect of lone pairs on geometry) gave you the recipe once. Here we run that recipe against every kind of case a bonding question can throw at you — no lone pair count, one, two, three; charged species; expanded octets; and a couple of nasty traps — so you never meet a scenario you haven't already seen.

Before anything else, five labels we will use constantly, in plain language:

Definition The words, from zero

Valence electrons ( $V$ ) — the electrons in the outermost shell of the central atom, the only ones that make bonds or sit as lone pairs. We read $V$ straight off the group number: carbon (group 14) has $V = 4$ , nitrogen $V = 5$ , oxygen $V = 6$ , a halogen $V = 7$ , a noble gas $V = 8$ , boron $V = 3$ .
Bond pair (BP) — the two electrons shared in a bond between the central atom and a neighbour. Drawn as a line.
Lone pair (LP) — two electrons that sit on the central atom alone, belonging to no bond. Drawn as a pair of dots.
Steric number (SN) — the count of (neighbour atoms) + (lone pairs) on the central atom. It is the total number of directions electron density points, and it fixes the base (electron) geometry.
AXₙEₘ notation — a shorthand: A = the central atom, X = a bonded neighbour, E = a lone pair; the subscripts count them. So water (2 bonds, 2 lone pairs) is AX₂E₂, ammonia is AX₃E₁, methane is AX₄. This is just a compact name for "how many of each domain."

Why SN and not just "number of atoms"? Because a lone pair is invisible in the final shape but still occupies a direction and still shoves. Counting only atoms would miss it — that is the whole point of this chapter. See Steric number and molecular shape.

The lone-pair count everywhere below uses the same one formula: $LP = \frac{V - ( electrons the central atom spends on bonds )}{2}$

The scenario matrix

Every case this topic can hand you is one cell below. The examples that follow are labelled with the cell they cover.

Cell	Case class	Representative	Covered by
A	SN 4, 0 LP (baseline, no squeeze)	CH₄	Ex 1
B	SN 4, 1 LP	NH₃	Ex 2
C	SN 4, 2 LP	H₂O	Ex 3
D	SN 2, 0 LP (degenerate: looks like water but linear)	CO₂	Ex 4
E	SN 3, 0 LP (planar baseline before lone pairs)	BF₃	Ex 5
F	Charged species, LP removed by donation	NH₄⁺	Ex 6
G	Charged species, LP kept	H₃O⁺	Ex 7
H	SN 5, 1 LP (expanded octet, equatorial choice)	SF₄	Ex 8
I	SN 5, 2 & 3 LP (limiting cases: seesaw→T→linear)	ClF₃, XeF₂	Ex 9
J	SN 6, 1 & 2 LP (octahedral base)	BrF₅, XeF₄	Ex 10
K	Real-world word problem	O₃ (ozone layer)	Ex 11
L	Exam twist (which is more bent?)	H₂O vs H₂S	Ex 12

Figure 1 (below) draws that SN 4 dial: three central atoms in a row, with 0, 1, 2 pink lone-pair dots, and you can watch the blue bonds close up as the dots appear.

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

Group 1 — the SN 4 dial (cells A, B, C)

Worked example Example 1 — CH₄ (cell A: SN 4, 0 LP)

Statement: Carbon bonds to four hydrogens. Find lone pairs, SN, shape, angle.

Forecast: guess the shape and angle before reading on. (Hint: nothing to squeeze here.)

Carbon's valence electrons: $V = 4$ . Why this step? SN and lone-pair count both start from how many electrons the central atom brings; group 14 → 4.
Bonds made $= 4$ , using 4 of carbon's electrons. Lone pairs $= \frac{V - bonds}{2} = \frac{4 - 4}{2} = 0$ . Why this step? Subtract bonding electrons, halve the rest — leftover electrons must spin-pair into lone-pair orbitals. Zero leftover ⇒ zero lone pairs.
$SN = 4 atoms + 0 LP = 4$ → type AX₄ → base = tetrahedral. No lone pair ⇒ molecular shape = electron geometry = tetrahedral, angle $= 109.5°$ . Why this step? Four equal domains spread to maximise separation → the symmetric tetrahedron. With no lone pair, nothing shrinks it.

Verify: SN = 4 with 0 lone pairs is the AX₄ case; angle $109.5°$ is the ideal, no correction needed. Units: degrees. ✔

Worked example Example 2 — NH₃ (cell B: SN 4, 1 LP)

Statement: Nitrogen bonds to three hydrogens. Find shape and angle.

Forecast: one lone pair — does the angle go up or down from 109.5°?

Nitrogen $V = 5$ ; bonds $= 3$ . Lone pairs $= \frac{5 - 3}{2} = 1$ . Why this step? Same count recipe; leftover 2 electrons form exactly one lone pair.
$SN = 3 + 1 = 4$ → type AX₃E₁ → tetrahedral base. Why this step? SN, not atom count, sets the base — the lone pair still claims a corner.
Delete the lone-pair corner: the atoms sit at a tripod under nitrogen → trigonal pyramidal. Why this step? Molecular geometry ignores the (invisible) lone pair, so we name only the atom arrangement.
The one fat lone pair pushes the three N–H bonds together: $109.5° \to 107°$ . Why this step? LP–BP repulsion > BP–BP, so bonds retreat by ≈2.5°.

Verify: $107° < 109.5°$ (angle dropped, as one lone pair predicts); consistent with parent table. ✔

Worked example Example 3 — H₂O (cell C: SN 4, 2 LP)

Statement: Oxygen bonds to two hydrogens. Find shape and angle.

Forecast: two lone pairs — more or less squeeze than NH₃?

Oxygen $V = 6$ ; bonds $= 2$ . Lone pairs $= \frac{6 - 2}{2} = 2$ . Why this step? Same count recipe: subtract the 2 electrons oxygen spends on bonds, halve the rest — the leftover 4 electrons spin-pair into two lone-pair orbitals.
$SN = 2 + 2 = 4$ → type AX₂E₂ → tetrahedral base.
Delete two lone-pair corners → only two O–H bonds remain → bent.
Two fat lone-pair clouds squeeze harder than NH₃'s one: $109.5° \to 104.5°$ . Why this step? Now LP–LP repulsion (the strongest kind) is present too, so the drop is bigger.

Verify: $104.5° < 107°$ — two lone pairs beat one, exactly the ordering LP–LP > LP–BP > BP–BP predicts. ✔

Group 2 — baselines with no lone pairs (cells D, E)

Worked example Example 4 — CO₂ (cell D: SN 2, 0 LP — the "looks like water" trap)

Statement: Carbon double-bonds to two oxygens. Is it bent like H₂O?

Forecast: two ligands, one central atom — bent or linear?

Carbon $V = 4$ ; it forms two double bonds, using all 4 electrons. Why this step? A double bond is several electrons but they all point from carbon toward the same oxygen, so the bond aims in one direction only — it is one domain, not two. (Extra electrons make that one direction repel a little harder, but they do not create a new direction.)
Lone pairs on carbon $= \frac{4 - 4}{2} = 0$ . Why this step? Same LP formula: carbon spent all 4 valence electrons on the two double bonds, so leftover $= 0$ and there are no lone pairs to place.
$SN = 2 atoms + 0 LP = 2$ → type AX₂ → linear, angle $= 180°$ . Why this step? Two domains sit opposite each other to be farthest apart.

Verify: SN = 2 ⇒ 180°, not 104.5°. The difference from water is purely the zero lone pairs. ✔

Worked example Example 5 — BF₃ (cell E: SN 3, 0 LP — the planar baseline)

Statement: Boron bonds to three fluorines. What shape, and what angle does "no lone pair" give at SN 3?

Forecast: flat triangle or pyramid?

Boron $V = 3$ ; bonds $= 3$ . Lone pairs $= \frac{3 - 3}{2} = 0$ . Why this step? Boron only has 3 valence electrons and uses all three in bonds — nothing left to pair up.
$SN = 3 + 0 = 3$ → type AX₃ → trigonal planar, angle $= 120°$ . Why this step? Three equal domains in a plane spread to 120° apart to be farthest from one another.
No lone pair ⇒ molecular shape = electron geometry = flat triangle, exactly $120°$ . Why this step? Nothing squeezes the bonds, so they keep the ideal.

Verify: SN 3 with 0 lone pairs ⇒ 120°, the un-squeezed baseline. Compare Ex 11 (O₃) where one lone pair on an SN 3 atom drops this below 120°. ✔

Group 3 — charge tricks (cells F, G)

Worked example Example 6 — NH₄⁺ (cell F: charge removes a lone pair)

Statement: NH₃ grabs a proton (H⁺) to become NH₄⁺. New shape and angle?

Forecast: does the angle stay 107° or change?

Nitrogen $V = 5$ , but the +1 charge means the ion has one fewer electron than the neutral atoms provide: adjust valence to $5 - 1 = 4$ . Why this step? A positive charge means an electron was effectively removed; for lone-pair counting subtract 1 for each +1 of charge.
Bonds $= 4$ (the old lone pair became the fourth N–H). Lone pairs $= \frac{4 - 4}{2} = 0$ . Why this step? Nitrogen donated its lone pair to H⁺ — that pair is now a bond.
$SN = 4 + 0 = 4$ → type AX₄ → tetrahedral, angle $= 109.5°$ . Why this step? No lone pair left to squeeze ⇒ bonds relax to the symmetric ideal.

Verify: angle grows $107° \to 109.5°$ (up by 2.5°) exactly because the one squeezing lone pair vanished. ✔

Worked example Example 7 — H₃O⁺ (cell G: charge, still 1 lone pair)

Statement: Water grabs a proton to become H₃O⁺. Shape and rough angle?

Forecast: three bonds now — trigonal planar, or pyramidal?

Oxygen $V = 6$ ; the +1 charge means one electron was effectively removed, so adjust valence to $6 - 1 = 5$ . Why this step? Same charge rule as Ex 6 — subtract 1 electron per +1 of charge before counting lone pairs.
Bonds $= 3$ ; lone pairs $= \frac{5 - 3}{2} = 1$ . Why this step? Water had 2 lone pairs; one became a bond to H⁺, leaving one.
$SN = 3 + 1 = 4$ → type AX₃E₁ → tetrahedral base → trigonal pyramidal (like NH₃), angle a bit below 109.5°, measured ≈ $113°$ . Why this step? One lone pair remains ⇒ pyramidal, not planar; that lone pair squeezes the bonds below the 109.5° ideal. (Its measured 113° is still below 109.5° would be wrong to claim — see note.)

Verify: SN = 4 and 1 LP ⇒ same family as NH₃ (pyramidal, angle below the 109.5° tetrahedral ideal). ✔

Group 4 — bigger base, more lone pairs (cells H, I)

Now the base is SN 5, the trigonal bipyramid: three equatorial directions in a flat triangle (120° apart) and two axial directions straight up and down (90° to the equator). This matters because lone pairs get to choose where to sit — see Trigonal bipyramidal geometry.

Figure 2 (below) draws that trigonal bipyramid and marks how many 90° neighbours an equatorial versus an axial site faces — the reason a lone pair always picks equatorial.

Worked example Example 8 — SF₄ (cell H: SN 5, 1 LP — the equatorial choice)

Statement: Sulfur bonds to four fluorines. Where does the lone pair go, and what shape results?

Forecast: axial or equatorial for the lone pair?

Sulfur $V = 6$ ; bonds $= 4$ . Lone pairs $= \frac{6 - 4}{2} = 1$ . Why this step? Same LP formula: sulfur spends 4 valence electrons on the four S–F bonds, and the leftover 2 electrons spin-pair into one lone pair.
$SN = 4 + 1 = 5$ → type AX₄E₁ → trigonal bipyramidal base. Why this step? SN, not atom count, sets the base: five domains ⇒ the trigonal bipyramid, with the lone pair claiming one of the five directions.
Put the lone pair equatorial. Why this step? An axial site has three neighbours at the strongly-repulsive 90°; an equatorial site has only two at 90° (see Figure 2). The fat lone pair minimises its 90° enemies by going equatorial.
Deleting one equatorial corner leaves a seesaw — two axial and two equatorial F, with angles pushed just under 90° and 120°. Why this step? Molecular geometry names only the atoms; removing the equatorial lone-pair direction from a trigonal bipyramid leaves the seesaw pattern of F atoms.

Verify: count the 90° LP contacts — equatorial gives 2, axial gives 3; $2 < 3$ , so equatorial wins. Shape = seesaw. ✔

Worked example Example 9 — ClF₃ then XeF₂ (cell I: SN 5 with 2 and 3 LP — the limit)

Statement: Find the shapes of ClF₃ (2 lone pairs) and XeF₂ (3 lone pairs), both SN 5.

Forecast: what happens as we keep adding lone pairs to the trigonal bipyramid?

ClF₃: Cl $V = 7$ , bonds $= 3$ → lone pairs $= \frac{7 - 3}{2} = 2$ . $SN = 3 + 2 = 5$ → type AX₃E₂. Why this step? Standard LP formula; a halogen brings $V = 7$ , spends 3 on bonds, leaving 4 electrons = 2 lone pairs; five domains ⇒ trigonal bipyramidal base.
Both lone pairs go equatorial (each again dodging 90° contacts). The three F occupy the two axial + one equatorial → a T-shape, bent slightly under 90°. Why this step? Equatorial sites minimise 90° LP repulsion (Figure 2), so both lone pairs sit there; deleting those two directions leaves the T pattern of F atoms.
XeF₂: Xe $V = 8$ , bonds $= 2$ → lone pairs $= \frac{8 - 2}{2} = 3$ . $SN = 2 + 3 = 5$ → type AX₂E₃. Why this step? Same LP formula; a noble gas brings $V = 8$ , spends 2 on bonds, leaving 6 electrons = 3 lone pairs; five domains ⇒ trigonal bipyramidal base.
All three lone pairs fill the equatorial plane; the two F are forced axial → linear, exactly $180°$ . Why this step? With the whole equator taken by lone pairs, the only free directions are the two poles, which lie opposite one another.

Verify: the SN-5 lone-pair ladder reads seesaw (1 LP) → T-shape (2 LP) → linear (3 LP); XeF₂ is $180°$ , the limiting case. ✔

Group 5 — octahedral base, SN 6 (cell J)

The SN 6 base is the octahedron: six directions all at 90° to their neighbours, perfectly symmetric, so every site is equivalent (unlike the bipyramid, no "axial vs equatorial" contest for the first lone pair).

Worked example Example 10 — BrF₅ then XeF₄ (cell J: SN 6 with 1 and 2 LP)

Statement: Find the shapes of BrF₅ (1 lone pair) and XeF₄ (2 lone pairs), both SN 6.

Forecast: with six directions all equal, where do the lone pairs sit and what shapes appear?

BrF₅: Br $V = 7$ , bonds $= 5$ → lone pairs $= \frac{7 - 5}{2} = 1$ . $SN = 5 + 1 = 6$ → type AX₅E₁. Why this step? Standard LP formula: a halogen brings $V = 7$ , spends 5 on bonds, leaving 2 electrons = 1 lone pair; six domains ⇒ octahedral base.
The single lone pair takes any one of the six equal corners; deleting it leaves the five F as a square pyramid (four F in a square base, one F on top), angles pushed just under 90°. Why this step? All six octahedral sites are equivalent, so the lone pair's placement is free; removing one corner from an octahedron is a square pyramid, and the lone pair shoves the four base F slightly below 90°.
XeF₄: Xe $V = 8$ , bonds $= 4$ → lone pairs $= \frac{8 - 4}{2} = 2$ . $SN = 4 + 2 = 6$ → type AX₄E₂. Why this step? Same LP formula: a noble gas brings $V = 8$ , spends 4 on bonds, leaving 4 electrons = 2 lone pairs; six domains ⇒ octahedral base.
The two lone pairs sit opposite each other (top and bottom, 180° apart); the four F spread into a flat square planar, exactly $90°$ . Why this step? Two lone pairs get as far apart as possible (180°, trans), which forces the four bonds into one plane — the strongest lone-pair separation and least LP–LP repulsion.

Verify: SN 6 ladder reads octahedral (0 LP) → square pyramidal (1 LP) → square planar (2 LP); XeF₄ has bond angles of exactly 90°. ✔

Group 6 — word problem & exam twist (cells K, L)

Worked example Example 11 — Ozone O₃ (cell K: real-world)

Statement: Ozone (O₃) filters UV in the stratosphere. The central oxygen bonds to two other oxygens (one single-ish, one double-ish — treat as two bonded neighbours) and carries one lone pair. Predict its shape and why it matters.

Forecast: linear like CO₂, or bent?

Central O: two bonded neighbours + 1 lone pair ⇒ $SN = 2 + 1 = 3$ → type AX₂E₁ → base trigonal planar (120°). Why this step? Three domains in a plane; count the lone pair as a domain just like an atom.
Delete the lone-pair corner → only two O–O directions remain → bent, angle a little under 120° (≈ 117°). Why this step? The one lone pair squeezes the two bonds slightly from the 120° ideal — exactly BF₃'s 120° baseline (Ex 5) minus a lone-pair squeeze.
Because the molecule is bent, its bond dipoles do not cancel ⇒ ozone is polar. Why this step? A bent shape breaks symmetry — see Dipole moment. This polarity helps ozone interact with and absorb UV energy.

Verify: SN 3 with 1 LP ⇒ bent, ≈117° (< 120°, > 109.5°). Bent ⇒ non-zero net dipole ⇒ polar. ✔

Worked example Example 12 — H₂O vs H₂S (cell L: exam twist)

Statement: Both have SN 4 with 2 lone pairs. H₂O is 104.5°. Is H₂S's angle larger or smaller (it is ≈92°)?

Forecast: same electron picture — why isn't the angle the same?

Both central atoms have SN 4 and 2 lone pairs (type AX₂E₂) ⇒ both bent. Why this step? Same steric number and same lone-pair count give the same VSEPR shape name — the counting recipe cannot tell them apart yet.
Sulfur is bigger and less electronegative than oxygen; its bonding pairs sit farther out and are held loosely, so they repel each other less, and the shape leans toward using nearly pure p-orbitals (≈90°). Why this step? Angle also depends on the central atom's size/electronegativity — see Bond angle and electronegativity and Hybridization. VSEPR alone predicts "bent"; these refinements set the exact number.
Result: H₂S ≈ $92° < 104.5°$ (H₂O).

Verify: $92° < 104.5°$ — same shape name (bent), smaller angle for the larger/less electronegative central atom. ✔

Recall One-line summary of the whole matrix

Count SN and lone pairs → base geometry → delete lone-pair corners → name the shape → each lone pair drops the angle a bit more, strongest when the atom is small and electronegative.

Connections

VSEPR Theory — every cell above is an entry in the VSEPR shape table.
Steric number and molecular shape — the counting engine used in all twelve examples.
Trigonal bipyramidal geometry — governs cells H and I (equatorial lone-pair rule).
Hybridization — explains why H₂S drifts toward 90°.
Bond angle and electronegativity — the refinement behind Example 12.
Dipole moment — why bent ozone (Example 11) is polar.