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

We assume you know only this: atoms have a middle (the nucleus, positively charged) and electrons around it (negatively charged). That is enough. Everything else — even the words "lone pair" — we build below.

Step 1 — What an "electron domain" actually is

WHAT: We stop counting individual electrons and start counting blobs. WHY: Electrons repel each other (same charge → they push apart). What matters for the shape is not each electron but each blob's direction — a blob points somewhere, and blobs shove each other away. Counting blobs is the whole trick. PICTURE: In the figure, the amber blob is a lone pair (fat, hugging the atom); the two cyan blobs are bonding pairs (thin, stretched toward the H nuclei).

Step 2 — Why blobs spread out: the "balloons on a stick" law

WHAT: We convert "electrons repel" into a geometric rule: maximise the minimum angle between blobs. WHY this rule and not another? Because repulsion energy grows when blobs get close. Nature settles into the lowest energy, which means the widest spread. There is no other force in play at this level — just push-apart. PICTURE: Two blobs → they go to opposite sides ( $180°$ ). Three → a flat triangle ( $120°$ ). Four → they cannot stay flat; they pop into 3-D as a tetrahedron ( $109.5°$ ). Watch the fourth balloon lift out of the page.

Recall Why does the fourth blob leave the plane?

If all four stayed flat you'd get only $90°$ between them (a cross). Lifting two of them out of the plane lets every pair reach $109.5°$ — bigger than $90°$ . Bigger minimum angle = less repulsion = what nature picks.

Step 3 — Counting the blobs: the steric number

To get the lone pairs we count the central atom's electrons:

$LP = \frac{V - ( electrons used in bonds )}{2}$

$V$ — the valence electrons the central atom brings (outer-shell electrons).
electrons used in bonds — one from the central atom per single bond.
We divide by 2 because leftover electrons must pair up (two per blob) to form a lone pair.

WHAT: A bookkeeping formula turning "how many electrons" into "how many blobs." WHY: SN alone fixes the electron geometry (Step 2). No chemistry left to guess. PICTURE: Two ledgers side by side — nitrogen ( $V = 5$ ) and oxygen ( $V = 6$ ) — showing electrons flowing into bonds (cyan) and the remainder pairing into lone pairs (amber).

For NH₃: $V = 5$ , three N–H bonds use $3$ → leftover $5 - 3 = 2$ → $LP = 2/2 = 1$ . So $SN = 3 + 1 = 4$ . For H₂O: $V = 6$ , two O–H bonds use $2$ → leftover $6 - 2 = 4$ → $LP = 4/2 = 2$ . So $SN = 2 + 2 = 4$ .

Both give SN = 4 → tetrahedral base (from Step 2). See Steric number and molecular shape for the full generalisation and Hybridization for why SN 4 means $s p^{3}$ .

Step 4 — The base tetrahedron, all four blobs equal

WHAT: Both CH₄, NH₃ and H₂O start from the same tetrahedron of four blobs at $109.5°$ . WHY $109.5°$ ? It is the exact angle four points on a sphere reach when spread as far as possible — the tetrahedral angle. (Verified numerically at the bottom of this page.) PICTURE: One shared tetrahedron drawn three times. In CH₄ all four blobs are cyan bonds. In NH₃ one corner turns amber (a lone pair). In H₂O two corners turn amber. Same skeleton, different paint.

Step 5 — Why the lone pair pushes harder (the "greedy blob")

A fatter, closer cloud sweeps out a wider angle and therefore shoves its neighbours harder. This gives the repulsion ladder:

$LP-LP > LP-BP > BP-BP$

LP–LP — two fat clouds fighting: strongest push.
LP–BP — one fat, one thin: medium.
BP–BP — two thin clouds: gentlest.

WHAT: We rank the four kinds of pushes by cloud fatness. WHY it must be this order: more electron density near the centre = more overlap with neighbours = more repulsion. Fatness tracks how many nuclei are pulling the cloud outward: fewer nuclei → fatter → closer. PICTURE: Side-by-side clouds — the amber lone-pair cloud drawn visibly fatter and hugging the nucleus; the cyan bond cloud drawn slim and stretched toward the neighbour. The amber wedge angle is wider.

See VSEPR Theory — this ladder is the heart of VSEPR.

Step 6 — Watch the angle collapse: 109.5° → 107° → 104.5°

WHAT: We turn the fat-cloud ladder into actual numbers on the drawn shape. WHY the bonds move and not the lone pairs? We report the angle between atoms (bonds). A fat lone pair spreads out and squeezes the bonds toward each other, so the reported H–X–H angle shrinks below $109.5°$ . PICTURE: Three overlaid molecules from the same centre. CH₄'s bonds sit at the wide $109.5°$ . NH₃'s single amber lone pair presses the three bonds inward to $107°$ . H₂O's two amber lone pairs press even harder (LP–LP is the strongest push) down to $104.5°$ . The amber wedge visibly widens as the cyan wedge narrows.

This angle-shrinking is why bent H₂O and pyramidal NH₃ are polar — the lone pairs break the symmetry (see Dipole moment). Ligand pull also nudges angles slightly, covered in Bond angle and electronegativity.

Step 7 — Edge cases: what happens at the extremes

Case A — Zero lone pairs (CH₄, NH₄⁺): No amber blob → nothing squeezes the bonds → they rest at the perfect $109.5°$ . This is why $N H_{4}^{+}$ (the lone pair donated away to $H^{+}$ ) jumps back up from $107°$ to $109.5°$ .

Case B — All atoms, no lone pair, but only 2 blobs (CO₂): SN $= 2$ → the two blobs go to opposite sides → $180°$ → linear, not bent. Same "two ligands" as water, but zero lone pairs, so no squeeze. This is the classic H₂O-vs-CO₂ trap.

Case C — A lone pair with a bigger base (SF₄, SN 5): The base is trigonal bipyramidal. The lone pair chooses the equatorial belt, where it meets only two $90°$ neighbours instead of three — dodging the strongest LP–BP pushes. Result: a seesaw.

WHAT: Three boundary molecules, each stressing a different variable (LP $= 0$ ; ligands $= 2$ ; SN $= 5$ ). WHY these three: they check the lower limit (no lone pairs), the "looks the same but isn't" limit (CO₂), and a higher-SN limit (SF₄). PICTURE: A three-panel strip: linear CO₂ ( $180°$ ), tetrahedral $N H_{4}^{+}$ ( $109.5°$ ), and seesaw SF₄ with the amber lone pair sitting in the equatorial plane.

The one-picture summary

Everything above, compressed: count blobs → tetrahedron → paint lone pairs amber → watch them squeeze the bonds.

Recall Feynman retelling — say it back in plain words

Around the central atom are little blobs of negative charge. Some blobs are bonds (pointing at a neighbour), some are lone pairs (pointing at nobody). All blobs push each other away, so four of them spread into a tetrahedron with $109.5°$ between them. A lone pair is held by just one nucleus, nothing pulls it away, so it stays fat and close and shoves the bonds harder than the bonds shove each other. One fat lone pair (ammonia) nudges the bonds in to $107°$ ; two fat lone pairs (water) shove even harder — because lone-pair-vs-lone-pair is the fiercest push — down to $104.5°$ . Take the lone pair away (as in $N H_{4}^{+}$ ) and the bonds relax straight back to the symmetric $109.5°$ . That's the whole story: count the blobs, spread them out, and let the greedy lone pairs squeeze.

Recall Quick self-check

Why is H₂O 104.5° and not 109.5°? ::: Its oxygen has two fat lone pairs (SN 4); their LP–LP and LP–BP repulsion squeezes the two O–H bonds inward from the tetrahedral ideal. Why does NH₄⁺ sit at exactly 109.5° while NH₃ is 107°? ::: NH₄⁺ has 0 lone pairs — all four blobs are equal bonds, so nothing squeezes and the bonds rest at the symmetric tetrahedral angle. Why is CO₂ linear but H₂O bent, though both are "AB₂"? ::: CO₂'s carbon has SN 2 (no lone pairs) → 180°; water's oxygen has SN 4 (two lone pairs) → bent.

Connections

VSEPR Theory — the repulsion ladder of Step 5 is its core.
Steric number and molecular shape — the counting of Step 3, generalised.
Hybridization — SN 4 → $s p^{3}$ orbitals on N and O.
Bond angle and electronegativity — the extra tweak on top of lone-pair squeezing.
Dipole moment — why the squeezed, asymmetric shapes are polar.
Trigonal bipyramidal geometry — the SN 5 base behind the SF₄ edge case.