VSEPR theory — geometry from electron pairs (linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral, etc: Worked examples

This page is the drill-ground for VSEPR theory. The parent note built the machinery: count domains, get the steric number $S N$ , place domains at maximum separation, then subtract lone pairs to read off the atom shape. Here we hit every case class — no lone pairs, one, two, three lone pairs; every steric number from 2 to 6; a molecule with no central atom to lean on; a real-world word problem; and a nasty exam twist.

The scenario matrix

Cell	Case class	Representative	$(b, L)$	Why it's a distinct case
A	$S N = 2$ , no lone pairs	$C O_{2}$	$(2, 0)$	linear baseline; double bonds = one domain
B	$S N = 3$ , no lone pairs	$B F_{3}$	$(3, 0)$	electron geom = molecular geom
C	$S N = 3$ , one lone pair	$S O_{2}$	$(2, 1)$	first angle-squeeze in a flat family
D	$S N = 4$ , lone pairs 0→1→2	$C H_{4}, N H_{3}, H_{2} O$	$(4, 0), (3, 1), (2, 2)$	the shrinking-angle series
D+	$S N = 4$ , three lone pairs	$H C l$ (on Cl)	$(1, 3)$	one bond, three lone pairs — degenerate but same count
E0	$S N = 5$ , no lone pairs	$P F_{5}$	$(5, 0)$	trigonal bipyramidal baseline (two seat types)
E	$S N = 5$ , one lone pair	$S F_{4}$	$(4, 1)$	the "where does the lone pair go" case
E2	$S N = 5$ , two lone pairs	$C l F_{3}$	$(3, 2)$	both LPs equatorial → T-shaped
F	$S N = 5$ , three lone pairs	$X e F_{2}$	$(2, 3)$	lone-pair-heavy → linear
G	$S N = 6$ , no lone pairs	$S F_{6}$	$(6, 0)$	octahedral baseline
H	$S N = 6$ , one lone pair	$B r F_{5}$	$(5, 1)$	square pyramidal
I	$S N = 6$ , two lone pairs	$X e F_{4}$	$(4, 2)$	trans lone pairs → square planar
J	Word problem (real world)	$H_{2} O$ vs $C O_{2}$ polarity	$(2, 2)$ vs $(2, 0)$	shape decides polarity (no new domain count)
K	Exam twist	$I_{3}^{-}$	$(2, 3)$	an ion — charge changes the electron count

Worked Examples

Worked example Cell A —

$C O_{2}$ (linear, $S N = 2$ , no lone pairs) Statement: Predict the shape and bond angle of carbon dioxide, $C O_{2}$ .

Forecast: Guess the shape before reading on. Two oxygens on a carbon — a bent "smile" or a straight rod?

Central atom = C; count bonded atoms. C is double-bonded to each O, so $b = 2$ . Why this step? Each O is one atom joined to C; a double bond is one domain (one direction of density), so two oxygens give two domains, not four.
Count lone pairs on C. Carbon's 4 valence electrons are all used in the two double bonds, so $L = 0$ . Why this step? Only lone pairs on the central atom bend the shape; here there are none.
Steric number: $S N = b + L = 2 + 0 = 2$ . Why this step? $S N$ alone fixes the electron geometry.
Two domains spread maximally = opposite ends of a line, angle $18 0^{\circ}$ . Linear. Why this step? Two points on a sphere are farthest apart when diametrically opposite.

Figure walk-through: The violet sphere is carbon, the two magenta spheres are the oxygens, and the whole molecule lies on one straight navy line. The two orange arrows are the C=O bond dipoles; they point in exactly opposite directions along that line, so head-to-tail they cancel to zero — that is why the caption reads "linear, SN = 2, L = 0" and why $C O_{2}$ carries no net dipole.

Verify: $b + L = 2 = S N$ ; the two opposite dipoles cancel → $C O_{2}$ is nonpolar, matching lab fact. See Bond Polarity and Dipole Moment.

Worked example Cell B —

$B F_{3}$ (trigonal planar, $S N = 3$ , no lone pairs) Statement: Predict shape and angle of boron trifluoride.

Forecast: Boron has only 3 outer electrons. Flat triangle or pyramid?

Bonds: B forms 3 single bonds to 3 F atoms → $b = 3$ . Why this step? We count directions of electron density (three separate B–F bonds), not the raw electron total.
Lone pairs on B: boron brings 3 valence electrons, all three go into the three B–F bonds → $L = 0$ . Why this step? No leftover pairs means nothing bends the ideal shape, so molecular geometry will equal electron geometry.
$S N = 3 + 0 = 3$ → three domains at $12 0^{\circ}$ in a plane. Trigonal planar. Why this step? Three points maximally spread on a sphere sit at the corners of a flat equatorial triangle, $12 0^{\circ}$ apart.
Molecular geometry = electron geometry (no lone pairs), so the atoms also form a flat triangle. Why this step? With $L = 0$ there is no invisible lone-pair seat to erase, so what you draw for electrons is what you see for atoms.

Figure walk-through: The violet centre is boron and the three magenta corners are the fluorines, joined by navy bonds. Notice the three bonds fan out evenly; the orange $12 0^{\circ}$ label marks the angle between neighbouring bonds. Because $3 \times 12 0^{\circ} = 36 0^{\circ}$ , the bonds exactly sweep out the full plane — proof the molecule is flat, with no room (or reason) to pucker.

Verify: $3 \times 12 0^{\circ} = 36 0^{\circ}$ confirms coplanarity. Bond dipoles at $12 0^{\circ}$ sum to zero → nonpolar.

Worked example Cell C —

$S O_{2}$ (bent, $S N = 3$ , one lone pair) Statement: Predict the shape and estimate the bond angle of sulfur dioxide.

Forecast: Same "3 domains" family as $B F_{3}$ — but sulfur keeps a lone pair. Still flat-triangle straight, or bent?

Bonds: S bonds to 2 O atoms (treat as one $S = O$ + one $S - O$ resonance; either way $b = 2$ ). Why this step? Two oxygens = two directions of bonding density, regardless of single/double resonance.
Lone pairs on S: sulfur has 6 valence electrons; the bonds use 4, leaving one lone pair (one LP) → $L = 1$ . Why this step? Leftover pairs on the centre count into $S N$ and will distort the angle.
$S N = 2 + 1 = 3$ → electron geometry trigonal planar, ideal $12 0^{\circ}$ . Why this step? Three domains (2 bonds + 1 LP) still arrange as a flat triangle before distortion.
Erase the LP's seat, keep its shove. Two atoms + one LP → bent. The fat LP repels harder, so the ideal $12 0^{\circ}$ is dented downward; the measured O–S–O angle is $\approx 11 9^{\circ}$ (real molecules are fit to spectroscopy, so we quote the lab value rather than a derived one — VSEPR only predicts the direction of the shift, "below $12 0^{\circ}$ ", not the exact number). Why this step? Repulsion order lone–bond $>$ bond–bond squeezes the two S–O bonds together.

Verify: Electron geometry (trigonal planar) $\neq =$ molecular geometry (bent) exactly because $L \neq = 0$ . Angle $< 12 0^{\circ}$ : consistent with one LP squeezing.

Worked example Cell D — the

$C H_{4} \to N H_{3} \to H_{2} O$ series ( $S N = 4$ , $L = 0, 1, 2$ ) Statement: Show how the bond angle changes across methane, ammonia, water — all with $S N = 4$ .

Forecast: All three have four electron domains. Do their bond angles stay at $109. 5^{\circ}$ ?

$C H_{4}$ : C bonds 4 H, $L = 0$ . So $b = 4$ , $S N = 4 + 0 = 4$ → tetrahedral, angle $109. 5^{\circ}$ . Molecular geometry = electron geometry. Why this step? Carbon's 4 valence electrons all go into the 4 C–H bonds, leaving no LP; four domains spread to a tetrahedron ( $arccos (- 1/3) = 109.4 7^{\circ}$ , checked below).
$N H_{3}$ : N bonds 3 H, has $L = 1$ . So $b = 3$ , $S N = 3 + 1 = 4$ → tetrahedral electron geometry; atoms form a trigonal pyramid, H–N–H $\approx 10 7^{\circ}$ . Why this step? Nitrogen has 5 valence electrons; 3 make bonds and 2 form one LP, so $S N$ is still 4 but one seat is now an invisible, fat LP that pushes the bonds inward.
$H_{2} O$ : O bonds 2 H, has $L = 2$ . So $b = 2$ , $S N = 2 + 2 = 4$ → tetrahedral electron geometry; atoms form a bent shape, H–O–H $\approx 104. 5^{\circ}$ . Why this step? Oxygen has 6 valence electrons; 2 make bonds and 4 form two LPs, so two fat seats now squeeze the bonds even harder.
Read the trend. $109. 5^{\circ} \to 10 7^{\circ} \to 104. 5^{\circ}$ as $L$ goes $0 \to 1 \to 2$ . Why this step? Same $S N$ means the same tetrahedral skeleton; only the count of fat LPs changes how much the bonds get pushed in. These $10 7^{\circ}$ and $104. 5^{\circ}$ are experimental values — VSEPR predicts only that each extra LP shrinks the angle further, which the numbers confirm.

Figure walk-through: Three panels, same tetrahedral skeleton. Left ( $C H_{4}$ ): symmetric $109. 5^{\circ}$ , no lone pair. Middle ( $N H_{3}$ ): one orange "LP" arrow (LP = lone pair) sticks up from the centre, and the two navy bonds have closed slightly to $10 7^{\circ}$ . Right ( $H_{2} O$ ): the bonds have closed further to $104. 5^{\circ}$ — watch the V get narrower left-to-right exactly as $L$ climbs.

Verify: Angles decrease monotonically $109.5 > 107 > 104.5$ as $L$ increases $0 < 1 < 2$ . Exact tetrahedral value $arccos (- 1/3) = 109.4 7^{\circ}$ (checked below).

Worked example Cell D+ —

$H C l$ on chlorine ( $S N = 4$ , one bond, three lone pairs) Statement: Apply the same $(b, L)$ machinery to the chlorine atom in $H C l$ . Is a two-atom molecule an exception?

Forecast: Only two atoms — does VSEPR break, or does the framework still return an answer?

Look at the chlorine centre. Cl has 7 valence electrons; 1 goes into the H–Cl bond, leaving 6 electrons = 3 lone pairs (3 LP). So $b = 1$ , $L = 3$ , $S N = 1 + 3 = 4$ . Why this step? The framework never asked for a big molecule — it asks for bonds and lone pairs on one atom, and those are perfectly defined here.
$S N = 4$ → tetrahedral electron geometry around Cl (one bond seat + three LP seats). Why this step? Four domains always spread to a tetrahedron; three of the four seats simply happen to be lone pairs.
Read the atom shape. With only two atoms (H and Cl) there is no third atom to define an angle, so the molecule is trivially linear — a single bond axis. Why this step? Molecular geometry reports atom positions; two atoms always lie on a line, so the elaborate tetrahedral electron cloud collapses to a plain rod when you erase the LP seats. This shows the degenerate case is inside the same map, not outside it.

Verify: $S N = b + L = 1 + 3 = 4$ — the same tetrahedral electron geometry as water, but with one bond instead of two; atom shape = line (no angle to measure).

Worked example Cell E0 —

$P F_{5}$ (trigonal bipyramidal, $S N = 5$ , no lone pairs) Statement: Predict the shape of phosphorus pentafluoride — the baseline for $S N = 5$ .

Forecast: Five bonds and nothing else. Can five arrows share a sphere as evenly as three or six can?

Bonds & lone pairs: P bonds 5 F; phosphorus's 5 valence electrons are all used, so $L = 0$ . $b = 5$ , $S N = 5 + 0 = 5$ . Why this step? All valence electrons are spent on bonds, so there is no LP to bend anything — molecular geometry = electron geometry.
Five domains spread → trigonal bipyramid: 3 in an equatorial triangle ( $12 0^{\circ}$ apart) + 2 axial poles (top and bottom, $9 0^{\circ}$ to the equator). Why this step? Unlike $S N = 2, 3, 4, 6$ , five points cannot all be equivalent on a sphere — the best compromise splits them into two axial + three equatorial seats. This is the one geometry with two different kinds of seat, which is why the later $S N = 5$ cases have a "where does the LP go" question.
Molecular geometry = electron geometry (no LP). Trigonal bipyramidal. Why this step? With $L = 0$ every seat holds an atom, so the atom shape is the full bipyramid.

Verify: $S N = 5$ , two seat types (axial $9 0^{\circ}$ , equatorial $12 0^{\circ}$ ); no LP ⇒ undistorted trigonal bipyramid. This is the baseline for Cells E, E2, F.

Worked example Cell E —

$S F_{4}$ (see-saw, $S N = 5$ , one lone pair) Statement: Where does the lone pair sit in $S F_{4}$ , and what shape results?

Forecast: In a trigonal bipyramid there are two kinds of seat: axial (top/bottom) and equatorial (the triangle around the waist). Which does the LP grab?

Bonds & lone pairs: S bonds 4 F, has $L = 1$ . $S N = 4 + 1 = 5$ . Why this step? $S N = 5$ gives trigonal bipyramidal electron geometry — the first shape with two different kinds of seat.
Count $9 0^{\circ}$ neighbours per seat. An axial seat faces 3 equatorial atoms at $9 0^{\circ}$ ; an equatorial seat faces only 2 axial atoms at $9 0^{\circ}$ (the other two equatorials are at $12 0^{\circ}$ ). Why this step? $9 0^{\circ}$ contacts are the closest and therefore worst repulsions; whichever seat has fewer of them is the "roomier" one.
LP → equatorial (fewer $9 0^{\circ}$ clashes). Erase that seat: atoms form a see-saw. Why this step? The fat LP minimises its worst repulsions by taking the equatorial seat with only two $9 0^{\circ}$ neighbours instead of three.

Figure walk-through: Two vertical navy bonds are the axial S–F pairs, two side navy bonds are equatorial S–F. The thick orange arrow points into the third equatorial seat — the lone pair (LP) — sitting in the roomy waist. Erase that orange seat and the four fluorines trace the tilting "see-saw" plank the shape is named after.

Verify: Equatorial LP has $2$ ninety-degree neighbours vs axial's $3$ ; $2 < 3$ , so equatorial wins. Shape "see-saw" ✔.

Worked example Cell E2 —

$C l F_{3}$ (T-shaped, $S N = 5$ , two lone pairs) Statement: Predict the shape of chlorine trifluoride — the $S N = 5$ case with two lone pairs.

Forecast: Three F atoms, two lone pairs on a trigonal-bipyramidal skeleton. Do the fluorines make a flat triangle, a see-saw, or a "T"?

Bonds & lone pairs: Cl bonds 3 F; chlorine has 7 valence electrons; 3 go to bonds, leaving 4 electrons = 2 lone pairs (2 LP). $b = 3$ , $L = 2$ , $S N = 3 + 2 = 5$ . Why this step? Two leftover pairs sit on the centre, so $S N = 5$ and we again choose axial-vs-equatorial for the LPs.
Both LPs take equatorial seats. Why this step? Each LP wants the roomy equatorial seat (only two $9 0^{\circ}$ neighbours, from Cell E); with two LPs, the two best equatorial seats are filled, keeping the strong lone–lone repulsion at the wide $12 0^{\circ}$ equatorial separation rather than a cramped $9 0^{\circ}$ .
The three F atoms occupy the remaining seats: the two axial poles + one equatorial seat. Erase the two equatorial LP seats → the three F make a flat T-shape (two axial F at top/bottom, one equatorial F out to the side), with the two straight arms bent slightly inward by the fat LPs. Why this step? Reading atom positions after removing the LP seats leaves a "T": the two axial bonds and one equatorial bond, all in one plane.

Figure walk-through: The violet centre is chlorine. One navy bond goes straight up, one straight down (the axial F atoms), and one navy bond goes out to the right (the equatorial F). The two orange arrows point left into the two empty equatorial seats — the two lone pairs (LP), both at the roomy waist and $12 0^{\circ}$ apart. Erase those two orange seats and the three fluorines trace the capital letter "T".

Verify: Two LPs both equatorial → three F at the "T" positions; axial arms bent under $9 0^{\circ}$ by LP push. Shape "T-shaped" ✔.

Worked example Cell F —

$X e F_{2}$ (linear, $S N = 5$ , three lone pairs) Statement: Predict the shape of xenon difluoride — the "lone-pair-heavy" trigonal bipyramid.

Forecast: With three lone pairs and only two bonds, will the two F atoms end up bent, T-shaped, or straight?

Bonds & lone pairs: Xe bonds 2 F. Xe has 8 valence electrons; 2 go to bonds, leaving 6 electrons = 3 lone pairs (3 LP). $S N = 2 + 3 = 5$ . Why this step? Even a noble-gas centre obeys the $S N$ count; the extra electrons must go somewhere, and here they form three lone pairs.
All 3 LPs take the equatorial seats. Why this step? From Cell E, equatorial seats have the fewest $9 0^{\circ}$ neighbours; the three fat pairs each want that roominess, and there are exactly three equatorial seats to fill.
The two F atoms are forced to the two axial positions, $18 0^{\circ}$ apart → linear. Why this step? Once the three equatorial seats are all occupied by LPs, the only seats left for the two bonded fluorines are the two axial poles, which lie on one straight line at $18 0^{\circ}$ — so the molecule is linear.

Verify: Two atoms $18 0^{\circ}$ apart = linear ✔. LP count $3$ = number of equatorial seats, forcing atoms axial.

Worked example Cell G —

$S F_{6}$ (octahedral, $S N = 6$ , no lone pairs) Statement: Predict the shape of sulfur hexafluoride.

Forecast: Six identical bonds and nothing else — how do six arrows share a sphere?

Bonds & lone pairs: S bonds 6 F; sulfur's 6 valence electrons are all used, so $L = 0$ . $b = 6$ , $S N = 6 + 0 = 6$ . Why this step? All valence electrons are spent on bonds, leaving no LP to distort anything.
Six domains spread maximally → octahedron, every neighbour $9 0^{\circ}$ apart. Why this step? Six points farthest apart on a sphere sit at the vertices of an octahedron (two square pyramids base-to-base); this is the pure-geometry answer.
Molecular geometry = electron geometry (no lone pairs). Perfectly octahedral, undistorted. Why this step? With $L = 0$ there is no fat seat to erase, so atoms and domains coincide.

Verify: $S N = 6$ , all angles $9 0^{\circ}$ , no distortion (baseline case for the octahedral family H and I below).

Worked example Cell H —

$B r F_{5}$ (square pyramidal, $S N = 6$ , one lone pair) Statement: Predict the shape of bromine pentafluoride.

Forecast: Octahedral skeleton, but one seat is a lone pair. Do the five fluorines stay flat, or pucker into a pyramid?

Bonds & lone pairs: Br has 7 valence electrons; 5 go to F bonds, leaving 2 electrons = 1 lone pair (1 LP). $b = 5$ , $L = 1$ , $S N = 5 + 1 = 6$ . Why this step? Counting leftover electrons on Br gives exactly one LP, which will occupy one octahedral vertex.
Put the LP at one vertex of the octahedron (all six vertices are equivalent in a bare octahedron). Why this step? With only one LP there is no "trans vs cis" choice yet — any single vertex is the same by symmetry.
Erase that vertex; the remaining five F form a square base plus one apex → square pyramidal. The LP below the base pushes the four basal F slightly up, so the F–Br–F angles dip just under $9 0^{\circ}$ . Why this step? Reading atom positions after removing the LP seat leaves a square pyramid, and the fat LP (repelling harder than bonds) bends the base fluorines away from it.

Figure walk-through: The violet centre is bromine; four magenta fluorines sit in a square around the waist and one magenta fluorine sits at the top apex. The orange arrow points straight down to the empty sixth vertex where the lone pair (LP) lives. Because that LP shoves upward, the four square-base bonds tilt up a touch — that is the "just under $9 0^{\circ}$ " note.

Verify: $S N = 6$ , $L = 1$ ⇒ octahedral electron geometry, square-pyramidal atom shape; angle slightly below $9 0^{\circ}$ from the single LP.

Worked example Cell I —

$X e F_{4}$ (square planar, $S N = 6$ , two lone pairs) Statement: Predict the shape of xenon tetrafluoride.

Forecast: Six domains → octahedral skeleton. Two lone pairs — do they sit next to each other ( $9 0^{\circ}$ ) or opposite ( $18 0^{\circ}$ )?

Bonds & lone pairs: Xe bonds 4 F; 8 valence − 4 bonding = 4 electrons = 2 lone pairs (2 LP). $S N = 4 + 2 = 6$ → octahedral electron geometry. Why this step? Six domains spread to the six vertices of an octahedron, all $9 0^{\circ}$ ; two of those seats are lone pairs.
Place the two LPs trans (opposite), $18 0^{\circ}$ apart. Why this step? Lone–lone repulsion is the strongest of all; putting the two fat pairs opposite maximises their separation ( $18 0^{\circ}$ is the farthest two vertices can be on an octahedron) and so minimises that strongest repulsion.
Erase the two axial LP seats; the four F atoms stay in the equatorial square → square planar. Why this step? Reading atom positions only (LPs invisible) leaves a flat square of four fluorines.

Figure walk-through: The violet centre is xenon; the four magenta fluorines form a flat square in the plane of the page, each bond $9 0^{\circ}$ from its neighbour (orange label). The two text notes above and below mark the lone pairs (LP) sticking straight out of the page front and back — they are trans, $18 0^{\circ}$ apart, which is why the four visible atoms are perfectly flat.

Verify: Trans LPs $= 18 0^{\circ}$ (the max possible on an octahedron) ✔. Four F at $9 0^{\circ}$ in a plane = square planar; opposite dipoles cancel → nonpolar.

Worked example Cell J — Word problem: why is water a great solvent but

$C O_{2}$ is not? (polarity from shape) Statement: A student notices water is an excellent solvent (polar) while $C O_{2}$ is not. Both have two identical bonds to a central atom. Use VSEPR to explain the difference. (This cell adds no new $(b, L)$ — it re-uses Cells A and D to show why shape matters.)

Forecast: Same "two bonds" — why is one polar and one not?

$C O_{2}$ : $S N = 2$ , $L = 0$ → linear (from Cell A). The two C=O bond dipoles point exactly opposite ( $18 0^{\circ}$ ) and cancel → net dipole $= 0$ → nonpolar. Why this step? Shape decides whether the bond dipoles cancel; a straight line puts them head-to-head so they annihilate — see Bond Polarity and Dipole Moment.
$H_{2} O$ : $S N = 4$ , $L = 2$ → bent at $104. 5^{\circ}$ (from Cell D). The two O–H dipoles are not opposite; they add to a net downward dipole → polar. Why this step? A bent frame leaves the two dipole vectors at $104. 5^{\circ}$ , so they cannot cancel; their vector sum points down through the oxygen.
Conclusion: water's bent shape (caused by its two lone pairs) is precisely why it is polar and a good solvent, while $C O_{2}$ 's linear shape makes it nonpolar. Why this step? This ties the whole page together: the same domain-counting that gave the shape now, one step later, hands us the polarity.

Verify: Two opposite unit vectors sum to zero; two vectors at $104. 5^{\circ}$ sum to a nonzero vector (checked in VERIFY). Geometry $\Rightarrow$ polarity ✔.

Worked example Cell K — Exam twist:

$I_{3}^{-}$ (an ion, $S N = 5$ ) Statement: Predict the shape of the triiodide ion, $I_{3}^{-}$ . (The trap: it is charged.)

Forecast: Three iodines and a $- 1$ charge. T-shaped? Bent? Linear?

Pick the central atom = middle I; count electrons carefully. The central I brings 7 valence electrons, plus 1 extra from the ion's $- 1$ charge → 8 electrons available for domains. Why this step? The charge changes the electron count — forgetting it is the classic trap that gives the wrong $L$ .
Bonds: central I bonds 2 terminal I atoms → $b = 2$ , using 2 electrons. Remaining $8 - 2 = 6$ electrons = 3 lone pairs (3 LP) → $L = 3$ . Why this step? Domain counting is identical to a neutral molecule once the charge's electron is added in.
$S N = 2 + 3 = 5$ → trigonal bipyramidal electron geometry. All 3 LPs go equatorial, the two bonded I atoms go axial → linear ( $18 0^{\circ}$ ). Why this step? Exactly the $X e F_{2}$ logic (Cell F): three equatorial seats swallow the three LPs, leaving only the two axial poles for the bonded atoms — a straight line.

Verify: $S N = 5$ , $L = 3$ ⇒ same class as Cell F ⇒ linear ✔. Adding the $- 1$ charge's electron was essential: without it you'd wrongly get 7 electrons and mis-count.

Recall Rapid self-test across all cells

$C O_{2}$ shape and why ::: Linear; $S N = 2$ , no lone pairs, double bonds count as one domain each. $S O_{2}$ vs $B F_{3}$ — same $S N$ , different shape? ::: Both $S N = 3$ ; $B F_{3}$ trigonal planar ( $L = 0$ ), $S O_{2}$ bent ( $L = 1$ squeezes it). $H C l$ in the $(b, L)$ framework? ::: On Cl: $b = 1$ , $L = 3$ , $S N = 4$ (tetrahedral electron geometry); atom shape is a trivial line. $P F_{5}$ shape and why two seat types? ::: Trigonal bipyramidal ( $S N = 5$ , $L = 0$ ); five points can't all be equal on a sphere, so 2 axial + 3 equatorial. Why does the LP go equatorial in $S F_{4}$ ? ::: Equatorial has only two $9 0^{\circ}$ neighbours vs axial's three, so it minimises the worst repulsion. $C l F_{3}$ shape? ::: T-shaped; $S N = 5$ , both lone pairs equatorial, three F at the "T" positions. $B r F_{5}$ shape? ::: Square pyramidal; $S N = 6$ , one lone pair takes one octahedral vertex. $X e F_{4}$ shape and lone-pair placement? ::: Square planar; two lone pairs sit trans ( $18 0^{\circ}$ ) on an octahedron. How does the $- 1$ charge change $I_{3}^{-}$ ? ::: It adds one electron to the central atom, giving $L = 3$ , $S N = 5$ → linear. Why is water polar but $C O_{2}$ not? ::: Water is bent so O–H dipoles don't cancel; $C O_{2}$ is linear so C=O dipoles cancel.

Related: Lewis Structures (to get the domains), Hybridization (the orbital picture of the same shapes), Formal Charge (to pick the best Lewis structure), Molecular Orbital Theory (the deeper bonding model).