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

This page is a self-test ladder. Cover the solutions, work each problem yourself, then reveal. Every problem builds on the parent topic. Before you start, make sure you can already draw Lewis Structures (you need the electron count), and it helps to have seen Hybridization and Bond Polarity and Dipole Moment.

Level 1 — Recognition

L1·Q1

State the electron geometry, molecular shape and ideal bond angle for a central atom with steric number 2 and no lone pairs.

Recall Solution — L1·Q1

WHAT: $S N = 2$ means two electron domains. WHY: Two clouds push to opposite ends of a line — that is the farthest two points can sit on a sphere. Answer: Electron geometry = linear, molecular shape = linear, angle $= 18 0^{\circ}$ .

L1·Q2

State electron geometry, molecular shape and angle for a central atom with steric number 3 and no lone pairs.

Recall Solution — L1·Q2

WHAT: Three domains, all bonding. WHY: Three points spread maximally on a sphere sit at the corners of a flat triangle, $12 0^{\circ}$ apart. Answer: Electron geometry = trigonal planar, molecular shape = trigonal planar, angle $= 12 0^{\circ}$ .

L1·Q3

A molecule has $S N = 4$ with zero lone pairs. Name the shape and angle.

Recall Solution — L1·Q3

Four domains spread to the corners of a tetrahedron. Shape = tetrahedral, angle $= 109. 5^{\circ}$ (exactly $arccos (- \frac{1}{3}) = 109.4 7^{\circ}$ ).

L1·Q4

Give electron geometry, molecular shape and characteristic angles for $S N = 5$ and $S N = 6$ , both with no lone pairs.

Recall Solution — L1·Q4

$S N = 5$ : five points → trigonal bipyramidal; molecular shape same (no LP). Angles $= 12 0^{\circ}$ (equatorial–equatorial) and $9 0^{\circ}$ (axial–equatorial).
$S N = 6$ : six points → octahedral; molecular shape same. Angle $= 9 0^{\circ}$ between all adjacent bonds.

Level 2 — Application

L2·Q1

Find $S N$ , electron geometry and molecular shape of $BeC l_{2}$ . (Be has 2 valence electrons.)

Recall Solution — L2·Q1

Step 1 — count electrons. Be valence $= 2$ . WHY: we always start from the central atom's own electrons, because those are the ones that arrange into bonds and lone pairs. Both electrons go into the two Be–Cl bonds → 0 lone pairs (LP). Step 2 — get $S N$ . $S N = 2 bonds + 0 LP = 2$ . WHY: the steric number counts every direction of electron density; here that is just the two bonds. Step 3 — read off the shape. Two domains push to opposite ends of a line. WHY: with no lone pairs there is nothing to bend the arrangement, so molecular = electron geometry. Answer: Electron geometry linear, molecular shape linear, angle $18 0^{\circ}$ .

L2·Q2

Find $S N$ and the molecular shape of $N H_{3}$ . (N has 5 valence electrons.)

Recall Solution — L2·Q2

Step 1: N valence $= 5$ . Three go to N–H bonds. Remaining $5 - 3 = 2$ electrons = 1 LP. Step 2: $S N = 3 bonds + 1 LP = 4$ . Step 3: Electron geometry = tetrahedral. Remove the lone-pair vertex (it is a "ghost", no atom there) → the three H atoms form a tripod. Answer: Molecular shape = trigonal pyramidal, angle $\approx 10 7^{\circ}$ (squeezed below $109. 5^{\circ}$ by the fatter lone pair).

What the figure shows: the cyan N sits above the tripod of white H atoms; the amber lobe is the lone pair pushing straight up and down on the bonds, which is why the H–N–H angle closes from $109. 5^{\circ}$ to $10 7^{\circ}$ . This is the picture behind the words "the lone pair squeezes the bonds."

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

L2·Q3

Find $S N$ and shape of $S O_{2}$ . (S central, two S=O bonds, S has 6 valence electrons.)

Recall Solution — L2·Q3

Step 1: Two double bonds to O. A double bond = one domain (one direction), so 2 bonding domains, using 4 of S's electrons (2 per double bond). Remaining $6 - 4 = 2$ electrons = 1 LP. Step 2: $S N = 2 bonded atoms + 1 LP = 3$ . Step 3: Electron geometry = trigonal planar. Remove the ghost lone-pair corner → two atoms left. Answer: Molecular shape = bent, angle a little under $12 0^{\circ}$ ( $\approx 11 9^{\circ}$ ).

Level 3 — Analysis

L3·Q1

Explain, using the axial-vs-equatorial argument, why the single lone pair in $S F_{4}$ ( $S N = 5$ ) sits equatorial, and name the resulting shape.

Recall Solution — L3·Q1

Step 1: $S N = 5$ → trigonal bipyramidal electron geometry: 2 axial positions ( $9 0^{\circ}$ from every equatorial neighbour) and 3 equatorial positions. Step 2 — WHY it matters: count the crowded $9 0^{\circ}$ neighbours.

Axial site: 3 neighbours at $9 0^{\circ}$ .
Equatorial site: only 2 neighbours at $9 0^{\circ}$ (the other two equatorials are $12 0^{\circ}$ away — much gentler). Step 3: The fat lone pair causes the worst repulsion, so it takes the least-crowded seat: equatorial. Answer: Lone pair equatorial → the four F atoms form a see-saw shape.

What the figure shows: the amber lobe (the LP) occupies one equatorial seat where it has only two $9 0^{\circ}$ white-F neighbours; the cyan note on the diagram contrasts this with the axial seat's three $9 0^{\circ}$ neighbours — this side-by-side count is why equatorial wins.

L3·Q2

$Cl F_{3}$ has $S N = 5$ with 2 lone pairs. Predict the shape and justify the lone-pair placement.

Recall Solution — L3·Q2

Step 1: Cl valence $= 7$ . Three go to Cl–F bonds; $7 - 3 = 4$ left = 2 LP. $S N = 3 + 2 = 5$ → trigonal bipyramidal. Step 2: Both lone pairs choose equatorial seats (each avoids the crowded 3-neighbour axial site). Placing both equatorial puts them $12 0^{\circ}$ apart, minimising the strong lone–lone repulsion. Step 3: The three F atoms then occupy the 2 axial + 1 equatorial positions → they line up in a slightly bent vertical column. Answer: Molecular shape = T-shaped. The two axial F–Cl bonds bend slightly toward the remaining equatorial F (lone pairs push them down), angle $\approx 87. 5^{\circ}$ .

L3·Q3

Compare the H–X–H bond angle order for $C H_{4}$ , $N H_{3}$ , $H_{2} O$ and explain the trend.

Recall Solution — L3·Q3

All three have $S N = 4$ (tetrahedral electron geometry) but different lone-pair counts: 0, 1, 2.

$C H_{4}$ : 0 LP → $109. 5^{\circ}$ .
$N H_{3}$ : 1 LP squeezes bonds → $10 7^{\circ}$ .
$H_{2} O$ : 2 LP squeeze harder → $104. 5^{\circ}$ . WHY: repulsion order is $LP-LP > LP-BP > BP-BP$ (lone pairs are fatter and closer, so they push harder). Each added lone pair pushes bonding pairs closer together. Answer (order): $C H_{4} (109. 5^{\circ}) > N H_{3} (10 7^{\circ}) > H_{2} O (104. 5^{\circ})$ .

Level 4 — Synthesis

L4·Q1

Derive the shape of $Xe F_{4}$ from scratch, including why it is square planar and not tetrahedral or "see-saw."

Recall Solution — L4·Q1

Step 1: Xe valence $= 8$ . Four go to Xe–F bonds; $8 - 4 = 4$ left = 2 LP. Step 2: $S N = 4 bonds + 2 LP = 6$ → electron geometry octahedral (six positions, all $9 0^{\circ}$ apart). Step 3 — where do the 2 lone pairs go? In an octahedron every position is equivalent, but two lone pairs want to be as far apart as possible. Placing them trans (opposite, $18 0^{\circ}$ ) removes all lone–lone $9 0^{\circ}$ contact — the minimum-repulsion choice. Step 4: Remove those two ghost positions (top and bottom). The four F atoms remain in one flat plane at $9 0^{\circ}$ . Answer: Square planar, F–Xe–F $= 9 0^{\circ}$ .

What the figure shows: the four white F atoms lie flat around cyan Xe at exact $9 0^{\circ}$ (amber corner marker), while the two dashed amber arrows are the trans lone pairs pointing above and below the plane — this makes visible why removing those two ghost positions leaves a flat square.

L4·Q2

$Xe F_{2}$ : predict the shape and state its polarity.

Recall Solution — L4·Q2

Step 1: Xe valence $= 8$ . Two go to bonds; $8 - 2 = 6$ left = 3 LP. $S N = 2 + 3 = 5$ → trigonal bipyramidal. Step 2: All three lone pairs take the roomy equatorial seats ( $12 0^{\circ}$ apart, no $9 0^{\circ}$ lone–lone crowding). The two F atoms are forced to the axial positions. Step 3: Two atoms on a straight axial line → shape = linear, F–Xe–F $= 18 0^{\circ}$ . Polarity: the two Xe–F bond dipoles point in exactly opposite directions and are equal, so they cancel → molecule is nonpolar (see Bond Polarity and Dipole Moment).

Level 5 — Mastery

L5·Q1

A central atom shows an experimental bond angle of exactly $109.4 7^{\circ}$ between two identical bonds. Using vectors, prove this is the tetrahedral angle (do not just quote it).

Recall Solution — L5·Q1

WHY vectors? A bond angle is the angle between two direction vectors from the central atom. The dot product answers "what is the angle between two vectors?" directly: $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$ Step 1: Put four tetrahedral vertices at alternating cube corners: $a = (1, 1, 1)$ , $b = (1, - 1, - 1)$ . Step 2: Dot product $a \cdot b = (1) (1) + (1) (- 1) + (1) (- 1) = 1 - 1 - 1 = - 1$ . Step 3: Magnitudes $∣ a ∣ = ∣ b ∣ = 1^{2} + 1^{2} + 1^{2} = 3$ . Step 4: $cos θ = \frac{- 1}{3 \cdot 3} = - \frac{1}{3}$ . Step 5: $θ = arccos (- \frac{1}{3}) = 109.4 7^{\circ}$ . ∎ The negative cosine tells us the angle is obtuse (>90°) — the bonds lean away from each other, exactly as repulsion demands.

L5·Q2

Rank the F–S–F "cis" angle (adjacent F atoms) of $S F_{6}$ , the axial–equatorial angle of $P F_{5}$ , and the H–O–H angle of $H_{2} O$ from smallest to largest, and justify each value.

Recall Solution — L5·Q2

$H_{2} O$ : $S N = 4$ , 2 lone pairs → $104. 5^{\circ}$ .
$S F_{6}$ : $S N = 6$ , octahedral, adjacent bonds $= 9 0^{\circ}$ .
$P F_{5}$ : $S N = 5$ , trigonal bipyramidal; axial–equatorial angle $= 9 0^{\circ}$ (equatorial–equatorial would be $12 0^{\circ}$ ). Order (smallest → largest): $S F_{6} (9 0^{\circ}) = P F_{5} ax-eq (9 0^{\circ}) < H_{2} O (104. 5^{\circ})$ . Justification: the two $9 0^{\circ}$ values come purely from octahedral/bipyramidal geometry; water's $104. 5^{\circ}$ is a tetrahedral $109. 5^{\circ}$ reduced by two lone pairs. So water's angle is the largest of the three.

L5·Q3

$I_{3}^{-}$ (triiodide anion): the central I bonds to two I atoms and carries the extra negative charge. Determine $S N$ and shape.

Recall Solution — L5·Q3

Step 1: Central I valence $= 7$ ; the $- 1$ charge adds 1 more electron → 8 electrons to place. Two are used in the two I–I bonds; $8 - 2 = 6$ remain = 3 LP. Step 2: $S N = 2 bonds + 3 LP = 5$ → trigonal bipyramidal electron geometry. Step 3: All three lone pairs take equatorial seats (as in $X e F_{2}$ ); the two bonded I atoms go axial. Answer: Shape = linear, I–I–I $= 18 0^{\circ}$ .

Recall Self-check ladder (reveal only after all 12)

L1: linear/180°; trigonal planar/120°; tetrahedral/109.5°; SN5 trig-bipyramidal (90° & 120°); SN6 octahedral 90°. L2: $B e C l_{2}$ linear; $N H_{3}$ pyramidal 107°; $S O_{2}$ bent ~119°. L3: $S F_{4}$ see-saw; $C l F_{3}$ T-shaped; angle order $C H_{4} > N H_{3} > H_{2} O$ . L4: $X e F_{4}$ square planar 90°; $X e F_{2}$ linear, nonpolar. L5: $arccos (- 1/3) = 109.47°$ ; $S F_{6} = P F_{5} = 90° < H_{2} O = 104.5°$ ; $I_{3}^{-}$ linear.