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

Before we start, one shared toolkit — every problem below runs on these three lines, so we define them once, in plain words, anchored to a picture.

Look at the figure below: it is the map every answer walks across — count → SN → electron geometry → erase lone pairs → shape.

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

The SN→geometry dictionary we will reuse:

SN	Electron geometry	Ideal angle
2	linear	180°
3	trigonal planar	120°
4	tetrahedral	109.5°
5	trigonal bipyramidal	120° & 90°
6	octahedral	90°

Level 1 — Recognition

L1.1 For each species, state the steric number of the central atom (bold): CO₂, CH₄, NH₃, OH₂.

Recall Solution

Count atoms bonded + lone pairs.

CO₂: C double-bonds two O. Two domains, no LP → SN = 2.
CH₄: C bonds 4 H, $V = 4$ , all 4 electrons used, LP $= (4 - 4) /2 = 0$ → SN = 4.
NH₃: N bonds 3 H, $V = 5$ , LP $= (5 - 3) /2 = 1$ → SN = 4.
H₂O: O bonds 2 H, $V = 6$ , LP $= (6 - 2) /2 = 2$ → SN = 4.

L1.2 Which of these has the most lone pairs on the central atom: CH₄, NH₃, H₂O, HF?

Recall Solution

CH₄: 0 LP. NH₃: 1 LP. H₂O: 2 LP.
HF: F has $V = 7$ , one bond → LP $= (7 - 1) /2 = 3$ . HF wins with 3 lone pairs (though it is diatomic, so the "shape" idea is moot — the electron count still works).

Level 2 — Application

L2.1 Find the number of lone pairs, SN, and molecular shape of $PCl_{3}$ .

Recall Solution

P is group 15, $V = 5$ . It makes 3 P–Cl bonds → uses 3 electrons. LP $= (5 - 3) /2 = 1$ . SN $= 3 + 1 = 4$ → tetrahedral base → delete one corner → trigonal pyramidal, angle a touch under 109.5° (≈100°, since Cl is large/electronegative). Same family as NH₃.

L2.2 Find SN and shape of $H_{3} O^{+}$ (hydronium).

Recall Solution

Central O: $V = 6$ , but the + charge removes one electron → effective $V = 5$ . Bonds: 3 O–H → 3 electrons used. LP $= (5 - 3) /2 = 1$ . SN $= 3 + 1 = 4$ → tetrahedral base → trigonal pyramidal (just like NH₃, angle ≈ 107°).

L2.3 Find SN and shape of $BF_{3}$ , and explain why it is not pyramidal.

Recall Solution

B is group 13, $V = 3$ . Three B–F bonds use all 3 electrons. LP $= (3 - 3) /2 = 0$ . SN $= 3 + 0 = 3$ → trigonal planar, exactly 120°. No lone pair means nothing pushes the bonds out of plane, so unlike NH₃ it stays flat.

Level 3 — Analysis

L3.1 Rank the H–X–H (or F–X–F) bond angle from largest to smallest: $CH_{4}$ , $NH_{3}$ , $H_{2} O$ . Justify with lone-pair repulsion.

Recall Solution

All three have SN 4 (tetrahedral base). The difference is lone-pair count: 0, 1, 2. Each lone pair is a fatter, closer cloud (held by one nucleus) so it repels harder than a bond pair, squeezing the remaining bonds together. $CH_{4} (109.5°) > NH_{3} (107°) > H_{2} O (104.5°)$ Roughly −2.5° per lone pair added. See the figure — the wedge of bonds narrows as lone pairs pile on.

L3.2 $NF_{3}$ and $NH_{3}$ are both pyramidal (SN 4, 1 LP). Their measured angles are 102.5° (NF₃) and 107° (NH₃). Explain the difference using Bond angle and electronegativity.

Recall Solution

Fluorine is far more electronegative than hydrogen. It pulls the bonding electrons away from N, toward itself. A bonding pair dragged out toward F occupies less angular space near the central N. Thinner bond clouds near N repel each other less → the bonds can be squeezed closer → smaller angle (102.5°). With H (low electronegativity), the bond electrons sit nearer N, take more room, and hold the bonds farther apart → larger angle (107°).

Level 4 — Synthesis

L4.1 Work out the full story for $SF_{4}$ : LP count, SN, electron geometry, where the lone pair sits, and the final shape name. Use Trigonal bipyramidal geometry.

Recall Solution

S is group 16, $V = 6$ . Four S–F bonds use 4 electrons. LP $= (6 - 4) /2 = 1$ . SN $= 4 + 1 = 5$ → trigonal bipyramidal electron geometry (two axial + three equatorial sites). Where does the LP go? In a TBP, axial sites have three 90° neighbours; equatorial sites have only two 90° neighbours. Since 90° repulsions are the strongest strain, the fat lone pair takes the equatorial seat to minimize the number of close 90° elbowings. Delete that equatorial lone pair → the remaining four F atoms form a seesaw (angles slightly under 120° and 90°). See figure: the lone pair (red) sits in the equatorial belt; the removed corner leaves the seesaw of F atoms.

L4.2 Predict LP, SN, and shape of $ClF_{3}$ and $XeF_{2}$ .

Recall Solution

ClF₃: Cl group 17, $V = 7$ . Three Cl–F bonds → 3 used. LP $= (7 - 3) /2 = 2$ . SN $= 3 + 2 = 5$ → TBP base. Both lone pairs go equatorial (fewest 90° neighbours) → the three F atoms make a T-shape. XeF₂: Xe group 18, $V = 8$ . Two Xe–F bonds → 2 used. LP $= (8 - 2) /2 = 3$ . SN $= 2 + 3 = 5$ → TBP base. All three lone pairs go equatorial → the two F atoms sit axially → linear.

Level 5 — Mastery

L5.1 A central atom X (main-group) forms $b$ single bonds and has $l$ lone pairs, SN $= b + l = 4$ . Show that its molecular shape depends only on $l$ , and name the shape for $l = 0, 1, 2, 3$ .

Recall Solution

With SN 4, the electron geometry is always tetrahedral. The molecular shape = the tetrahedron with $l$ corners erased:

$l = 0$ : 4 atoms → tetrahedral (CH₄).
$l = 1$ : 3 atoms → trigonal pyramidal (NH₃).
$l = 2$ : 2 atoms → bent (H₂O).
$l = 3$ : 1 atom → linear (any diatomic like HF — trivially). Because $b = 4 - l$ is fixed by $l$ , the shape is a pure function of $l$ . This is exactly the Steric number and molecular shape recipe generalized.

L5.2 Explain, using this page's tools, why $XeF_{4}$ is square planar even though it has 2 lone pairs.

Recall Solution

Xe: $V = 8$ , four Xe–F bonds → 4 used. LP $= (8 - 4) /2 = 2$ . SN $= 4 + 2 = 6$ → octahedral base (six sites, all 90° apart). In an octahedron the sites are symmetric, so the two lone pairs minimize LP–LP repulsion by sitting opposite each other (180° apart, one "top," one "bottom"). Erase those two axial lone pairs → the four F atoms lie in one plane at 90° → square planar. The lone pairs are hidden above and below, keeping the visible shape perfectly flat.

Connections

VSEPR Theory — the framework every exercise applies.
Steric number and molecular shape — the counting engine used throughout.
Trigonal bipyramidal geometry — L4's equatorial-lone-pair rule.
Hybridization — SN 4 ↔ sp³, SN 5 ↔ sp³d, SN 6 ↔ sp³d².
Bond angle and electronegativity — the L3.2 NF₃/NH₃ contrast.
Dipole moment — square-planar XeF₄ and linear XeF₂ end up nonpolar despite lone pairs.