Chapter 4 · Chemical Bonding and Molecular Structure

4.2 VSEPR Theory, Molecular Shapes and Polarity

Electron pairs around a central atom repel each other and spread out as far as possible. From that single idea follow the shapes of molecules — linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral, and the bent, pyramidal, see-saw, T-shaped and square shapes created by lone pairs — and whether a molecule has a dipole moment.

Carbon dioxide and water are both molecules of the type $\mathrm{AB_{2}}$: a central atom with two atoms bonded to it. Yet carbon dioxide is straight, $\mathrm{O{=}C{=}O}$ with an angle of $180^{\circ}$, and has no dipole moment, while water is bent at $104.5^{\circ}$ and is strongly polar — which is why water dissolves salt, climbs up plant stems and heats up in a microwave oven, and carbon dioxide does none of these things. Lewis structures count electrons, but they say nothing about the arrangement of atoms in space.

In 1940 Nevil Sidgwick and Herbert Powell suggested a simple rule, which Ronald Gillespie and Ronald Nyholm developed in 1957 into the valence shell electron pair repulsion (VSEPR) theory: the electron pairs in the valence shell of the central atom repel one another and arrange themselves as far apart as possible. Counting the pairs, and knowing that lone pairs push harder than bond pairs, lets you predict the shape of almost any small molecule or ion of the main-group elements — and from the shape, its polarity.

The VSEPR Model

📘 Definition 4.8 — Postulates of VSEPR theory
  1. The shape of a molecule depends on the number of valence-shell electron pairs, bonded and non-bonded, around the central atom.
  2. Electron pairs repel each other because their electron clouds are negatively charged; they take up positions as far apart as possible, which minimises repulsion.
  3. The valence shell is treated as a sphere, with the electron pairs localised on its surface at maximum distance from one another.
  4. A multiple bond is treated as if it were a single electron pair (one "super pair"), and the two or three pairs of a multiple bond as a single domain.
  5. The VSEPR model applies to any structure that can be described by two or more resonance structures.

A lone pair is held by one nucleus only, so its cloud spreads out more widely near the central atom; a bond pair is shared between two nuclei and is pulled out along the bond. Lone pairs therefore occupy more space and repel more strongly.

📐 Rule 4.5 — Order of electron-pair repulsions
$$\text{lone pair–lone pair}\ \gt\ \text{lone pair–bond pair}\ \gt\ \text{bond pair–bond pair}$$ When lone pairs are present, the molecule takes the shape that keeps them as far as possible from other pairs, and the bond angles shrink below the ideal values of the arrangement. Repulsions at $90^{\circ}$ are far larger than those at $120^{\circ}$ or more, so only the $90^{\circ}$ contacts need to be compared when deciding between arrangements.
🧭 Problem-solving strategy — Predicting a shape with VSEPR
  1. Find the number of electron pairs (domains) around the central atom A. Either draw the Lewis structure and count the atoms bonded to A plus its lone pairs, or use the shortcut $$\text{pairs}=\tfrac12\,(V+M-c+a)$$ with $V$ the valence electrons of A, $M$ the number of monovalent atoms bonded to A (H, F, Cl, Br, I; doubly bonded O or S count $0$), $c$ the positive charge and $a$ the negative charge of an ion.
  2. Bond pairs = number of atoms bonded to A; lone pairs = pairs − bond pairs.
  3. The total number of pairs fixes the electron-pair geometry (2 linear, 3 trigonal planar, 4 tetrahedral, 5 trigonal bipyramidal, 6 octahedral).
  4. The shape of the molecule is described by the positions of the atoms only — lone pairs are invisible to the shape. Place lone pairs where they suffer the fewest $90^{\circ}$ repulsions.
Examples: $\mathrm{SO_{2}}$: $\tfrac12(6+0)=3$ pairs, 2 bonded O → 1 lone pair, bent. $\mathrm{NH_{4}^{+}}$: $\tfrac12(5+4-1)=4$, tetrahedral. $\mathrm{I_{3}^{-}}$: $\tfrac12(7+2+1)=5$, 2 bonded → 3 lone pairs.

When every pair is a bond pair (all the outer atoms identical), the molecule simply has the geometry of the pairs, with ideal angles:

Electron pairs23456
Geometry / shapelineartrigonal planartetrahedraltrigonal bipyramidaloctahedral
Bond angle(s)$180^{\circ}$$120^{\circ}$$109.5^{\circ}$$120^{\circ}$ (equatorial), $90^{\circ}$ (axial–equatorial)$90^{\circ}$
Example$\mathrm{BeCl_{2}}$$\mathrm{BF_{3}}$$\mathrm{CH_{4}}$, $\mathrm{NH_{4}^{+}}$$\mathrm{PCl_{5}}$$\mathrm{SF_{6}}$
✏️ Example 1 — Counting electron pairs
Use the shortcut to find the number of electron pairs, bond pairs and lone pairs on the central atom, and name the electron-pair geometry, for (a) $\mathrm{BCl_{3}}$; (b) $\mathrm{SiCl_{4}}$; (c) $\mathrm{NO_{3}^{-}}$; (d) $\mathrm{H_{3}O^{+}}$; (e) $\mathrm{SF_{6}}$; (f) $\mathrm{XeF_{4}}$.
(a), (b) $\mathrm{BCl_{3}}$: $\tfrac12(3+3)=3$; 3 bonded Cl → 3 bp, 0 lp; trigonal planar. $\mathrm{SiCl_{4}}$: $\tfrac12(4+4)=4$; 4 bp, 0 lp; tetrahedral.
(c) $\tfrac12(5+0+1)=3$ (the O atoms count 0, the charge adds 1); 3 bonded O → 3 bp, 0 lp; trigonal planar.
(d), (e) $\mathrm{H_{3}O^{+}}$: $\tfrac12(6+3-1)=4$; 3 bp, 1 lp; tetrahedral geometry. $\mathrm{SF_{6}}$: $\tfrac12(6+6)=6$; 6 bp; octahedral.
(f) $\tfrac12(8+4)=6$; 4 bp, 2 lp; octahedral electron-pair geometry (the shape is decided in the next subsection).
Check Lewis count for $\mathrm{XeF_{4}}$: Xe has 8 valence electrons, 4 are used in the four $\mathrm{Xe{-}F}$ bonds, 4 remain as 2 lone pairs ✓.

Shapes of Molecules with Lone Pairs

Lone pairs take part in the electron-pair geometry but are not "seen" in the shape, which describes the positions of the nuclei only. Ammonia has four pairs arranged tetrahedrally, but one of them is a lone pair: the three hydrogen atoms form a trigonal pyramid with nitrogen at the apex. Because the lone pair pushes harder than the bond pairs, the $\mathrm{H{-}N{-}H}$ angle closes from $109.5^{\circ}$ to $107^{\circ}$; with two lone pairs, water closes further to $104.5^{\circ}$.

PairsBond pairsLone pairsShapeExamples (angle)
321bent (V-shaped)$\mathrm{SO_{2}}$ ($119^{\circ}$), $\mathrm{O_{3}}$ ($117^{\circ}$)
431trigonal pyramidal$\mathrm{NH_{3}}$ ($107^{\circ}$), $\mathrm{H_{3}O^{+}}$
422bent (V-shaped)$\mathrm{H_{2}O}$ ($104.5^{\circ}$)
541see-saw$\mathrm{SF_{4}}$
532T-shaped$\mathrm{ClF_{3}}$
523linear$\mathrm{XeF_{2}}$, $\mathrm{I_{3}^{-}}$ ($180^{\circ}$)
651square pyramidal$\mathrm{BrF_{5}}$
642square planar$\mathrm{XeF_{4}}$, $\mathrm{ICl_{4}^{-}}$ ($90^{\circ}$)
📐 Rule 4.6 — Where the lone pairs go
  1. In a trigonal bipyramid the two axial positions each have three neighbours at $90^{\circ}$, the three equatorial positions only two. Lone pairs therefore occupy equatorial positions: one lone pair gives a see-saw, two a T-shape, three a linear molecule.
  2. In an octahedron all six positions are equivalent. One lone pair may go anywhere (square pyramid); a second lone pair goes opposite (trans to) the first, at $180^{\circ}$, giving a square planar molecule.
In every case the arrangement chosen is the one with the fewest lone pair–lone pair repulsions at $90^{\circ}$, and then the fewest lone pair–bond pair repulsions at $90^{\circ}$.
✏️ Example 2 — Comparing bond angles
(a) Explain why the bond angles fall in the order $\mathrm{CH_{4}}\ (109.5^{\circ})\gt\mathrm{NH_{3}}\ (107^{\circ})\gt\mathrm{H_{2}O}\ (104.5^{\circ})$. (b) The $\mathrm{F{-}N{-}F}$ angle in $\mathrm{NF_{3}}$ is $102^{\circ}$, smaller than the $\mathrm{H{-}N{-}H}$ angle in $\mathrm{NH_{3}}$. Explain. (c) In methanal, $\mathrm{H_{2}C{=}O}$ (3 electron domains on C), the $\mathrm{H{-}C{-}H}$ angle is about $116^{\circ}$ rather than $120^{\circ}$. Why?
(a) All three have 4 electron pairs (tetrahedral geometry). $\mathrm{CH_{4}}$ has 4 bond pairs: perfect tetrahedron, $109.5^{\circ}$. $\mathrm{NH_{3}}$ has 1 lone pair, whose stronger lp–bp repulsion squeezes the bond pairs together. $\mathrm{H_{2}O}$ has 2 lone pairs: the lp–lp repulsion is the strongest of all and squeezes the bond pairs further.
(b) Both have 3 bp + 1 lp. Fluorine is far more electronegative than hydrogen and pulls the $\mathrm{N{-}F}$ bond pairs away from nitrogen; bond pairs that are further from the central atom, and from each other, repel less, so the angle closes more. In $\mathrm{NH_{3}}$ nitrogen is the more electronegative atom and holds the bond pairs close to itself.
(c) The $\mathrm{C{=}O}$ double bond is one electron domain but contains two pairs; its larger electron cloud repels the $\mathrm{C{-}H}$ bond pairs more than they repel each other, so $\angle\mathrm{HCH}$ closes below $120^{\circ}$ (and each $\angle\mathrm{HCO}$ opens to about $122^{\circ}$; check: $116+2(122)=360$).
✏️ Example 3 — Five and six electron pairs
Predict the shapes of (a) $\mathrm{SF_{4}}$; (b) $\mathrm{ClF_{3}}$; (c) $\mathrm{XeF_{2}}$; (d) $\mathrm{BrF_{5}}$; (e) $\mathrm{ICl_{4}^{-}}$; (f) $\mathrm{I_{3}^{-}}$.
(a), (b) $\mathrm{SF_{4}}$: $\tfrac12(6+4)=5$ pairs, 4 bp + 1 lp (equatorial) → see-saw. $\mathrm{ClF_{3}}$: $\tfrac12(7+3)=5$, 3 bp + 2 lp (both equatorial) → T-shaped.
(c) $\tfrac12(8+2)=5$, 2 bp + 3 lp: all three lone pairs equatorial, the two F atoms axial → linear, $180^{\circ}$.
(d), (e) $\mathrm{BrF_{5}}$: $\tfrac12(7+5)=6$, 5 bp + 1 lp → square pyramidal. $\mathrm{ICl_{4}^{-}}$: $\tfrac12(7+4+1)=6$, 4 bp + 2 lp (trans) → square planar.
(f) $\tfrac12(7+2+1)=5$, 2 bp + 3 lp → linear, like $\mathrm{XeF_{2}}$ (the two species are isoelectronic in their valence shells).
✏️ Example 4 — Why is ClF₃ T-shaped?
$\mathrm{ClF_{3}}$ has 3 bond pairs and 2 lone pairs in a trigonal bipyramid. Consider the three ways of placing the lone pairs: (i) both axial; (ii) one axial, one equatorial; (iii) both equatorial. For each, count the lp–lp, lp–bp and bp–bp repulsions at $90^{\circ}$, and decide the shape.
Geometry In a trigonal bipyramid the $90^{\circ}$ contacts are exactly the axial–equatorial pairs: $2\times3=6$ of them. Equatorial–equatorial contacts are at $120^{\circ}$, axial–axial at $180^{\circ}$.
(i) lp–lp: $0$ (they are at $180^{\circ}$). lp–bp: each axial lp touches 3 equatorial bp → $6$. bp–bp: $0$. Planar triangle of F atoms.
(ii) lp–lp: $1$. lp–bp: axial lp with 2 equatorial bp, equatorial lp with 1 axial bp → $3$. bp–bp: $2$.
(iii) lp–lp: $0$ ($120^{\circ}$). lp–bp: each equatorial lp with 2 axial bp → $4$. bp–bp: $2$. (Totals: $0+6+0$, $1+3+2$, $0+4+2$ — each is $6$, as it must be.)
Decide (ii) has a $90^{\circ}$ lp–lp contact, the worst repulsion — rejected. Between (i) and (iii), (iii) has fewer lp–bp contacts ($4$ against $6$). So both lone pairs are equatorial and the molecule is T-shaped; the lone pairs push the axial F atoms back slightly, so $\angle\mathrm{F_{ax}ClF_{eq}}\approx87.5^{\circ}$, just under $90^{\circ}$.
Interactive 1 — VSEPR Shape Explorer

Choose the number of bond pairs and lone pairs on the central atom. The lab places the pairs as far apart as possible (lone pairs in the positions of Rule 4.6), draws the 3-D model — atoms in blue, lone pairs as purple lobes — and names the electron-pair geometry and the shape. It also adds the bond directions as vectors: if they cancel, a molecule with identical outer atoms is non-polar. Press ▶ Play to rotate the model.

electron-pair geometry—
shape—
bond angle(s)—
identical outer atoms →—

Dipole Moment and Molecular Shape

Section 4.1 showed that a bond between atoms of different electronegativity has a dipole moment, a vector pointing towards the more electronegative atom. A molecule with several polar bonds has a dipole moment equal to the vector sum of its bond moments (and of the small moments of its lone pairs). The shape decides whether the bond moments add up or cancel.

📐 Rule 4.7 — Molecular polarity from shape
  1. A molecule $\mathrm{AB}_{n}$ with identical outer atoms B arranged symmetrically with no lone pairs on A — linear $\mathrm{AB_{2}}$, trigonal planar $\mathrm{AB_{3}}$, tetrahedral $\mathrm{AB_{4}}$, trigonal bipyramidal $\mathrm{AB_{5}}$, octahedral $\mathrm{AB_{6}}$ — has $\mu=0$, however polar its bonds. The same is true of the symmetric lone-pair shapes linear $\mathrm{AB_{2}E_{3}}$ ($\mathrm{XeF_{2}}$) and square planar $\mathrm{AB_{4}E_{2}}$ ($\mathrm{XeF_{4}}$).
  2. Bent, pyramidal, see-saw, T-shaped and square pyramidal molecules, and molecules with different outer atoms ($\mathrm{CHCl_{3}}$), are generally polar.
  3. For a bent molecule $\mathrm{AB_{2}}$ with bond moment $\mu_{b}$ and bond angle $\theta$, the two bond moments add to $$\mu=2\mu_{b}\cos\tfrac{\theta}{2}$$ which is $0$ at $\theta=180^{\circ}$ and grows as the molecule bends.
Molecule$\mathrm{CO_{2}}$$\mathrm{BF_{3}}$$\mathrm{CCl_{4}}$$\mathrm{XeF_{4}}$$\mathrm{H_{2}O}$$\mathrm{H_{2}S}$$\mathrm{SO_{2}}$$\mathrm{NH_{3}}$$\mathrm{NF_{3}}$$\mathrm{CHCl_{3}}$
Shapelineartrig. planartetrahedralsq. planarbentbentbentpyramidalpyramidaltetrahedral
$\mu$ / D00001.850.951.631.470.231.04

Ammonia and nitrogen trifluoride are both pyramidal with one lone pair on nitrogen, yet $\mu(\mathrm{NH_{3}})=1.47\ \mathrm{D}$ and $\mu(\mathrm{NF_{3}})=0.23\ \mathrm{D}$. The difference lies in the directions of the vectors. The lone pair on N contributes a moment pointing away from the nucleus, towards the lone pair. In $\mathrm{NH_{3}}$ nitrogen is more electronegative than hydrogen, so the three $\mathrm{N{-}H}$ bond moments point towards N — the same way as the lone-pair moment — and they add. In $\mathrm{NF_{3}}$ fluorine is more electronegative than nitrogen, so the bond moments point away from N, towards the F atoms, and largely cancel the lone-pair moment.

✏️ Example 5 — Adding bond moments
(a) Water has $\mu=1.85\ \mathrm{D}$ and $\angle\mathrm{HOH}=104.5^{\circ}$. Find the $\mathrm{O{-}H}$ bond moment, treating the molecular moment as the vector sum of the two bond moments. (b) Repeat for $\mathrm{SO_{2}}$: $\mu=1.63\ \mathrm{D}$, angle $119^{\circ}$. (c) An $\mathrm{AB_{2}}$ molecule has bond moment $1.50\ \mathrm{D}$. Find $\mu$ if the bond angle is $180^{\circ}$, $120^{\circ}$ and $90^{\circ}$.
(a) $\mu=2\mu_{b}\cos\tfrac{\theta}{2}$, so $\mu_{b}=\dfrac{1.85}{2\cos52.25^{\circ}}=\dfrac{1.85}{2\times0.6122}=1.51\ \mathrm{D}$.
(b) $\mu_{b}=\dfrac{1.63}{2\cos59.5^{\circ}}=\dfrac{1.63}{2\times0.5075}=1.61\ \mathrm{D}$.
(c) $180^{\circ}$: $2(1.50)\cos90^{\circ}=0$. $120^{\circ}$: $2(1.50)\cos60^{\circ}=1.50\ \mathrm{D}$. $90^{\circ}$: $2(1.50)\cos45^{\circ}=2.12\ \mathrm{D}$.
Check At $120^{\circ}$ the two vectors and their sum form an equilateral triangle, so $\mu=\mu_{b}$ exactly ✓. (The lone pairs also contribute, so bond moments found this way are effective values.)
✏️ Example 6 — Polar or non-polar?
(a) Decide whether each molecule has a dipole moment: $\mathrm{BeCl_{2}}$, $\mathrm{SO_{2}}$, $\mathrm{BF_{3}}$, $\mathrm{PCl_{3}}$, $\mathrm{CCl_{4}}$, $\mathrm{CH_{2}Cl_{2}}$, $\mathrm{XeF_{4}}$, $\mathrm{SF_{4}}$, $\mathrm{PCl_{5}}$. (b) Why is $\mathrm{CHCl_{3}}$ polar while $\mathrm{CCl_{4}}$ is not? (c) $\mathrm{BF_{3}}$ has $\mu=0$ but $\mathrm{NF_{3}}$ has $\mu=0.23\ \mathrm{D}$. Explain.
(a) zero $\mathrm{BeCl_{2}}$ (linear), $\mathrm{BF_{3}}$ (trigonal planar), $\mathrm{CCl_{4}}$ (tetrahedral), $\mathrm{XeF_{4}}$ (square planar), $\mathrm{PCl_{5}}$ (trigonal bipyramidal): the bond moments cancel by symmetry.
(a) polar $\mathrm{SO_{2}}$ (bent), $\mathrm{PCl_{3}}$ (pyramidal: $\tfrac12(5+3)=4$ pairs, 1 lp), $\mathrm{CH_{2}Cl_{2}}$ (tetrahedral but with two different kinds of bond), $\mathrm{SF_{4}}$ (see-saw).
(b) In $\mathrm{CCl_{4}}$ four identical $\mathrm{C{-}Cl}$ moments point to the corners of a regular tetrahedron and cancel. In $\mathrm{CHCl_{3}}$ one corner holds H, whose bond moment is small and different; the three $\mathrm{C{-}Cl}$ moments are no longer balanced, leaving $1.04\ \mathrm{D}$.
(c) $\mathrm{BF_{3}}$ is trigonal planar with no lone pair on B: three equal moments at $120^{\circ}$ cancel. $\mathrm{NF_{3}}$ has a lone pair and is pyramidal: the bond moments do not cancel each other, and their resultant only partly cancels the lone-pair moment.
Interactive 2 — Adding Bond Dipoles in a Bent Molecule

Choose a molecule $\mathrm{AB_{2}}$, or set the bond angle and the bond moment yourself. The two blue bond-moment arrows are added by the parallelogram rule; the red arrow is the molecular dipole $\mu=2\mu_{b}\cos(\theta/2)$. The graph on the right shows how $\mu$ changes with the angle. Press ▶ Play to bend the molecule from $180^{\circ}$ to $60^{\circ}$ and back.

104.5°1.51 D
bond angle θ—
bond moment μ_b—
μ = 2μ_b cos(θ/2)—
measured μ—

Exercises

I
The VSEPR model and molecules with up to four electron pairs.
  • 1 State three postulates of VSEPR theory.
  • 2 Arrange lp–lp, bp–bp and lp–bp repulsions in decreasing order, and explain why a lone pair repels more than a bond pair.
  • 3 Find the number of electron pairs, bond pairs and lone pairs on the central atom of $\mathrm{BeH_{2}}$, $\mathrm{NO_{2}^{-}}$, $\mathrm{NH_{2}^{-}}$ and $\mathrm{PF_{5}}$.
  • 4 Predict the shapes of $\mathrm{PH_{3}}$, $\mathrm{H_{2}S}$, $\mathrm{BCl_{3}}$ and $\mathrm{SiH_{4}}$.
  • 5 What are the electron-pair geometry and the shape of $\mathrm{H_{3}O^{+}}$? Is its bond angle more or less than $109.5^{\circ}$?
  • 6 Why is $\mathrm{CO_{2}}$ linear but $\mathrm{SO_{2}}$ bent?
  • 7 Predict the shape of the carbonate ion and its $\mathrm{O{-}C{-}O}$ angle.
  • 8 Gaseous $\mathrm{SnCl_{2}}$ is a molecule. Predict its shape (Sn is in group 14).
1 Any three of: shape depends on the number of valence-shell pairs (bonded and lone); pairs repel and get as far apart as possible; the valence shell is a sphere with pairs on its surface; a multiple bond counts as one pair; lone pairs repel more than bond pairs.
2 lp–lp $\gt$ lp–bp $\gt$ bp–bp. A lone pair is attracted by one nucleus only, so its cloud is fatter and closer to the central atom; a bond pair is drawn out between two nuclei.
3 $\mathrm{BeH_{2}}$: $\tfrac12(2+2)=2$; 2 bp, 0 lp. $\mathrm{NO_{2}^{-}}$: $\tfrac12(5+0+1)=3$; 2 bp, 1 lp. $\mathrm{NH_{2}^{-}}$: $\tfrac12(5+2+1)=4$; 2 bp, 2 lp. $\mathrm{PF_{5}}$: $\tfrac12(5+5)=5$; 5 bp, 0 lp.
4 $\mathrm{PH_{3}}$: 4 pairs, 1 lp → trigonal pyramidal. $\mathrm{H_{2}S}$: 4 pairs, 2 lp → bent. $\mathrm{BCl_{3}}$: trigonal planar. $\mathrm{SiH_{4}}$: tetrahedral.
5 $\tfrac12(6+3-1)=4$ pairs (3 bp, 1 lp): tetrahedral electron-pair geometry, trigonal pyramidal shape; angle less than $109.5^{\circ}$ (lp–bp repulsion), like $\mathrm{NH_{3}}$, with which it is isoelectronic.
6 $\mathrm{CO_{2}}$: C has 4 valence electrons, all used in the two double bonds — 2 domains, no lone pair → linear. $\mathrm{SO_{2}}$: S has 6; 4 are used in bonding, leaving 1 lone pair — 3 domains → bent.
7 $\tfrac12(4+0+2)=3$ domains, no lone pair: trigonal planar, $120^{\circ}$ (all three C–O bonds are equivalent by resonance).
8 $\tfrac12(4+2)=3$ pairs: 2 bp + 1 lp → bent (angle below $120^{\circ}$).
II
Five and six electron pairs.
  • 1 Predict the shape of $\mathrm{SF_{4}}$ and state where its lone pair sits.
  • 2 Predict the shape of $\mathrm{BrF_{3}}$.
  • 3 Predict the shapes of $\mathrm{XeF_{2}}$ and $\mathrm{I_{3}^{-}}$. Why are they the same?
  • 4 Predict the shape of $\mathrm{BrF_{5}}$ and of $\mathrm{XeOF_{4}}$ (O doubly bonded to Xe).
  • 5 Why is $\mathrm{XeF_{4}}$ square planar rather than see-saw?
  • 6 Predict the shape of $\mathrm{ICl_{4}^{-}}$ and of $\mathrm{XeO_{3}}$.
  • 7 In $\mathrm{PCl_{5}}$ the axial bonds ($240\ \mathrm{pm}$) are longer than the equatorial bonds ($202\ \mathrm{pm}$). Explain with VSEPR.
  • 8 For $\mathrm{SF_{4}}$, count the lp–bp repulsions at $90^{\circ}$ when the lone pair is axial and when it is equatorial. Which is preferred?
1 5 pairs, 4 bp + 1 lp; the lone pair is equatorial → see-saw.
2 $\tfrac12(7+3)=5$, 3 bp + 2 lp (both equatorial) → T-shaped, like $\mathrm{ClF_{3}}$.
3 Both linear: $\mathrm{XeF_{2}}$ $\tfrac12(8+2)=5$ and $\mathrm{I_{3}^{-}}$ $\tfrac12(7+2+1)=5$, each 2 bp + 3 lp, the three lone pairs equatorial. The central atoms have the same number of valence-shell electrons (10).
4 $\mathrm{BrF_{5}}$: $\tfrac12(7+5)=6$, 5 bp + 1 lp → square pyramidal. $\mathrm{XeOF_{4}}$: $\tfrac12(8+4)=6$ (O counts 0), 5 atoms bonded + 1 lp → square pyramidal.
5 It has 6 pairs (4 bp + 2 lp), so the geometry is octahedral, not trigonal bipyramidal; the two lone pairs go trans ($180^{\circ}$ apart) to avoid a $90^{\circ}$ lp–lp contact, leaving the four F atoms in a square plane.
6 $\mathrm{ICl_{4}^{-}}$: $\tfrac12(7+4+1)=6$, 4 bp + 2 lp → square planar. $\mathrm{XeO_{3}}$: $\tfrac12(8+0)=4$, 3 bp + 1 lp → trigonal pyramidal.
7 Each axial bond pair has three bond pairs at $90^{\circ}$; each equatorial bond pair has only two. The axial bonds suffer more repulsion and lengthen.
8 Axial lp: $90^{\circ}$ contacts with the 3 equatorial bp → $3$. Equatorial lp: with the 2 axial bp → $2$. Equatorial is preferred (fewer strong repulsions).
III
Bond angles.
  • 1 Arrange $\mathrm{H_{2}O}$, $\mathrm{CH_{4}}$ and $\mathrm{NH_{3}}$ in order of increasing bond angle.
  • 2 Why is the $\mathrm{H{-}N{-}H}$ angle in $\mathrm{NH_{4}^{+}}$ larger than in $\mathrm{NH_{3}}$?
  • 3 Which has the larger bond angle, $\mathrm{H_{2}O}$ ($\angle\mathrm{HOH}$) or $\mathrm{OF_{2}}$ ($\angle\mathrm{FOF}$)? Explain.
  • 4 Explain why $\angle\mathrm{HSH}$ in $\mathrm{H_{2}S}$ ($92^{\circ}$) is much smaller than $\angle\mathrm{HOH}$ in water.
  • 5 Compare the bond angles in $\mathrm{BF_{3}}$ and $\mathrm{NF_{3}}$.
  • 6 Why is the $\mathrm{F_{ax}{-}S{-}F_{ax}}$ angle in $\mathrm{SF_{4}}$ about $173^{\circ}$ instead of $180^{\circ}$?
  • 7 Arrange $\mathrm{NO_{2}^{+}}$, $\mathrm{NO_{2}}$ and $\mathrm{NO_{2}^{-}}$ in order of decreasing $\mathrm{O{-}N{-}O}$ angle. ($\mathrm{NO_{2}^{+}}$ has no non-bonding electrons on N; $\mathrm{NO_{2}}$ has one; $\mathrm{NO_{2}^{-}}$ has a lone pair.)
  • 8 In methanal, $\angle\mathrm{HCH}\approx116^{\circ}$. Find the two $\angle\mathrm{HCO}$ angles if the molecule is planar and symmetrical.
1 $\mathrm{H_{2}O}\ (104.5^{\circ})\lt\mathrm{NH_{3}}\ (107^{\circ})\lt\mathrm{CH_{4}}\ (109.5^{\circ})$.
2 In $\mathrm{NH_{4}^{+}}$ the former lone pair has become a bond pair: four equal bond pairs give a regular tetrahedron, $109.5^{\circ}$. In $\mathrm{NH_{3}}$ the lone pair squeezes the angle to $107^{\circ}$.
3 $\mathrm{H_{2}O}$ ($104.5^{\circ}$, against $103^{\circ}$ for $\mathrm{OF_{2}}$). In $\mathrm{OF_{2}}$ the very electronegative F atoms pull the bond pairs away from O, so the bond pairs repel each other less.
4 S is larger and less electronegative than O; the $\mathrm{S{-}H}$ bond pairs lie further from the central atom and from each other, repel less, and are pushed together by the two lone pairs.
5 $\mathrm{BF_{3}}$: trigonal planar, $120^{\circ}$. $\mathrm{NF_{3}}$: pyramidal with a lone pair, $102^{\circ}$. $\mathrm{BF_{3}}$ has the larger angle.
6 The equatorial lone pair repels the two axial S–F bond pairs (lp–bp $\gt$ bp–bp), bending them back away from itself.
7 $\mathrm{NO_{2}^{+}}$ (2 domains, linear, $180^{\circ}$) $\gt$ $\mathrm{NO_{2}}$ (a single non-bonding electron repels weakly, about $134^{\circ}$) $\gt$ $\mathrm{NO_{2}^{-}}$ (a full lone pair, about $115^{\circ}$).
8 The angles round C sum to $360^{\circ}$: $2x+116=360$, $x=122^{\circ}$.
IV
Shape and dipole moment.
  • 1 Which of these molecules are polar: $\mathrm{CO_{2}}$, $\mathrm{SO_{2}}$, $\mathrm{BF_{3}}$, $\mathrm{NF_{3}}$, $\mathrm{CH_{4}}$, $\mathrm{CHCl_{3}}$, $\mathrm{XeF_{4}}$, $\mathrm{SF_{4}}$?
  • 2 The B–F bond is very polar. Why is $\mu(\mathrm{BF_{3}})=0$?
  • 3 Using an $\mathrm{O{-}H}$ bond moment of $1.51\ \mathrm{D}$ and $\angle\mathrm{HOH}=104.5^{\circ}$, calculate the dipole moment of water.
  • 4 Explain why $\mu(\mathrm{NH_{3}})\gt\mu(\mathrm{NF_{3}})$.
  • 5 A molecule $\mathrm{AB_{2}}$ has polar bonds but zero dipole moment. What is its shape?
  • 6 An $\mathrm{AB_{2}}$ molecule has bond moment $1.5\ \mathrm{D}$ and dipole moment $1.5\ \mathrm{D}$. Find the bond angle.
  • 7 $\mathrm{H_{2}S}$ has $\mu=0.95\ \mathrm{D}$ and a bond angle of $92^{\circ}$. Find the effective $\mathrm{S{-}H}$ bond moment.
  • 8 Applied. A microwave oven heats food because polar molecules try to line up with the rapidly changing electric field. Would it heat a cup of liquid tetrachloromethane, $\mathrm{CCl_{4}}$, quickly? Explain.
1 Polar: $\mathrm{SO_{2}}$, $\mathrm{NF_{3}}$, $\mathrm{CHCl_{3}}$, $\mathrm{SF_{4}}$. Non-polar: $\mathrm{CO_{2}}$, $\mathrm{BF_{3}}$, $\mathrm{CH_{4}}$, $\mathrm{XeF_{4}}$.
2 It is trigonal planar: three equal bond moments at $120^{\circ}$ add to zero (any two add to a vector equal and opposite to the third).
3 $\mu=2(1.51)\cos52.25^{\circ}=2(1.51)(0.6122)=1.85\ \mathrm{D}$.
4 In $\mathrm{NH_{3}}$ the N–H bond moments point towards N, in the same direction as the lone-pair moment, so they add. In $\mathrm{NF_{3}}$ the N–F moments point towards F, against the lone-pair moment, and largely cancel it.
5 Linear ($\theta=180^{\circ}$, $\cos90^{\circ}=0$).
6 $1.5=2(1.5)\cos\tfrac{\theta}{2}$ → $\cos\tfrac{\theta}{2}=0.5$ → $\tfrac{\theta}{2}=60^{\circ}$, $\theta=120^{\circ}$.
7 $\mu_{b}=\dfrac{0.95}{2\cos46^{\circ}}=\dfrac{0.95}{2\times0.6947}=0.68\ \mathrm{D}$.
8 No: $\mathrm{CCl_{4}}$ is tetrahedral and non-polar ($\mu=0$), so its molecules have no permanent dipole for the field to turn; it warms only slowly. Water ($1.85\ \mathrm{D}$) heats rapidly.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. NEET The shape of $\mathrm{ClF_{3}}$ is (A) trigonal planar (B) T-shaped (C) trigonal pyramidal (D) see-saw

Idea $\tfrac12(7+3)=5$ pairs: 3 bp + 2 lp, both lone pairs equatorial.
Answer (B).

E2. JEE Main Which pair of species has the same shape? (A) $\mathrm{XeF_{2}}$ and $\mathrm{CO_{2}}$ (B) $\mathrm{NH_{3}}$ and $\mathrm{BF_{3}}$ (C) $\mathrm{SF_{4}}$ and $\mathrm{CH_{4}}$ (D) $\mathrm{H_{2}O}$ and $\mathrm{BeCl_{2}}$

Shapes $\mathrm{XeF_{2}}$ linear (2 bp + 3 lp), $\mathrm{CO_{2}}$ linear; $\mathrm{NH_{3}}$ pyramidal vs $\mathrm{BF_{3}}$ planar; $\mathrm{SF_{4}}$ see-saw vs $\mathrm{CH_{4}}$ tetrahedral; $\mathrm{H_{2}O}$ bent vs $\mathrm{BeCl_{2}}$ linear.
Answer (A).

E3. Board · 3 marks $\mathrm{CH_{4}}$, $\mathrm{NH_{3}}$ and $\mathrm{H_{2}O}$ all have four electron pairs around the central atom, yet their bond angles are $109.5^{\circ}$, $107^{\circ}$ and $104.5^{\circ}$. Explain using VSEPR theory.

1 mark Four pairs arrange themselves tetrahedrally; with four bond pairs ($\mathrm{CH_{4}}$) the angle is the ideal $109.5^{\circ}$.
1 mark Repulsions decrease in the order lp–lp $\gt$ lp–bp $\gt$ bp–bp. $\mathrm{NH_{3}}$ has one lone pair, which pushes the three bond pairs closer: $107^{\circ}$.
1 mark $\mathrm{H_{2}O}$ has two lone pairs; the strong lp–lp repulsion compresses the bond angle further: $104.5^{\circ}$.

E4. JEE Main Numerical answer. How many of the following species are square planar? $\mathrm{XeF_{4}}$, $\mathrm{SF_{4}}$, $\mathrm{ICl_{4}^{-}}$, $\mathrm{BrF_{4}^{-}}$, $\mathrm{CCl_{4}}$, $\mathrm{XeO_{4}}$

Count $\mathrm{XeF_{4}}$: $\tfrac12(8+4)=6$, 4 bp + 2 lp → square planar. $\mathrm{SF_{4}}$: 5 pairs → see-saw. $\mathrm{ICl_{4}^{-}}$: $\tfrac12(7+4+1)=6$, 4 bp + 2 lp → square planar. $\mathrm{BrF_{4}^{-}}$: $\tfrac12(7+4+1)=6$ → square planar. $\mathrm{CCl_{4}}$: tetrahedral. $\mathrm{XeO_{4}}$: $\tfrac12(8+0)=4$, 4 bp → tetrahedral.
Answer 3.

E5. JEE Advanced One or more options correct. (A) $\mathrm{NF_{3}}$ has a smaller dipole moment than $\mathrm{NH_{3}}$. (B) In $\mathrm{SF_{4}}$ the lone pair occupies an axial position. (C) $\mathrm{BF_{3}}$ and $\mathrm{NF_{3}}$ both have zero dipole moment. (D) $\mathrm{XeF_{2}}$ is linear with three lone pairs on xenon.

(A) True ($0.23$ vs $1.47\ \mathrm{D}$): in $\mathrm{NF_{3}}$ the bond moments oppose the lone-pair moment.
(B) False: it is equatorial (2 lp–bp contacts at $90^{\circ}$ instead of 3).
(C) False: $\mathrm{NF_{3}}$ is pyramidal and polar.
(D) True: 5 pairs, the 3 lone pairs equatorial.
Answer (A), (D).

E6. Board · Case study Carbon dioxide and water are both triatomic. Their dipole moments are $0$ and $1.85\ \mathrm{D}$; the $\mathrm{H{-}O{-}H}$ angle is $104.5^{\circ}$. (i) Explain why $\mathrm{CO_{2}}$ is non-polar although each $\mathrm{C{=}O}$ bond is polar, while water is polar. (2 marks) (ii) Treating the dipole of water as the sum of two equal bond moments, find the $\mathrm{O{-}H}$ bond moment. (1 mark) (iii) Is $\mathrm{SO_{2}}$ polar? Give a reason. (1 mark)

(i) $\mathrm{CO_{2}}$ has 2 electron domains on C and no lone pair: it is linear, and the two equal bond moments point in opposite directions and cancel. Water has 2 bp + 2 lp: it is bent, so the two bond moments add to a resultant along the bisector of the angle.
(ii) $\mu_{b}=\dfrac{1.85}{2\cos52.25^{\circ}}=1.51\ \mathrm{D}$.
(iii) Yes: S has a lone pair (3 domains), so $\mathrm{SO_{2}}$ is bent and its bond moments do not cancel ($1.63\ \mathrm{D}$).
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Nitrogen and oxygen make up 99% of dry air, yet they play almost no part in the greenhouse effect, while traces of $\mathrm{CO_{2}}$, $\mathrm{CH_{4}}$, $\mathrm{N_{2}O}$ and water vapour do. What has molecular shape and polarity to do with it?

Why it matters. The Earth radiates heat as infrared (IR) light. A molecule can absorb IR only if one of its vibrations changes its dipole moment. The Intergovernmental Panel on Climate Change (IPCC) bases its assessments of global warming on how strongly each gas absorbs in the IR — so this simple rule underpins international climate policy, from the Paris Agreement to national emission inventories.

What to do.

  1. Use VSEPR to give the shape of $\mathrm{N_{2}}$, $\mathrm{O_{2}}$, $\mathrm{CO_{2}}$, $\mathrm{H_{2}O}$, $\mathrm{CH_{4}}$ and $\mathrm{SF_{6}}$, and decide which have a permanent dipole moment.
  2. A homonuclear diatomic molecule stays non-polar however much it stretches. Explain why $\mathrm{N_{2}}$ and $\mathrm{O_{2}}$ therefore cannot absorb IR by vibrating.
  3. $\mathrm{CO_{2}}$ is non-polar, but it absorbs strongly at about $15\ \mu\mathrm{m}$ when it bends. With Interactive 2, find the dipole created when $\mathrm{CO_{2}}$ bends to $170^{\circ}$, as a fraction of one bond moment ($\mu=2\mu_{b}\cos85^{\circ}$). Do the same for $160^{\circ}$.
  4. Look up the "global warming potential" of $\mathrm{CH_{4}}$, $\mathrm{N_{2}O}$ and $\mathrm{SF_{6}}$ (100-year values) in an IPCC report summary, and suggest why $\mathrm{SF_{6}}$, an octahedral molecule with six polar bonds, is such a powerful greenhouse gas.

Tools: Rule 4.7, Interactive 1 and Interactive 2, and an IPCC or national meteorological agency summary.

Hand in: A table of shapes and permanent dipoles, your two bent-$\mathrm{CO_{2}}$ calculations, and a paragraph explaining in your own words which molecules are greenhouse gases and why.

Going further. Argon (about 0.9% of air) is a single atom. Can it absorb IR by vibrating? What does that tell you about the IR activity of monatomic gases?

Instructor note: permanent dipoles only for $\mathrm{H_{2}O}$; $\mathrm{N_{2}}$, $\mathrm{O_{2}}$, $\mathrm{CO_{2}}$ (linear), $\mathrm{CH_{4}}$ (tetrahedral), $\mathrm{SF_{6}}$ (octahedral) have $\mu=0$, but the last three have vibrations (bending, asymmetric stretching) that create a temporary dipole. $2\cos85^{\circ}=0.174$ (17% of $\mu_{b}$); $2\cos80^{\circ}=0.347$ (35%). $\mathrm{SF_{6}}$: six very polar S–F bonds give intense IR absorption, and it survives thousands of years in the atmosphere; its 100-year GWP is over 20 000.

🧪 Self-check quiz

0/5 correct

Q1. The shape of $\mathrm{XeF_{4}}$ is

Q2. The correct order of bond angles is

Q3. Which molecule has zero dipole moment?

Q4. $\mathrm{SF_{4}}$ has four bond pairs and one lone pair. Its shape is

Q5. In a trigonal bipyramidal arrangement, lone pairs occupy equatorial positions because

← 4.1 Lewis Structures, Ionic Bonds and Bond Parameters4.3 Valence Bond Theory and Hybridisation →