5th Year OL · Chemistry
Structure of Matter · Bonding and properties
Geometry, hybridisation and polarity
VSEPR theory: bonding pairs and lone pairs Lone pairs also take up space Sigma and pi bonds, and hybridisation Polarity of molecules Distorted angles and finer comparisons Summary
A Lewis diagram tells you which atoms are joined, but not the shape of the molecule. Electron pairs repel each other, so they spread out in space as far apart as they can. That one idea lets you predict the 3D shape of a molecule and then work out whether it is polar. It also explains why a stream of water from a tap bends towards a charged comb, while a stream of tetrachloromethane would not.
This page uses the AB_n notation of the specification, with lone pairs counted separately. Hybridisation (sp, sp², sp³) is not named in the Leaving Certificate specification: it is an extra model used here to explain the shapes. The sigma (σ) and pi (π) bonds from orbital overlap are on the course.
VSEPR theory: bonding pairs and lone pairs
Experiments show that beryllium fluoride, BeF₂, is a perfectly straight molecule (180°), while boron trifluoride, BF₃, is a flat equilateral triangle (120°). What rule decides how far apart the bonds sit?
Valence Shell Electron Pair Repulsion (VSEPR) theory says that the pairs of electrons around a central atom arrange themselves in space to be as far apart as possible. Molecules are written as AB_n, where A is the central atom and B the atoms bonded to it, with up to four regions of electrons around A. Here we also write the number of lone pairs as E, so a molecule is AB_nE_m: n bonded atoms and m lone pairs. For shape, a double or triple bond counts as one region of electrons.
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Bonding pairs
Lone pairs
Shape of molecule
Bond angle
Example
2
0
linear
180°
BeF₂
3
0
trigonal planar
120°
BF₃
4
0
tetrahedral
109.5°
CH₄
Beryllium in BeF₂ has 2 bonding pairs and no lone pair: AB₂. Boron in BF₃ has 3 bonding pairs and no lone pair: AB₃.
Two pairs are furthest apart at 180°, so BeF₂ is linear . Three pairs in a plane are furthest apart at 120°, so BF₃ is trigonal planar .
Neither Be nor B completes an octet, but that does not matter for the shape. Only the number of electron regions counts.
Next step (1/3)
Lone pairs also take up space
A lone pair is held by one nucleus only, so its electron cloud is spread out closer to the central atom. It repels neighbouring pairs more strongly than a bonding pair does: $$\text{lone pair–lone pair} > \text{lone pair–bonding pair} > \text{bonding pair–bonding pair}$$ Each lone pair squeezes the bond angles to a little below the ideal value.
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C, N and O each have 4 pairs of electrons around them, arranged tetrahedrally. They differ in lone pairs: CH₄ is AB₄, NH₃ is AB₃E and H₂O is AB₂E₂.
The shape of the molecule describes only the atoms: tetrahedral for CH₄, pyramidal (trigonal pyramid) for NH₃ and bent (V-shaped) for H₂O.
The bond angle falls along the series: 109.5° (no lone pair), 107.3° (one) and 104.5° (two). Each extra lone pair pushes the bonding pairs closer together.
Next step (1/3)
The four pairs around oxygen in water point to the corners of a tetrahedron, but the molecule is bent , because only atoms count when naming a shape. In an exam, give the shape of the molecule unless asked otherwise.
Sigma and pi bonds, and hybridisation
Ground-state carbon is 1s² 2s² 2p² with two unpaired electrons in p orbitals at 90°. Yet in methane it forms four identical bonds at 109.5°. How?
Hybridisation is a model in which the s and p orbitals of an atom mix to form new, equivalent hybrid orbitals pointing in the directions VSEPR predicts: sp (linear, 2 directions), sp² (trigonal planar, 3) and sp³ (tetrahedral, 4). In carbon, one 2s electron is promoted to the empty 2p orbital, giving four unpaired electrons to share.
A σ bond forms by head-on overlap of orbitals along the line joining the nuclei. A π bond forms by sideways overlap of two unhybridised p orbitals, above and below the bond axis. A single bond is one σ bond, a double bond is σ + π and a triple bond is σ + 2π.
Carbon hybridisation: four sp³ bonds in methane, three sp² bonds with a double bond in ethene, and two sp bonds with a triple bond in ethyne.
sp³ (CH₄)
sp² (ethene, C=C)
sp (ethyne, C≡C)
No experiment shows an sp³ orbital. It is a mathematical device that reproduces the observed shapes. It is a useful working model, not a measurable quantum state, and it has limitations: for example, it does not explain the spectra of molecules.
Polarity of molecules
Hold a charged comb near a thin stream of water and it bends towards the comb. With tetrachloromethane, CCl₄, nothing happens, even though its bonds are very polar. Why do some molecules respond to an electric field and others do not?
A bond is polar if the two atoms differ in electronegativity (Pauling scale). Draw it as an arrow pointing towards the more electronegative atom. A molecule is polar only if these bond dipoles do not cancel . A symmetrical shape can cancel polar bonds and give a non-polar molecule.
Why? In a symmetrical molecule the centres of positive and negative charge are at the same point, so there is no overall separation of charge. On this course you work with arrows and symmetry only: dipole moment values are not required.
Bond dipoles: in linear carbon dioxide the dipoles are opposite and cancel, so the molecule is non-polar; in bent water they add up to a net dipole.
CO₂ (linear)
dipoles cancel: non-polar
H₂O (bent)
dipoles add: polar
C
O
net dipole
O
O
H
H
A molecule with no lone pairs on the central atom is non-polar only if all the bonded atoms are the same. In chloromethane, CH₃Cl, the shape is tetrahedral, but C–Cl is much more polar than C–H, so the dipoles do not cancel and the molecule is polar.
★★☆ Not started
Carbon dioxide, $\mathrm{CO_2}$, and water, $\mathrm{H_2O}$, are triatomic molecules with polar bonds due to the high electronegativity of oxygen compared with that of… Carbon dioxide, $\mathrm{CO_2}$, and water, $\mathrm{H_2O}$, are triatomic molecules with polar bonds due to the high electronegativity of oxygen compared with that of carbon and hydrogen. However, their behaviour in an electrostatic field is radically different. The following illustration shows the spatial orientations and bond dipole moments for both substances:
Carbon dioxide (CO₂)
O
C
O
μ
μ
Water (H₂O)
μ
μ
O
H
H
a) Justify the molecular geometry of both species using VSEPR theory (specify the $\mathrm{AX}_n\mathrm{E}_m$ notation).b) Deduce, from the vector sum of the bond dipole moments, whether or not each molecule has a resultant molecular dipole moment. Solve
Distorted angles and finer comparisons
Phosgene, COCl₂: a trigonal planar molecule with a C=O double bond and two C–Cl bonds, where the Cl–C–Cl angle is 111.8°.
C
O
Cl
Cl
111.8°
Carbon has three regions of electrons (one C=O and two C–Cl) and no lone pair: AB₃, trigonal planar, sp² hybridisation.
The double bond region holds more electron density than a single bond, so it repels the C–Cl bonds more strongly, much as a lone pair does. The two C–Cl bonds close up a little below 120°, and the measured angle is 111.8°.
Next step (1/2)
Each carbon is sp². The C=C is a σ bond plus a π bond. The π bond stops the two ends of the molecule rotating, so the cis and trans forms are stable and can be separated.
In the cis isomer both C–Cl dipoles point to the same side and add up: the molecule is polar. In the trans isomer they point in opposite directions and cancel: it is non-polar.
Polar molecules attract each other by permanent dipole–dipole forces, so cis has the higher boiling point (60.3 °C) and trans the lower (47.5 °C).
Next step (1/3)
NF₃ has a much smaller overall dipole than NH₃, although fluorine is more electronegative than hydrogen. In NF₃ the N–F dipoles point towards the fluorine atoms, opposite to the effect of the lone pair, so they partly cancel. In NH₃ the N–H dipoles point towards nitrogen, the same way as the lone pair, so they add up. Always think about direction as well as size.
Aoife says NF₃ must be more polar than NH₃ because the electronegativity difference between F and N is bigger than between N and H. Is she right?
Show answer No. In NF₃ the bond dipoles point towards F, against the lone pair, so they largely cancel. In NH₃ the bond dipoles and the lone pair act in the same direction and add up.
Back to the start: water bends towards a charged comb because H₂O is bent, so its bond dipoles add up. CCl₄ is tetrahedral, so its dipoles cancel and it is not affected.
Summary
VSEPR theory predicts shape from the number of electron regions around the central atom (a multiple bond counts as one), written AB_n with m lone pairs: linear (180°), trigonal planar (120°), tetrahedral (109.5°).
Lone pairs repel more than bonding pairs and squeeze bond angles: CH₄ 109.5°, NH₃ 107.3°, H₂O 104.5°. Name the shape of the molecule, not the electron arrangement.
Head-on overlap gives σ bonds and sideways overlap of p orbitals gives π bonds: double = σ + π, triple = σ + 2π. Hybridisation (sp³, sp², sp) is a model that explains the shapes.
A molecule is polar only if its bond dipoles do not cancel. Symmetry can make a molecule with polar bonds non-polar, as in CO₂, BF₃ and CCl₄.
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