VSEPR and Hybridization: Study Guide (Chemistry 2.7)

Why electron pairs spread out

A Lewis diagram tells you which atoms are bonded and where the lone pairs are. It does not tell you the shape. Water drawn as H–O–H on paper is not a straight molecule. To get from the diagram to the shape, you use valence shell electron pair repulsion (VSEPR) theory.

The idea is Coulombic. Every bonding pair and every lone pair around an atom is a region of negative charge. Negative regions repel one another, so they arrange themselves around the atom to be as far apart as possible. That arrangement fixes the bond angles and so the shape.

Each region is called an electron domain. To count them on an atom:

• each lone pair is one domain;
• each bond to another atom is one domain, whether it is single, double or triple.

The second rule matters. In CO₂ the carbon has two double bonds, but those are only two domains, because each double bond points in one direction, towards one oxygen atom.

From domains to shape

Step 1: draw the best Lewis diagram (Topics 2.5 and 2.6). Step 2: count the domains on the central atom. Step 3: the number of domains gives the electron-domain geometry. Step 4: name the molecular geometry from the positions of the atoms only, ignoring the lone pairs.

Domains	Electron-domain geometry	Lone pairs	Molecular geometry	Ideal angles	Hybridization	Example
2	linear	0	linear	180°	sp	CO₂
3	trigonal planar	0	trigonal planar	120°	sp²	BF₃
3	trigonal planar	1	bent	just under 120°	sp²	SO₂
4	tetrahedral	0	tetrahedral	109.5°	sp³	CH₄
4	tetrahedral	1	trigonal pyramidal	under 109.5°	sp³	NH₃
4	tetrahedral	2	bent	under 109.5°	sp³	H₂O
5	trigonal bipyramidal	0	trigonal bipyramidal	90°, 120°, 180°	shape only	AsF₅
5	trigonal bipyramidal	1	seesaw	about 90° and 120°	shape only	SF₄
5	trigonal bipyramidal	2	T-shaped	about 90°	shape only	ClF₃
5	trigonal bipyramidal	3	linear	180°	shape only	I₃⁻
6	octahedral	0	octahedral	90°, 180°	shape only	SF₆
6	octahedral	1	square pyramidal	about 90°	shape only	IF₅
6	octahedral	2	square planar	90°, 180°	shape only	ICl₄⁻

You need all eleven molecular geometry names in this table. Two names appear twice: “linear” (2 domains, or 5 domains with 3 lone pairs) and “bent” (3 or 4 domains). Always say how many domains the bent shape comes from, because the angles differ.

Lone pairs push harder

A lone pair is held by only one nucleus, so it stays closer to the central atom and spreads out more widely than a bonding pair does. It repels neighbouring domains more strongly. The order of repulsion is:

lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair

So the more lone pairs on a central atom, the more the bonds are pushed together. Methane, ammonia and water all have 4 domains, but their angles fall as lone pairs are added (Figure 1).

Multiple bonds push harder too. A double or triple bond holds more electron density in its domain than a single bond. In methanal, H₂C=O, the carbon has three domains, but the H–C–H angle is a little less than 120° because the C=O domain pushes the two C–H bonds together. On the exam, “about 120°” or “slightly less than 120°” are both reasonable; a number such as 90° is not.

Five and six domains

With five domains, the positions are not all the same. The three equatorial positions lie in a triangle at 120°; the two axial positions point up and down. An axial domain has three neighbours at 90°; an equatorial domain has only two. Lone pairs take the roomier equatorial positions first. That is why one lone pair gives a seesaw, two give a T-shape and three give a linear molecule.

With six domains, all positions are equivalent and 90° apart. The first lone pair can go anywhere (square pyramidal). A second lone pair goes opposite the first, 180° away, giving a square planar shape.

For five or six domains you only need the shape and approximate angles. The course does not ask for the hybridization of these atoms.

Worked example 1: two oxides of sulfur

Question. For sulfur dioxide, SO₂, and sulfur trioxide, SO₃, give the molecular geometry, the approximate O–S–O angle, the hybridization of sulfur, and whether the molecule has a dipole moment.

SO₂. Electrons: 6 + 2(6) = 18. An octet diagram is O=S–O with one lone pair on S (S has 1 double bond, 1 single bond and 1 lone pair). Resonance (Topic 2.6) moves the double bond, but in every structure S has 2 bonded atoms + 1 lone pair = 3 domains.

• Electron-domain geometry: trigonal planar. Molecular geometry: bent.
• Angle: just under 120° (the lone pair pushes the bonds together).
• Hybridization: 3 domains → sp².
• Dipole moment: yes. Each S–O bond is polar (O is more electronegative). The bent shape means the two bond dipoles both point partly in the same direction, so they add up instead of cancelling.

SO₃. Electrons: 6 + 3(6) = 24. Whether you draw one S=O and two S–O bonds (octet on S) or three S=O bonds (formal charges all zero), S has 3 bonded atoms and no lone pairs = 3 domains.

• Molecular geometry: trigonal planar, O–S–O 120°.
• Hybridization: sp².
• Dipole moment: no. Resonance makes the three S–O bonds identical. Three equal bond dipoles 120° apart cancel exactly.

Interpretation. Both sulfur atoms are sp², yet one molecule is polar and the other is not. The lone pair makes the difference. Notice also that the choice between Lewis diagrams for SO₃ does not change the domain count: shape depends on how many directions the electrons point in, not on how many pairs are in each bond.

Worked example 2: a central atom with six domains

Question. Predict the shape of iodine pentafluoride, IF₅, the approximate F–I–F bond angles, and whether IF₅ has a dipole moment.

1. Electrons: 7 + 5(7) = 42.
2. Five I–F single bonds use 10. Three lone pairs on each F use 5 × 6 = 30. That leaves 42 − 40 = 2 electrons, one lone pair on iodine (an expanded octet: I has 12 electrons).
3. Domains on I: 5 bonded atoms + 1 lone pair = 6. Electron-domain geometry: octahedral.
4. One position holds the lone pair, so the atoms form a square pyramid: four F atoms in a square, the fifth F above the middle of the square, and the lone pair pointing down.
5. Angles: the four F atoms of the square are pushed slightly up and away from the lone pair, so the angles between the top F and the square F atoms are a little less than 90°.
6. Dipole moment: yes. The bonds are polar and the lone pair on one side makes the shape unsymmetrical, so the bond dipoles do not cancel.

No hybridization is asked for. With six domains, you are only responsible for the shape.

What if there were two lone pairs? With 6 domains and 2 lone pairs, the lone pairs sit opposite each other and the four atoms form a square plane. That shape is symmetrical, so a molecule with four identical polar bonds in a square plane has no dipole moment.

Hybridization: naming the atom

Chemists describe an atom with 2, 3 or 4 electron domains by its hybridization. You only need the names and what they tell you about the angles:

Electron domains on the atom	Hybridization	Ideal angle
2	sp	180°
3	sp²	120°
4	sp³	109.5°

Count the domains on the atom you are asked about, lone pairs included. The oxygen in water has 2 bonds and 2 lone pairs, so it is sp³ even though only two atoms are attached. The rule applies to every atom in a molecule, not only to the “central” one.

How hybrid orbitals are derived from s and p orbitals, and how they are drawn, is not assessed. Focus on the link between domains, name and angle.

Sigma and pi bonds

A covalent bond forms where orbitals on two atoms overlap. There are two kinds of overlap (Figure 2):

• A sigma (σ) bond has its overlap on the line between the two nuclei (head-on). Every single bond is a σ bond.
• A pi (π) bond has its overlap above and below that line (side-on). A π bond is only ever formed in addition to a σ bond.

Bond	σ bonds	π bonds
single	1	0
double	1	1
triple	1	2

Pi bonds are weaker. Side-on overlap is less effective than head-on overlap, so a π bond has a lower bond energy than a σ bond. The course bond energies from Topic 2.2 show this: C–C is 347 kJ mol⁻¹ and C=C is 614 kJ mol⁻¹. The extra π bond adds only 614 − 347 = 267 kJ mol⁻¹, less than the σ bond. A double bond is stronger than a single bond, but less than twice as strong (2 × 347 = 694 kJ mol⁻¹).

Pi bonds stop rotation. Atoms joined by only a σ bond can rotate about the bond, because head-on overlap is the same at any angle. Twisting a π bond would pull its side-on lobes out of line and break it. So groups on either side of a double bond are locked in place. This creates geometric isomers. In 1,2-dichloroethene, ClHC=CHCl, the two Cl atoms can be on the same side of the C=C bond (cis) or on opposite sides (trans). These are two different compounds. The cis isomer has a dipole moment; in the trans isomer the two C–Cl bond dipoles point in opposite directions and cancel.

Worked example 3: hybridization and bonds in a chain molecule

Question. Acrylonitrile has the structure CH₂=CH–C≡N. (a) Count the σ and π bonds. (b) Give the hybridization of each carbon and of nitrogen. (c) Give the approximate C–C–C and C–C≡N angles.

Check the Lewis diagram. Electrons: 3(4) + 3(1) + 5 = 20. C=C (4) + three C–H (6) + C–C (2) + C≡N (6) + lone pair on N (2) = 20. ✓

(a) Every bond contains one σ bond: C=C, three C–H, C–C and C≡N give 6 σ bonds. The C=C adds 1 π bond and the C≡N adds 2: 3 π bonds.

(b) Count domains on each atom:

Atom	Domains	Hybridization
C of CH₂=	3 (two H, one C)	sp²
C of =CH–	3 (one H, two C)	sp²
C of –C≡N	2 (one C, one N)	sp
N	2 (one C, one lone pair)	sp

(c) The middle carbon is sp², so C–C–C is about 120°. The nitrile carbon is sp, so C–C≡N is 180°: those three atoms lie in a straight line.

Interpretation. Hybridization is a property of each atom, so one molecule can contain sp, sp² and sp³ atoms side by side. The fastest route is always: count domains on that atom, then name it.

Bond order, bond length and bond energy

Two factors decide how long and how strong a bond is:

• Bond order. More shared pairs between the same two atoms pull the nuclei closer and hold them more tightly. C≡C is shorter and stronger than C=C, which is shorter and stronger than C–C. With resonance, use the average bond order from Topic 2.6.
• Atomic radius. Larger atoms have their bonding electrons further from the nucleus, so their bonds are longer and usually weaker. Down group 17, H–F < H–Cl < H–Br < H–I in length.

Compare like with like: bond order alone decides only when you compare bonds between the same pair of elements, such as C–O with C=O.

Dipole moments: polarity plus shape

A molecule has a dipole moment when one end is overall partly negative and the other end partly positive. You need two things to decide:

1. Are any bonds polar? If the bonded atoms differ in electronegativity, the bond has a bond dipole pointing towards the more electronegative atom.
2. Do the bond dipoles cancel? Treat each bond dipole as an arrow. If identical polar bonds are arranged symmetrically around the central atom, the arrows cancel.

Shapes that cancel when all the outer atoms are the same: linear (2 domains), trigonal planar, tetrahedral, trigonal bipyramidal, octahedral, square planar, and linear from 5 domains. Shapes with lone pairs that leave one side different (bent, trigonal pyramidal, seesaw, T-shaped, square pyramidal) do not cancel, so polar bonds give a dipole moment. Changing one outer atom also breaks the symmetry: CCl₄ has no dipole moment, but CH₂Cl₂ does.

Common misconceptions

• “A double bond counts as two domains.” It is one domain. CO₂ has 2 domains on C and is linear.
• “Lone pairs are part of the shape name.” The molecular geometry describes atoms only. NH₃ has a tetrahedral electron-domain geometry but a trigonal pyramidal shape.
• “All bent molecules have 109.5° or 104.5° angles.” A bent shape from 3 domains (SO₂) is near 120°; from 4 domains (H₂O) it is near 104.5°.
• “Lone pairs in a trigonal bipyramid go on the axis.” They go in the equatorial positions, where they have fewer neighbours at 90°.
• Forgetting lone pairs when naming hybridization. Count every domain on the atom: N in C≡N is sp, O in water is sp³.
• “A double bond is twice as strong as a single bond.” The π part is weaker than the σ part.
• “Polar bonds always make a polar molecule.” Symmetrical shapes with identical bonds, such as SO₃ and CCl₄, have no dipole moment.
• Giving sp³d or sp³d² labels. For more than four domains, give the shape only.

Where this leads

This topic completes the Lewis-diagram toolkit of Topic 2.5, Lewis Diagrams and Topic 2.6, Resonance and Formal Charge. Next, Topic 3.1, Intermolecular and Interparticle Forces, uses the dipole moments you predicted here to explain how molecules attract one another, and so their boiling points and solubilities. Try the practice questions now, then use the revision notes and the checklist to consolidate.