Advanced Theories of Covalent Bonding
Lewis dots tell you which atoms share pairs. They do not tell you why a shared pair holds, why a double bond refuses to twist, or why liquid oxygen sticks to a magnet. For that you need orbitals. In this chapter you will build two pictures of the same bond. The first keeps each atom's own orbitals and lets them overlap. The second mixes orbitals across the whole molecule. Each one pays off somewhere the other falls short. Together they explain shape, stiffness, bond strength and magnetism with the same small set of rules.
By the end of this chapter you can
- Explain a covalent bond as two orbitals that overlap and share a spin‑paired pair of electrons.
- Read an energy curve to find a bond length and a bond energy.
- Count regions of electron density and name the hybrid set: sp, sp2, sp3, sp3d or sp3d2.
- Tell a sigma bond from a pi bond, and count both in a molecule.
- Say why a double bond locks rotation while a single bond spins.
- Fill a molecular orbital diagram for any period 1 or period 2 diatomic.
- Work out bond order, then predict bond length and paramagnetic or diamagnetic behaviour.
Section 1Valence bond theory and orbital overlap
A bond is a shared pair in a shared space
Valence bond theory keeps the atoms you already know. Each atom brings its own orbitals to the meeting. A bond forms when an orbital on one atom overlaps an orbital on the other. Two electrons then move into that shared space. Both nuclei pull on that pair at once. The double pull is the glue.
The key idea is orbital overlap. Overlap means two orbitals share some of the same space. More overlap piles more electron density between the nuclei. More density there makes a stronger bond. Overlap also has a direction, because orbitals point. A p orbital points along one axis. The bond it builds points along that axis too. That is why this theory predicts shapes so well.
One rule limits every bond to a pair. The two shared electrons must have opposite spins. Spin‑paired electrons in overlapping orbitals lower the energy. The same two electrons with matching spins raise it instead. So a full orbital has nothing to offer a new partner. That is why two He atoms meet and bounce apart.
Why atoms stop at one best distance
Slide two H atoms together and track the energy. Far apart they barely notice each other, so we call that energy zero. As they close in, each nucleus starts to pull on the other atom's electron. The energy falls. Lower energy means a steadier pair.
Keep pushing and the fall stops. Both nuclei carry positive charge, so they begin to shove. The energy climbs fast. Between the two trends sits a dip. The bottom of the dip is the bond length, the distance the atoms really keep. The depth of the dip is the bond energy, the price of pulling them apart again.
For H2 that dip sits at 74 pm and runs 436 kJ/mol deep. A picometre is 10−12 m, so 74 pm is a very short reach. Note that 436 kJ/mol is a mole of bonds, not one bond. The next example turns it into the cost of a single bond.
Worked example · the energy of one bond
The H–H bond energy is 436 kJ/mol. How much energy does it take to break just one H–H bond?
Step 1 · put the energy in joules436 kJ/mol × 1000 J/kJ = 436,000 J/mol.
Step 2 · divide by the number of bonds in a moleOne mole holds NA = 6.022 × 1023 bonds.
436,000 J/mol ÷ (6.022 × 1023 /mol) = 7.24 × 10−19 J
Step 3 · check it backwards7.24 × 10−19 J × 6.022 × 1023 = 4.36 × 105 J, which is 436 kJ. The units cancel to joules per bond, so the answer is the energy of one bond.
That is a tiny number, and it should be. One bond is one pair of atoms. The big number only shows up because a mole is such a huge count.
- A covalent bond is two orbitals overlapping with a spin‑paired pair of electrons inside.
- More overlap means more density between the nuclei, so a stronger bond.
- Orbitals point, so bonds point. Overlap sets the shape of the molecule.
- The dip in an energy curve gives the bond length at its bottom and the bond energy as its depth.
- For H2 those values are 74 pm and 436 kJ/mol.
Section 2Hybrid orbitals
The shape problem that forces mixing
Carbon has one 2s orbital and three 2p orbitals in its valence shell. The 2s is round. The three 2p orbitals point along x, y and z, at 90° to each other. Yet CH4 has four identical bonds at 109.5°. Plain s and p orbitals cannot give that. Something has to change.
The fix is hybridization. Orbitals on the same atom are waves, and waves on the same atom can add together. Mix one s with three p orbitals and you get four new hybrid orbitals. Each one has the same shape, the same energy, and one fat lobe pointing out. Four fat lobes spread as far apart as they can. That lands them at 109.5°, which is exactly what CH4 shows.
Mixing never creates or destroys orbitals. Put four orbitals in and you take four out. The names just record the recipe. An sp3 set comes from one s plus three p. An sp2 set comes from one s plus two p, leaving one p orbital untouched. An sp set comes from one s plus one p, leaving two p orbitals untouched. Those leftover p orbitals matter a lot in section 3.
Counting your way to the answer
You never have to guess a hybrid set. Count regions of electron density on the central atom. A region is one sigma bond or one lone pair. A double or triple bond still counts as one region, because the whole group points one way. The number of regions equals the number of hybrid orbitals you need.
Two regions give sp. Three give sp2. Four give sp3. Five and six need more room, so the bookkeeping adds d orbitals: sp3d for five and sp3d2 for six. The s character of a hybrid is the share of s in the mix. For sp it is 1 of 2, or 50%. For sp2 it is 1 of 3, near 33%. For sp3 it is 1 of 4, or 25%. More s character pulls the pair in closer, so those bonds run shorter and stronger.
Worked example · naming two hybrid sets
Find the hybrid set on the oxygen in H2O and on the sulfur in SF4.
Step 1 · count regions on OTwo O–H sigma bonds plus two lone pairs gives 2 + 2 = 4 regions.
Step 2 · match the countFour regions need four hybrids, so oxygen is sp3. Its s character is 1/4 = 25%.
Step 3 · count regions on SFour S–F sigma bonds plus one lone pair gives 4 + 1 = 5 regions.
Step 4 · match againFive regions need five hybrids, so sulfur is sp3d. Its s character is 1/5 = 20%.
Water bends to 104.5° rather than 109.5°. A lone pair spreads wider than a bonding pair, so it squeezes the two O–H bonds together. The hybrid set still tells you where to start.
- Hybrid orbitals are mixes of s and p orbitals on one atom.
- Orbitals in equals orbitals out: 1 s + 3 p gives 4 sp3 hybrids.
- Count regions of electron density, then read off the set: 2 → sp, 3 → sp2, 4 → sp3, 5 → sp3d, 6 → sp3d2.
- A multiple bond counts as one region, not two or three.
- s character runs 50%, 33% and 25% for sp, sp2 and sp3. More s means shorter bonds.
Section 3Sigma and pi bonds
Two ways for orbitals to meet
Overlap can happen in two shapes. Orbitals can meet end to end, along the line joining the nuclei. That line is the internuclear axis. End‑to‑end overlap builds a sigma bond, written σ. The electron density piles up right on the axis, between the two nuclei. Look down the axis and the bond looks like a plain round tube. Nothing changes as you spin it.
Orbitals can also meet side by side. Two p orbitals that stay parallel overlap above the axis and below it at the same time. That is a pi bond, written π. A pi bond puts no density on the axis at all. The axis sits inside a nodal plane, a flat sheet where the chance of finding the electron is zero. So a pi bond is really one bond in two slabs, one over and one under.
Sigma overlap is the better deal. Pointing two lobes straight at each other gets them as close as they can get. Side‑on lobes have to lean past one another, so they share less space. Less overlap means a weaker bond. Every single bond you have ever drawn is a sigma bond. Pi bonds only show up as extras, on top of a sigma bond that is already there.
| Feature | Sigma (σ) bond | Pi (π) bond |
|---|---|---|
| Overlap | end to end, along the axis | side by side, parallel lobes |
| Density on the axis | highest there | zero, a nodal plane runs through it |
| Built from | s, p or hybrid orbitals | unhybridized p orbitals |
| Relative strength | stronger, more overlap | weaker, less overlap |
| Rotation about the axis | free | blocked |
| Where it appears | in every covalent bond | only on top of a σ bond |
Counting sigma and pi bonds
The counting rule is short. A single bond is 1 σ. A double bond is 1 σ + 1 π. A triple bond is 1 σ + 2 π. The second pi bond in a triple bond uses the other pair of p orbitals, at right angles to the first. The two pi bonds wrap the sigma bond from four sides.
Worked example · sigma and pi in acetic acid
Acetic acid gives vinegar its bite. Its formula is CH3COOH. Count every σ and π bond.
Step 1 · list the bondsThree C–H bonds, one C–C bond, one C=O bond, one C–O bond and one O–H bond.
Step 2 · count sigmaEvery bond on that list holds one sigma bond: 3 + 1 + 1 + 1 + 1 = 7 σ.
Step 3 · count piOnly the C=O bond is a double bond, so it adds 1 π. Nothing else does.
Step 4 · check the total7 σ + 1 π = 8 bonding pairs, which is 16 bonding electrons. A Lewis structure of CH3COOH shows 8 lines, and the count agrees.
- Sigma bonds come from head‑on overlap and hold their density on the internuclear axis.
- Pi bonds come from side‑on overlap of parallel p orbitals, above and below a nodal plane.
- Sigma overlap is larger, so a sigma bond is stronger than a pi bond.
- Single = 1 σ. Double = 1 σ + 1 π. Triple = 1 σ + 2 π.
- A pi bond never stands alone. It always rides on a sigma bond.
Section 4Multiple bonds and rotation
Why a double bond will not twist
Take ethene, C2H4. Each carbon is sp2, so each one keeps one unhybridized p orbital. Those two p orbitals stand up at right angles to the flat part of the molecule. When they line up parallel, they overlap side by side and make the pi bond. The whole molecule goes flat and stays flat.
Now try to twist one CH2 group. The sigma bond does not care, because it is round about the axis. The pi bond cares a great deal. Twisting tips one p orbital out of line with the other. Their overlap shrinks. It falls roughly as the cosine of the twist angle. At 90° the two p orbitals sit at right angles and share nothing. The pi bond is simply gone.
That is restricted rotation. To spin a double bond you must first pay for breaking a pi bond. For ethene that bill is near 270 kJ/mol. Room temperature cannot cover it, so the molecule stays locked. A single bond, as in ethane, only has to slide past a small bump of about 12 kJ/mol. Ethane spins millions of times a second.
What the lock buys you
Restricted rotation is not a nuisance. It is the reason shape can carry information. Two compounds can share a formula and still differ, because the groups on a double bond are pinned in place. Cis means two chosen groups sit on the same side. Trans means they sit on opposite sides. Cis‑2‑butene and trans‑2‑butene are different substances, with different melting points and different boiling points.
Your eyes run on this. A molecule called retinal sits in the back of your eye with one cis double bond. Light hits it and that bond flips to trans. The shape change starts the nerve signal. Without restricted rotation the molecule would wobble all day and you would see nothing.
Bond data backs the picture. A C–C single bond is 154 pm long. A C=C double bond is 134 pm. A C≡C triple bond is 120 pm. More shared pairs pull the nuclei closer. The bonds get stronger too, but not in equal steps, as the next example shows.
Worked example · how strong is a pi bond?
A C–C bond takes 347 kJ/mol to break. A C=C bond takes 614 kJ/mol. A C≡C bond takes 839 kJ/mol. Find the strength of each pi bond.
Step 1 · first pi bondA double bond is one sigma plus one pi. So the first pi bond is the extra:
614 kJ/mol − 347 kJ/mol = 267 kJ/mol
Step 2 · second pi bondA triple bond adds one more pi on top of the double bond:
839 kJ/mol − 614 kJ/mol = 225 kJ/mol
Step 3 · read the resultBoth pi bonds come out weaker than the 347 kJ/mol sigma bond. Side‑on overlap really is the poorer deal. Double is not twice single: 2 × 347 = 694 kJ/mol, well above the measured 614 kJ/mol.
Step 4 · compare with heatAt 298 K the thermal energy RT is (8.314 J/mol·K)(298 K) = 2477 J/mol, or 2.48 kJ/mol. The pi bond costs 267 ÷ 2.48 ≈ 108 times that. Room temperature has no chance.
- A pi bond needs its two p orbitals parallel, so twisting destroys it.
- Pi overlap falls off with the cosine of the twist angle and hits zero at 90°.
- Rotating a C=C bond costs near 270 kJ/mol. Rotating a C–C bond costs about 12 kJ/mol.
- The lock makes cis and trans isomers real, separate substances.
- More shared pairs means shorter bonds: 154 pm, 134 pm and 120 pm for single, double and triple carbon bonds.
Section 5Molecular orbital theory
Spread the electrons over the whole molecule
Valence bond theory has one blind spot. It says O2 has a tidy double bond with every electron paired. Hang liquid oxygen between the poles of a magnet and it clings there. Only unpaired electrons do that. So O2 must have some, and the shared‑pair picture misses them.
Molecular orbital theory fixes this by letting go of the idea that an electron belongs to one bond. Atomic orbitals are waves, so add them across the whole molecule. Adding two atomic orbitals in step builds up density between the nuclei. That gives a bonding molecular orbital, lower in energy than either atomic orbital. Adding them out of step cancels density between the nuclei. That gives an antibonding molecular orbital, higher in energy, with a node right between the atoms. The method is called the linear combination of atomic orbitals.
Two atomic orbitals always give two molecular orbitals, one down and one up. Electrons then fill those molecular orbitals by the same rules you already use. Lowest energy first. Two electrons per orbital, with opposite spins. Equal‑energy orbitals each take one electron before any of them takes a second. Orbitals at the same energy are called degenerate.
Period 1 and period 2 diatomics
For H2 and He2 only 1s orbitals matter. They give σ1s and σ*1s. The star marks antibonding. H2 puts both electrons in σ1s and holds together. He2 has to fill the star as well, which undoes the gain, so He2 does not exist.
Period 2 brings the 2s and 2p orbitals. The 2s pair gives σ2s and σ*2s. The six 2p orbitals give six molecular orbitals: σ2p and σ*2p from the pair pointing along the axis, plus two degenerate π2p and two degenerate π*2p from the two side‑on pairs.
Here is the twist that trips people up. The order of those levels is not the same across the row. For B2, C2 and N2 the π2p pair sits below σ2p. For O2, F2 and Ne2 the σ2p sits below the π2p pair. The reason is mixing. The 2s and 2p orbitals lie close in energy on the left of the row, so the sigma orbitals from 2s and 2p push on each other and σ2p gets shoved up. Moving right, the nuclear charge grows and the 2s drops far below the 2p. The mixing fades and the normal order returns.
| Order of levels, low to high | Holds for | Why |
|---|---|---|
| σ2s, σ*2s, π2p, σ2p, π*2p, σ*2p | Li2, Be2, B2, C2, N2 | strong 2s–2p mixing lifts σ2p above π2p |
| σ2s, σ*2s, σ2p, π2p, π*2p, σ*2p | O2, F2, Ne2 | 2s sits far below 2p, so mixing is small |
Worked example · filling O2
Fill the diagram for O2 and state the magnetism.
Step 1 · count valence electronsOxygen is in group 16, so each atom brings 6. That makes 6 + 6 = 12 electrons to place.
Step 2 · pick the orderOxygen uses the O/F order: σ2s, σ*2s, σ2p, π2p, π*2p, σ*2p.
Step 3 · fill from the bottomσ2s takes 2. σ*2s takes 2. σ2p takes 2. The two π2p orbitals take 4. That is 10 placed, with 2 left.
Step 4 · obey the half‑filling ruleThe last 2 go into the two degenerate π*2p orbitals, one each, with the same spin. Two unpaired electrons, so O2 is paramagnetic.
Step 5 · bond orderBonding = 8, antibonding = 4, so the bond order is (8 − 4)/2 = 2. The magnet and the double bond both come out of one diagram.
- Molecular orbitals spread over the whole molecule, not over one bond.
- Two atomic orbitals give one bonding orbital (lower) and one antibonding orbital (higher, marked with a star).
- Fill lowest first, two per orbital with opposite spins, and spread singly across degenerate orbitals first.
- B2, C2 and N2 put π2p below σ2p. O2 and F2 do the opposite.
- That ordering is why O2 ends up with two unpaired electrons.
Section 6Bond order and magnetism
One number that ties it together
Bond order is the net number of bonding pairs holding two atoms together. Count the electrons in bonding molecular orbitals. Count the electrons in antibonding ones. Subtract, then halve.
bond order = (bonding electrons − antibonding electrons) ÷ 2
Antibonding electrons cancel bonding ones, so they carry a minus sign. A bond order of 0 means no bond at all. That is the whole story of He2, Be2 and Ne2. They never form, because their stars fill as fast as their bonding orbitals do. Bond order can also land on a half, such as 2.5. Nothing is wrong. It just means one lone electron is pulling its weight in a bonding or antibonding orbital.
Bond order sets length and strength
Bond order is not just bookkeeping. It predicts two things you can measure. Raise the bond order and the two nuclei pull closer, so the bond length shrinks. Raise the bond order and the bond also gets harder to break, so the bond energy climbs. The row from N2 to F2 shows it clearly. N2 has bond order 3, a length of 110 pm, and needs 945 kJ/mol. O2 has bond order 2, a length of 121 pm, and needs 498 kJ/mol. F2 has bond order 1, a length of 142 pm, and needs only 159 kJ/mol.
Reading the magnet
Magnetism is the other prize. A substance with one or more unpaired electrons is paramagnetic. Each unpaired electron acts like a tiny magnet, so the sample is drawn into a magnetic field. A substance with every electron paired is diamagnetic. Paired spins cancel, and the sample is pushed out of the field, very weakly. So the test is simple. Finish the diagram, then look for a lone arrow. One or more lone arrows means paramagnetic.
This is where molecular orbital theory earns its keep. Pour liquid oxygen between the poles of a strong magnet and it hangs there in a bridge. Liquid nitrogen in the same gap pours straight through. The diagrams say why. O2 keeps two lone arrows in its π* level. N2 has none.
Worked example · the whole O2 family
Find the bond order and magnetism of O2+, O2, O2− and O22−, then rank them by bond length.
Step 1 · count electronsO2 has 12 valence electrons. Taking one away gives 11. Adding one gives 13. Adding two gives 14.
Step 2 · place themEvery one of these fills σ2s2 σ*2s2 σ2p2 π2p4 first. That is 10 electrons and 8 of them are bonding. Whatever is left goes into the two π*2p orbitals.
Step 3 · do the four sumsBonding stays at 8 every time. Antibonding is 2 from σ*2s plus whatever lands in π*.
O2+: π* holds 1, so antibonding = 3. (8 − 3)/2 = 2.5, with 1 unpaired.
O2: π* holds 2, so antibonding = 4. (8 − 4)/2 = 2, with 2 unpaired.
O2−: π* holds 3, so antibonding = 5. (8 − 5)/2 = 1.5, with 1 unpaired.
O22−: π* holds 4, so antibonding = 6. (8 − 6)/2 = 1, with 0 unpaired.
Step 4 · rank the lengthsBond order falls in the order O2+ > O2 > O2− > O22−, so bond length must rise in that same order. Measurement agrees: 112 pm, 121 pm, 133 pm and 149 pm.
Step 5 · magnetismThe first three carry at least one lone arrow, so all three are paramagnetic. Only O22−, the peroxide ion, is diamagnetic.
- Bond order = (bonding electrons − antibonding electrons) ÷ 2.
- Bond order 0 means the molecule does not form. He2, Be2 and Ne2 all land there.
- Half‑integer bond orders are fine and show up in ions such as O2+.
- Higher bond order gives a shorter and stronger bond.
- Any unpaired electron makes a substance paramagnetic. All pairs means diamagnetic.
- N2, O2 and F2 run 110, 121 and 142 pm, and 945, 498 and 159 kJ/mol.
The chapter in one breath
- Valence bond theory builds a bond from two overlapping orbitals holding one spin‑paired pair.
- The dip in an energy curve gives the bond length at its lowest point and the bond energy as its depth.
- Hybrid orbitals mix s and p on one atom so the lobes point where the molecule needs them.
- Count regions of electron density: 2 → sp, 3 → sp2, 4 → sp3, 5 → sp3d, 6 → sp3d2.
- Sigma bonds come from head‑on overlap, pi bonds from side‑on overlap of leftover p orbitals.
- A pi bond needs parallel p orbitals, so a double bond cannot twist. That is what makes cis and trans real.
- Molecular orbital theory spreads electrons over the molecule in bonding and antibonding levels.
- B2, C2 and N2 keep π2p below σ2p. O2, F2 and Ne2 swap them.
- Bond order = (bonding − antibonding)/2, and it predicts length, strength and magnetism in one go.
ReferenceWords worth knowing
These are the words this chapter leans on. Each one is defined the way it is used here.
- antibonding molecular orbital
- A molecular orbital that is higher in energy than the atomic orbitals that built it. It has a node between the nuclei and weakens the bond. Its symbol carries a star, as in σ*2p.
- bond energy
- The energy needed to break one mole of a given bond. It is the depth of the dip in the energy curve. For H2 it is 436 kJ/mol.
- bond length
- The average distance between two bonded nuclei. It sits at the bottom of the energy dip. For H2 it is 74 pm.
- bond order
- Half the difference between bonding and antibonding electrons. It gives the net number of bonds and can be a half number.
- bonding molecular orbital
- A molecular orbital lower in energy than the atomic orbitals that built it. It piles electron density between the nuclei and holds the atoms together.
- degenerate orbitals
- Two or more orbitals with exactly the same energy, such as the pair of π2p orbitals. Electrons spread out singly across them before pairing up.
- diamagnetic
- Having every electron paired. A diamagnetic sample is pushed out of a magnetic field, and only very weakly.
- hybrid orbital
- A new orbital on one atom built by mixing that atom's own s and p orbitals. It has one large lobe and one small back lobe.
- hybridization
- The mixing of atomic orbitals on the same atom to make an equal number of hybrid orbitals that point where the bonds go.
- linear combination of atomic orbitals
- The method of adding and subtracting atomic orbital waves to build molecular orbitals. Adding in step gives a bonding orbital, out of step gives an antibonding one.
- molecular orbital
- An orbital that spreads over a whole molecule rather than one atom. Electrons in it belong to the molecule.
- molecular orbital theory
- The model that builds molecular orbitals from atomic orbitals and then fills them with all the electrons of the molecule.
- node
- A place where the chance of finding the electron is zero. A pi bond has a nodal plane through both nuclei.
- orbital overlap
- Two orbitals sharing some of the same space. More overlap puts more density between the nuclei and makes a stronger bond.
- paramagnetic
- Having one or more unpaired electrons. A paramagnetic sample is pulled into a magnetic field. Liquid O2 does this.
- pi bond
- A bond from side‑on overlap of two parallel p orbitals. Its density sits above and below the internuclear axis, never on it.
- region of electron density
- One sigma bond or one lone pair on an atom. A double or triple bond still counts as a single region.
- restricted rotation
- The block on twisting about a double or triple bond. Twisting would break a pi bond, which costs far too much energy.
- sigma bond
- A bond from head‑on overlap along the internuclear axis, the straight line joining the two nuclei. Its density is highest on that axis and it spins freely.
- valence bond theory
- The model in which a bond is two overlapping atomic or hybrid orbitals sharing one pair of electrons with opposite spins.