Difficulty: Intermediate | Prerequisites: Lewis structures, VSEPR geometry, valence electrons.
Once you can draw Lewis structures and predict geometry, the next layer is understanding how atomic orbitals combine to form bonds. Hybridisation explains why carbon can make four equivalent bonds, and sigma/pi bond theory explains what happens in single, double, and triple bonds at the orbital level. Bond energies then let you estimate whether a reaction releases or absorbs energy, without needing full thermodynamic data. These topics tie molecular structure to energetics and sit at the heart of this exam.
Hybridisation (sp, sp2, sp3) is determined by the number of electron groups on an atom. Every single bond and every first bond in a double/triple bond is a sigma (σ) bond; the second bond in a double and the second and third in a triple are pi (π) bonds. Bond energies let you estimate ΔH for a reaction: energy in (breaking reactant bonds) minus energy out (forming product bonds).
Hybridisation
The mixing of atomic orbitals on an atom to form new, equivalent hybrid orbitals that better explain observed bonding geometry. Think of it as blending different-shaped orbitals into a matching set.
sp3 hybridisation
One s orbital mixes with three p orbitals to produce four equivalent sp3 hybrid orbitals arranged tetrahedrally (109.5°). Occurs when an atom has 4 electron groups.
sp2 hybridisation
One s orbital mixes with two p orbitals to produce three equivalent sp2 hybrid orbitals in a trigonal planar arrangement (120°), leaving one unhybridised p orbital for pi bonding. Occurs when an atom has 3 electron groups.
sp hybridisation
One s orbital mixes with one p orbital to produce two equivalent sp hybrid orbitals in a linear arrangement (180°), leaving two unhybridised p orbitals. Occurs when an atom has 2 electron groups.
Sigma (σ) bond
A bond formed by head-on (end-to-end) overlap of orbitals along the internuclear axis. Every single bond is a sigma bond. The first bond in any double or triple bond is also sigma.
Pi (π) bond
A bond formed by sideways (lateral) overlap of unhybridised p orbitals above and below the internuclear axis. The second bond in a double bond and the second and third bonds in a triple bond are pi bonds.
Bond energy (bond dissociation energy)
The energy required to break one mole of a particular bond in the gas phase, in kJ/mol. Higher bond energy means a stronger bond.
Enthalpy change from bond energies (ΔH)
The estimated energy change for a reaction, calculated as the sum of bond energies broken (reactants) minus the sum of bond energies formed (products). A positive result means the reaction is endothermic; negative means exothermic.
The hybridisation of any atom is set by its number of electron groups (same count used in VSEPR):
2 electron groups → sp (linear, 180°)
3 electron groups → sp2 (trigonal planar, 120°)
4 electron groups → sp3 (tetrahedral, 109.5°)
Lone pairs count as electron groups. A double or triple bond counts as one electron group.
A single bond = 1 σ bond
A double bond = 1 σ + 1 π bond
A triple bond = 1 σ + 2 π bonds
Sigma bonds are formed by head-on overlap of hybrid orbitals (or hybrid with s/p on hydrogen/halogens). Pi bonds are formed by sideways overlap of unhybridised p orbitals.
The exam uses the notation σ: C(sp2) - Cl(p), meaning a sigma bond formed between a carbon atom using an sp2 hybrid orbital and a chlorine atom using a p orbital.
For each bond, identify:
The type (σ or π)
Each atom's hybridisation (the orbital it contributes to that bond)
Terminal atoms (H, halogens) typically contribute:
H contributes its 1s orbital (written as "s")
Halogens (F, Cl, Br, I) contribute a p orbital (written as "p")
First, determine the hybridisation of each non-hydrogen atom:
N (in N≡C): 2 electron groups (triple bond + lone pair) → sp
C (in N≡C): 2 electron groups (triple bond + single bond) → sp
C (in CF=): 3 electron groups (single bond + single bond + double bond) → sp2
C (in CHBr): 3 electron groups (double bond + single bond + single bond) → sp2
Bond-by-bond labels:
N≡C: one σ bond: N(sp) - C(sp), plus two π bonds: N(p) - C(p)
C-C (single bond between the two central carbons): σ: C(sp) - C(sp2)
C-F: σ: C(sp2) - F(p)
C=C: one σ bond: C(sp2) - C(sp2), plus one π bond: C(p) - C(p)
C-H: σ: C(sp2) - H(s)
C-Br: σ: C(sp2) - Br(p)
Reaction: 2 CO2 + N2 → 2 CO + 2 NO
Step 1: Identify bond types. CO2 has two C=O double bonds (use the special value 799 kJ/mol for C=O in CO2, as noted in the table). N2 has one N≡N triple bond. CO has one C≡O triple bond. NO has one N=O double bond.
Step 2: Sum bonds broken (reactants).
2 CO2: 2 molecules x 2 C=O bonds = 4 bonds at 799 kJ/mol = 3196 kJ
1 N2: 1 N≡N bond = 946 kJ
Total energy in = 4142 kJ
Step 3: Sum bonds formed (products).
2 CO: 2 C≡O bonds at 1072 kJ/mol = 2144 kJ
2 NO: 2 N=O bonds at 590 kJ/mol = 1180 kJ
Total energy out = 3324 kJ
Step 4: ΔH = bonds broken - bonds formed = 4142 - 3324 = +818 kJ/mol.
The reaction is endothermic (positive ΔH).
\Delta H_{\text{rxn}} = \sum (\text{bond energies of bonds broken}) - \sum (\text{bond energies of bonds formed})A positive result means the reaction is endothermic (absorbs energy). A negative result means it is exothermic (releases energy).
Remember: breaking bonds costs energy (endothermic step), forming bonds releases energy (exothermic step).
Bond energies are how chemists estimate the energy content of fuels and the feasibility of industrial reactions without running expensive calorimetry on every possibility. Hybridisation and sigma/pi bond theory underpin all of organic chemistry: the difference between a C-C single bond (free rotation, sp3) and a C=C double bond (restricted rotation, sp2) explains why fats can be saturated or unsaturated.
Students often think that hybridisation is determined by the number of bonds. It is not. It is determined by the number of electron groups, which includes lone pairs. An atom with 3 bonds and 1 lone pair is sp3, not sp2.
Students often forget to use the special C=O bond energy for CO2 (799 kJ/mol) given in the table footnote. Using the generic C=O value (736 kJ/mol) instead will give the wrong answer for Question 3.
Students often label all bonds in a triple bond as sigma. Only the first bond is sigma; the other two are pi bonds.
Students often assume terminal halogens use hybrid orbitals. In this course's notation, terminal halogens (F, Cl, Br, I) contribute a p orbital to their sigma bond with the central atom.
⚠️ The bond labelling question (Q5) requires you to label every bond with the notation σ: atom1(orbital) - atom2(orbital) or π: atom1(p) - atom2(p). Draw the structure large enough to fit the labels.
⚠️ Pay close attention to bond types in the enthalpy calculation. CO2 has double bonds (and a special bond energy value in the table), while CO has a triple bond and NO has a double bond.
⚠️ The exam provides a bond energy table. You do not need to memorise values, but you do need to read the table carefully, including footnotes.
⚠️ For bond labelling, first assign hybridisation to every atom, then label each bond. Doing it in two passes reduces mistakes.
Fill in the blank: An atom with 3 electron groups is ____ hybridised.
Answer: sp2.
True or False: A triple bond consists of 3 sigma bonds.
Answer: False. A triple bond is 1 sigma + 2 pi bonds.
Fill in the blank: In the bond energy method, ΔH = Σ(bonds broken) ____ Σ(bonds formed).
Answer: minus.
True or False: Breaking bonds is an exothermic process.
Answer: False. Breaking bonds requires energy (endothermic). Forming bonds releases energy.
Fill in the blank: A terminal bromine atom contributes a ____ orbital to its sigma bond.
Answer: p orbital.
Q: For the molecule N≡C-CF=CHBr, what is the hybridisation of each carbon atom?
A: The carbon in N≡C has 2 electron groups (triple bond + single bond), so it is sp. The carbon bonded to F has 3 electron groups (single + single + double), so it is sp2. The carbon bonded to H and Br has 3 electron groups (double + single + single), so it is sp2.
Q: How many sigma bonds and how many pi bonds are in N≡C-CF=CHBr?
A: Sigma bonds: N≡C (1) + C-C (1) + C-F (1) + C=C (1) + C-H (1) + C-Br (1) = 6 sigma bonds. Pi bonds: N≡C (2) + C=C (1) = 3 pi bonds. Total: 6σ + 3π.
Q: Using bond energies, calculate ΔH for 2 CO2 + N2 → 2 CO + 2 NO. The table gives C=O in CO2 as 799 kJ/mol, N≡N as 946, C≡O as 1072, and N=O as 590.
A: Bonds broken: 4(799) + 1(946) = 4142 kJ. Bonds formed: 2(1072) + 2(590) = 3324 kJ. ΔH = 4142 - 3324 = +818 kJ/mol. The reaction is endothermic.
Q: Label the C-F bond in N≡C-CF=CHBr using the σ/π notation with orbital types.
A: σ: C(sp2) - F(p). The carbon is sp2 hybridised (3 electron groups), and fluorine contributes a p orbital.
Hybridisation is directly linked to VSEPR geometry: the same electron group count that determines geometry also determines hybridisation (4 groups = tetrahedral = sp3, etc.). This connects to intermolecular forces as well, because molecular geometry determines polarity, and polarity determines the strength of dipole-dipole interactions.
Bond energies connect to thermochemistry. The bond energy method gives an estimate of ΔH that complements Hess's law and standard enthalpies of formation. In practice, you use whichever data set the problem provides.
Hybridisation, hybridization, sp sp2 sp3, orbital hybridisation, sigma bond, pi bond, sigma and pi bonds, bond order, single double triple bond, bond dissociation energy, average bond energy, bond enthalpy, enthalpy change from bond energies, Hess's law bond energy method, bond labelling notation, orbital overlap, head-on overlap, lateral overlap, CHEM1000, NYU general chemistry, molecular orbital basics, organic bonding