IMAT Chemistry · Organic Chemistry

How Carbon Bonds, and How We Write Organic Molecules

Carbon always makes four bonds; how many atoms it bonds to sets its hybridisation and shape, and every formula is a way of counting those atoms.

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Why this matters

The 2025 paper asked for the hybridisation and shape of the carbons in benzene, and older papers asked for the molecular formula of a molecule drawn as a skeleton. Almost every other question in the chapter also needs you to turn a drawing or a condensed formula into atoms quickly.

Concept 1 of 2: Carbon's four bonds and its hybridisation (sp³, sp², sp)

A carbon atom has four electrons to share, so it always makes four bonds. Those bonds point as far apart as possible. If the carbon is joined to four atoms, they spread into a tetrahedron. A double or triple bond holds extra electrons between just two atoms, so fewer directions are used and the shape becomes flat or straight.

Definition

Hybridisation describes how carbon's outer orbitals mix before bonding. Count the atoms joined to the carbon:

4 atoms (all single bonds): sp³, tetrahedral, angles about 109.5°.
3 atoms (one double bond): sp², trigonal planar, angles 120°.
2 atoms (a triple bond, or two double bonds): sp, linear, angle 180°.
Every single bond is one σ (sigma) bond. A double bond is one σ plus one π (pi) bond; a triple bond is one σ plus two π bonds.
Atoms can rotate freely about a single bond, but not about a double bond, because the π bond would break.

Hybridisation	Atoms joined to the carbon	Shape and bond angle	Example
$\mathrm{sp^3}$	4 (four single bonds: 4 σ)	Tetrahedral, about 109.5°	Both carbons of ethane $\mathrm{CH_3CH_3}$ ; the carbon of $\mathrm{CH_4}$
$\mathrm{sp^2}$	3 (one double bond: 3 σ and 1 π)	Trigonal planar, 120°	Both carbons of ethene $\mathrm{CH_2{=}CH_2}$ ; the C of a C=O group; every carbon of benzene
$\mathrm{sp}$	2 (a triple bond, or two double bonds: 2 σ and 2 π)	Linear, 180°	Both carbons of ethyne $\mathrm{HC{\equiv}CH}$ ; the carbon of $\mathrm{CO_2}$

Count atoms, not bonds: a carbon in a C=C still has four bonds, but it is joined to only three atoms, so it is sp².

Practice this conceptself-check · 4 quick reps

A double bond is one σ and one π, not two π bonds

The first bond between two atoms is always σ. Only the second and third bonds are π. So a double bond has 1 σ + 1 π and a triple bond 1 σ + 2 π. Options that call a double bond two π bonds are wrong.

Hybridisation follows the number of atoms joined, not the number of bonds

Every carbon makes four bonds, so counting bonds tells you nothing. Count the atoms attached: 4 gives sp³, 3 gives sp², 2 gives sp.

Concept 2 of 2: Molecular, condensed and skeletal formulas, and counting the hydrogens

Chemists draw the same molecule in several ways, from a full picture of every bond to a bare zigzag of lines. The shorter drawings hide the hydrogens on carbon, because you can always work them out: every carbon must end up with four bonds. Whatever is missing is hydrogen.

Definition

Molecular formula: the number of each atom only, e.g. $\mathrm{C_3H_8O}$ .
Condensed (structural) formula: atoms carbon by carbon, e.g. $\mathrm{CH_3CH_2CH_2OH}$ . Brackets show a branch: $\mathrm{CH_3CH(CH_3)CH_3}$ .
Displayed formula: every atom and every bond drawn.
Skeletal formula: only the carbon skeleton as lines. Each corner and each line end is a carbon. Hydrogens on carbon are not drawn. Every other atom (O, N, Cl) is drawn, and so are the hydrogens on it (OH, $\mathrm{NH_2}$ ).
A ring drawn as a polygon has one carbon at each corner: a hexagon is six carbons.

Hidden hydrogens on a skeletal carbon

n_{\mathrm{H}} = 4 - (\text{number of bonds drawn to that carbon})

$n_{\mathrm{H}}$ hydrogens on that carbon
bonds drawna double line counts 2, a triple line counts 3

Worked example

Write the molecular formula of the compound

\mathrm{CH_3CH_2CH(OH)CH_2CHO}

Practice this conceptself-check · 4 quick reps

The hydrogens you cannot see in a skeletal formula are still there

A skeletal formula hides the hydrogens on carbon but shows the ones on O and N. Forgetting the hidden ones gives a formula short of hydrogen; forgetting the drawn OH hydrogen loses one more. Count carbon by carbon with

4 - \text{bonds drawn}

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Molecular, condensed and skeletal formulas, and counting the hydrogens
Hidden hydrogens on a skeletal carbon
$n_{\mathrm{H}} = 4 - (\text{number of bonds drawn to that carbon})$

Reference tables (1)

Carbon's four bonds and its hybridisation (sp³, sp², sp)3 rows

Hybridisation	Atoms joined to the carbon	Shape and bond angle	Example
$\mathrm{sp^3}$	4 (four single bonds: 4 σ)	Tetrahedral, about 109.5°	Both carbons of ethane $\mathrm{CH_3CH_3}$ ; the carbon of $\mathrm{CH_4}$
$\mathrm{sp^2}$	3 (one double bond: 3 σ and 1 π)	Trigonal planar, 120°	Both carbons of ethene $\mathrm{CH_2{=}CH_2}$ ; the C of a C=O group; every carbon of benzene
$\mathrm{sp}$	2 (a triple bond, or two double bonds: 2 σ and 2 π)	Linear, 180°	Both carbons of ethyne $\mathrm{HC{\equiv}CH}$ ; the carbon of $\mathrm{CO_2}$

Count atoms, not bonds: a carbon in a C=C still has four bonds, but it is joined to only three atoms, so it is sp².

Watch out for (3)

A double bond is one σ and one π, not two π bonds
→ Carbon's four bonds and its hybridisation (sp³, sp², sp)
Hybridisation follows the number of atoms joined, not the number of bonds
→ Carbon's four bonds and its hybridisation (sp³, sp², sp)
The hydrogens you cannot see in a skeletal formula are still there
→ Molecular, condensed and skeletal formulas, and counting the hydrogens

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