Tetravalence of Carbon and Shapes

Why carbon is tetravalent. Carbon's electron configuration is $1s^2\,2s^2\,2p^2$: two electrons fill the inner shell, four sit in the valence shell. Gaining four electrons to complete an octet, or losing four to empty the shell, both cost far too much energy for an atom this small and weakly electronegative. Sharing is the only workable route, so carbon contributes all four valence electrons to four separate covalent bonds — a fixed valency of four, called tetravalence. Because these bonds are strong and similar in size to a carbon atom itself, carbon also bonds readily to other carbon atoms, chaining into long backbones and rings; this self-linking property is called catenation, the reason organic chemistry has millions of known compounds built from one element's skeleton.

Why hybridisation is needed at all. In the ground state, carbon's $2s$ orbital is spherical and lower in energy, while the three $2p$ orbitals are dumbbell-shaped and mutually perpendicular ($90^\circ$ apart) — nothing about them predicts methane's real, measured bond angle of $109.5^\circ$, with all four C–H bonds identical. Hybridisation resolves this: the $s$ orbital and some number of $p$ orbitals of similar energy mix into a matching number of new, energy-equivalent hybrid orbitals, each pointing as far from its neighbours as geometry allows (minimising electron-pair repulsion).

The three hybridisation states of carbon.

• $sp^3$: one $s$ + three $p$ orbitals mix into four equivalent $sp^3$ orbitals, arranged tetrahedrally at $109.5^\circ$ ($109^\circ28'$ precisely) to keep four electron pairs as far apart as possible in three dimensions. Methane, $\text{CH}_4$, is the textbook example: four identical C–H sigma bonds.
• $sp^2$: one $s$ + two $p$ orbitals mix into three equivalent $sp^2$ orbitals, arranged in a trigonal planar pattern at $120^\circ$; the one $p$ orbital left unhybridised sticks out perpendicular to that plane and forms a pi bond. Ethene, $\text{CH}_2{=}\text{CH}_2$, is the example: each carbon is $sp^2$ with three sigma-bond regions (two C–H, one C=C).
• $sp$: one $s$ + one $p$ orbital mix into two equivalent $sp$ orbitals, arranged linearly at $180^\circ$; the two $p$ orbitals left unhybridised are mutually perpendicular and form two pi bonds. Ethyne, $\text{HC}{\equiv}\text{CH}$, is the example: each carbon is $sp$ with only two sigma-bond regions.

The bond-angle pattern can be written as one relation linking the number of hybrid orbitals to the ideal angle between them:

where: $n$ is the number of hybrid orbitals formed at that carbon (equivalently, the number of sigma-bond regions around it — a double or triple bond still counts as ONE region), and $\theta$ is the ideal angle between them. The $120^\circ$ and $180^\circ$ cases follow from simple planar/linear geometry ($360^\circ/3$ and half of $360^\circ/2$); the $109.5^\circ$ tetrahedral angle is the three-dimensional case and is normally quoted as a fixed geometric result rather than computed from the same $360^\circ/n$ formula, since four points cannot be spread evenly in a single plane.

Sigma and pi bonds. Every single covalent bond, C–C or C–H, is a sigma ($\sigma$) bond, formed by head-on, axial overlap of orbitals — unaffected by rotation. A double bond is one $\sigma$ bond plus one pi ($\pi$) bond, and a triple bond is one $\sigma$ bond plus two $\pi$ bonds; $\pi$ bonds come from sideways overlap of unhybridised $p$ orbitals above and below the bond axis. Because sideways overlap breaks on twisting, $\pi$ bonds restrict free rotation while $\sigma$ bonds allow it freely — why an ethane C–C single bond rotates easily but an ethene C=C double bond is rigid.

There is no single algebraic derivation of "which hybridisation a carbon adopts" — it is a counting rule grounded in orbital theory: count sigma-bond regions, then read off the hybridisation and shape. Every atom or group directly attached to a carbon by a sigma bond counts as one region, and a double or triple bond to one neighbour still counts as only ONE region (it just uses extra $p$ orbitals for the $\pi$ part). Four regions force $sp^3$ mixing because all four valence orbitals are needed to hold four separate sigma bonds, and the tetrahedral arrangement keeps four repelling electron pairs as far apart as physically possible. Three regions force $sp^2$ mixing because only three sigma bonds are needed, freeing one $p$ orbital for a $\pi$ bond; the three hybrid orbitals settle into the flattest, most separated planar arrangement, $120^\circ$ apart. Two regions force $sp$ mixing for the same reason, leaving two $p$ orbitals free for two $\pi$ bonds, and two orbitals separate maximally by pointing in exactly opposite directions, $180^\circ$ apart. This is also why bond length shortens from $sp^3$ to $sp^2$ to $sp$: $sp$ orbitals have the most "s character" (50%, closest to the nucleus), pulling bonded atoms in tighter, while $sp^3$ orbitals have the least (25%) and give the longest, weakest carbon bonds.

Predict the hybridisation, shape, and bond angle at each carbon in propene, $\text{CH}_3{-}\text{CH}{=}\text{CH}_2$, and state which carbon has the shortest bonds to its neighbours.

Step 1. List the regions around each carbon. C1 ($\text{CH}_3$) is bonded to 3 H atoms and 1 carbon (C2) — four single-bond regions, no multiple bond. C2 is bonded to C1 (single bond), C3 (double bond), and 1 H — three regions. C3 is bonded to C2 (double bond) and 2 H atoms — three regions.

Step 2. Assign hybridisation using the region count from the derivation above. C1 has four single-bond regions, so it is $sp^3$. C2 and C3 each have three regions with one double bond, so both are $sp^2$.

Step 3. Read off shape and angle. Around C1: tetrahedral, $\approx 109.5^\circ$. Around C2 and C3: trigonal planar, $\approx 120^\circ$.

Step 4. Compare bond character. The C2=C3 double bond (both $sp^2$) is shorter and stronger than the C1–C2 single bond ($sp^3$ joined to $sp^2$), because the $sp^2$ orbitals have more s character and the bond includes an extra $\pi$ overlap. Answer: C1 is $sp^3$/tetrahedral/$109.5^\circ$; C2 and C3 are both $sp^2$/trigonal planar/$120^\circ$; the C2=C3 double bond is the shortest bond in the molecule.

Competitive stretch

The same method resolves why the H–C–H angle in methane ($109.5^\circ$, all four groups identical, $sp^3$) differs from the H–C–H angle in ethene ($\approx 117^\circ$, not the "ideal" $120^\circ$). Both ethene carbons are $sp^2$, so the baseline is $120^\circ$, but the three groups around each carbon are not identical: two are C–H single bonds and one is the C=C double bond. The doubly-bonded region carries extra electron density and repels the two C–H bonds slightly more than they repel each other, squeezing the H–C–H angle a little below $120^\circ$ while opening the H–C=C angles slightly above it — a small, testable deviation, not a contradiction.

• Counting sigma-bond regions, not bond lines — a double or triple bond to one neighbour is still ONE region, so miscounting a $\text{C}{=}\text{C}$ as "two regions" wrongly predicts $sp^3$ instead of $sp^2$.
• Confusing bond angle with bond length — $sp$ bonds are shortest and strongest, $sp^3$ are longest and weakest; not every C–C bond in a molecule has the same length.
• Believing a $\pi$ bond permits free rotation like a $\sigma$ bond does — sideways $p$-orbital overlap breaks on twisting, why double bonds are rigid (the basis of cis-trans isomerism studied later).
• Treating catenation as an unrelated fact rather than a direct consequence of strong, similarly sized carbon-carbon $\sigma$ bonds.
• Assuming a whole molecule has one hybridisation — most organic molecules mix $sp^3$, $sp^2$ and $sp$ carbons at different positions, as propene shows.

Ask the tutor: Want to work through the hybridisation of every carbon in a bigger molecule like isoprene or benzene, or see how bond angles shift further with different substituents?