Tetravalency of carbon; hybridization recap (sp, sp², sp³): Worked examples

Before anything, we define the one quantity everything rests on:

Reminder of what a " $σ$ bond" and a " $π$ bond" are, in plain words — we lean on Sigma and Pi bonds — orbital overlap here:

A == $σ$ (sigma) bond== is the first bond between two atoms, formed by orbitals pointing head-on at each other. Every single bond is one $σ$ . Every double or triple bond contains exactly one $σ$ .
A == $π$ (pi) bond== is any extra bond in a double/triple, formed by leftover $p$ orbitals overlapping sideways. A double bond $= 1 σ + 1 π$ ; a triple bond $= 1 σ + 2 π$ .

The trick to counting fast:

The scenario matrix

Here is every "cell" — every distinct situation this topic can produce. Each worked example below is tagged with the cell(s) it covers.

#	Case class	What's tricky about it	Example
C1	Pure single bonds ( $s p^{3}$ , all $σ$ )	baseline, no $π$	Ex 1 (propane)
C2	One double bond ( $s p^{2}$ )	$π$ must be ignored in SN	Ex 2 (ethene)
C3	Triple / cumulated doubles ( $s p$ )	two $π$ on one atom	Ex 3 (allene — mixed centres!)
C4	Central atom with lone pairs	LP counts toward SN	Ex 4 (H₂O, NH₃)
C5	Multiple different carbons in one molecule	each C judged separately	Ex 5 (acrylonitrile)
C6	Degenerate / smallest cases	single atom (SN 0), radical	Ex 6 (C atom, :CH₂, ·CH₃)
C7	Limiting s-character → a physical trend	numbers, not just labels	Ex 7 (bond length & acidity)
C8	Real-world word problem	translate words → structure	Ex 8 (dry ice)
C9	Exam twist — looks like one thing, is another	resonance / charged atoms	Ex 9 (carbocation, carbanion)
C10	Expanded octet (SN 5, 6)	needs $d$ orbitals, $s p^{3} d$ / $s p^{3} d^{2}$	Ex 10 (PCl₅, SF₆)

Example 1 — Propane (Cell C1: all single bonds)

Worked example Assign hybridization to every carbon in propane,

$CH_{3} -CH_{2} -CH_{3}$ . Forecast: All three carbons are joined only by single bonds. Guess their hybridization before reading on.

Step 1 — Count neighbours of each carbon. Why this step? Number of $σ$ bonds = number of atoms directly attached.

End carbon: bonded to 3 H + 1 C $= 4$ neighbours $\Rightarrow 4 σ$ .
Middle carbon: bonded to 2 H + 2 C $= 4$ neighbours $\Rightarrow 4 σ$ .

Step 2 — Add lone pairs. Why? Carbon in a neutral hydrocarbon has no lone pairs, so add $0$ .

Every carbon: $SN = 4 + 0 = 4$ .

Step 3 — Read the table. $SN = 4 \Rightarrow s p^{3}$ , tetrahedral, $109. 5^{\circ}$ . Why this step? The steric number is just a count; it becomes chemistry only when we map it to a hybridization/geometry. The table is that dictionary — it turns the number $4$ into the physical prediction "four equivalent hybrids arranged tetrahedrally." Without it, $SN = 4$ tells us nothing about shape. The figure below shows those four $s p^{3}$ hybrids spreading to the corners of a tetrahedron.

Figure: a single $s p^{3}$ carbon (red) with its four hybrid orbitals pointing to the corners of a tetrahedron; every H–C–H angle is $109. 5^{\circ}$ — the maximally-spread arrangement of four directions in space.

Verify: Propane has no double or triple bonds, so there can be no $π$ orbitals and no leftover $p$ — all $s$ and $p$ orbitals are used in hybrids. That is exactly what $s p^{3}$ means (0 unhybridized p). ✔

Example 2 — Ethene (Cell C2: one double bond)

Worked example Hybridization of each carbon in ethene,

$CH_{2} = CH_{2}$ , and its bond angle. Forecast: There is a C=C double bond. Does "double" push it to $s p$ ? (No — watch.)

Step 1 — Count neighbours. Why? The double bond connects each C to only one other C, so it counts as one neighbour, not two.

Each C: 2 H + 1 C $= 3$ neighbours $\Rightarrow 3 σ$ bonds.

Step 2 — Lone pairs = 0, so $SN = 3 + 0 = 3 \Rightarrow s p^{2}$ .

Step 3 — Account for the leftover $p$ . Why? $s p^{2}$ uses one $s$ + two $p$ , leaving one unhybridized $p$ on each carbon. These two $p$ orbitals overlap sideways → the $π$ bond of the double bond. Look at the red sideways lobes in the figure.

Figure: two $s p^{2}$ carbons joined by a $σ$ bond; the red $p$ orbitals stick up and down out of the molecular plane and overlap sideways to make the $π$ bond of the C=C double bond.

Verify: The measured H–C–H / H–C–C angles in ethene are $\approx 11 7^{\circ} - 12 1^{\circ}$ , clustering around the predicted $12 0^{\circ}$ for $s p^{2}$ . The molecule is flat and cannot rotate about C=C — a signature of a real $π$ bond locking the plane. ✔

Example 3 — Allene (Cell C3: cumulated double bonds, mixed centres)

Worked example

$H_{2} C = C = CH_{2}$ (allene / propadiene). Give the hybridization of all three carbons. Forecast: The two ends look like ethene; the middle carbon has two double bonds. Same or different? Guess.

Step 1 — Terminal carbons. Why? Each end C is bonded to 2 H + 1 C $= 3$ neighbours $\Rightarrow SN = 3 \Rightarrow s p^{2}$ .

Step 2 — Central carbon. Why? It is bonded to two carbons only (one on each side). Two neighbours $\Rightarrow 2 σ$ , $0$ LP, $SN = 2 \Rightarrow s p$ , linear.

Step 3 — Where do the two $π$ bonds live? Why? An $s p$ centre keeps two unhybridized $p$ orbitals, perpendicular to each other. One makes the left C=C, the other makes the right C=C. Because the two $p$ orbitals are at $9 0^{\circ}$ , the two end $CH_{2}$ groups are twisted $9 0^{\circ}$ apart — allene is non-planar. See the figure: the central atom's two red $p$ orbitals point in perpendicular directions.

Figure: the linear C=C=C backbone; the central $s p$ carbon carries two red $p$ orbitals at $9 0^{\circ}$ to each other, forcing the left CH₂ group into a vertical plane and the right CH₂ into a horizontal one.

Verify: Same rule as CO₂ (two double bonds on the middle atom $\to s p$ , linear). Here the ends are $s p^{2}$ and the middle is $s p$ — proof that you must judge each atom separately, never the whole molecule at once. ✔

Example 4 — Water & Ammonia (Cell C4: lone pairs count)

Worked example Hybridization of O in

$H_{2} O$ and N in $NH_{3}$ . Forecast: Both have only single bonds. Are they $s p^{3}$ like carbon in methane, even though they have lone pairs?

Step 1 — Water, oxygen. Why include lone pairs? Lone pairs occupy hybrid orbitals too, so they add to SN.

O bonded to 2 H $\Rightarrow 2 σ$ ; O has $2$ lone pairs. $SN = 2 + 2 = 4 \Rightarrow s p^{3}$ .

Step 2 — Predicted vs real angle. Why less than $109. 5^{\circ}$ ? A lone pair is a cloud of two electrons held by only one nucleus, whereas a bonding pair is shared between two nuclei that pull it thin and tuck it between them. So the lone-pair cloud is fatter and sits closer to the central atom; it therefore takes up more angular room and shoves the neighbouring bonding pairs together. That is why the ideal $109. 5^{\circ}$ shrinks. (This is the electron-pair-repulsion idea of VSEPR theory and molecular geometry.) Measured H–O–H $= 104. 5^{\circ}$ .

Step 3 — Ammonia, nitrogen.

N bonded to 3 H $\Rightarrow 3 σ$ ; $1$ lone pair. $SN = 3 + 1 = 4 \Rightarrow s p^{3}$ . One lone pair, so measured H–N–H $= 10 7^{\circ}$ (only slightly squeezed).

Verify: More lone pairs $\Rightarrow$ more squeeze. Water ( $2$ LP, $104. 5^{\circ}$ ) $<$ ammonia ( $1$ LP, $10 7^{\circ}$ ) $<$ methane ( $0$ LP, $109. 5^{\circ}$ ). The ordering matches lone-pair count exactly. ✔

Example 5 — Acrylonitrile (Cell C5: several different carbons)

Worked example

$CH_{2} = CH - C \equiv N$ (acrylonitrile). Hybridization of each carbon and the nitrogen. Forecast: Four heavy atoms, three different bond situations. Predict a different label for the $s p^{2}$ region and the $s p$ region before reading.

Step 1 — The $= CH_{2}$ and $= CH -$ carbons. Why? Each is part of one double bond and bonded to a total of 3 neighbours.

$CH_{2}$ : 2 H + 1 C $= 3 σ \Rightarrow s p^{2}$ .
$= CH -$ : 1 H + 2 C $= 3 σ \Rightarrow s p^{2}$ .

Step 2 — The nitrile carbon $- C \equiv N$ . Why? It is bonded to the middle C (single) and to N (triple). Two neighbours only $\Rightarrow 2 σ$ , $0$ LP $\Rightarrow SN = 2 \Rightarrow s p$ .

Step 3 — The nitrogen. Why include its lone pair? N is bonded to just C (the triple bond = one neighbour) and carries $1$ lone pair. $SN = 1 + 1 = 2 \Rightarrow s p$ , and that lone pair sits in an $s p$ hybrid pointing straight out.

Verify: One molecule, carbons labelled $s p^{2}, s p^{2}, s p$ and N labelled $s p$ . Every atom judged on its own $σ$ +LP count — the theme of this page. The two $s p^{2}$ carbons connect via a $π$ system that can conjugate into the nitrile (Resonance and delocalization (sp² systems)). ✔

Example 6 — Bare C atom, carbene, and radical (Cell C6: degenerate/smallest cases)

Worked example Give the hybridization state of (i) a bare carbon atom in the gas phase, (ii) singlet carbene

$: CH_{2}$ , (iii) the methyl radical $\cdot CH_{3}$ . Forecast: With almost nothing attached, does "hybridization" even apply? And does a single unpaired electron count like a lone pair?

Step 1 — A single carbon atom, no neighbours ( $SN = 0$ ). Why does hybridization vanish? Plug into the formula literally: $0 σ$ bonds $+ 0$ lone pairs $= SN = 0$ . Hybridization only mixes orbitals to point at the things an atom bonds to or holds a pair for; with $SN = 0$ there are zero such directions, so there is nothing to hybridize. The atom keeps its plain $2 s$ and $2 p$ orbitals. So $SN = 0$ is a real, allowed answer meaning "unhybridized," not a failure of the rule.

Step 2 — Singlet carbene $: CH_{2}$ . Why? C is bonded to 2 H $\Rightarrow 2 σ$ ; it keeps $1$ lone pair. $SN = 2 + 1 = 3 \Rightarrow s p^{2}$ , bent (angle $\approx 10 0^{\circ} - 10 3^{\circ}$ ). One unhybridized $p$ is empty.

Step 3 — Methyl radical $\cdot CH_{3}$ . Why the formula and experiment disagree here. First do the strict SN count: C is bonded to 3 H $\Rightarrow 3 σ$ . The odd electron is not a lone pair — a lone pair is two electrons — so it contributes $0$ to the lone-pair term, giving $SN = 3 + 0 = 3 \Rightarrow s p^{2}$ . Now the experiment: methyl radical is indeed found to be (near-)planar $s p^{2}$ , and its single unpaired electron sits in the leftover unhybridized $p$ orbital. So the strict SN count and the measured geometry agree: one electron is not a full pair, so it neither demands its own hybrid orbital nor pushes the molecule pyramidal. (Contrast the carbene in Step 2, whose full lone pair does raise SN to 3 with a filled — not empty — extra orbital, and bends the molecule.)

Verify: Three limiting cases: $SN = 0$ (bare atom, unhybridized), carbene $SN = 3$ ( $s p^{2}$ , empty $p$ ), and radical $SN = 3$ ( $s p^{2}$ , singly-filled $p$ ). The SN rule survives the degenerate end (zero bonds) and the odd-electron twist — you just remember that a single electron is not a full pair, so it never enters the lone-pair count. ✔

Example 7 — s-character trend (Cell C7: limiting/numeric behaviour)

Worked example Show numerically why the C–H bond gets shorter and its H more acidic as we go

$s p^{3} \to s p^{2} \to s p$ . Forecast: Which of ethane, ethene, ethyne has the most acidic hydrogen? Guess.

Step 1 — Write the s-character fractions. Why fractions, and what is $n$ ? The label $s p^{n}$ means the hybrid is built from one $s$ orbital and $n$ $p$ orbitals — so $n = 1$ for $s p$ , $n = 2$ for $s p^{2}$ , $n = 3$ for $s p^{3}$ (the little exponent literally counts the $p$ orbitals mixed in). The total number of orbitals mixed is $1 + n$ , and just one of them is the $s$ orbital, so the fraction of $s$ -character is $\frac{1}{1 + n}$ .

$s p$ ( $n = 1$ ): $\frac{1}{2} = 0.50$ (50%). $s p^{2}$ ( $n = 2$ ): $\frac{1}{3} \approx 0.333$ (33%). $s p^{3}$ ( $n = 3$ ): $\frac{1}{4} = 0.25$ (25%).

Step 2 — Link s-character to the C–H bond length. Why does more $s$ shorten the C–H bond? An $s$ orbital hugs the nucleus tightly; higher $s$ -fraction pulls the shared C–H electrons in, so the H sits closer. Measured C–H bond lengths follow exactly this: $s p (1.06 \overset{˚}{A}) < s p^{2} (1.08 \overset{˚}{A}) < s p^{3} (1.09 \overset{˚}{A})$ — shortest for the highest- $s$ carbon, matching the stated goal.

Step 3 — Link s-character to acidity. Why? Losing $H^{+}$ leaves a lone pair on carbon; a high- $s$ orbital holds that pair close to the nucleus and stabilizes the anion. So terminal alkyne ( $s p$ , $p K_{a} \approx 25$ ) $≪$ alkene ( $s p^{2}$ , $\approx 44$ ) $≪$ alkane ( $s p^{3}$ , $\approx 50$ ). Detail in Acidity of terminal alkynes.

Verify: Order of s-character $0.50 > 0.333 > 0.25$ matches (a) decreasing C–H bond length ( $1.06 < 1.08 < 1.09$ Å) and (b) increasing acidity of the attached H. Most acidic = ethyne. The same $s$ -character trend also shortens the C–C frame ( $s p$ 1.20 Å $< s p^{2}$ 1.34 Å $< s p^{3}$ 1.54 Å), covered in Bond length and bond strength trends and Inductive effect and electronegativity of hybrid carbons. ✔

Example 8 — Dry ice (Cell C8: real-world word problem)

Worked example "Dry ice" is solid carbon dioxide used to make fog. In one CO₂ molecule, what is the hybridization of carbon, and is the molecule bent or straight?

Forecast: Two oxygens, two double bonds. Bent like water, or linear? Guess.

Step 1 — Translate words to structure. Carbon dioxide is $O = C = O$ . Why this step? The SN rule needs a bonding picture, not a name. Half of exam mistakes happen before any counting — students apply the formula to a structure they drew wrong. Committing to an explicit Lewis structure ( $O = C = O$ , carbon central, two double bonds, no lone pairs on C) first makes the counting reliable.

Step 2 — Count on the central C. Why only $σ$ ? Each C=O is one neighbour (one $σ$ + one $π$ ). Carbon has $0$ lone pairs.

$SN = 2 σ + 0 = 2$ .

Step 3 — Map SN → hybridization → geometry. Why this step? SN is only a count; the table turns it into a shape. $SN = 2 \Rightarrow s p$ hybridization $\Rightarrow$ the two $s p$ hybrids point exactly opposite each other $\Rightarrow$ linear, bond angle $18 0^{\circ}$ . The two leftover $p$ orbitals on the $s p$ carbon (perpendicular to the axis) each make one $π$ bond, one to each oxygen. See the figure.

Figure: the linear O=C=O molecule; the red central $s p$ carbon sits between two oxygens at $18 0^{\circ}$ , and the two C=O bond dipoles (arrows) point in exactly opposite directions.

Verify: Because the molecule is linear and symmetric, the two C=O bond dipoles point opposite and cancel — CO₂ is non-polar, which is why it sublimes to a gas so readily. The $s p$ /linear ( $18 0^{\circ}$ ) prediction explains the real property. ✔

Example 9 — Carbocation vs carbanion (Cell C9: exam twist on charged carbon)

Worked example Give the hybridization of the charged carbon in (i) methyl cation

$CH_{3}^{+}$ and (ii) methyl anion $CH_{3}^{-}$ . Forecast: Both are $CH_{3}$ with 3 C–H bonds. Same hybridization? Charge changes nothing? Watch carefully.

Step 1 — Methyl cation $CH_{3}^{+}$ . Why? C bonded to 3 H $\Rightarrow 3 σ$ ; it lost its 4th electron, so it has $0$ lone pairs. $SN = 3 + 0 = 3 \Rightarrow s p^{2}$ , trigonal planar, $12 0^{\circ}$ , with an empty $p$ orbital.

Step 2 — Methyl anion $CH_{3}^{-}$ . Why different? C bonded to 3 H $\Rightarrow 3 σ$ ; but it gained an electron pair, so it has $1$ lone pair. $SN = 3 + 1 = 4 \Rightarrow s p^{3}$ , pyramidal (like ammonia).

Step 3 — Read the punchline. Same $CH_{3}$ skeleton, opposite charge $\Rightarrow$ different SN $\Rightarrow$ different shape ( $s p^{2}$ flat vs $s p^{3}$ pyramidal). Why this step? The whole point of the example is to build a transferable reflex: never read hybridization off the visible bonds alone — always count the electrons the charge added or removed. Stating the contrast explicitly cements that the lone-pair term, driven purely by charge here, changed the geometry. That is the exam trap this cell exists to defuse.

Verify: The empty $p$ of the planar $s p^{2}$ cation is what lets carbocations be stabilized by neighbouring $π$ systems (hyperconjugation/resonance). The rule "count $σ$ + LP" survives even when carbon is charged. ✔

Example 10 — PCl₅ and SF₆ (Cell C10: expanded octet, SN 5 and 6)

Worked example Give the hybridization and shape of the central atom in (i) phosphorus pentachloride

$PCl_{5}$ and (ii) sulfur hexafluoride $SF_{6}$ . Forecast: Both central atoms make more than four bonds. Which hybrid labels appear when SN climbs past 4? Guess.

Step 1 — PCl₅, phosphorus. Why? P is bonded to 5 Cl $\Rightarrow 5 σ$ ; P has $0$ lone pairs. $SN = 5 + 0 = 5$ .

Step 2 — Map SN 5. Why a new label? Four hybrids ( $s p^{3}$ ) is the most you can build from one $s$ + three $p$ . To point in a fifth direction you must borrow one empty $d$ orbital, giving $s p^{3} d$ . Five directions spread out as a trigonal bipyramid (three in a plane at $12 0^{\circ}$ , two axial at $9 0^{\circ}$ ). So $SN = 5 \Rightarrow s p^{3} d$ , trigonal bipyramidal.

Step 3 — SF₆, sulfur. Why? S is bonded to 6 F $\Rightarrow 6 σ$ , $0$ lone pairs. $SN = 6 + 0 = 6$ . Now two $d$ orbitals are borrowed $\Rightarrow s p^{3} d^{2}$ , and six directions spread out as a perfect octahedron (all F–S–F neighbours at $9 0^{\circ}$ ).

Verify: Same rule as every carbon example — count $σ$ + lone pairs — extended past 4. $SN = 5 \Rightarrow s p^{3} d$ (trigonal bipyramidal, borrows 1 $d$ ); $SN = 6 \Rightarrow s p^{3} d^{2}$ (octahedral, borrows 2 $d$ ). Carbon never reaches these because it has no low-energy $d$ orbitals — only period-3-and-beyond atoms (P, S, …) can expand their octet. ✔

Recall

Recall Quick self-test

What does SN stand for and how do you compute it? ::: Steric number; SN = number of σ bonds (neighbouring atoms) + lone pairs on that atom. In allene, why is the central carbon sp but the ends sp²? ::: Central C has 2 σ (0 LP) → SN 2 → sp; each end C has 3 σ → SN 3 → sp². CH₃⁺ vs CH₃⁻ — which is planar and why? ::: CH₃⁺ (sp², no lone pair, empty p) is planar; CH₃⁻ (sp³, one lone pair) is pyramidal. Why is CO₂ linear though it has two double bonds? ::: Central C has only 2 σ bonds + 0 LP → SN 2 → sp → 180°. Which hydrogen is most acidic: ethane, ethene, or ethyne, and why? ::: Ethyne; its sp carbon has 50% s-character, holding the conjugate-base lone pair close to the nucleus. Does hybridization apply to a lone gas-phase carbon atom? ::: No — with SN = 0 there are no bonding directions to mix, so it stays in unhybridized 2s/2p orbitals. What hybridization is the methyl radical ·CH₃ and why? ::: sp² and (near-)planar; the single unpaired electron is not a lone pair (SN = 3) and sits in the leftover p orbital, repelling too weakly to force sp³. Which hybridizations correspond to SN 5 and SN 6? ::: SN 5 → sp³d (trigonal bipyramidal, e.g. PCl₅); SN 6 → sp³d² (octahedral, e.g. SF₆).