Organic Chemistry · Structure and Bonding

sp Hybrid Orbitals and the Structure of Acetylene

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

Acetylene (ethyne, HC≡CH) carries the most extreme bonding in this chapter: a . Each carbon has just two electron groups (one C–H σ bond and one C–C σ bond), so hybridization mixes one 2s orbital with only one 2p orbital, producing two sp hybrid orbitals that point in opposite directions at 180° — the signature of . Two 2p orbitals remain unhybridized on each carbon, both perpendicular to the molecular axis and perpendicular to each other. Head-on overlaps build the σ framework (an sp–sp C–C bond and two sp–1s C–H bonds); the two sets of parallel p orbitals then overlap side-by-side to give two π bonds lying in perpendicular planes. The triple bond is therefore one σ plus two π bonds. The molecule is linear (H–C≡C–H, 180°), the C≡C bond is very short (1.20 Å), and the 50% s character of the sp carbons makes the terminal C–H bond uniquely acidic: ≈ 25, compared with ≈ 50 for an alkane C–H.

Why this matters

Alkynes (Chapter 9) are linear building blocks used throughout synthesis. The C–H is acidic enough to be deprotonated by strong bases, generating anions (RC≡C⁻) — powerful carbon nucleophiles that form new C–C bonds, one of the most valuable operations in organic synthesis. Triple bonds also appear in natural products, pharmaceuticals, and materials, and their linear geometry can lock molecular fragments into extended, rigid arrangements. Understanding also explains why sp carbons form the shortest, strongest bonds in organic chemistry and why acetylene C–H bonds behave so differently from alkane C–H bonds in acid–base chemistry.

The college version

Core Concepts

Two electron groups: the sp recipe

Count electron groups on one carbon of acetylene: one C–H σ bond plus one C–C σ bond = two groups. Two groups → sp: mix one s with one p orbital. Two equivalent sp orbitals result, each 50% s and 50% p, pointing in opposite directions at 180°.

Two leftover p orbitals and two π bonds

Because only one p orbital was used in the mixing, each carbon keeps two unhybridized p orbitals. They are perpendicular to the molecular axis and to each other — imagine one p orbital in the page plane and the other coming out of the page. Each leftover p orbital overlaps side-by-side with its partner on the other carbon, giving two π bonds in mutually perpendicular planes. Together with the sp–sp σ bond, this gives the triple bond its full strength.

The σ framework and linear geometry

The two sp orbitals of each carbon overlap head-on: sp–sp for the C–C bond and sp–1s for each C–H bond. Because the sp orbitals point in exactly opposite directions, all four atoms of acetylene lie on one straight line: H–C≡C–H with 180° angles. This is the defining shape of sp-hybridized carbon.

Short bonds, high s character

The C≡C bond is 1.20 Å — dramatically shorter than the C=C double bond (1.33 Å) and the C–C single bond (1.54 Å). The C–H bond of acetylene (1.06 Å) is also the shortest C–H bond of the series. Higher s character (50% vs. 33% vs. 25%) pulls electrons closer to the nucleus, strengthening and shortening every bond on the sp carbon.

Terminal alkyne acidity

The same s character concentrates the C–H bonding electrons, and the resulting carbon anion is strongly stabilized by the electronegative sp carbon: the conjugate base (the acetylide ion) is more stable, so the terminal alkyne C–H is a stronger acid. With pKa ≈ 25 it is far more acidic than an alkane C–H (pKa ≈ 50), although still much weaker than water (pKa ≈ 15.7) or carboxylic acids (pKa ≈ 4–5).

How It Works / Step-by-Step Process

Draw the Lewis structure of acetylene: H–C≡C–H.
Count electron groups per carbon: two (one C–H σ + one C–C σ) → sp.
Mix one s + one p → two sp orbitals at 180°; leave two p orbitals unhybridized on each carbon.
Build the σ framework: sp–sp (C–C) and sp–1s (C–H).
Overlap the leftover p orbitals side-by-side in two perpendicular planes to form two π bonds.

Common Confusions

Do Not Confuse	With	The Difference
Triple bond = three σ bonds	Triple bond = one σ + two π	Only one bond can lie on the internuclear axis; the other two must be π
The two π bonds in one plane	The two π bonds in perpendicular planes	The two leftover p orbitals are perpendicular to each other
Acetylene C–H acidity = alkane C–H	Terminal alkyne is more acidic	50% s character stabilizes the acetylide conjugate base (pKa 25 vs. 50)
sp geometry = bent	sp geometry = linear	Two groups repel to exactly 180°
Every carbon in an alkyne is linear	Only sp carbons are linear	An sp³ carbon adjacent to a triple bond stays tetrahedral

Eli explains

The same idea, in plain words

Explain it like I’m 10

Take one arrow and let it point both ways at once — that is the sp hybrid, and it makes everything lie on a straight line (180°). Around that arrow, two crossed paper fans (the leftover p orbitals) stick out at right angles to each other. When two such arrows meet, each pair of fans grabs its partner in its own direction, making two π grabs wrapped around one strong σ shaft. The double grab locks the whole molecule into a straight line.

Worked example

Example 1: Counting σ and π bonds in acetylene

In HC≡CH, count the bonds: one C–C σ bond, two C–H σ bonds, and two π bonds. Total: 3 σ bonds + 2 π bonds = 5 bonding interactions. Each carbon's two electron groups (one C–H σ + one C–C σ) confirm sp hybridization, and the two leftover p orbitals per carbon account for the two π bonds.

Example 2: Converting the C≡C bond length (dimensional analysis)

The C≡C bond in acetylene is 1.20 Å. Convert to picometers using 1 Å = 100 pm:

$1.20 Å \times 100 pm 1 Å = 120 pm$

Then to nanometers using 1 nm = 1000 pm:

$120 pm \times 1 nm 1000 pm = 0.120 nm$

The trend is now complete: C–C 154 pm → C=C 133 pm → C≡C 120 pm.

Example 3: Percent s character across hybrids

Compute the s character of sp with the formula % s = (s orbitals mixed ÷ total orbitals mixed) × 100%:

$% s character = 12 \times 100 % = 50 %$

Compare the full series — sp³: 25%, sp²: 33%, sp: 50% — and note the consequences: as s character rises, bonds shorten and strengthen, and the C–H bond becomes more acidic (alkane pKa ≈ 50 → alkene ≈ 44 → terminal alkyne ≈ 25).

Key takeaways

2 electron groups → sp, linear geometry, 180°.
sp = one s + one p; 50% s character, the most of any carbon hybrid.
A triple bond = one σ + two π bonds, with the two π bonds in perpendicular planes.
Each sp carbon keeps two unhybridized p orbitals.
C≡C ≈ 1.20 Å; C–H ≈ 1.06 Å — the shortest bonds in the sp³/sp²/sp series.
Terminal alkyne C–H: pKa ≈ 25 (vs. ≈ 50 for alkane C–H) because the acetylide anion is stabilized by the sp carbon.
Acetylide anions (RC≡C⁻) are strong carbon nucleophiles for building new C–C bonds.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

How many electron groups surround each carbon in acetylene?
Show answer
Two electron groups per carbon (one C–H σ + one C–C σ).
What is the hybridization of each carbon, and what is the overall molecular geometry?
Show answer
Each carbon is sp-hybridized; the molecule is linear (H–C≡C–H, 180°).
How many σ and π bonds does HC≡CH contain?
Show answer
Three σ bonds (one C–C + two C–H) and two π bonds.
Why are the two π bonds of acetylene perpendicular to each other?
Show answer
Each carbon keeps two unhybridized p orbitals, which are perpendicular to each other; each pair forms its own π bond in its own plane.
What is the approximate C≡C bond length?
Show answer
About 1.20 Å (120 pm).
Why is a terminal alkyne C–H more acidic than an alkane C–H?
Show answer
The sp carbon's 50% s character stabilizes the acetylide anion (the conjugate base), so the C–H proton is lost more readily: pKa ≈ 25 vs. ≈ 50 for an alkane.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

sp hybridization: Mixing one s + one p orbital → two linear orbitals at 180°
triple bond: One σ + two π bonds between two atoms
terminal alkyne: Alkyne with a C–H on a triply bonded carbon
acetylide: Anion RC≡C⁻ formed by deprotonating a terminal alkyne
pKa: Logarithmic measure of acid strength (lower = stronger acid)
linear geometry: All atoms arranged on a straight line at 180°
sp³ hybridization: Mixing one s + three p orbitals to make four equivalent orbitals

Sources & references

openstax.org — Organic Chemistry

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