What you'll be able to do: Combine hybridization, geometry and s character to explain and compare real bond angles.
Best after: Hybrid Orbitals from Domain Count
The neat table breaks down in real molecules, and the deviations are themselves predictable.
These are recommended, not required. You can start this lesson at any time.
The ideal sp, sp² and sp³ sets have 50%, about 33% and 25% s character, respectively. Their symmetric reference geometries are linear, trigonal planar and tetrahedral, with angles 180°, 120° and about 109.5°. These labels describe related local models; s character is not an independent dial that calculates every real bond angle.
Within comparable carbon bonds, greater s character places more electron density closer to the nucleus and can contribute to shorter, stronger bonds. Ethyne's C–H bond differs from ethane's in this way. But bond order, atom identity, lone pairs and substituents can also matter. Do not use 50% versus 25% alone to predict an exact angle or bond energy in unrelated molecules.
NH₃ and PH₃ each have three bonds and one lone pair, so the elementary domain-count model starts from a tetrahedral framework. Their measured angles differ substantially: about 107° for NH₃ and 93° for PH₃. PH₃ can be described with P–H bonds having much more p character and a lone pair with more s character than an equal sp³-hybrid picture suggests. The nearly perpendicular p directions provide a useful account of angles near 90°.
This is a limitation of assigning identical ideal hybrids from domain count, not an arithmetic failure of electron counting. The extent of mixing depends on the electronic structure and bonding partners; a simple claim about one fixed s–p energy gap does not establish every comparison. Use the measured angle or a stated bonding model when explaining a heavy-atom hydride, and do not predict 107° for every AX₃E species.
In allene, H₂C=C=CH₂, the central carbon has two domains and is sp, while the two terminal carbons have three domains each and are sp². The central sp carbon uses two perpendicular p orbitals for its two pi bonds, so the two CH₂ groups end up in planes rotated 90 degrees from one another and the molecule is not planar.
Label the atoms of allene C1=C₂=C₃ before assigning them. C₂ has two carbon neighbors and no lone pair, so it is linear and sp in the localized description. C1 and C₃ each have two H neighbors plus the central C, giving three domains and sp². The two unhybridized p orbitals on C₂ are perpendicular; one overlaps with C1 and the other with C₃. Each terminal CH₂ plane must orient to its own π system, leaving the terminal planes perpendicular even though the C–C–C skeleton is linear.
| Site in CH₃–C≡N | σ directions | Lone-pair directions | Local label and geometry |
|---|---|---|---|
| Methyl C | 4 | 0 | sp³; tetrahedral |
| Nitrile C | 2 | 0 | sp; linear |
| Nitrile N | 1 | 1 | sp description; bond and lone-pair directions |
A molecule has several local geometries rather than one global hybridization. In propene, C1 in CH₃–CH=CH₂ is sp³ and the two alkene carbons are sp²; specify which three atoms define any angle you compare. Ideal references of 109.5° and 120° are useful, but different substituents can distort measured angles within the same hybridization class. Count the local connections, then explain the angle using the appropriate model and its limits.
Ideal sp, sp² and sp³ sets have reference angles 180°, 120° and 109.5°. Within comparable bonds, greater s character can favor shorter, stronger bonds; it is not a universal exact angle rule.
The fraction of s orbital in a hybrid: 50 percent for sp, 33 percent for sp², 25 percent for sp³.
PH₃ has H-P-H angles near 93 degrees because the bonds are nearly pure 3p.
sp central carbon and sp² terminal carbons give two perpendicular CH₂ planes.
s fraction = 1 / (1 + number of p orbitals used)
For propene CH₃–CH=CH₂, label the carbons C1–C₂=C₃. Assign each carbon’s local hybridization, then compare the ideal reference angles H–C1–H, C1–C₂–C₃ and H–C₃–H.
Try it first: Count domains on each carbon separately.
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Explain why the H-P-H angle in PH₃ (about 93 degrees) is smaller than the H-N-H angle in NH₃ (about 107 degrees).
Try it first: Compare the measured angles with the 90° p-orbital reference and the equal-hybrid tetrahedral model.
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In acetonitrile CH₃-C≡N, compare the two carbon centers.
Try it first: Count the local σ-bond directions around each C.
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| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Assuming every AX3E molecule has 107 degree angles. | Heavier central atoms hybridize poorly and approach 90 degrees. | Consider the period of the central atom before quoting a value. |
| Assigning one hybridization to a whole molecule. | Hybridization is a property of each atom, and different atoms can differ. | Count domains separately at every non-terminal atom. |
| Thinking allene is planar like ethene. | The central sp carbon uses two perpendicular p orbitals for two pi bonds. | Track which p orbitals each pi bond uses. |
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After the lesson · Question bank
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Assign each atom its own local bonding description, then compare ideal geometry with measured angles. Ideal sp/sp²/sp³ sets differ in s character, but lone pairs, substituents and atom identity can alter real angles. PH₃ illustrates mostly p-like bonds near 90°, and allene illustrates perpendicular π systems.
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