Worked example 1
Assign the hybridization of each carbon in propene, CH₃-CH=CH₂, and rank the C-C bond angles.
Try it first: Count domains on each carbon separately.
0 of 4 steps revealed.
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.
An sp hybrid is half s and half p, an sp² hybrid is one third s, and an sp³ hybrid is one quarter s. s orbitals are spherical and hold electron density closer to the nucleus, so more s character pulls the hybrids into a wider arrangement. That ordering gives 180, 120 and 109.5 degrees respectively, and it also explains why the C-H bond in ethyne is shorter and stronger than the one in ethane.
Ammonia has angles near 107 degrees, consistent with sp³, but phosphine has angles near 93 degrees. Phosphorus is larger and its 3s and 3p orbitals differ more in energy and size, so mixing is poor. The bonds use nearly pure 3p orbitals, which are mutually perpendicular, and the lone pair stays in a mostly s orbital. Heavier group 15 and 16 hydrides show the same pattern.
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.
Higher s character gives a wider bond angle, a shorter bond and a more electronegative atom.
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)
Assign the hybridization of each carbon in propene, CH₃-CH=CH₂, and rank the C-C bond angles.
Try it first: Count domains on each carbon separately.
0 of 4 steps revealed.
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 size and energy match of the valence orbitals of N and P.
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Why it's wrong: Heavier central atoms hybridize poorly and approach 90 degrees.
Check instead: Consider the period of the central atom before quoting a value.
Why it's wrong: Hybridization is a property of each atom, and different atoms can differ.
Check instead: Count domains separately at every non-terminal atom.
Why it's wrong: The central sp carbon uses two perpendicular p orbitals for two pi bonds.
Check instead: Track which p orbitals each pi bond uses.
No practice questions are available for this topic yet. You can still practice the whole unit.
Hybridization, electron geometry and bond angle are three views of the same domain count. More s character in a hybrid orbital gives a wider bond angle, so sp (50 percent s) is wider than sp² (33 percent) which is wider than sp³ (25 percent). Heavy central atoms such as phosphorus use nearly pure p orbitals, which drives their angles toward 90 degrees.
This lesson is original Chem Help content. No external sources were adapted.