2.3.10 · D5Chemical Bonding
Question bank — Valence Bond Theory (VBT) — hybridization (sp, sp², sp³, sp³d, sp³d²)
Before we start, one reminder of the only formula you need — the steric number:
The picture below is the visual dictionary for the whole bank — every shape a steric number can produce, drawn once so you can point to it as you read.
And here is the geometry that forces the famous tetrahedral angle, referenced in the "Why questions" section — so the number is earned, not asserted.
True or false — justify
Every "true/false" below wants the reason, not the verdict.
True or false: A double bond adds 2 to the steric number.
False. A double bond is one + one ; only the counts toward SN, so a double bond contributes exactly 1. The part rides on an unhybridized orbital, which is invisible to the geometry. See Sigma and Pi Bonds — the is end-on overlap along the bond axis, the is sideways overlap above and below it.
True or false: Lone pairs can be ignored when finding hybridization.
False. Lone pairs sit inside hybrid orbitals, so they occupy a geometric position and count in SN. Ignore water's 2 lone pairs and you'd wrongly predict linear instead of bent ().
True or false: hybridization always gives exactly .
False. is the ideal when all four positions are identical bonding pairs. Lone pairs (water , ammonia ) and electronegativity differences distort the real angle — Bond Angle and Bent's Rule says more s-character concentrates in bonds to less electronegative atoms, widening those angles.
True or false: Hybridization changes the total number of orbitals available.
False. Orbitals in = orbitals out. Mix 1 + 3 and you get exactly 4 hybrids — never 3 or 5. It's a redistribution, not creation.
True or false: More s-character in a hybrid orbital gives a wider bond angle.
True. More %s pulls the orbital closer to the nucleus with a fatter lobe, so the orbitals splay apart more: (, ) (, ) (, ).
True or false: Two molecules with the same hybridization must have the same shape.
False. Same hybridization means same electron geometry, but lone pairs change the molecular shape. , , are all yet are tetrahedral, pyramidal and bent respectively.
True or false: A hybrid orbital can form a bond.
False. Hybrid orbitals point at the partner atom (end-on overlap) and make bonds or hold lone pairs. bonds need sideways overlap, which only leftover unhybridized orbitals can provide.
Spot the error
Each line states a wrong claim; the reveal says what broke and why. (Recall = valence electrons of the central atom, defined above.)
" has two double bonds, so its SN is 4 and carbon is ."
Error: double bonds are counted as 1 each. C has 2 bonds + 0 lone pairs SN , linear (), not .
"Water is linear because oxygen only bonds to two hydrogens."
Error: it forgot the 2 lone pairs. SN , so oxygen is and the molecule is bent (), not linear.
" has SN 3 and one lone pair on boron pushing it into a bent shape."
Error: boron has (three valence electrons), all three go into bonds, so LP . It's , trigonal planar with an empty orbital — flat, not bent.
"In ethene, the C=C bond comes from overlap of two hybrids."
Error: hybrids make the framework. The bond comes from the unhybridized left on each carbon, overlapping sideways.
" needs 6 unhybridized orbitals to make 6 bonds."
Error: sulfur has only 3 orbitals. The label is — one , three , two hybridized. Also note (nuance) real d-participation is minimal; it's a bookkeeping model.
"Since promotion of an electron in carbon costs energy, hybridization is energetically unfavourable."
Error: the promotion cost is repaid with interest — forming two extra strong C–H bonds releases far more energy than promotion consumed. Net: favourable.
" has 4 bonds so SN is 4 and it is tetrahedral."
Error: Xe () keeps 2 lone pairs: LP . SN ; the two lone pairs go trans, giving square planar.
Why questions
Why do we hybridize orbitals at all instead of using pure and ?
Pure and orbitals would give unequal, wrongly-angled bonds. Experiment shows 's four bonds are identical at , which only a set of four equivalent hybrids can explain.
Why does more s-character widen the bond angle?
The orbital is spherical and held tight to the nucleus. Blending in more of it makes the hybrid fatter and closer in, so the lobes must splay farther apart to minimise repulsion — hence a larger angle.
Why do lone pairs squeeze bond angles below the ideal value?
A lone pair is held by only one nucleus, so its cloud spreads wider and pushes harder than a bonding pair (shared by two nuclei). This extra shove compresses the neighbouring bonding angles — water drops to .
Why do only orbitals of comparable energy on the same atom hybridize?
Mixing requires the orbitals to "average" cleanly. Orbitals far apart in energy or on different atoms won't blend into a single, uniform set — the model only averages nearby-energy, same-atom orbitals.
Why is the leftover orbital in chemically important?
It is empty and perpendicular to the plane, so it can accept an electron pair — this is exactly what makes a Lewis acid. See Formal Charge and Lewis Structures.
Why can VBT/hybridization predict shape but not always explain magnetism or bond order like ?
Hybridization is a localized-pair picture built for geometry. Spread-out electron behaviour (e.g. 's two unpaired electrons) needs Molecular Orbital Theory (MOT) instead.
Why does the angle come out as and not a round number?
Put the four bonds at alternate corners of a cube — say and from the centre (see the second figure). The angle between them uses , so . It's forced by geometry, not chosen — the same balloon-repulsion picture as VSEPR Theory.
Edge cases
What hybridization does a terminal atom (like each H in ) have?
Hydrogen has only one orbital and nothing of similar energy to mix with, so it cannot hybridize. Hybridization is asked about the central atom.
What does the steric-number rule say for ?
SN means a single bond and no lone pairs — a lone diatomic fragment. There is nothing to arrange, so no hybridization is meaningfully assigned; the geometry is trivially a point-to-point line. The useful domain of the SNhybridization table is to .
What happens for (say , as in )?
The simple scheme runs out; is described as giving a pentagonal bipyramid. Beyond that (, e.g. , ) VBT hybrid labels become pure bookkeeping and are rarely required — real bonding is better handled by Molecular Orbital Theory (MOT).
but the molecule has a lone pair on the central atom — is it still linear?
With SN there is no room for a lone pair and two bonds and still be SN 2; if a lone pair existed SN would rise. A true SN-2 species (, ) has no central lone pair and is linear.
In a trigonal bipyramid (), where does the first lone pair go — axial or equatorial?
Equatorial. An equatorial position has only two neighbours at (vs three for axial), so the fat lone pair suffers less repulsion there. This is why is see-saw, not trigonal.
Does a single unpaired electron (radical) count in the steric number?
Yes — treat it like a lone "half" pair: it occupies a hybrid orbital and takes up a geometric position, so it counts as one toward SN (e.g. behaves near ).
Is evidence that d orbitals fully participate in bonding?
No. For exam VBT we label it , but modern theory shows d-orbital contribution is minimal. It's a convenient bookkeeping label, not literal heavy d-mixing.
Can a molecule have SN that changes when it ionizes or gains a lone pair?
Yes. Adding or removing electrons alters the lone-pair count, hence SN and shape — e.g. (, pyramidal) becomes (, tetrahedral) when the lone pair becomes a bond.
Recall One-line survival summary
Count only bonds + lone pairs on the central atom SN hybridization shape. Never count bonds; never forget lone pairs; and remember the table gives ideal angles that lone pairs then distort.