Question bank — Tetravalency of carbon; hybridization recap (sp, sp², sp³)

False as stated — it is $s{p}^2$ only if that C has exactly 3 σ bonds. The central C in O=C=O forms two double bonds yet is $sp$, because it has only 2 σ bonds. Count σ, not double bonds.

False — the opposite. π bonds use unhybridized p orbitals, so more π bonds means fewer hybrid orbitals ($sp$ has 2 hybrids and 2 leftover p; $s{p}^3$ has 4 hybrids and 0 leftover p — exactly the trend drawn in s02).

False — carbon has no desires. Promotion costs energy, but that cost is repaid by the two extra bonds formed afterward. It happens purely because the refund exceeds the bill, not out of "want."

True — that identity is the reason hybridization must exist. One $s$ + three unequal $p$ orbitals mix into four equivalent $s{p}^3$ hybrids, so every bond has the same length (109 pm) and strength.

True — $sp$ has 50% s-character vs 25% for $s{p}^3$. More s-character holds shared electrons closer to the nucleus, so the $sp$ carbon pulls electron density harder.

True — recall $\text{SN}=\sigma \text{ bonds}+\text{lone pairs on central atom}$. In methanol's oxygen, 2 σ + 2 lone pairs gives SN 4 → $s{p}^3$, even though O forms only two bonds.

True — higher s-character shortens bonds: $sp(1.20\text{A˚})<s{p}^2(1.34\text{A˚})<s{p}^3(1.54\text{A˚})$. This is a consequence of s-character, not a memorized list.

False — only orbitals of nearly equal energy on the same atom mix. That is why $2s$ mixes with $2p$ but not, say, with $1s$.

The triple bond is 1 σ + 2 π (see the head-on/sideways split in s01). Each carbon has only 2 σ bonds (one C–H, one C–C) and 0 lone pairs → SN 2 → ==$sp$==, linear $180^{\circ }$. The "3" of $s{p}^3$ was mistakenly read off the "triple."

Only lone pairs on the central atom count. The oxygens' lone pairs are irrelevant to carbon. Carbon has 2 σ + 0 lone pairs → SN 2 → ==$sp$==, linear (O=C=O).

The pairing observation is right for the ground state ($1s^{2}2s^{2}2p_x^{1}2p_y^1$), but the conclusion is wrong: promotion unpairs the $2s$ electron into $2p_z$, giving four unpaired electrons and true tetravalency.

The $120^{\circ }$ angle comes from the three $s{p}^2$ hybrid orbitals (the head-on σ framework) spreading as far apart as possible in a plane. The π bond is the sideways overlap sitting perpendicular above and below — it locks planarity but does not set the angle. Compare the σ vs π geometry in s01.

Every ring carbon has 3 σ bonds (2 ring C–C, 1 C–H) + 0 lone pairs → all are ==$s{p}^2$==. The leftover p orbitals (one per carbon, the grey lobe in s02's $s{p}^2$ panel) overlap into a single delocalized π system.

Promotion alone costs energy, but you must account for the whole process: the two additional bonds formed afterward release far more energy, so the final molecule is much more stable. Never judge an intermediate step in isolation.

A hybrid orbital is needed for each σ bond and each lone pair (they define the directional skeleton). π bonds form from pure unhybridized p orbitals that lie sideways on top of that skeleton (the orange lobes in s01), so they never demand their own hybrid.

An $s$ orbital is spherical and centred right on the nucleus, so more s-character keeps the bonding electrons closer to the positive nucleus — a tighter, shorter, stronger bond.

The $sp$ carbon (50% s-character) holds the C–H electron pair very close to the nucleus, so when the H⁺ leaves, the resulting negative charge on carbon is well-stabilized. That stability makes losing H⁺ favourable.

Each orbital is a mathematical wavefunction, and hybrids are just weighted sums of the originals; you cannot build more independent sum-functions than you started with, nor fewer without discarding one. So $n$ atomic wavefunctions produce exactly $n$ hybrid wavefunctions ($1s+3p\to 4s{p}^3$), same as $n$ input vectors spanning an $n$-dimensional space.

Oxygen is $s{p}^3$ (SN 4) but two of the four positions are lone pairs. A lone pair is held by only one nucleus, so its cloud is fatter and spreads out more than a bonding pair (which is pinched between two nuclei); it therefore pushes the two bonding orbitals closer, dropping the angle below the ideal tetrahedral value.

The double bond includes a π bond from sideways p-orbital overlap (the orange lobes in s01); twisting by $90^{\circ }$ would slide the lobes apart and break the bond. A single bond is only σ (end-on, cylindrically symmetric), so rotation leaves the overlap unchanged and costs almost nothing.

The central carbon has 2 σ bonds + 0 lone pairs → SN 2 → ==$sp$==. Its two π bonds use two perpendicular p orbitals, forcing the two end CH₂ groups into perpendicular planes. The two end carbons are $s{p}^2$.

SN = 2 σ + 1 lone pair = 3 → ==$s{p}^2$== (bent, angle < $120^{\circ }$). This shows the divalent carbon of the ground state can still hybridize once you count its lone pair.

The single unpaired electron sits in one orbital, so it counts like a lone pair: SN = 3 σ + 1 = 4 → nominally $s{p}^3$, but the shallow energy cost makes it nearly ==planar $s{p}^2$== with the odd electron in a p orbital. It hovers between the two, which is why radicals are floppy and easily flipped.

The carbon has 3 σ bonds + 0 lone pairs (the "missing" electrons leave an empty orbital, which is not counted) → SN 3 → ==$s{p}^2$==, trigonal planar, with the empty orbital being the leftover unhybridized p perpendicular to the plane. Contrast this with the carbanion below.

Yes — the carbanion has 3 σ bonds + 1 lone pair → SN 4 → $s{p}^3$, pyramidal. The extra filled pair behaves exactly like any lone pair in the count (unlike the cation's empty orbital, which does not).

All three carbons are $s{p}^2$ throughout; the "moving" π electrons are delocalized over the unhybridized p orbitals, but the σ skeleton and hybridization stay fixed.

The N has 1 σ bond (to C) + 1 lone pair = SN 2 → ==$sp$==, and it carries two π bonds through its two leftover p orbitals. Same $sp$ logic as carbon, applied to nitrogen.

No — carbon's valence shell (n=2) has no $d$ orbitals, so it maxes out at $s{p}^3$ (SN 4). Expanded octets requiring $d$ orbitals appear only from period 3 onward.