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

WHAT we count: carbon has 4 C–H single bonds → 4 σ bonds, 0 lone pairs. SN $=4+0=4$. $\text{SN}=4\Rightarrow \boxed{s{p}^3}$, tetrahedral, angle $109.5^{\circ }$.

Each carbon: 2 C–H σ + 1 C–C σ (the σ part of the double bond) = 3 σ, 0 lone pairs. The second stroke of the "=" is a π bond → it does not count. SN $=3+0=3\Rightarrow \boxed{s{p}^2}$ on each carbon. Planar, $120^{\circ }$.

Each carbon: 1 C–H σ + 1 C–C σ = 2 σ, 0 lone pairs. The triple bond is 1σ + 2π; the two π's don't count. SN $=2+0=2\Rightarrow \boxed{sp}$. Linear, $180^{\circ }$.

The carbon is double-bonded to each oxygen. A double bond = 1σ + 1π, so the carbon has 1 σ to each O = 2 σ total, and 0 lone pairs on carbon. SN $=2+0=2\Rightarrow \boxed{sp}$, linear $180^{\circ }$. The two leftover p orbitals on carbon make the two π bonds (one to each oxygen). See the sideways lobes in figure s01.

Oxygen has 2 O–H single bonds → 2 σ, and 2 lone pairs. SN $=2+2=4\Rightarrow \boxed{s{p}^3}$. Geometry is bent, angle $\approx 104.5^{\circ }$ — slightly less than $109.5^{\circ }$ because the two lone pairs push harder than bonding pairs and squeeze the H–O–H angle. (This links to VSEPR theory and molecular geometry.)

Nitrogen: 3 N–H σ + 1 lone pair. SN $=3+1=4\Rightarrow \boxed{s{p}^3}$, trigonal pyramidal, angle $\approx 107^{\circ }$ (one lone pair squeezes it below $109.5^{\circ }$).

Carbon: 2 C–H σ + 1 C–O σ (the σ of the C=O) = 3 σ, 0 lone pairs. SN $=3\Rightarrow \boxed{s{p}^2}$, planar, $\approx 120^{\circ }$.

• C1 (${\text{CH}}_3$): 3 C–H σ + 1 C–C σ = 4 σ, 0 LP → SN 4 → $s{p}^3$.
• C2 (inner triple-bond C): 1 σ to C1 + 1 σ to C3 = 2 σ, 0 LP → SN 2 → $sp$.
• C3 (terminal $\equiv$CH): 1 C–H σ + 1 σ to C2 = 2 σ, 0 LP → SN 2 → $sp$. Answer: $s{p}^3,sp,sp$. A single molecule can carry different hybridizations on different carbons.

WHY s-character matters: an $s$ orbital hugs the nucleus more tightly than a $p$ orbital. More s-character in a hybrid means its electrons sit closer to the positive nucleus.

• $sp$ = $\frac{1}{2}$ s-character (50%), $s{p}^3$ = $\frac{1}{4}$ (25%).
• When the acidic proton leaves, the electron pair it left behind lives in the carbon hybrid orbital. In an $sp$ carbon that lone pair is held close and low in energy → the resulting anion (carbanion) is stabilized.
• A stable conjugate base ⇒ the acid gives up its proton more readily ⇒ more acidic. So the 50% s-character of $sp$ carbon is exactly what makes terminal alkyne H acidic. (Details in Acidity of terminal alkynes.)

• Ethane C–C: $s{p}^3$–$s{p}^3$, $\approx 1.54\text{A˚}$.
• Ethene C=C: $s{p}^2$–$s{p}^2$, $\approx 1.34\text{A˚}$.
• Ethyne C≡C: $sp$–$sp$, $\approx 1.20\text{A˚}$. Order (longest → shortest): ethane $>$ ethene $>$ ethyne. Why: two effects both shorten as we go $s{p}^3\to s{p}^2\to sp$: (i) higher s-character pulls the shared electrons closer to each nucleus, and (ii) more π bonds add extra sideways binding that clamps the atoms together. (See Bond length and bond strength trends.)

• Terminal carbons: each has 2 C–H σ + 1 C–C σ = 3 σ, 0 LP → SN 3 → $s{p}^2$.
• Central carbon: two C=C double bonds → 1 σ to each neighbour = 2 σ, 0 LP → SN 2 → $sp$, linear. Shape: the central $sp$ carbon has two leftover perpendicular p orbitals. One makes the π bond to the left terminal carbon, the other (at $90^{\circ }$) makes the π bond to the right. Because each terminal $s{p}^2$ carbon lies in the plane of its π bond, and the two π planes are perpendicular, the two ${\text{CH}}_2$ ends are twisted $90^{\circ }$ to each other. So the four H atoms are NOT coplanar. See figure s02.

Structure: ${\text{CH}}_3-\stackrel{=O}{C}-O-H$.

• Methyl C: 3 C–H σ + 1 C–C σ = 4 σ, 0 LP → SN 4 → $s{p}^3$.
• Carbonyl C: 1 σ to methyl C + 1 σ (of C=O) + 1 σ (C–O of the –OH) = 3 σ, 0 LP → SN 3 → $s{p}^2$.
• Carbonyl O (the $=O$): 1 σ (to carbonyl C) + 2 LP → SN 3 → $s{p}^2$.
• Hydroxyl O (the $-O-H$): 2 σ (to C and to H) + 2 LP → SN 4 → $s{p}^3$. Answer: $s{p}^3,s{p}^2,s{p}^2,s{p}^3$. Note the hydroxyl O is often drawn/argued as $s{p}^2$ so its lone pair can overlap with the carbonyl π system (resonance) — see Resonance and delocalization (sp² systems). By the bare SN count it is $s{p}^3$; that's the answer expected here.

Each ring carbon: 1 C–H σ + 2 C–C σ (to its two ring neighbours) = 3 σ, 0 LP → SN 3 → $s{p}^2$, so every ring angle is $120^{\circ }$ and the ring is a flat regular hexagon. Each carbon keeps one un-hybridized p orbital, perpendicular to the ring plane. All six line up parallel and overlap sideways all the way around → a continuous ring of π electron density above and below the plane. That delocalized loop is what makes benzene aromatic and unusually stable. (See Aromaticity and benzene (all-sp² ring).)

Linear $180^{\circ }$ ⇒ SN $=2$ ⇒ $\boxed{sp}$. $sp$ has SN 2 (so σ + LP $=2$) and two un-hybridized p orbitals. Two π bonds use exactly those two p orbitals — consistent. To have SN $=2$ with the geometry linear and no lone pairs bending it, the atom has 2 σ bonds, 0 lone pairs (e.g. central C of ${\text{CO}}_2$ or an alkyne carbon).

The cationic carbon: 3 C–H σ + 0 lone pairs → SN $=3\Rightarrow \boxed{s{p}^2}$, trigonal planar, $120^{\circ }$. Edge subtlety: the leftover un-hybridized p orbital is empty (it holds the "missing" electron pair). An empty p orbital does not count in SN — SN only counts σ bonds and filled lone pairs. So the shape is flat, and that empty perpendicular p orbital is exactly where a nucleophile attacks. Compare with the methyl radical ${\text{CH}}_3^{\bullet }$ (also $s{p}^2$, nearly planar, one electron in the p).

Carbon: 3 C–H σ + 1 lone pair → SN $=4\Rightarrow \boxed{s{p}^3}$, trigonal pyramidal (like ${\text{NH}}_3$), angle a bit under $109.5^{\circ }$. Contrast the trio at the same ${\text{CH}}_3$ centre:

• ${\text{CH}}_3^{+}$: 3 σ + 0 LP → SN 3 → $s{p}^2$, flat.
• ${\text{CH}}_3^{\bullet }$: 3 σ + (1 unpaired e⁻, not a full pair) → treated as SN 3 → $s{p}^2$, near-flat.
• ${\text{CH}}_3^{-}$: 3 σ + 1 LP → SN 4 → $s{p}^3$, pyramidal. Same three bonds, three different shapes — decided entirely by what fills the fourth region.

Label: ${\text{C}}_a{\text{H}}_2={\text{C}}_b\text{H}-{\text{C}}_c\equiv \text{N}$.

• ${\text{C}}_a$ ($={\text{CH}}_2$): 2 C–H σ + 1 σ (of C=C) = 3 σ, 0 LP → SN 3 → $s{p}^2$.
• ${\text{C}}_b$ ($=\text{CH}-$): 1 C–H σ + 1 σ (of C=C) + 1 σ (to ${\text{C}}_c$) = 3 σ, 0 LP → SN 3 → $s{p}^2$.
• ${\text{C}}_c$ ($-\text{C}\equiv$): 1 σ (to ${\text{C}}_b$) + 1 σ (of C≡N) = 2 σ, 0 LP → SN 2 → $sp$.
• N ($\equiv \text{N}$): 1 σ (of C≡N) + 1 lone pair → SN 2 → $sp$. Answer: ${\text{C}}_a=s{p}^2,{\text{C}}_b=s{p}^2,{\text{C}}_c=sp,\text{N}=sp$. The nitrogen's lone pair sits in an $sp$ hybrid pointing straight out along the axis.