Valence Bond Theory (VBT) — hybridization (sp, sp², sp³, sp³d, sp³d²): Worked examples

Four short words and symbols we lean on the whole page — earn them once:

σ (sigma) bond ::: a bond where the two atoms' orbitals overlap head-on, straight along the line joining the nuclei. Every single bond is one σ. See Sigma and Pi Bonds.
π (pi) bond ::: a bond where two unhybridized p orbitals overlap sideways, above and below the bond line. A double bond = 1 σ + 1 π; a triple bond = 1 σ + 2 π. π bonds are invisible to SN.
BP = bonding pair ::: a pair of electrons shared between two atoms — i.e. a bond. Its cloud is pulled out between two nuclei, so it is relatively thin.
LP = lone pair ::: a pair of electrons that belongs to one atom only, forming no bond. With just one nucleus tugging it, its cloud is fatter and closer in, so it pushes harder than a BP.

The figure below is your master key for this whole page: its left panel shows the electron-geometry-vs-shape idea on $N H_{3}$ (the lone-pair domain is counted, but the named shape uses only the atoms), and its right panel shows why a lone pair squeezes angles — the fat, one-nucleus LP cloud vs the thin, two-nucleus BP cloud. Keep glancing back at the right panel whenever an example squeezes an angle (Ex 2, 6, 7, 8, 9).

Figure — Valence Bond Theory (VBT) — hybridization (sp, sp², sp³, sp³d, sp³d²)

The scenario matrix

Every hybridization problem is one of these case classes. Each row is a distinct trap or behaviour; the worked examples below are tagged with the cell they cover.

#	Case class	What makes it tricky	Example
C1	SN 4, zero lone pairs	baseline, ideal angle	Ex 1 ( $C H_{4}$ )
C2	Lone pairs present, angle squeezed	LP repels harder than BP	Ex 2 ( $N H_{3}$ )
C3	Multiple bonds — the π trap	double/triple counts as one in SN	Ex 3 ( $C O_{2}$ )
C4	Charged ion	must adjust valence electrons for charge	Ex 4 ( $N H_{4}^{+}$ )
C5	Expanded octet, SN 5	axial vs equatorial, non-equivalent positions	Ex 5 ( $P C l_{5}$ )
C6	Expanded octet with lone pairs	lone pair steals an equatorial slot	Ex 6 ( $S F_{4}$ )
C7	SN 6 with 2 lone pairs — degenerate-looking shape	octahedron collapses to square planar	Ex 7 ( $X e F_{4}$ )
C8	Real-world word problem	translate description → SN	Ex 8 (dry ice / $S O_{2}$ )
C9	Exam twist — same formula, two answers	odd electron / resonance-free counting	Ex 9 ( $N O_{2}$ vs $N O_{2}^{-}$ )
C10	Limiting / smallest case, SN 2	two σ only, linear boundary	Ex 10 ( $B e C l_{2}$ , $C_{2} H_{2}$ )

Formula The one machine, restated (with the π caveat built in)

$SN = (atoms σ -bonded to centre) + (lone pairs on centre)$ To find the lone pairs, use $LP = \frac{V - S}{2}$ where $V$ = valence electrons of the central atom (adjust for ionic charge) and $S$ = the number of the centre's own electrons it has spent making bonds. Key rule: each σ bond the centre makes uses 1 of its electrons, and each π bond uses 1 more. So a single bond costs 1, a double bond costs 2, a triple bond costs 3. $SN = 2 \to s p, 3 \to s p^{2}, 4 \to s p^{3}, 5 \to s p^{3} d, 6 \to s p^{3} d^{2}$

Newcomer note: for molecules with only single bonds, $S$ = number of σ partners, so the older shortcut $LP = \frac{V - B}{2}$ (with $B$ = σ partners) gives the same answer. The $S$ -version above is the general one that also handles double and triple bonds correctly — that is why Ex 3 counts the π electrons.

Ex 1 — Baseline: no lone pairs (cell C1)

Worked example Methane

$C H_{4}$ Forecast: four identical bonds to H — guess the SN, the label, and the angle before reading on.

Step 1 — Central atom & valence. C has $V = 4$ . Why this step? SN always starts from the central atom's electron budget; H is never central (it has only one orbital).

Step 2 — Count σ partners. C is single-bonded to 4 H, so it has 4 σ partners and spends $S = 4$ of its own electrons. Why this step? Each C–H is one head-on overlap = one σ = one unit of SN, and each single bond uses exactly one of carbon's electrons.

Step 3 — Lone pairs. $LP = \frac{4 - 4}{2} = 0$ . Why this step? With zero leftover electrons there is nothing to squeeze the angle — this is the ideal case.

Step 4 — SN and label. $SN = 4 + 0 = 4 \Rightarrow s p^{3}$ , tetrahedral, $109.5°$ . Why this step? SN 4 has one and only one entry in the table; converting SN → label is the whole point of the machine.

Verify: Place the four bonds at alternate cube corners $(1, 1, 1), (1, - 1, - 1), (- 1, 1, - 1), (- 1, - 1, 1)$ . Then $cos θ = \frac{- 1}{3} \Rightarrow θ = 109.47°$ . Geometry, not guesswork.

Ex 2 — Lone pair squeeze (cell C2)

Worked example Ammonia

$N H_{3}$ Forecast: N has 5 valence electrons and makes 3 bonds — where do the other 2 electrons go, and what does that do to the angle?

Step 1 — Valence. N: $V = 5$ . Why this step? Everything downstream is a budget calculation on these 5 electrons. Step 2 — σ partners. 3 single bonds to H, so $S = 3$ . Why this step? Three single bonds use three of nitrogen's electrons; that leaves $5 - 3 = 2$ to account for. Step 3 — Lone pairs. $LP = \frac{5 - 3}{2} = 1$ . Why this step? Those two leftover electrons pair up into one hybrid orbital — invisible in the shape's name but very visible in the angle. Step 4 — SN. $3 + 1 = 4 \Rightarrow s p^{3}$ . Why this step? SN counts σ plus lone pairs, so the lone pair earns a slot; the electron geometry is tetrahedral, but the molecular shape (atoms only) is trigonal pyramidal.

Step 5 — The angle. Ideal $s p^{3}$ is $109.5°$ , but the fat lone pair pushes the three N–H bonds together, so the observed angle is $\approx 107°$ . Why this step? An LP sits closer to the nucleus (no second nucleus pulling it out), so its cloud is fatter and repels bonds harder than a BP does. See Bond Angle and Bent's Rule.

Verify: $107° < 109.5°$ ✅ — one lone pair, one small squeeze. Compare water ( $104.5°$ , two lone pairs, bigger squeeze). The trend $C H_{4} > N H_{3} > H_{2} O$ matches lone-pair count $0 < 1 < 2$ .

Ex 3 — The π trap (cell C3)

Worked example Carbon dioxide

$C O_{2}$ (O=C=O) Forecast: two double bonds "look like a lot of bonding." Will SN be 4? Guess, then check.

Step 1 — Central atom. C, $V = 4$ . Why this step? The budget always starts from the central atom's own valence electrons — here carbon's four — because it is those electrons we track through every bond. Step 2 — Count σ only for SN. Each C=O is one σ + one π. C has two double bonds ⇒ 2 σ, so 2 counts toward SN. Why this step? SN counts directions in space. A double bond points in exactly one direction — its π is stacked on the σ, adding no new direction. Step 3 — Lone pairs on C. Now use the general formula. Each double bond spends 2 of carbon's electrons (1 σ + 1 π), so $S = 2 \times 2 = 4$ , giving $LP = \frac{4 - 4}{2} = 0$ . Why this step? This is exactly why the LP formula counts π electrons: carbon really has used up all four of its own electrons (2 on the σ framework, 2 on the π's), so none are left to form a lone pair. Step 4 — SN. $2 + 0 = 2 \Rightarrow s p$ , linear, $180°$ . Why this step? Only the σ count (2) and the LP count (0) enter SN — the π electrons were needed to prove LP=0, but they never add to SN themselves.

Step 5 — Where do the π's live? C is $s p$ , leaving two unhybridized p orbitals ( $p_{y}, p_{z}$ ), one for each π bond. Why this step? π always comes from unhybridized p — never from a hybrid. Two leftover p ⇒ room for exactly two π bonds. ✅ consistent.

Verify: $O = C = O$ is linear at $180°$ — measured value confirms. If you had wrongly counted π toward SN you'd get $4 \Rightarrow s p^{3}$ and predict a bent molecule — wrong.

Ex 4 — Charged ion: fix the valence first (cell C4)

Worked example Ammonium ion

$N H_{4}^{+}$ Forecast: it's ammonia plus a proton — did the lone pair survive?

Step 1 — Adjust valence for charge. N normally has $V = 5$ ; the $+ 1$ charge means the ion is short one electron, so use $V = 5 - 1 = 4$ . Why this step? A positive charge = electrons removed; a negative charge = electrons added. Skipping this is the classic ion error. Step 2 — σ partners. 4 single bonds to H, so $S = 4$ . Why this step? Only after the charge fix is the budget correct; each N–H still costs one nitrogen electron, so four bonds spend four. Step 3 — Lone pairs. $LP = \frac{4 - 4}{2} = 0$ . Why this step? The lone pair that ammonia had is now donating to the extra $H^{+}$ — it became a bond, so no lone pair remains. Step 4 — SN. $4 + 0 = 4 \Rightarrow s p^{3}$ , tetrahedral, $109.5°$ . Why this step? SN 4 with no lone pairs means electron geometry = molecular shape = tetrahedral, no angle squeeze.

Verify: $N H_{4}^{+}$ is a perfect tetrahedron with all four N–H bonds identical (the coordinate N–H bond is indistinguishable once formed). $109.5°$ ✅ — same as $C H_{4}$ , which is exactly why a charge adjustment was needed to get there. See Formal Charge and Lewis Structures.

Ex 5 — Expanded octet, SN 5 (cell C5)

Worked example Phosphorus pentachloride

$P C l_{5}$ Forecast: five bonds — can P even hold that many? What shape shares 5 directions fairly?

Step 1 — Valence. P, $V = 5$ . Why this step? We track phosphorus's five valence electrons through the bonds; knowing the starting budget is what lets us test at the end whether any electrons are left over as a lone pair. Step 2 — σ partners. 5 single bonds to Cl, so $S = 5$ . Why this step? Five single bonds spend all five of phosphorus's own electrons — a clue that no lone pair can remain. Step 3 — Lone pairs. $LP = \frac{5 - 5}{2} = 0$ . Why this step? The budget is exhausted ( $V - S = 0$ ), so there are zero leftover electrons and hence no lone pair to steal a slot — every SN unit will be a real bond. Step 4 — SN. $5 + 0 = 5 \Rightarrow s p^{3} d$ , trigonal bipyramidal. Why this step? SN 5 is the first row that needs a d-orbital in the mix; five domains cannot all be equal, so the table hands us the trigonal bipyramid. Geometry: Five balloons can't sit at equal angles; nature splits them into 3 equatorial (in a flat triangle, $120°$ apart) and 2 axial (straight up/down, $90°$ to the equator).

Step 5 — Axial vs equatorial. Axial bonds face three neighbours at $90°$ ; equatorial bonds face only two at $90°$ . More close neighbours = more repulsion = longer, weaker axial bonds.

Verify: Measured P–Cl: axial $\approx 219$ pm $>$ equatorial $\approx 204$ pm ✅. Angles present: $90°, 120°, 180°$ — all three, exactly what a trigonal bipyramid must contain.

Ex 6 — Expanded octet with a lone pair (cell C6)

Worked example Sulfur tetrafluoride

$S F_{4}$ Forecast: four bonds and leftover electrons on S. Where does the lone pair go in a trigonal bipyramid?

Step 1 — Valence. S, $V = 6$ . Why this step? Sulfur brings six valence electrons; because six is more than the four bonds it makes, we should expect a leftover pair — the budget is what makes that prediction precise. Step 2 — σ partners. 4 single bonds to F, so $S = 4$ . Why this step? Four single bonds spend four of sulfur's six electrons, leaving two — a lone pair is coming. Step 3 — Lone pairs. $LP = \frac{6 - 4}{2} = 1$ . Why this step? The two unspent electrons ( $V - S = 2$ ) pair into exactly one lone pair; that lone pair will claim a real domain in the SN count and reshape the molecule. Step 4 — SN. $4 + 1 = 5 \Rightarrow s p^{3} d$ . Electron geometry trigonal bipyramidal, but one slot is a lone pair, so the molecular shape is a see-saw. Why this step? The lone pair earns a real domain in SN even though we don't see it — deleting its vertex from the trigonal bipyramid leaves the see-saw of atoms.

Step 5 — Which slot for the lone pair? The lone pair takes an equatorial position, not axial. Why this step? Equatorial has only 2 close ( $90°$ ) neighbours vs axial's 3. A fat LP wants the least crowded slot — putting it equatorial minimises repulsion. This is the same "least-crowded" logic as VSEPR Theory / Bond Angle and Bent's Rule.

Verify: With the lone pair equatorial, the two axial S–F bonds bend slightly toward the far side; observed axial angle $\approx 173°$ (below $180°$ ) and equatorial $\approx 102°$ (below $120°$ ) — both squeezed by the lone pair, exactly as predicted. ✅

Ex 7 — SN 6, two lone pairs (cell C7)

Worked example Xenon tetrafluoride

$X e F_{4}$ Forecast: a noble gas making four bonds plus leftover pairs — what happens to an octahedron when two vertices are lone pairs?

Step 1 — Valence. Xe, $V = 8$ . Why this step? A noble gas normally bonds to nothing, so its full eight valence electrons are the entire budget; four bonds cannot use them all, which foreshadows two leftover lone pairs. Step 2 — σ partners. 4 single bonds to F, so $S = 4$ . Why this step? Four single bonds spend four of xenon's eight electrons, leaving four — that is two lone pairs. Step 3 — Lone pairs. $LP = \frac{8 - 4}{2} = 2$ . Why this step? The four unspent electrons ( $V - S = 4$ ) pair into two lone pairs; both will claim octahedral slots, and where they sit decides the final flat shape. Step 4 — SN. $4 + 2 = 6 \Rightarrow s p^{3} d^{2}$ , electron geometry octahedral. Why this step? Six domains (4 σ + 2 LP) fill all six octahedral slots; the shape name will come from deleting the two lone-pair slots.

Step 5 — Place the 2 lone pairs. In an octahedron the two lone pairs sit trans (opposite each other, top and bottom), so they're $180°$ apart — maximally separated, minimum repulsion. Why this step? If both lone pairs were adjacent ( $90°$ ) they'd repel strongly; opposite poles avoids that. What's left of the four F is a flat square in the middle → square planar.

Verify: The four Xe–F bonds are coplanar and equal, all $90°$ to each other. So a molecule that forms from an octahedron looks flat — the degeneracy of the two axial lone-pair slots collapses the visible shape to a square. ✅

Ex 8 — Real-world word problem (cell C8)

Worked example "The stinging gas from burning sulfur"

A student burns sulfur in air and smells a sharp gas. Chemistry tells her it is $S O_{2}$ , drawn as one S=O double bond, one S–O single bond, and a lone pair on S. Predict its hybridization, shape and roughly its bond angle.

Forecast: two O attached, plus something extra on S — SN 2, 3, or 4?

Step 1 — Translate words to a count. S is bonded to two oxygens (one single, one double) and carries one lone pair. Why this step? Real problems hide the numbers in prose — first extract "σ partners" and "lone pairs." Step 2 — σ partners. Both S–O links (single and double) give one σ each ⇒ 2 σ toward SN. Why this step? Even the double bond adds only one direction, so it contributes one σ, not two. Step 3 — Lone pairs. Given as 1. Check with the general formula: S has $V = 6$ ; the single bond spends 1 electron and the double bond spends 2, so $S = 3$ , giving $LP = \frac{6 - 3}{2} = 1.5$ ... which flags that $S O_{2}$ has resonance and an expanded-octet Lewis form. Using the standard exam Lewis structure (one S=O, one S–O, one LP on S) the accepted lone-pair count is 1. Why this step? Cross-checking the prose against the formula catches resonance/expanded-octet subtleties; here it tells us to trust the given "one lone pair" for the SN count. Step 4 — SN. $2 + 1 = 3 \Rightarrow s p^{2}$ , electron geometry trigonal planar; with one slot a lone pair the molecule is bent. Why this step? SN 3 → $s p^{2}$ ; deleting the lone-pair vertex from the flat triangle leaves a bent, two-atom shape.

Step 5 — Angle. Ideal $s p^{2}$ is $120°$ ; the lone pair squeezes it down to $\approx 119°$ (only slightly, because S is large).

Verify: Measured O–S–O angle $\approx 119°$ ✅, and $S O_{2}$ is a known bent polar molecule (that polarity is why it dissolves in rain to make acid). Contrast dry ice $C O_{2}$ : also two O, but no lone pair ⇒ SN 2 ⇒ linear. Same "two oxygens," opposite shape — the lone pair decides.

Ex 9 — Exam twist: neutral vs anion (cell C9)

Worked example

$N O_{2}$ (radical) vs $N O_{2}^{-}$ (nitrite) Forecast: both have N bonded to two O. Same hybridization? Guess before reading — the twist is what happens to the third thing on nitrogen.

The neutral radical $N O_{2}$ : Step 1. N, $V = 5$ ; bonded to 2 O ⇒ 2 σ toward SN. Why this step? Two oxygens fix the σ count at 2; the interesting part is the leftover electron. Step 2 — leftover electrons. After the σ framework there is 1 electron left over — an odd, unpaired electron, not a full lone pair. Why this step? A single unpaired electron still occupies a hybrid orbital and repels bonds, so it counts like "half a lone pair" for the SN slot: SN = 2 σ + 1 (single-electron domain) = 3 ⇒ $s p^{2}$ , bent. But the single electron repels less than a full pair, so the angle opens up to $\approx 134°$ (well above $120°$ ).

The anion $N O_{2}^{-}$ : Step 1 — adjust charge. $- 1$ adds one electron ⇒ effective $V = 6$ . Why this step? The extra electron completes the odd electron into a genuine pair — the whole difference between the two species. Step 2 — lone pairs. Now N carries a full lone pair. SN $= 2 σ + 1 LP = 3 \Rightarrow s p^{2}$ , bent, angle squeezed to $\approx 115°$ (below $120°$ ).

Verify: Both are $s p^{2}$ and bent, yet the angles differ sharply: $N O_{2} (\approx 134°) > N O_{2}^{-} (\approx 115°)$ . The lone-pair count (½ vs 1) explains the $19°$ gap ✅. The exam twist: same hybridization label, very different angle — the charge changed the electron domain, not the SN.

Ex 10 — The limiting case, SN 2 (cell C10)

Worked example

$B e C l_{2}$ and acetylene $C_{2} H_{2}$ Forecast: the smallest non-trivial SN. What's the most spread-out two bonds can get?

$B e C l_{2}$ (electron-deficient): Step 1. Be, $V = 2$ ; bonded to 2 Cl ⇒ $S = 2$ ; $LP = \frac{2 - 2}{2} = 0$ . Why this step? Beryllium's two electrons are both spent on bonds, so nothing is left for a lone pair. Step 2 — SN. $2 + 0 = 2 \Rightarrow s p$ , linear, $180°$ . Why this step? Two balloons on a string push to exact opposite ends — $180°$ is the maximum possible separation of two directions. This is the boundary of the whole SN table.

$C_{2} H_{2}$ (H–C≡C–H): Step 1. Each C: bonded to 1 H + 1 C ⇒ 2 σ; the C≡C triple = 1 σ + 2 π. So 2 σ toward SN, and $S = 1 (C-H) + 3 (C\equivC) = 4$ , giving $LP = \frac{4 - 4}{2} = 0$ . Why this step? The triple bond spends three of carbon's electrons, so together with the C–H bond all four are used — no lone pair, as expected. Step 2 — SN. $2 \Rightarrow s p$ , linear, $180°$ . Each C keeps two unhybridized p orbitals, and those two leftover p orbitals overlap sideways to form the two π bonds of the triple bond (1 σ + 2 π = triple). Why this step? Confirms the triple bond still gives SN 2 — the extra two π's are stacked on the single σ direction, invisible to SN, and they must come from unhybridized p, never from a hybrid.

Verify: $B e C l_{2}$ (gas) and $C_{2} H_{2}$ are both linear at $180°$ ✅. Two unhybridized p on each acetylene carbon ⇒ exactly 2 π ⇒ a triple bond, consistent with H–C≡C–H.

Recall Self-test: name the cell, then answer

Neutral molecule, two double bonds to central atom, no lone pair — hybridization? ::: $s p$ (SN 2), linear — this is cell C3, e.g. $C O_{2}$ . Cation with 4 bonds and no lone pair after charge adjustment? ::: $s p^{3}$ tetrahedral — cell C4, e.g. $N H_{4}^{+}$ . SN 6 electron geometry but square-planar molecular shape means how many lone pairs? ::: Two, sitting trans — cell C7, e.g. $X e F_{4}$ . Same "two oxygens" but one is bent and one is linear — what decides? ::: A lone pair on the central atom ( $S O_{2}$ bent, $C O_{2}$ linear) — cells C8/C3. A single unpaired electron on the centre counts as how much in SN? ::: One domain (like a lone pair for counting), but it repels less, so it widens the angle — cell C9, $N O_{2}$ . Difference between electron geometry and molecular shape? ::: Electron geometry counts all domains (bonds + lone pairs); molecular shape names only the atoms you see — identical when there are no lone pairs.