# Hybridisation and Molecular Orbitals

Chemistry I · Atoms, Bonds and Reaction Rates · https://tryals.app/learn/chemistry-i/hybridisation-and-molecular-orbitals

## Two Ways to Explain a Bond

**Valence bond theory** says a bond is the overlap of two atomic orbitals, each contributing one electron. Head-on overlap along the internuclear axis makes a **sigma** bond; sideways overlap of parallel $p$ orbitals makes a **pi** bond, which has a nodal plane through the axis.

Pure atomic orbitals give the wrong shapes, carbon's $2s^2 2p^2$ would predict two bonds at $90^\circ$, not four at $109.5^\circ$. **Hybridisation** mixes them into equivalent hybrids:

| Hybrid | Orbitals mixed | Geometry | Angle |
|---|---|---|---|
| $sp$ | one s, one p | Linear | $180^\circ$ |
| $sp^2$ | one s, two p | Trigonal planar | $120^\circ$ |
| $sp^3$ | one s, three p | Tetrahedral | $109.5^\circ$ |

The count is simple: **the number of hybrid orbitals equals the number of electron domains.** So a single bond is one sigma; a double bond is one sigma plus one pi; a triple bond is one sigma plus two pi. In ethene each carbon is $sp^2$, leaving one unhybridised $p$ orbital on each to form the pi bond, and because pi overlap is destroyed by twisting, the double bond cannot rotate.

**Molecular orbital theory** takes a different route: atomic orbitals combine into orbitals belonging to the whole molecule. Two atomic orbitals give one **bonding** MO (lower energy, electron density between the nuclei) and one **antibonding** MO (higher energy, with a node between them). Electrons fill these by the same rules as atoms, and

$$\text{bond order} = \tfrac{1}{2}(N_{\text{bonding}} - N_{\text{antibonding}})$$

A bond order of zero means no bond, which is exactly why $\mathrm{He_2}$ does not exist. MO theory also predicts what valence bond theory cannot: $\mathrm{O_2}$ has two unpaired electrons in degenerate antibonding orbitals and is therefore paramagnetic, it sticks to a magnet, which a Lewis structure with a tidy double bond would never suggest.

> **Common pitfall:** thinking hybridisation is something an atom physically does before bonding. It is a mathematical recombination of orbitals chosen to match the observed geometry, the shape is the evidence, and the hybrid is the description.

## Practice questions

8 of this lesson's 12 practice questions, with answers. The full set is in the app.

### 1. A diatomic molecule has 8 electrons in bonding molecular orbitals and 4 in antibonding orbitals. Compute its bond order.

**Answer:** 2

**Why:** Bond order $= (8 - 4)/2 = 2$, a double bond. Each pair of antibonding electrons cancels the effect of one bonding pair.

Page: https://tryals.app/practice/chemistry-i/hybridisation-and-molecular-orbitals/a-diatomic-molecule-has-8-electrons-in-bonding-molecular-orbitals-and

### 2. The helium molecule $\mathrm{He_2}$ is unstable because its bond order is zero.

**Answer:** True

**Why:** True, four electrons fill both the bonding and the antibonding orbital, giving $(2-2)/2 = 0$. With no net bonding there is nothing to hold the atoms together, which is why helium is monatomic.

Page: https://tryals.app/practice/chemistry-i/hybridisation-and-molecular-orbitals/the-helium-molecule-he2-is-unstable-because-its-bond-order-is-zero

### 3. Hybridisation is a mathematical construct to explain observed geometry rather than a physical event occurring prior to bonding. What follows from this distinction when analysing a newly synthesised molecule?

A. The experimental geometry dictates which hybridisation scheme should be applied
B. Spectroscopy detects the physical excitation of electrons into hybrid states
C. Molecules must dynamically unhybridise whenever their bonds undergo cleavage
D. Hybridisation provides the thermodynamic driving force for assembling a molecule

**Answer:** A. The experimental geometry dictates which hybridisation scheme should be applied

**Why:** Assuming hybrids exist as physical precursors confuses descriptive models with physical steps. Hybrid schemes are chosen retrospectively to reflect measured bond angles, while overall molecular stability emerges from total energy minimisation across all nuclei and electrons.

Page: https://tryals.app/practice/chemistry-i/hybridisation-and-molecular-orbitals/hybridisation-is-a-mathematical-construct-to-explain-observed

### 4. A carbon-carbon triple bond is drawn between two atoms. How many of its three bonds are pi bonds?

**Answer:** 2

**Why:** The first bond formed between two atoms is always sigma; every additional one is pi. A triple bond is therefore one sigma and **two** pi bonds.

Page: https://tryals.app/practice/chemistry-i/hybridisation-and-molecular-orbitals/a-carbon-carbon-triple-bond-is-drawn-between-two-atoms-how-many-of

### 5. Which species have an $sp^3$-hybridised central atom?

A. Water
B. Ammonia
C. Methane
D. Ethyne

**Answer:** A. Water; B. Ammonia; C. Methane

**Why:** Methane (4 bonds), ammonia (3 bonds + 1 lone pair) and water (2 bonds + 2 lone pairs) all have four domains and are $sp^3$. Ethyne’s carbons carry only two domains each and are $sp$.

Page: https://tryals.app/practice/chemistry-i/hybridisation-and-molecular-orbitals/which-species-have-an-sp-hybridised-central-atom

### 6. Why can an alkene not rotate freely about its carbon-carbon double bond?

A. The head-on overlap of the sigma bond is too rigid to turn
B. Twisting would break the sideways overlap of the pi bond
C. The sp2 hybridised carbon atoms are held by electrostatic lock
D. Rotating the bond forces both carbon nuclei into direct contact

**Answer:** B. Twisting would break the sideways overlap of the pi bond

**Why:** The sigma bond is cylindrically symmetric and would permit rotation on its own. The pi bond requires the two $p$ orbitals to stay parallel, so rotating by $90^\circ$ destroys the overlap entirely, a real energy barrier of roughly 260 kJ/mol.

Page: https://tryals.app/practice/chemistry-i/hybridisation-and-molecular-orbitals/why-can-an-alkene-not-rotate-freely-about-its-carbon-carbon-double

### 7. Hybridisation increases the total number of orbitals available to an atom.

**Answer:** False

**Why:** False, hybridisation conserves the number of orbitals. Mixing one $s$ and three $p$ orbitals gives exactly four $sp^3$ hybrids; it redistributes shape and direction, never quantity.

Page: https://tryals.app/practice/chemistry-i/hybridisation-and-molecular-orbitals/hybridisation-increases-the-total-number-of-orbitals-available-to-an

### 8. Match each model or feature to what it accounts for.

**Answer:**

- Bonding molecular orbital → Electron density concentrated between the nuclei
- Antibonding molecular orbital → A node between the nuclei that weakens binding
- $sp^3$ hybridisation → Four equivalent bonds at 109.5 degrees
- Unhybridised p orbital → Supplies the second component of a double bond

**Why:** Bonding MOs pile electron density between the nuclei; antibonding MOs put a node there; $sp^3$ explains tetrahedral geometry; and the leftover unhybridised $p$ orbital supplies pi overlap.

Page: https://tryals.app/practice/chemistry-i/hybridisation-and-molecular-orbitals/match-each-model-or-feature-to-what-it-accounts-for