Bond Energy, Bond Length and Intermolecular Forces: PECTAA Class 11 Chemistry, Punjab Board

Why does hydrogen fluoride hold together so firmly while hydrogen iodide falls apart easily enough to act as a strong acid? Two numbers answer it: bond energy and bond length. Chapter 3 of the book uses them to compare every kind of bond and force.

Chapter 3 of the PECTAA Chemistry 11 book is Chemical Bonding. Sections 3.3 to 3.5 cover intermolecular forces, bond energy and bond length, and a comparison of ionic, covalent and metallic bonds with intermolecular forces. This note is for FSc Part 1 (1st year, Class 11) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

Intermolecular forces (section 3.3). Molecules attract one another when they are close enough. These attractions are much weaker than chemical bonds, but they hold molecules together and give gases, liquids and solids their physical properties. They are called van der Waals forces. The book names three:

1. instantaneous dipole–induced dipole forces, also called London dispersion forces;
2. permanent dipole–permanent dipole forces;
3. hydrogen bonding.

Their full treatment comes in Chapter 5 on states and phases of matter.

Bond energy (section 3.4). Bond energy is the average energy required to break all bonds of a particular type in one mole of a substance. It is found experimentally from the heat involved in a reaction, in kJ mol⁻¹. It measures how strong a bond is, which in turn affects how reactive the substance is. Bond strength depends on:

• the electronegativity difference of the bonded atoms;
• the size of the atoms.

Electronegativity and bond energy. In the H–X series (X = F, Cl, Br, I), bond energy rises as the electronegativity difference rises. A bigger difference makes the bond more polar, which adds attraction between the atoms. So HF (565 kJ mol⁻¹) has a much stronger bond than HI (295 kJ mol⁻¹).

Multiple bonds. Triple bonds are stronger than double bonds, which are stronger than single bonds: C≡C > C=C > C–C.

Bond length. Bond length is the distance between the nuclei of two covalently bonded atoms. It is measured by electron diffraction, X-ray diffraction or spectral studies. It depends on electronegativity, atomic size and bond type (single, double or triple).

• Size: larger atoms give longer bonds. C–Cl is about 180 pm; C–F is 135 pm, because Cl is much larger than F.
• Electronegativity difference: a larger difference gives the bond some ionic character, and the extra attraction shortens it. In SiF₄ the measured Si–F length is near 155 pm, while adding the covalent radii (Si 117 pm + F 64 pm) predicts 181 pm.

Comparing bonds and forces (section 3.5). Ionic, covalent and metallic bonds are true chemical bonds: they form new species by transfer or sharing of electrons and affect chemical properties. Intermolecular forces bring molecules closer and affect physical properties. The clearest difference is strength:

The book explains the lower strength of metallic bonds by the extensive delocalisation of electrons in the metal crystal.

Key points

• Higher bond energy means a stronger, less easily broken bond.
• Bond energy rises with electronegativity difference and with bond multiplicity.
• Bond length rises with atomic size and falls with electronegativity difference.
• Chemical bonds are far stronger than intermolecular forces.

Worked examples

Example 1 (book Quick Check 3.5 a). Why is HI a stronger acid and a strong reducing agent, while HF is a weaker acid?

1. The H–I bond energy (295 kJ mol⁻¹) is much lower than the H–F bond energy (565 kJ mol⁻¹).
2. The weaker H–I bond breaks easily, releasing H⁺ readily, so HI is the stronger acid.
3. The strong H–F bond resists breaking, so HF is the weaker acid.

Example 2 (book Quick Check 3.5 b). Why is acetylene (HC≡CH) more stable than ethene (H₂C=CH₂)?

1. Acetylene has a carbon–carbon triple bond; ethene has a double bond.
2. Bond energy order: C≡C > C=C. More energy is needed to break the triple bond, so the book treats acetylene as the more stable of the two.

Example 3 (from the book's SiF₄ data). Predicted Si–F length from covalent radii = 117 + 64 = 181 pm. Measured ≈ 155 pm. The bond is about 26 pm shorter because the large electronegativity difference adds ionic character.

Common mistakes in the exam

1. Calling bond energy the energy to break one bond. The book defines it per mole of bonds.
2. Thinking higher electronegativity difference means a longer bond. It shortens the bond.
3. Ranking metallic bonds above covalent bonds. The book places metallic bonds below ionic and covalent.
4. Mixing intermolecular forces with chemical bonds. Hydrogen bonds act between molecules and are much weaker.
5. Leaving out units. Bond energy is in kJ mol⁻¹, bond length in pm.

Exam technique

• Quote one pair of numbers (HF 565, HI 295) when explaining the electronegativity trend.
• For "explain" questions, link cause to effect: difference → polarity → extra attraction → stronger, shorter bond.
• Learn the strength order from the table above.
• Check your board's notice for the sections in your paper.

Practice questions

1. Define bond length. Answer: The distance between the nuclei of two covalently bonded atoms.
2. Why is the C–Cl bond longer than the C–F bond? Answer: The Cl atom is much larger than the F atom.
3. Arrange C–C, C=C and C≡C in order of increasing bond energy. Answer: C–C < C=C < C≡C.
4. Name the three intermolecular forces listed in section 3.3. Answer: London dispersion forces, dipole–dipole forces and hydrogen bonding.
5. Which is weaker on average, a metallic bond or a covalent O–H bond? Answer: The metallic bond (100 to 150 kJ mol⁻¹ against 464 kJ mol⁻¹).

Ready to practise this? Continue in 1st Year Chemistry.