IB DP Chemistry - S2.2.5 Bond polarity- Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – S2.2.5 Bond polarity – Study Notes – New Syllabus
IITian Academy excellent Study Notes and effective strategies will help you prepare for your IB DP Chemistry exam.
- IB DP Chemistry 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Chemistry 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Chemistry 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Chemistry 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Structure 2.2.5 — Bond Polarity and Electronegativity
Structure 2.2.5 — Bond Polarity and Electronegativity
Electronegativity
Electronegativity (\( \chi \)) is a measure of an atom’s ability to attract a shared pair of electrons in a covalent bond. It is a relative scale introduced by Linus Pauling.
- Fluorine is the most electronegative element: \( \chi = 4.0 \).
- Electronegativity increases across a period (left to right) and decreases down a group in the periodic table.
Bond Polarity
Bond polarity arises when there is a difference in electronegativity between two atoms in a covalent bond.
- The bonding electrons are pulled more towards the more electronegative atom.
- This creates a partial negative charge (\( \delta^- \)) on the more electronegative atom and a partial positive charge (\( \delta^+ \)) on the less electronegative one.
This unequal sharing leads to a bond dipole a separation of charge across the bond.
Bond Dipoles Can Be Represented in Two Ways:
Using Partial Charges:
\( \delta^+ \text{H} – \text{Cl} \delta^- \)
Using Dipole Vectors:
An arrow points from the partial positive to the partial negative end, e.g. \( \overset{\delta^+}{\text{H}} \rightarrow \overset{\delta^-}{\text{Cl}} \)
How to Deduce Bond Polarity
- Use electronegativity values from the IB Data Booklet (Table 8).
- Calculate the difference: \( \Delta \chi = \chi_{\text{more}} – \chi_{\text{less}} \)
- Interpret the result as follows:
| \( \Delta \chi \) | Bond Type | Bond Polarity |
|---|---|---|
| \( < 0.4 \) | Pure Covalent | Electrons shared equally; non-polar bond (e.g. Cl2) |
| \( 0.4 – 1.7 \) | Polar Covalent | Electrons shared unequally; bond has dipole (e.g. HCl) |
| \( > 1.7 \) | Ionic Bond | Electron transferred completely; ions form (e.g. NaCl) |
Important: Even if a molecule has polar bonds, the overall molecule may be non-polar if the molecular geometry causes the dipoles to cancel out. Examples: CO2, CCl4.
Example 1: HCl (Hydrogen Chloride)
- \( \chi_{\text{Cl}} = 3.0 \), \( \chi_{\text{H}} = 2.1 \)
- \( \Delta \chi = 3.0 – 2.1 = 0.9 \)
This is a polar covalent bond. Dipole representation:
- Partial charges: \( \delta^+ \text{H} – \text{Cl} \delta^- \)
- Vector: \( \overset{\delta^+}{\text{H}} \rightarrow \overset{\delta^-}{\text{Cl}} \)
Example 2: NaCl (Sodium Chloride)
- \( \chi_{\text{Cl}} = 3.0 \), \( \chi_{\text{Na}} = 0.9 \)
- \( \Delta \chi = 2.1 \)
Since \( \Delta \chi > 1.7 \), this bond is ionic — electron transfer from Na to Cl forms Na+ and Cl−.
Example 3: O2 (Oxygen molecule)
- \( \chi_{\text{O}} = 3.5 \)
- \( \Delta \chi = 0 \)
This is a non-polar covalent bond. No dipole is formed.
Example
Deduce whether the bond in hydrogen fluoride (HF) is polar, and show the bond dipole using both partial charges and dipole vector notation.
▶️Answer/Explanation
Electronegativity values (from IB data booklet):
- \( \chi_{\text{F}} = 4.0 \)
- \( \chi_{\text{H}} = 2.1 \)
- \( \Delta \chi = 4.0 – 2.1 = 1.9 \)
Since \( \Delta \chi = 1.9 \), the bond is polar covalent (almost ionic).
Partial charges: \( \delta^+ \text{H} – \text{F} \delta^- \)
Dipole vector: \( \overset{\delta^+}{\text{H}} \rightarrow \overset{\delta^-}{\text{F}} \)
Example
The Si-Cl bond exists in several silicon compounds. Use electronegativity values to determine whether the Si-Cl bond is polar. If so, identify the more electronegative atom.
▶️Answer/Explanation
Electronegativity values:
- \( \chi_{\text{Si}} = 1.8 \)
- \( \chi_{\text{Cl}} = 3.0 \)
- \( \Delta \chi = 3.0 – 1.8 = 1.2 \)
The bond is polar covalent.
Chlorine is more electronegative, so:
Dipole vector: \( \overset{\delta^+}{\text{Si}} \rightarrow \overset{\delta^-}{\text{Cl}} \)
Example
Explain why carbon dioxide (CO2) has polar bonds but is a non-polar molecule overall.
▶️Answer/Explanation
Electronegativity difference:
- \( \chi_{\text{O}} = 3.5 \), \( \chi_{\text{C}} = 2.5 \)
- \( \Delta \chi = 3.5 – 2.5 = 1.0 \) → polar covalent bonds
CO2 has two polar bonds pointing in opposite directions (linear geometry).
The dipoles cancel due to symmetry, so the molecule is non-polar overall.
Shape: linear — bond angle \( 180^\circ \)
Conclusion: Bond polarity does not always lead to molecular polarity.