Electronegativity & polarity
Electronegativity & polarity
Electronegativity and polar bonds
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Electronegativity tells you how strongly an atom attracts the shared pair in a covalent bond. Unequal attraction makes the bond polar: the more electronegative atom is δ− and its partner is δ+.
Electronegativity generally increases across a period and up a group. Fluorine is the most electronegative element.
Calculate a difference by subtracting the smaller value from the larger one. Always use the numbers supplied in that question: some practice questions use H = 2.1, so they need not give the same numerical difference as a different notes example.
A partial charge is not an ionic charge. The atoms are still sharing the bonding pair.
- Electronegativity
- The ability of an atom to attract the bonding electrons in a covalent bond.
- Polar bond
- A covalent bond with an unequal distribution of bonding electron density because the atoms have different electronegativities.
- Permanent dipole
- A permanent separation of partial charge, represented by δ+ and δ−.
| Atom | H | C | N | O | F | Cl |
|---|---|---|---|---|---|---|
| Approximate Pauling value for these notes | 2.2 | 2.5 | 3.0 | 3.5 | 4.0 | 3.0 |
Compare C–H and O–H
The Pauling electronegativities are C = 2.5, H = 2.2 and O = 3.5. Calculate the electronegativity difference for C–H and O–H. State which bond is more polar and identify the δ+ and δ− atoms in that bond.
Worked answer
- ΔEN(C–H) = 2.5 − 2.2 = 0.3.
- ΔEN(O–H) = 3.5 − 2.2 = 1.3.
- The larger difference gives more unequal sharing.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
(a) Define electronegativity using precise OCR A wording. [2 marks]
(b) State which element is the most electronegative and describe the general direction in which electronegativity increases in the periodic table. [2 marks]
(c) Pauling electronegativities are H 2.1, C 2.5, O 3.5 and F 4.0. Calculate the electronegativity difference for C-H and O-H, then identify the more polar bond. [3 marks]
Total: 7 marks
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(a) Electronegativity is the ability of an atom to attract the bonding electrons [1] in a covalent bond [1]. [2]
(b) Fluorine is the most electronegative element [1]; electronegativity generally increases towards fluorine, across a period and up a group [1]. [2]
(c) C-H: |2.5 - 2.1| = 0.4 [1]. O-H: |3.5 - 2.1| = 1.4 [1]. O-H is more polar because it has the larger electronegativity difference [1]. [3]
(a) Explain what is meant by a polar covalent bond and a permanent dipole. [2 marks]
(b) Use δ+ and δ- to show the permanent dipole in H-Cl. Add a bond-dipole arrow pointing in the correct direction. [2 marks]
(c) Distinguish bond polarity from the polarity of a whole molecule. [2 marks]
Total: 6 marks
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(a) A polar covalent bond has an unequal distribution of bonding electron density caused by different electronegativities [1], producing permanent partial charges / a permanent dipole [1]. [2]
(b) H is δ+ and Cl is δ- [1]; the dipole arrow points from H towards Cl, with the crossed tail at the H end [1]. [2]
(c) Bond polarity concerns one bond [1]. Molecular polarity is the vector result of all bond dipoles in the 3-D shape; they may reinforce or cancel [1]. [2]
Selected Pauling electronegativities are Mg 1.2, H 2.1, C 2.5, Cl 3.0, O 3.5 and F 4.0.
(a) Calculate the electronegativity difference for Mg-Cl, C-Cl and O-H. [3 marks]
(b) Rank these three bonds from least polar to most polar. [2 marks]
(c) Use the values to suggest why MgCl2 has mainly ionic bonding whereas CCl4 has covalent bonding. [2 marks]
(d) State which atom is δ- in a C=O bond. [1 mark]
Total: 8 marks
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(a) Mg-Cl: 3.0 - 1.2 = 1.8 [1]; C-Cl: 3.0 - 2.5 = 0.5 [1]; O-H: 3.5 - 2.1 = 1.4 [1]. [3]
(b) C-Cl < O-H < Mg-Cl [2]. [2]
(c) Mg-Cl has a much larger electronegativity difference, indicating much greater ionic character [1]; C-Cl has a smaller difference and the electrons are shared covalently [1]. [2]
(d) O is δ- because it is more electronegative than C [1]. [1]
Bond polarity and molecular polarity
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A molecule can contain polar bonds and still be non-polar overall. The answer depends on whether the bond dipoles cancel in its three-dimensional shape.
- Identify which bonds are polar and mark their dipole directions.
- Work out the molecular shape, including the effect of lone pairs.
- Check whether the bond dipoles balance in all directions.
- If they cancel completely, the molecule has no permanent dipole. If they do not, the molecule is polar.
CO₂ is linear, so its equal C=O dipoles cancel. H₂O is bent, so its O-H dipoles do not cancel. Draw the shape before deciding.
- Polar molecule
- A molecule with an overall dipole because its bond dipoles do not cancel in its molecular shape.
| Molecule | Shape / dipole arrangement | Overall polarity |
|---|---|---|
| HCl | One bond dipole | Polar |
| CO2 | Linear, equal opposite dipoles | Non-polar |
| BF3 | Symmetrical trigonal planar | Non-polar |
| CH4 | Symmetrical tetrahedral | Non-polar |
| NH3 | Pyramidal; dipoles do not cancel | Polar |
| H2O | Bent; dipoles do not cancel | Polar |
| SF6 | Symmetrical octahedral | Non-polar |
H2O versus CO2
H₂O and CO₂ both contain polar covalent bonds. State which molecule is polar overall and explain the difference using molecular shape and bond dipoles.
Worked answer
CO2 is linear, so the equal C=O dipoles point in opposite directions and cancel. H2O is bent, so its O–H dipoles do not cancel.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
(a) For each molecule, state whether it is polar or non-polar. [5]
| Molecule | Polar / non-polar |
|---|---|
| CO2 | … |
| BF3 | … |
| H2O | … |
| NH3 | … |
| CCl4 | … |
(b) Choose one non-polar molecule from part (a) that contains polar bonds. State its shape, then explain why it has no overall dipole. A drawing is not required. [3]
Total: 8 marks
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(a) CO2 non-polar [1]; BF3 non-polar [1]; H2O polar [1]; NH3 polar [1]; CCl4 non-polar [1]. [5]
(b) Example CO2: each C=O bond is polar [1], but the molecule is linear and symmetrical [1], so equal dipoles act in opposite directions and cancel [1]. [3]
Carbon dioxide, O=C=O, and hydrogen cyanide, H-C≡N, are both linear. Pauling electronegativities are H 2.1, C 2.5, N 3.0 and O 3.5.
(a) State the shape name and bond angle around carbon in each molecule. Names and angles only; diagrams are not required in part (a). [4 marks]
(b) Add bond-dipole arrows to both molecules. Include partial charges at the terminal atoms. [2 marks]
(c) Explain why CO2 is non-polar but HCN is polar. [2 marks]
Total: 8 marks
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(a) CO2 is linear, 180° [2]. HCN is linear around carbon, 180° [2]. [4]
(b) Correct terminal partial charges and dipole directions towards O in CO2 and towards C then N across H-C≡N [1 each molecule]. [2]
(c) CO2 has equal opposite C=O dipoles that cancel [1]. HCN is unsymmetrical, so its directed bond dipoles do not cancel and it has an overall dipole [1]. [2]