Molecular Polarity Practice Questions with Answers

A molecule can be full of polar bonds and still be non-polar overall. That single idea catches out more students than any other part of this topic, so these questions return to it again and again. Work through all fourteen, write your reasoning in full, and only then check the answer key. The reasoning is where the marks are.

What you need before you start

Polarity is decided in two stages.

Stage 1: are the bonds polar? Compare the Pauling electronegativities of the two bonded atoms. The more electronegative atom gets a partial negative charge (δ−), the other a partial positive charge (δ+). The values used in this set are:

Element	H	B	C	N	O	F	P	S	Cl
Pauling electronegativity	2.20	2.04	2.55	3.04	3.44	3.98	2.19	2.58	3.16

Stage 2: do the bond dipoles cancel? Work out the shape with VSEPR. Treat each bond dipole as an arrow pointing towards the δ− atom. If the arrows cancel by symmetry, the molecule is non-polar; if they leave a net arrow, it is polar.

Shortcuts that follow from this:

• A central atom with no lone pairs and identical outer atoms gives a symmetrical shape (linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral) and a non-polar molecule.
• Lone pairs on the central atom usually make the shape lopsided (bent, pyramidal, seesaw, T-shaped), and the molecule polar. The exceptions are the symmetrical lone-pair shapes: linear XeF₂ and square planar XeF₄.
• Different outer atoms on an otherwise symmetrical shape (for example CHCl₃) usually break the cancellation.

The step-by-step version of this method, with worked examples, is in how to tell if a molecule is polar.

Questions

Total: 36 marks.

1. Using the table above, rank these bonds in order of increasing polarity: C–H, N–H, O–H, F–H. Show the electronegativity differences. (2 marks)
2. Label the δ+ and δ− ends of each bond: (a) C–Cl, (b) O–H, (c) C–O, (d) P–Cl. (2 marks)
3. Carbon dioxide contains two polar C=O bonds, yet it is non-polar. Explain why. (2 marks)
4. Explain why water is polar. (2 marks)
5. Classify each as polar or non-polar, giving the shape: (a) BF₃, (b) NF₃. (3 marks)
6. Classify CH₄, CH₂Cl₂, CHCl₃ and CCl₄ as polar or non-polar, and explain the pattern. (4 marks)
7. SO₂ is polar but SO₃ is non-polar. Explain using shapes. (3 marks)
8. State whether each is polar: (a) PCl₃, (b) PCl₅. Give the shapes. (2 marks)
9. State whether each is polar: (a) SF₄, (b) SF₆. Explain your answers. (3 marks)
10. Classify XeF₂, XeF₄ and ClF₃ as polar or non-polar, with shapes. (3 marks)
11. 1,2-Dichloroethene exists as cis and trans isomers. Which isomer is polar? Explain. (2 marks)
12. NH₃ and NF₃ are both trigonal pyramidal, and the N–F bond is more polar than the N–H bond (compare the electronegativity differences). Yet NH₃ has a much larger dipole moment. Explain. (3 marks)
13. Ozone, O₃, contains only oxygen atoms, but it has a small dipole moment. Suggest why. (2 marks)
14. A thin stream of liquid from a burette is held next to a charged plastic rod. Predict which of these liquids would be deflected, and explain: tetrachloromethane (CCl₄), trichloromethane (CHCl₃), 1,2-dichlorobenzene and 1,4-dichlorobenzene (melted). (3 marks)

Answer key

1. Differences: C–H 2.55 − 2.20 = 0.35; N–H 3.04 − 2.20 = 0.84; O–H 3.44 − 2.20 = 1.24; F–H 3.98 − 2.20 = 1.78 (1). Increasing polarity: C–H < N–H < O–H < F–H (1). The C–H bond is only weakly polar, which is why hydrocarbons are treated as non-polar.

2. (a) C is δ+, Cl is δ− (3.16 > 2.55). (b) H is δ+, O is δ−. (c) C is δ+, O is δ−. (d) P is δ+, Cl is δ− (3.16 > 2.19). (2 for all four; 1 for three.)

3. Each C=O bond is polar, with O δ− (1). The molecule is linear (two domains, no lone pairs on C), so the two equal bond dipoles point in exactly opposite directions and cancel; there is no net dipole (1).

4. The O–H bonds are polar, with O δ− and H δ+ (1). Oxygen has two lone pairs, so the molecule is bent (about 104.5°); the two bond dipoles do not point in opposite directions, and they add to give a net dipole from the hydrogens towards the oxygen (1).

5. (a) BF₃: three bonds, no lone pair on B, trigonal planar (1). The three identical B–F dipoles at 120° cancel: non-polar (1). (b) NF₃: three bonds and one lone pair on N, trigonal pyramidal. The dipoles do not cancel: polar (1), although only weakly (see question 12).

6. All four are tetrahedral around carbon (1). CH₄ is non-polar: four identical, weakly polar C–H bonds cancel. CCl₄ is non-polar: four identical C–Cl dipoles cancel exactly by symmetry (1). CH₂Cl₂ and CHCl₃ are polar (1): mixing C–H and C–Cl bonds, which have different polarities, breaks the symmetry, so the dipoles no longer cancel (1). The lesson is that shape alone is not enough; the outer atoms must also be identical.

7. SO₂ has two S=O bonds and one lone pair on sulfur, so it is bent; the S–O dipoles (O is δ−, 3.44 vs 2.58) do not cancel, so SO₂ is polar (1). SO₃ has three S=O bonds and no lone pair, so it is trigonal planar (1); three identical dipoles at 120° cancel, making it non-polar (1).

8. (a) PCl₃: one lone pair, trigonal pyramidal, polar. (b) PCl₅: no lone pair, trigonal bipyramidal; the two axial dipoles cancel each other and the three equatorial ones cancel each other, so it is non-polar. (1 each.)

9. (a) SF₄ has four bonds and one lone pair on sulfur: seesaw shape (1). The dipoles do not cancel, so it is polar. (b) SF₆ has six bonds and no lone pairs: octahedral (1). Every S–F dipole is cancelled by the one directly opposite, so it is non-polar (1).

10. XeF₂: two bonds, three lone pairs; the lone pairs sit around the middle, leaving the fluorines at 180°: linear, non-polar (1). XeF₄: four bonds, two lone pairs opposite each other: square planar, non-polar; opposite Xe–F dipoles cancel in pairs (1). ClF₃: three bonds, two lone pairs: T-shaped, polar, because the three Cl–F dipoles cannot cancel in a T (1).

11. In the cis isomer both chlorines are on the same side of the C=C bond, so the two C–Cl dipoles add to give a net dipole: cis is polar (1). In the trans isomer the chlorines are on opposite sides and the molecule has a centre of symmetry, so the dipoles cancel: trans is non-polar (1). This is a neat way to tell the two isomers apart.

12. In NH₃, nitrogen is δ− (3.04 vs 2.20), so each N–H dipole points towards nitrogen, and the lone pair on nitrogen also pushes electron density away from the hydrogens. The bond dipoles and the lone pair reinforce each other (1). In NF₃, fluorine is δ− (3.98 vs 3.04), so the N–F dipoles point away from nitrogen towards the fluorines, while the lone pair still points the other way. The two effects oppose each other (1) and largely cancel, leaving NF₃ with a very small dipole moment (about 0.2 D) compared with NH₃ (about 1.5 D) (1). This shows that lone pairs contribute to a molecule’s dipole, not just bonds.

13. Ozone is bent, because the central oxygen carries a lone pair (1). The central oxygen and the two end oxygens are not in the same bonding situation: in the Lewis structure the central atom has a formal charge of +1 and the negative charge is shared between the ends, so the electron density is uneven. In a bent shape, this uneven distribution does not cancel, giving a small net dipole (1). Identical atoms do not guarantee a non-polar molecule; see dipole moments.

14. Only polar molecules are attracted to a charged rod, because their δ+ or δ− ends turn towards it (1). CHCl₃ (polar, see question 6) and 1,2-dichlorobenzene (both C–Cl dipoles on the same side of the ring, so they add) are deflected (1). CCl₄ (symmetrical tetrahedron) and 1,4-dichlorobenzene (the C–Cl dipoles point in opposite directions and cancel) are not (1).

Exam tips

• Always state the shape. Most mark schemes give a mark for the shape and another for the symmetry argument; “it has polar bonds” alone scores nothing.
• Write the word “cancel” or “do not cancel” and say why (symmetry, lone pairs, different outer atoms).
• Check the outer atoms. A tetrahedral or trigonal planar molecule with mixed substituents is usually polar.
• Remember the symmetrical lone-pair shapes, XeF₂ and XeF₄, which are non-polar despite their lone pairs.
• Polarity has consequences. Polar molecules have dipole–dipole attractions, and those with H bonded to N, O or F can also hydrogen bond. The intermolecular forces practice questions build on this set.

Key takeaways

• Bond polarity comes from an electronegativity difference; molecular polarity needs polar bonds plus an unsymmetrical shape.
• Symmetrical, lone-pair-free shapes with identical outer atoms (CO₂, BF₃, CH₄, CCl₄, SF₆) are non-polar.
• Lone pairs usually make molecules polar (H₂O, NH₃, SO₂, SF₄, ClF₃), except in linear XeF₂ and square planar XeF₄.
• Lone pairs can add to or oppose bond dipoles, which explains why NH₃ is far more polar than NF₃.
• For the electronegativity values behind every answer, compare fluorine with its neighbours on the periodic table.