The shape of the lesson
By the end of the lesson, learners can
- Colour and volatility: Compare the three halogens in the simulator at room temperature, then warm each one and watch for the vapour.
- Cl2 gas, Br2 liquid, I2 solid.
- Volatility falls: more electrons, stronger id-id.
- Bond energy falls Cl–Cl > Br–Br > I–I.
- F–F anomalously weak: lone-pair repulsion.
One picture every idea returns to
The halogens exist as diatomic molecules, X2. Down the group the molecules have more electrons, so the instantaneous dipole-induced dipole forces between them get stronger and the boiling points rise: volatility decreases from chlorine (a gas) to bromine (a liquid) to iodine (a solid). The colours deepen too. Within each molecule, the X–X bond energy decreases from Cl–Cl to I–I as the atoms get larger and the bonding pair is further from the nuclei; F–F is anomalously weak because the small atoms' lone pairs repel each other.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Darker, denser, less volatile.. Chlorine is a pale gas, bromine a heavy red-brown liquid, iodine a dark solid that turns to purple vapour. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 30 min | Teach and check: Colour and volatility | Compare the three halogens in the simulator at room temperature, then warm each one and watch for the vapour. Project tab 1 of the lab (Group 17 lab, colour and volatility) and let a student drive it while the class predicts. | Cooperative Concept Mapping |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 14 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Colour and volatility, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Cooperative Concept Mapping
Teams receive the key terms on cards and arrange them on a large sheet, drawing and labelling the links. Compare maps between teams before showing the model map.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Boiling a halogen breaks intermolecular forces, not the X–X covalent bond. | The rise in boiling point and the fall in bond energy are separate trends with separate explanations: one is about forces between molecules, the other about the bond inside each molecule. |
Physical properties of the halogens: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
Give the colour and physical state at room temperature of chlorine, bromine and iodine.
Explain the trend in volatility from chlorine to iodine.
Explain why the bond energy decreases from Cl–Cl to I–I.
Suggest why the F–F bond energy (158 kJ mol−1) is lower than that of Cl–Cl (242 kJ mol−1).
Predict the state and colour of astatine at room temperature and give a reason.
When iodine crystals are warmed gently, a purple vapour forms. State whether any covalent bonds are broken and explain.
Answer key and marking notes
- Chlorine: pale yellow-green gas.
- Bromine: red-brown liquid.
- Iodine: grey-black solid (purple vapour).
- Volatility decreases down the group.
- The molecules have more electrons, so the instantaneous dipole-induced dipole forces between them are stronger.
- More energy is needed to separate the molecules, so the boiling point is higher.
- The atoms get larger, so the bond is longer.
- The bonding pair is further from the nuclei and more shielded, so it is attracted less strongly.
- Fluorine atoms are very small, so the F–F bond is very short.
- The lone pairs on the two atoms are close together and repel, weakening the bond.
- A black solid.
- Colours darken and boiling points rise down the group; astatine has even more electrons than iodine.
- No covalent bonds are broken.
- Only the weak id-id forces between I2 molecules are overcome; the vapour is still I2.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Which row correctly describes bromine at room temperature and pressure?
- Aa red-brown liquid that gives off an orange-brown vapour
- Ba pale yellow-green gas
- Ca grey-black solid that gives a purple vapour when warmed
- Da colourless liquid that gives off a brown vapour
The chart shows the boiling points of the halogens. Which statement best explains the trend from F2 to I2?
- Athe molecules have more electrons down the group, so the instantaneous dipole-induced dipole forces between molecules get stronger and more energy is needed to separate them
- Bthe X-X covalent bond gets stronger down the group, so more energy is needed to boil the halogen
- Cthe molecules become more polar down the group, so permanent dipole-dipole forces increase
- Dhydrogen bonding between the molecules increases down the group as the atoms get larger
Solid iodine is warmed gently in a test tube. What is observed?
- Athe grey-black solid gives off a purple vapour
- Bthe white solid melts to a colourless liquid
- Cthe grey-black solid gives off an orange-brown vapour
- Dthe yellow solid gives off a pale green gas
The bond energies are Cl-Cl 242, Br-Br 193 and I-I 151 kJ mol−1. Which explanation of this trend is correct?
- Athe atoms get larger down the group, so the bonding pair is further from the nuclei and attracted less strongly
- Bthe atoms get smaller down the group, so the bonding pair is held more weakly
- Cthe number of bonding pairs decreases from two in Cl2 to one in I2
- Dthe bond gets weaker because the id-id forces between the molecules increase
How does the electronegativity of the halogens change from fluorine to iodine?
- Ait decreases, because the bonding electrons are further from the nucleus and more shielded
- Bit increases, because the nuclear charge increases down the group
- Cit stays the same, because every halogen needs one electron to fill its outer shell
- Dit decreases, because the nuclear charge decreases down the group
The F-F bond energy (158 kJ mol−1) is much lower than the Cl-Cl bond energy (242 kJ mol−1), breaking the trend down the group. What is the usual explanation?
- Athe F atoms are so small that the F-F bond is very short, and the lone pairs on the two atoms repel each other strongly, weakening the bond
- Bfluorine has fewer electrons, so its id-id forces are weaker
- Cfluorine is the least electronegative halogen, so it attracts the bonding pair weakly
- Dthe F-F bond is a double bond, which is weaker than a single bond
Astatine, At, lies below iodine in Group 17. Which set of predictions for astatine is most likely to be correct?
- Aa black solid with a boiling point above 184 °C, an At-At bond weaker than I-I, and a lower electronegativity than iodine
- Ba dark liquid with a boiling point below 184 °C, because the trend reverses for heavy atoms
- Ca black solid with an At-At bond stronger than I-I, because astatine has more electrons
- Da solid with a higher electronegativity than iodine, because astatine has the largest nuclear charge
How much energy is needed to break all the Br-Br bonds in 7.99 g of bromine, Br2? (Ar Br = 79.9; Br-Br bond energy = 193 kJ mol−1)
- A9.65 kJ
- B19.3 kJ
- C4.83 kJ
- D1540 kJ
Quiz key
Bromine boils at 59 °C, so at room temperature it is a dense red-brown liquid, and it is volatile enough to give off orange-brown fumes. A pale yellow-green gas describes chlorine (boiling point −34 °C). A grey-black solid with a purple vapour describes iodine (boiling point 184 °C). Bromine is strongly coloured, not colourless; the halogens all absorb visible light, and the colour deepens down the group.
Boiling points rise from −188 °C (F2) to 184 °C (I2), so volatility decreases down the group. The halogen molecules are non-polar, so the only intermolecular forces are instantaneous dipole-induced dipole (id-id) forces, which grow with the number of electrons (18 in F2, 106 in I2). Boiling does not break the covalent bond, and in any case the bond gets WEAKER from Cl-Cl to I-I. X2 molecules are symmetrical with identical atoms, so there is no permanent dipole. Hydrogen bonding needs H bonded to N, O or F; halogen molecules contain no hydrogen.
Iodine is a shiny grey-black solid at room temperature, and on gentle warming it readily forms a purple vapour. Iodine is not white, and its vapour is not colourless. An orange-brown vapour is bromine, not iodine. A pale yellow-green gas is chlorine; no halogen is a yellow solid at room temperature.
Down the group the atomic radius increases, so the shared pair of electrons is further from both nuclei and more shielded by inner shells; the attraction is weaker and the bond energy falls (242 > 193 > 151 kJ mol−1). The atoms get larger down the group, not smaller. Every halogen molecule has a single covalent bond, one bonding pair. id-id forces act BETWEEN molecules and do not change the strength of the covalent bond inside a molecule.
Electronegativity is the power of an atom to attract the bonding pair in a covalent bond. Fluorine (4.0) is the most electronegative element; the value falls to about 2.7 for iodine because the bonding electrons are further from the nucleus and more shielded. The nuclear charge does increase, but the extra protons are outweighed by the extra inner shells, so electronegativity does not increase. All halogens need one electron, but how strongly they attract it differs, so it is not constant. The nuclear charge increases down a group, it never decreases.
Fluorine atoms are very small, so the bond is short and the three lone pairs on each atom are close together. Their mutual repulsion weakens the bond, so F-F (158 kJ mol−1) is weaker than Cl-Cl and close to I-I. This weak bond is one reason fluorine is so reactive. id-id forces are intermolecular and affect the boiling point, not the covalent bond energy. Fluorine is the MOST electronegative element, not the least. F-F is a single bond, and in any case double bonds are stronger than single bonds.
Astatine has even more electrons than iodine, so its id-id forces are stronger: it should be a dark (black) solid with a boiling point above iodine's 184 °C. Its atoms are larger, so the At-At bond should be weaker than I-I (151 kJ mol−1) and its electronegativity lower. There is no reason for the volatility trend to reverse, so a boiling point below iodine's is not expected. More electrons strengthen the intermolecular forces, not the covalent bond; larger atoms give weaker bonds. The larger nuclear charge is outweighed by extra shielding and distance, so electronegativity keeps falling.
Mr(Br2) = 159.8, so n(Br2) = 7.99 / 159.8 = 0.0500 mol. Each molecule contains one Br-Br bond, so the energy needed is 0.0500 × 193 = 9.65 kJ. 19.3 kJ divides by the Ar (79.9), giving moles of atoms (0.100 mol) instead of moles of molecules. 4.83 kJ halves the correct answer, as if each bond were shared by two molecules. 1540 kJ multiplies the mass by the bond energy without converting to moles.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.