Physical properties of the Group 17 elements

1. Overview

The Group 17 elements, the halogens, are non-metals made of diatomic molecules, X₂. This section covers what chlorine, bromine and iodine look like, why they become less volatile down the group, and why the bond inside each molecule becomes weaker from chlorine to iodine. The two trends are easy to mix up: volatility depends on the forces between molecules, while bond strength depends on the covalent bond inside each molecule. The same reasoning about bond length returns in 11.2, for the bonds in the hydrogen halides.

Key Definitions

• Halogen: an element of Group 17 of the Periodic Table, such as fluorine, chlorine, bromine or iodine; as elements, fluorine to iodine exist as diatomic molecules, X₂, held together by a single covalent bond.
• Volatility: how readily a substance turns into a vapour; of two substances, the more volatile one has the lower boiling point.
• Sublimation: a change of state directly from solid to gas, without melting first; iodine sublimes when it is warmed gently.
• Van der Waals' forces: the general name for all intermolecular forces, the attractions between molecules (or noble gas atoms) other than those due to bond formation; the types are id-id forces and pd-pd forces, and hydrogen bonding is a special case of pd-pd forces.
• Instantaneous dipole-induced dipole (id-id) forces: attractions that arise when an uneven spread of electrons at one instant gives a molecule a temporary dipole, which induces a dipole in a neighbouring molecule; also called London dispersion forces, they are the only intermolecular forces between X₂ halogen molecules.
• Covalent bond: the electrostatic attraction between a shared pair of electrons and the nuclei of the two bonded atoms.
• Bond energy: the energy required to break one mole of a particular covalent bond in the gaseous state; for a halogen it is the energy for X₂(g) → 2X(g) per mole of X₂, and it is always positive.
• Bond length: the distance between the nuclei of two atoms joined by a covalent bond.

Content

The halogens at a glance

Each halogen atom has seven electrons in its outer shell. Two atoms share one pair of electrons to make a molecule, X₂, with a single covalent bond, X-X. Chlorine, bromine and iodine are simple molecular substances: separate X₂ molecules with only weak intermolecular forces between them.

The two atoms in X₂ are identical, so they attract the shared pair equally. The bond is non-polar and the molecule has no permanent dipole. The only forces between halogen molecules are therefore id-id forces.

Molecule	Electrons	State at room temperature
F₂	18	gas
Cl₂	34	gas
Br₂	70	liquid
I₂	106	solid

The number of electrons per molecule is twice the proton number: iodine is element 53, so I₂ has 2 × 53 = 106 electrons.

Bond	Bond energy / kJ mol⁻¹
F-F	158
Cl-Cl	242
Br-Br	193
I-I	151

The bond energies come from the Data Booklet, which calls them exact values: each is measured for one particular molecule, X₂, not averaged over many different compounds.

Colours and states

You need the colours of chlorine, bromine and iodine. Learn each colour together with the state at room temperature:

Element	At room temperature	Gas or vapour
chlorine, Cl₂	pale green (yellow-green) gas	pale green
bromine, Br₂	dark red-brown liquid	orange-brown
iodine, I₂	grey-black solid (shiny crystals)	purple

Dissolved, the colours change with the solvent and the concentration:

Solution of	In water	In hexane
chlorine	very pale green, almost colourless	very pale green
bromine	yellow to orange	orange to red
iodine	yellow to brown	purple (violet)

• Chlorine is pale green or yellow-green; "yellow" alone is a different colour.
• Purple is the colour of iodine vapour and of iodine dissolved in hexane. Describe solid iodine as grey-black, not purple.
• Liquid bromine is not bromine water, a dilute solution that looks yellow to orange. Any solution looks paler when more dilute.
• Bromine evaporates easily, so a bottle of it is full of orange-brown vapour above the liquid. Warm iodine crystals gently and a purple vapour forms straight from the solid: iodine sublimes. Stronger heating can melt it.

The colours get darker down the group, but "darker" is not a colour: name the colour and state of each element. You will use the solution colours in 11.2 to tell which halogen is present.

The trend in volatility

Volatility is how readily a substance turns into a vapour. A volatile substance has a low boiling point. Down Group 17, from chlorine to iodine:

• melting points and boiling points increase;
• volatility decreases;
• the state at room temperature changes from gas (Cl₂) to liquid (Br₂) to solid (I₂).

Of the three, chlorine is the most volatile and iodine the least. Fluorine fits the same pattern: it is a gas and is more volatile than chlorine. Following the trend, astatine, below iodine, is expected to be less volatile than iodine.

Boiling point and volatility move in opposite directions. "Boiling point increases" and "volatility decreases" describe the same trend.

Why volatility decreases: id-id forces

To boil a halogen, or to sublime iodine, you separate whole X₂ molecules from one another. Only the forces between the molecules are overcome. The X-X covalent bonds stay intact: bromine vapour is still made of Br₂ molecules.

id-id forces arise like this (3.6 describes them for all molecules):

1. Electrons move constantly. At one instant more of a molecule's electrons may be at one end than the other, giving an instantaneous dipole: δ− at one end, δ+ at the other.
2. This dipole repels or attracts the electrons of a neighbouring molecule and distorts its electron cloud, giving it an induced dipole.
3. The δ+ end of one molecule attracts the δ− end of the next.
4. The dipoles keep vanishing and forming again. Each induced dipole always lines up so that it is attracted to the dipole that caused it, so overall the molecules attract each other.

Down the group the number of electrons per molecule increases: 34 in Cl₂, 70 in Br₂ and 106 in I₂. A molecule with more electrons has a larger electron cloud, which is more easily distorted. Larger instantaneous dipoles form, and they induce larger dipoles in neighbouring molecules. So the chain is:

more electrons → larger instantaneous and induced dipoles → stronger id-id forces → more energy needed to separate the molecules → higher boiling point → lower volatility.

The cause is the number of electrons in each molecule, not the mass of the molecule. Electronegativity and the pull of a nucleus on its own outer electrons describe atoms; volatility depends on forces between molecules.

Worked example 1: bromine and iodine at room temperature

Question: At room temperature bromine is a liquid but iodine is a solid. Explain this difference in terms of the forces involved.

Step 1: identify the structure and the forces. Both elements are simple molecular, made of non-polar X₂ molecules. The only forces between the molecules are id-id forces.

Step 2: count the electrons. Bromine has proton number 35, so Br₂ has 2 × 35 = 70 electrons. Iodine has proton number 53, so I₂ has 2 × 53 = 106 electrons.

Step 3: link the electrons to the forces. I₂ has more electrons, so larger instantaneous and induced dipoles form. The id-id forces between I₂ molecules are stronger than those between Br₂ molecules.

Step 4: link the forces to energy and state. More energy is needed to overcome the id-id forces between I₂ molecules, so iodine has the higher melting and boiling points. Room temperature is below the melting point of iodine, so it is a solid. It is above the melting point of bromine but below its boiling point, so bromine is a liquid.

Answer: I₂ has more electrons than Br₂ (106 against 70), so the id-id forces between I₂ molecules are stronger and need more energy to overcome; iodine is a solid and bromine a liquid at room temperature. The covalent bonds play no part: I-I (151 kJ mol⁻¹) is actually weaker than Br-Br (193 kJ mol⁻¹).

The trend in bond strength

Bond energy measures the strength of a bond. For a halogen it is the energy needed to split one mole of gaseous X₂ molecules into atoms:

$$\text{X}_2(\text{g}) \rightarrow 2\text{X}(\text{g}) \qquad \Delta H = +E(\text{X-X})$$

Here $E$ is the bond energy in kJ mol⁻¹. From chlorine to iodine the bond energy decreases: Cl-Cl 242, Br-Br 193 and I-I 151 kJ mol⁻¹. The bonds get weaker down the group.

Explain the trend as a chain:

1. A covalent bond is the attraction between the shared pair of electrons and the nuclei of both atoms.
2. Down the group each atom has one more occupied shell, so the atomic radius increases and the bond gets longer.
3. The shared pair is further from each nucleus, and more filled inner shells lie between it and each nucleus, so there is more shielding.
4. The attraction between the nuclei and the shared pair is weaker, so less energy is needed to break the bond.

The nuclear charge also rises, from +17 to +35 to +53, but the extra inner electrons shield the shared pair from almost all of this extra charge, so the greater distance decides the trend.

This explanation is about the shared pair in a bond. Ionisation energy (removing an electron from an atom) and id-id forces (between molecules) do not explain it.

Fluorine: the exception

If the trend carried on upwards, F-F would be the strongest halogen bond. It is not: at 158 kJ mol⁻¹ it is weaker than Cl-Cl and Br-Br, and only a little stronger than I-I.

Fluorine atoms are very small, so the F-F bond is very short. Each F atom has three lone pairs, and in such a short bond the lone pairs on the two atoms are close together. They repel each other strongly, and this weakens the bond. In the larger halogens the lone pairs are further apart and this repulsion matters much less.

So bond energy decreases steadily from Cl-Cl to Br-Br to I-I, with F-F out of line. Bring in fluorine only when a question includes it.

Worked example 2: how far out of line is fluorine?

Question: Use bond energies from the Data Booklet: F-F 158, Cl-Cl 242, Br-Br 193, I-I 151 kJ mol⁻¹. (a) Calculate the average decrease in bond energy per step from Cl-Cl to I-I. (b) Use it to estimate the F-F bond energy that the trend predicts. (c) Compare your estimate with the real value and explain the difference.

Step 1: the fall per step. $$242 - 193 = 49\ \text{kJ mol}^{-1}$$ $$193 - 151 = 42\ \text{kJ mol}^{-1}$$ $$\text{average} = \frac{49 + 42}{2} = 45.5\ \text{kJ mol}^{-1}$$

Step 2: go one step up the group. Fluorine is one step above chlorine, so the trend predicts $$E(\text{F-F}) \approx 242 + 45.5 = 287.5 \approx 290\ \text{kJ mol}^{-1}$$ This is an estimate, not a measured value. The two steps are not equal, so give it to only 2 s.f. The steps grow going up the group (42, then 49), so if anything the trend points even higher. What matters is that the trend predicts a value well above 242.

Step 3: compare. $$\frac{158}{287.5} = 0.55$$ $$287.5 - 158 = 129.5 \approx 130\ \text{kJ mol}^{-1}$$

Answer: (a) 45.5 kJ mol⁻¹ per step. (b) An estimate of about 290 kJ mol⁻¹. (c) The real F-F bond energy, 158 kJ mol⁻¹, is only about half of the estimate, roughly 130 kJ mol⁻¹ lower. The F atoms are so small that the bond is very short, so the lone pairs on the two atoms are close together and repel strongly, which weakens the bond.

Volatility and bond strength: two different forces

	Volatility	Bond strength
What is overcome	id-id forces between X₂ molecules	the covalent X-X bond inside a molecule
Change	X₂(l) → X₂(g), or I₂(s) → I₂(g)	X₂(g) → 2X(g)
Trend from Cl₂ to I₂	decreases, because id-id forces get stronger	decreases, because the bonds get weaker
Main cause	more electrons per molecule	larger atoms, longer bond, more shielding

Both decrease from chlorine to iodine, but for different reasons. Iodine has the weakest bond of the three, yet it is the least volatile: boiling and subliming do not break X-X bonds.

Worked example 3: boiling chlorine versus breaking its bonds

Question: At its boiling point, 20.4 kJ is needed to boil one mole of liquid chlorine, Cl₂(l) → Cl₂(g). A sample of 4.26 g of liquid chlorine is boiled. ($A_{\text{r}}$: Cl 35.5; Cl-Cl bond energy 242 kJ mol⁻¹)
(a) Calculate the energy needed to boil the sample.
(b) Calculate the energy needed to break all the Cl-Cl bonds in the chlorine gas formed.
(c) How many times larger is (b) than (a), and what does this show?

Step 1: amount of chlorine. $M(\text{Cl}_2) = 2 \times 35.5 = 71.0$ g mol⁻¹. $$n = \frac{m}{M} = \frac{4.26}{71.0} = 0.0600\ \text{mol}$$

Step 2: energy to boil. $$0.0600 \times 20.4 = 1.224\ \text{kJ} = 1.22\ \text{kJ}$$

Step 3: energy to break the bonds. Each Cl₂ molecule has one Cl-Cl bond, so 0.0600 mol of Cl₂ contains 0.0600 mol of bonds. $$0.0600 \times 242 = 14.52\ \text{kJ} = 14.5\ \text{kJ}$$

Step 4: compare. $$\frac{14.52}{1.224} = 11.9$$

Answer: (a) 1.22 kJ. (b) 14.5 kJ. (c) About 12 times larger (11.9). Overcoming the id-id forces between Cl₂ molecules takes far less energy than breaking the covalent bonds inside them. Boiling separates the molecules and leaves every Cl-Cl bond intact.

Key Equations

Electrons in a halogen molecule ($N_{\text{e}}$ = number of electrons in X₂; $Z$ = proton number of the halogen): $$N_{\text{e}} = 2Z$$

Breaking the bond ($E$ = bond energy in kJ mol⁻¹, for gases, always positive): $$\text{X}_2(\text{g}) \rightarrow 2\text{X}(\text{g}) \qquad \Delta H = +E(\text{X-X})$$

Boiling (the molecules separate; the X-X bond does not break): $$\text{X}_2(\text{l}) \rightarrow \text{X}_2(\text{g})$$

Energy for a sample ($n$ = amount in mol; $m$ = mass in g; $M$ = molar mass in g mol⁻¹; the energy per mole in kJ mol⁻¹ gives the energy in kJ): $$n = \frac{m}{M}$$ $$\text{energy} = n \times \text{energy per mole}$$

Data Booklet bond energies (kJ mol⁻¹): F-F 158, Cl-Cl 242, Br-Br 193, I-I 151.

Common Mistakes to Avoid

• Mistake: describing chlorine as "yellow" or solid iodine as "purple". Fix: chlorine is a pale green (yellow-green) gas and iodine is a grey-black solid. Purple is the colour of iodine vapour and of iodine dissolved in hexane.
• Mistake: giving a colour without the state, or calling iodine a liquid. Fix: at room temperature chlorine is a gas, bromine a liquid and iodine a solid. Write "dark red-brown liquid" for bromine, not just "red-brown".
• Mistake: mixing up liquid bromine with bromine water, or writing only "the colours get darker down the group". Fix: liquid bromine is dark red-brown, while bromine water is yellow to orange. Name the colour of each element separately.
• Mistake: getting the direction of the volatility trend backwards. Fix: boiling points increase down the group, so volatility decreases: chlorine is the most volatile of the three and iodine the least.
• Mistake: explaining volatility with electronegativity, the attraction of a nucleus for its outer electrons, or the mass of the molecules. Fix: name the forces between molecules and their cause. Br₂ (70 electrons) has stronger id-id forces than Cl₂ (34 electrons), so more energy is needed to separate Br₂ molecules.
• Mistake: saying that X-X bonds break when a halogen boils, or explaining a state with bond strength. Fix: I-I (151 kJ mol⁻¹) is the weakest of the three bonds, yet iodine is the least volatile. In Worked example 3, breaking the Cl-Cl bonds needs about 12 times the energy needed to boil the chlorine.
• Mistake: explaining the bond-strength trend with ionisation energy or with id-id forces. Fix: talk about the shared pair. The Br-Br bond is longer than Cl-Cl, so the shared pair is further from both nuclei and more shielded, and the bond energy falls from 242 to 193 kJ mol⁻¹.
• Mistake: assuming F-F is the strongest halogen bond because fluorine is at the top. Fix: check the Data Booklet: F-F is 158 kJ mol⁻¹, well below Cl-Cl at 242 kJ mol⁻¹, because lone pairs repel in the very short bond.

Exam Tips

• Describe the colours by giving the colour and the state of each element at room temperature, for example "bromine: dark red-brown liquid". For a solution, say which solvent.
• State and explain the volatility trend as a chain: volatility decreases from Cl₂ to I₂; the forces are id-id forces between molecules; the number of electrons increases; so more energy is needed to overcome the forces, and the boiling point is higher.
• Name the force in full the first time: "instantaneous dipole-induced dipole (id-id) forces". "Van der Waals' forces" alone is only the general name.
• Explain the bond-strength trend with the words "bond length", "shared pair", "nuclei" and "shielding". When a question says "use data", quote the Data Booklet bond energies.
• Keep the two trends apart. Volatility is about forces between molecules; bond strength is about the bond within one molecule. Never use one to explain the other.
• Predict for another halogen, such as astatine, by continuing the trend, and give the reason as well as the prediction. The one value that breaks its trend is the F-F bond energy.