# Group 7: The halogens

### Group 7: The halogens

#### Lessons

In this lesson, we will learn:
• To recall properties of the halogens
• To understand the trend in properties found in the halogens
• To apply knowledge of electronic structure and bonding to explain the trends in halogens.
• To know the key reactions of the halogens.

Notes:

• We have already seen that the Periodic Table is arranged, top-left to bottom-right, by proton number and number of outer shell electrons. The number of outer shell electrons dictates the chemical properties of an element.

• Just like with the alkali metals, the halogens are another example of a well-studied group of elements which display trends in their common properties as you go down the group. As they are reactive elements, they were known and studied in their compounds before being isolated in their reactive, toxic elemental forms.

• The halogens have the following properties:
• They are non-metals stable as diatomic molecules (this means at room temperature and pressure, they exist as molecules made of two atoms, e.g. Cl2).
• They have a valence of 1 and form covalent bonds with non-metals atoms, or ionic bonds with metal atoms. Halogen ions will usually have a single negative charge (X-), where they are known as halides.
• They are colored.
• They have relatively low melting and boiling points compared to other non-metals (except the noble gases).
• Halogens in elemental form are relatively toxic, reactive substances.
• They do not conduct heat or electricity.
• They are brittle as solids.

• As you go down the group, the properties of the elements change in the following ways:
• The melting and boiling point gets higher – starting as gases, bromine is a liquid while iodine is a solid.
• The color of the halogens gets darker – fluorine is pale yellow, followed by green chlorine, brown/purple bromine and purple iodine.
• Electronegativity decreases down the group. The smallest halogen, fluorine, is the most electronegative element in the periodic table.
• The halogens get less reactive – fluorine, top of the group, is the most reactive element known. Iodine is the least reactive halogen (besides astatine which is often ignored because it is extremely rare).

• As with group 1 and 2, the trends in properties and GENERAL reactivity in group 7 can be explained by their electronic configuration:
• The reason that melting and boiling points increase down the group is because the intermolecular forces between the halogen molecules (e.g. F2, Cl2, Br2) get stronger down the group.
The increasingly large halogen atoms with more electrons produce stronger van der Waals forces between the molecules, so more heat energy is needed to overcome them.
This is why fluorine and chlorine are gases at room temperature, bromine is a liquid and iodine a solid.
• The reason that electronegativity decreases down the group is because each halogen further down the group has an extra inner electron shell shielding the nucleus from the outer electron shell.
As explained in Periodic Trends: Electronegativity more inner shells make it more difficult to attract electrons into (and keep them held in) the outer shell, which is increasingly further away from the nucleus. This is why fluorine (with only one inner shell shielding the nucleus) is the single most electronegative element in the periodic table, and electronegativity drops from there.
• Electronegativity decreasing down the group is the reason why reactivity decreases down the group! Halogens react by ‘pulling in’ electrons to their outer shell to complete it. Electronegativity measures this; how easily does an atom attract bonding electrons into its outer shell and hold onto them?
Fluorine is the most electronegative halogen, so it is the most reactive halogen. Iodine is the least electronegative (except astatine), so it is the least reactive.

• As one of the more reactive groups of elements, there are a variety of reactions the halogens take part in:
• The halogens react well with group 1 and 2 metals because these have electron configurations that complement the halogens. The metals react by losing electrons; the halogens react by gaining them. These are vigorous, exothermic reactions.

• Group 1 with halogens:
2M + X2 $\,$$\,$ 2MX

Group 2 with halogens:
M + X2 $\,$$\,$ MX2

In both equations:
M = group 1 or group 2 metal.
X = halogen (F2, Cl2, Br2, I2).

• A more reactive halogen can displace a less reactive halogen in a salt. This is known as a displacement reaction.

• X2 + 2MY $\,$$\,$ 2MX + Y2

X = F, Cl, Br or I,
Y = halogen atom below X in the group.
X = For example:
Cl2 + 2NaBr $\,$$\,$ 2NaCl + Br2

This reaction works as long as the ‘X2’ halogen is higher up the group than the halogen in the ‘MY’ halide salt, where M is the metal and Y is the halide ion.
If this reaction happens in water (‘aqueous solution’), it’s more accurate to write this reaction ignoring the metal ions because they are just spectator ions. Spectator ions don’t take part in the reaction – this is just an oxidation of the halide (Cl-, Br- or I-) ions by a halogen molecule (see the point just below).

• The halogens are good oxidising agents, and more reactive halogens can oxidise halide ions further down the group.
A general equation:
X2 + 2Y- (aq) $\,$$\,$ 2X- (aq) + Y2

Where X is a halogen atom higher up the group than Y.

This reaction works as long as the halogen (X2) reactant is higher up the group than the halide ions (2Y-) they are oxidising. You can think of it as electrons (or negative charge) “moving up” the group.
For example, Cl2 will react with Br- to produce chloride ions and bromine:

Cl2 (g) + 2Br- (aq) $\,$$\,$ 2Cl- (aq) + Br2 (l)

• Fluorine is such a strong oxidising agent that this reaction doesn’t practically work for fluorine because it literally oxidises the water instead of the halide ions!

This reaction happens less readily going down the group. This is an effect of the enthalpy change of X2 $\,$$\,$ 2X- (aq) getting less exothermic (more positive). The reasons given for why this happens is misleading though; there is a ‘short answer’ and a more accurate long answer.
• The short answer is that in the halogen group, electronegativity is higher and there is less electron shielding of the nucleus at the top of the group. This means fluorine/chlorine is better at attracting electrons into its outer shell and forming an ion than the larger halogens like bromine/iodine below. Depending on your course/syllabus, this answer will probably be enough, but it isn’t completely true.

• The short answer is inaccurate because it talks about only one part of the reaction called the electron affinity, and the enthalpy data2 doesn’t back it up. Fluorine isn’t better than chlorine at ‘attracting electrons into its outer shell’. The total enthalpy change still shows fluorine as the most reactive because…

• The long answer is that this reaction has multiple enthalpy changes involved.
The first is the enthalpy of atomisation: the energy needed to break all bonds within a substance into separate atoms in the gas phase. For the halogens this change can be written as:

• X2 (g) $\,$$\,$ 2X (g)

In elemental fluorine and chlorine, this is half of the bond enthalpy1 because they are diatomic gases at room temperature. F-F is a weaker bond than Cl-Cl, so F is more readily atomised.
The second is the electron affinity: the energy released when an electron joins a gaseous atom to form a negative ion. This change can be written as:

2X (g) $\,$$\,$ 2X- (g)

Fluorine’s electron affinity is actually less than chlorine’s.2 Fluorine is a small atom with very high electron density where the bonding electrons are being attracted into. This decreases its electron affinity to being lower than that of Cl. Cl $\,$$\,$ Cl- releases slightly more energy than F $\,$$\,$ F-, so Cl is more readily ionized.
The third is the enthalpy of hydration: the energy released when an ionic substance dissolves in and interacts with water. This change can be written as:

2X- (g) $\,$$\,$ 2X- (aq)

The enthalpy of hydration is much higher when F- ions dissolve than when Cl- ions dissolve.3 In other words, F- interacting with water releases A LOT more energy than Cl- interacting with water.
In summary, when adding the three enthalpy values1,2,3 together to get total enthalpy change, you find fluorine is the most exothermic by far. The change gets less exothermic going down the group in a consistent manner.

• The halogens, particularly chlorine, also react in disproportionation reactions. These are reactions where atoms of the same substance are both oxidised and reduced. In a reaction with Cl2, one atom is reduced when forming one product and the other Cl atom is oxidised when forming part of another product.

The disproportionation of chlorine with water is below:

• Cl2 + H2O $\, \rightleftharpoons \,$ HClO + HCl

The disproportionation of chlorine with cold sodium hydroxide is below:

Cl2 + 2NaOH $\,$$\,$ NaClO + NaCl + H2O

This reaction with cold sodium hydroxide produces sodium hypochlorite (NaClO), which is the active ingredient in bleach products. Hypochlorite salts and elemental chlorine (in very small quantities) are both used in water treatment as disinfectants.

Disproportionation reactions are best understood using oxidation numbers (see Oxidation number for more):
• Like all elemental forms, Cl in Cl2 has an oxidation state of zero.
• Na, being in group 1, takes a +1 oxidation state; so does H when bonded to non-metals.
• Oxygen, being the most electronegative element besides fluorine, has an oxidation state of -2 in compounds.
• In the reactions above, no oxidation numbers change except for chlorine: chlorine goes from two atoms with an oxidation number of zero to one atom with an oxidation number of -1 and another with +1.

The reaction of chlorine with hot sodium hydroxide is slightly different, producing sodium chlorate instead:
3Cl2 + 6NaOH $\,$$\,$ 5NaCl + NaClO3 + 3H2O

Just like in the reaction with water and cold sodium hydroxide, the only oxidation state change here is with the chlorine atoms.
• 5 of the 6 Cl atoms in the reactants formed sodium chloride, going from an oxidation number of 0 to -1.
• The last Cl atom forms sodium chlorate, with a change in oxidation number of 0 to +5.
To balance this equation, memorize the products and then look at the oxidation number changes:
• An oxidation number just tells you about electrons being transferred. Electrons don’t get created or destroyed, so the changes in oxidation number in a reaction must cancel to zero.
• When Cl2 reacts with hot NaOH, NaClO3 forms which is an oxidation number change of 0 to +5 in the chlorine atom. If one atom of Cl changed 0 $\,$$\,$ +5, there must be a 0 $\,$$\,$ -5 change in the other chlorine atoms in total. In NaCl, chlorine has a 1- oxidation state, so it must be 5 moles of NaCl to make -5 overall change.
• This gives you 5NaCl for every NaClO3. To produce this, you would need 3Cl2 for the chlorine atoms, 6 NaOH for 6 Na atoms and the leftover H and O make 3 H2O molecules.
See below for the full equations:

This disproportionation reactions can happen with bromine and iodine as well. The reaction to produce the chlorate/bromate/iodate (NaXO3, X = Cl, Br, I) occurs more readily going down the group.

References:
1: https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf - source for bond enthalpy data.
2: http://hyperphysics.phy-astr.gsu.edu/hbase/Chemical/eleaff.html#c1 - source for electron affinity data.
3: http://www.wiredchemist.com/chemistry/data/enthalpies-hydration - source for enthalpy of hydration.
• Introduction
Halogens: A summary
a)
Introduction to halogens.

b)
Properties of the halogens.

c)
How do the properties change going down the group? (Trends in properties)

d)
Explaining the trends in properties.

e)
Reaction of halogens with group 1 and 2 metals.

f)
Halogen displacement reactions.

g)
Halogens as oxidising agents.

h)
Halogens and disproportionation reactions.

• 1.
Recall the properties of the halogens.
Sort the halogens according to the properties listed below:
a)
Reactivity; lowest to highest.

b)
State at room temperature; solids, then liquids, then gases.

• 2.
Explaining the trend in properties of the halogens.
A student in a school laboratory has a solution of sodium bromide (NaBr) in a beaker. To this, she adds an equal amount of chlorine water (chlorine dissolved in water).
a)
What observations would she make from the reaction?

b)
Write a chemical equation to describe this reaction, and explain why this observation and reaction occurred.

c)
Explain why no observed change would be seen if sodium bromide (NaBr) was reacted with iodine (I2) in solution.