Halogens

Welcome to the Halogens (Group 7 / Group 17)

Welcome to one of the most colourful and reactive groups in the periodic table! The halogens (Group 7, also known as Group 17) consist of non-metals that you encounter in daily life: from the chlorine purifying swimming pool water, to the iodine in first-aid antiseptics, to the fluoride in your toothpaste.

In this chapter, we will explore their physical appearances, trends in boiling points and electronegativity, their chemical reactivity (as oxidising agents), how halide ions behave as reducing agents, and standard laboratory tests used to identify them. Don't worry if chemical equations seem daunting at first—we will break down each reaction step by step!

1. Physical Properties and Trends of the Halogens

All halogen atoms have the outer electron configuration \(ns^2 np^5\). They need just one more electron to achieve a stable octet (noble gas configuration). Because of this, they exist naturally as diatomic molecules: \(\text{F}_2\), \(\text{Cl}_2\), \(\text{Br}_2\), and \(\text{I}_2\).

Appearance and State at Room Temperature

• Fluorine (\(\text{F}_2\)): Pale yellow gas (highly toxic and reactive).
• Chlorine (\(\text{Cl}_2\)): Pale green-yellow gas (pungent and choking smell).
• Bromine (\(\text{Br}_2\)): Red-brown liquid that readily vaporises into a dense orange-brown gas.
• Iodine (\(\text{I}_2\)): Shiny grey-black solid that sublimes on gentle heating to produce a vivid purple vapour.
• Astatine (\(\text{At}_2\)): Black solid (radioactive and rare).

Memory Trick: Notice how the colours get darker and the physical states get denser as you move down the group (Gas \(\rightarrow\) Gas \(\rightarrow\) Liquid \(\rightarrow\) Solid).

Trend in Boiling and Melting Points

Trend: Melting and boiling points increase as you go down Group 7.
Explanation:
1. Halogen molecules are simple non-polar covalent molecules held together by weak van der Waals forces (induced dipole-dipole or London dispersion forces).
2. Going down the group, the number of electrons in each molecule increases (\(\text{F}_2\) has 18 electrons, while \(\text{I}_2\) has 106 electrons).
3. More electrons lead to larger, more polarisable electron clouds, creating stronger van der Waals forces between molecules.
4. More thermal energy is required to overcome these stronger intermolecular forces, leading to higher melting and boiling points.

Trend in Electronegativity

Definition: Electronegativity is the ability of an atom to attract the bonding pair of electrons in a covalent bond.
Trend: Electronegativity decreases down Group 7.
Explanation:
• As you move down the group, each element has an extra quantum shell of electrons.
• The atomic radius increases and there is increased electron shielding from inner shells.
• Consequently, the positively charged nucleus has a weaker electrostatic attraction for the shared bonding pair of electrons.

Key Takeaway: Going down Group 7, the molecules get darker in colour, change from gas to solid, have higher boiling points due to stronger van der Waals forces, and show decreasing electronegativity.

2. Halogens as Oxidising Agents (Displacement Reactions)

A halogen atom acts as an oxidising agent by accepting an electron to form a halide ion (reduction):
\(\text{X}_2 + 2\text{e}^- \rightarrow 2\text{X}^-\)

Trend in Oxidising Ability: Decreases down the group (\(\text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2\)).
Why? Smaller atoms (like fluorine and chlorine) have their outer shells closer to the nucleus with less shielding, so they attract and capture an incoming electron much more strongly.

Halogen Displacement Reactions

A more reactive halogen (stronger oxidising agent) will displace a less reactive halide ion from its aqueous solution.

1. Chlorine displacing Bromide:
\(\text{Cl}_2\text{(aq)} + 2\text{NaBr(aq)} \rightarrow 2\text{NaCl(aq)} + \text{Br}_2\text{(aq)}\)
Ionic equation: \(\text{Cl}_2\text{(aq)} + 2\text{Br}^-\text{(aq)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{Br}_2\text{(aq)}\)
Observation: Colourless solution turns orange/yellow due to the formation of aqueous bromine (\(\text{Br}_2\)).

2. Chlorine displacing Iodide:
\(\text{Cl}_2\text{(aq)} + 2\text{NaI(aq)} \rightarrow 2\text{NaCl(aq)} + \text{I}_2\text{(aq)}\)
Ionic equation: \(\text{Cl}_2\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{I}_2\text{(aq)}\)
Observation: Colourless solution turns brown due to aqueous iodine (\(\text{I}_2\)).

3. Bromine displacing Iodide:
\(\text{Br}_2\text{(aq)} + 2\text{NaI(aq)} \rightarrow 2\text{NaBr(aq)} + \text{I}_2\text{(aq)}\)
Ionic equation: \(\text{Br}_2\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Br}^-\text{(aq)} + \text{I}_2\text{(aq)}\)
Observation: Orange solution turns brown.

Using a Non-Polar Organic Solvent (e.g., Hexane / Cyclohexane)

In water, bromine and iodine solutions can both look brownish-orange, making them hard to distinguish. To confirm which halogen is present, we add an organic solvent like hexane and shake the tube:

• Halogens are non-polar molecules and dissolve much better in the upper organic layer than in water.
• Chlorine in organic solvent: virtually colourless / pale green
• Bromine in organic solvent: distinct orange/red
• Iodine in organic solvent: distinct vibrant purple / violet

Key Takeaway: Chlorine displaces both bromide and iodide. Bromine displaces only iodide. Iodine cannot displace any of them. Organic solvents make identifying free halogens easy (especially purple for iodine!).

3. Halide Ions as Reducing Agents

While halogens gain electrons, halide ions (\(\text{X}^-\)) act as reducing agents by losing an electron:
\(2\text{X}^- \rightarrow \text{X}_2 + 2\text{e}^-\)

Trend in Reducing Ability: Increases down the group (\(\text{F}^- < \text{Cl}^- < \text{Br}^- < \text{I}^-\)).
Why? Iodide ions (\(\text{I}^-\)) are much larger than chloride ions. The outermost electrons in \(\text{I}^-\) are further from the nucleus and experience more shielding, so they are held less tightly and are lost most easily.

Reactions of Solid Sodium Halides with Concentrated Sulfuric Acid (\(\text{H}_2\text{SO}_4\))

Concentrated sulfuric acid is both a strong acid and an oxidising agent. This classic practical demonstrates the difference in reducing power between halide ions.

1. Sodium Chloride (\(\text{NaCl}\)) + Conc. \(\text{H}_2\text{SO}_4\)

• Chloride is a weak reducing agent; it cannot reduce concentrated sulfuric acid.
• Only an acid-base reaction occurs:
\(\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HCl(g)}\)
• Observations: Steamy/misty white fumes of hydrogen chloride (\(\text{HCl}\)) gas that turn damp blue litmus paper red.

2. Sodium Bromide (\(\text{NaBr}\)) + Conc. \(\text{H}_2\text{SO}_4\)

• Bromide is a stronger reducing agent. It first undergoes the acid-base reaction:
\(\text{NaBr(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HBr(g)}\) (produces misty fumes of \(\text{HBr}\))
• The \(\text{HBr}\) then reduces sulfuric acid to sulfur dioxide (\(\text{SO}_2\)), where sulfur is reduced from \(+6\) to \(+4\):
\(2\text{HBr(g)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{Br}_2\text{(g)} + \text{SO}_2\text{(g)} + 2\text{H}_2\text{O(l)}\)
• Observations: Misty fumes (\(\text{HBr}\)), orange-brown fumes/liquid (\(\text{Br}_2\)), and a colourless, choking gas (\(\text{SO}_2\)).

3. Sodium Iodide (\(\text{NaI}\)) + Conc. \(\text{H}_2\text{SO}_4\)

• Iodide is the strongest reducing agent of the three. It reduces sulfuric acid (\(+6\)) through multiple stages to \(\text{SO}_2\) (\(+4\)), elemental sulfur \(\text{S}\) (\(0\)), and hydrogen sulfide \(\text{H}_2\text{S}\) (\(-2\)):
1. Acid-base reaction: \(\text{NaI(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HI(g)}\)
2. Reduction to sulfur dioxide: \(2\text{HI} + \text{H}_2\text{SO}_4 \rightarrow \text{I}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\)
3. Reduction to sulfur: \(6\text{HI} + \text{H}_2\text{SO}_4 \rightarrow 3\text{I}_2 + \text{S} + 4\text{H}_2\text{O}\)
4. Reduction to hydrogen sulfide: \(8\text{HI} + \text{H}_2\text{SO}_4 \rightarrow 4\text{I}_2 + \text{H}_2\text{S} + 4\text{H}_2\text{O}\)
• Observations: Misty fumes (\(\text{HI}\)), purple fumes / dark solid (\(\text{I}_2\)), choking gas (\(\text{SO}_2\)), yellow solid (\(\text{S}\)), and a rotten-egg smell (\(\text{H}_2\text{S}\)).

Key Takeaway: Chloride cannot reduce sulfuric acid (acid-base only). Bromide reduces sulfur from \(+6\) to \(+4\) (\(\text{SO}_2\)). Iodide reduces sulfur all the way from \(+6\) to \(-2\) (\(\text{H}_2\text{S}\)).

4. Testing and Identifying Halide Ions

To identify unknown halide ions in aqueous solution, we carry out a two-step confirmation test using acidified silver nitrate followed by aqueous ammonia.

Step 1: Acidified Silver Nitrate Test

1. Add dilute nitric acid (\(\text{HNO}_3\)) to the unknown solution.
Why? To react with and remove any carbonate (\(\text{CO}_3^{2-}\)) or hydroxide (\(\text{OH}^-\)) impurities, which would otherwise form misleading silver precipitates (\(\text{Ag}_2\text{CO}_3\) or \(\text{Ag}_2\text{O}\)).
Common Mistake to Avoid: Never use hydrochloric acid (\(\text{HCl}\)) to acidify, because \(\text{HCl}\) contains chloride ions that give a false-positive white precipitate!
2. Add aqueous silver nitrate (\(\text{AgNO}_3\)) dropwise.

Precipitate Observations and Equations:
• Chloride (\(\text{Cl}^-\)): Forms a white precipitate of silver chloride.
\(\text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \rightarrow \text{AgCl(s)}\)
• Bromide (\(\text{Br}^-\)): Forms a cream precipitate of silver bromide.
\(\text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{AgBr(s)}\)
• Iodide (\(\text{I}^-\)): Forms a yellow precipitate of silver iodide.
\(\text{Ag}^+\text{(aq)} + \text{I}^-\text{(aq)} \rightarrow \text{AgI(s)}\)

Step 2: Confirming with Aqueous Ammonia (\(\text{NH}_3\))

Because cream and yellow precipitates can look very similar under laboratory lighting, we confirm the result using solubility in ammonia:

• Silver Chloride (\(\text{AgCl}\)): Dissolves in dilute \(\text{NH}_3\text{(aq)}\) to form a colourless solution.
• Silver Bromide (\(\text{AgBr}\)): Insoluble in dilute \(\text{NH}_3\text{(aq)}\), but dissolves in concentrated \(\text{NH}_3\text{(aq)}\).
• Silver Iodide (\(\text{AgI}\)): Insoluble in both dilute and concentrated \(\text{NH}_3\text{(aq)}\).

Quick Review:
\(\text{AgCl}\) = White \(\rightarrow\) Soluble in dilute \(\text{NH}_3\)
\(\text{AgBr}\) = Cream \(\rightarrow\) Soluble only in conc \(\text{NH}_3\)
\(\text{AgI}\) = Yellow \(\rightarrow\) Insoluble in conc \(\text{NH}_3\)

5. Disproportionation Reactions of Chlorine

Disproportionation is a specific redox reaction in which the same element is simultaneously oxidised and reduced.

Reaction 1: Chlorine with Water (Water Treatment)

When chlorine gas dissolves in water, it reacts reversibly:
\(\text{Cl}_2\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{HCl(aq)} + \text{HClO(aq)}\)

• The oxidation state of chlorine in \(\text{Cl}_2\) is \(0\).
• In hydrochloric acid (\(\text{HCl}\)), chlorine is reduced to \(-1\).
• In chloric(I) acid (\(\text{HClO}\), also called hypochlorous acid), chlorine is oxidised to \(+1\).
• Application: Chloric(I) acid acts as an antibacterial agent and sterilises drinking water and swimming pools. Chloric(I) ions kill microorganisms and prevent waterborne diseases like cholera.

Reaction 2: Chlorine with Cold, Dilute Sodium Hydroxide (Bleach Production)

When chlorine is mixed with cold, dilute aqueous \(\text{NaOH}\), household bleach is produced:
\(\text{Cl}_2\text{(g)} + 2\text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{NaClO(aq)} + \text{H}_2\text{O(l)}\)

• In \(\text{Cl}_2\), chlorine has an oxidation state of \(0\).
• In sodium chloride (\(\text{NaCl}\)), chlorine is reduced to \(-1\).
• In sodium chlorate(I) (\(\text{NaClO}\)), chlorine is oxidised to \(+1\).
• Application: Sodium chlorate(I) is the active bleaching and disinfectant agent found in standard household bleach solutions.

Key Takeaway: Both the water reaction and the cold dilute alkali reaction are classic disproportionation examples where chlorine's oxidation state splits from \(0\) into \(-1\) and \(+1\).

Chapter Summary & Quick Revision Checklist

• Volatility & Boiling Points: \(\text{F}_2 < \text{Cl}_2 < \text{Br}_2 < \text{I}_2\) (boiling point increases due to stronger van der Waals forces).
• Electronegativity: Decreases down the group due to larger atomic radius and more shielding.
• Oxidising power: \(\text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2\) (chlorine displaces bromide and iodide).
• Reducing power: \(\text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^-\) (iodide reduces conc. \(\text{H}_2\text{SO}_4\) all the way to \(\text{H}_2\text{S}\)).
• Silver Nitrate Test: Acidify with \(\text{HNO}_3\); \(\text{AgCl}\) (white, sol. in dilute \(\text{NH}_3\)), \(\text{AgBr}\) (cream, sol. in conc. \(\text{NH}_3\)), \(\text{AgI}\) (yellow, insoluble).
• Bleach equation: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\) (chlorine goes from \(0 \rightarrow -1\) and \(+1\)).