Chemistry (9620) | Group 7(17), the halogens

Chemistry 9620: Inorganic Chemistry Study Notes

Chapter 3.2.3: Group 7(17), The Halogens (International AS)

Welcome to Group 7! This group is one of the most exciting sections of Inorganic Chemistry. The elements here—Fluorine (F), Chlorine (Cl), Bromine (Br), and Iodine (I)—are known as the Halogens (meaning salt-former). They are highly reactive non-metals, and understanding their trends is crucial for success in this module!

In this chapter, we will explore why their properties change so dramatically as you move down the group, focusing on physical trends, redox chemistry, and how we identify them in the lab.

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1. General Properties and Physical Trends

The halogens exist as diatomic molecules, $X_2$ (e.g., $F_2$, $Cl_2$, $Br_2$, $I_2$). They all have 7 valence electrons, meaning they typically need to gain just one electron to achieve a stable noble gas configuration. This explains why they are potent oxidising agents.

1.1 Trend in Electronegativity (EN)

Electronegativity is the power of an atom to attract the pair of electrons in a covalent bond.

• Trend: Electronegativity decreases down Group 7 (Fluorine is the most electronegative element on the Periodic Table!).
• Explanation: As you go down the group:
  1. The number of electron shells increases (larger atomic radius).
  2. The outer electrons are further away from the positively charged nucleus.
  3. More inner shells means greater shielding.

  Result: The nucleus's attraction for bonding electrons is weaker, so the ability to attract electrons decreases.

1.2 Trend in Boiling Point (BP)

The state of the halogens at standard temperature and pressure reflects their boiling points:

• $F_2$ and $Cl_2$ are gases.
• $Br_2$ is a liquid.
• $I_2$ is a solid.

Trend: Boiling point increases down Group 7.

Explanation (Structure and Bonding):

1. Halogens exist as simple molecular substances ($X_2$).
2. These molecules are non-polar, so the only intermolecular forces acting between them are induced dipole-dipole forces (also called London forces or van der Waals forces).
3. As the atoms get larger going down the group, the molecules have:
   • More electrons.
   • A larger surface area.
4. This results in stronger London forces, requiring more energy to break them apart. Therefore, boiling point increases from $F_2$ to $I_2$.

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2. Redox Trends: Oxidising and Reducing Ability

Understanding redox chemistry is where the halogens really shine!

2.1 Halogens as Oxidising Agents

Halogens ($X_2$) are oxidising agents because they are easily reduced (they gain electrons):

$$X_2 + 2e^- \rightarrow 2X^-$$

• Trend in Oxidising Ability: Decreases down the group ($F_2 > Cl_2 > Br_2 > I_2$).
• Explanation: Since electronegativity decreases down the group, the tendency to attract and gain electrons decreases, making them poorer oxidising agents.

Displacement Reactions (Key Practical Test)

A more powerful oxidising halogen (higher up) will displace a halide ion (lower down) from its aqueous salt solution.

Analogy: Think of it as a competition for electrons. The more reactive halogen wins the electrons and becomes the halide ion.

Example 1: Chlorine displacing Bromine

If you bubble chlorine gas through a solution containing bromide ions ($KBr$):

$$Cl_2(aq) + 2Br^-(aq) \rightarrow 2Cl^-(aq) + Br_2(aq)$$

Chlorine (the better oxidising agent) steals the electrons from the bromide ions. The colour changes from colourless to orange/yellow-brown (due to the formation of aqueous $Br_2$).

Example 2: Bromine displacing Iodine

$$Br_2(aq) + 2I^-(aq) \rightarrow 2Br^-(aq) + I_2(aq)$$

Bromine displaces iodide. The colour changes from colourless to brown/purple (due to the formation of aqueous $I_2$).

Did you know? We often add an organic solvent (like hexane) to these displacement mixtures. The resulting halogen colour is much clearer in the organic layer: $Cl_2$ is pale green, $Br_2$ is orange/brown, and $I_2$ is violet/purple.

2.2 Halide Ions as Reducing Agents

Halide ions ($X^-$) are reducing agents because they are easily oxidised (they lose electrons):

$$2X^- \rightarrow X_2 + 2e^-$$

• Trend in Reducing Ability: Increases down the group ($I^- > Br^- > Cl^- > F^-$).
• Explanation: As the ion gets larger ($I^-$ is huge compared to $F^-$), the outer electrons are further from the nucleus and are shielded more effectively. This means they are held less tightly and are much easier to lose (oxidised).

Reaction of Solid Sodium Halides with Concentrated Sulfuric Acid ($H_2SO_4$)

Concentrated $H_2SO_4$ is both a strong acid and a moderate oxidising agent. The reaction proceeds in two stages:

Stage 1: Acid-Base Reaction (Occurs for all halides)

$$\text{NaX}(s) + H_2SO_4(l) \rightarrow NaHSO_4(s) + HX(g)$$

This forms the corresponding hydrogen halide gas ($HCl$, $HBr$, $HI$). This step is reversible.

Stage 2: Redox Reaction (Only for stronger reducing agents, $Br^-$ and $I^-$)

1. Chloride ($Cl^-$) and Fluoride ($F^-$): $Cl^-$ and $F^-$ are very weak reducing agents. Stage 2 does not occur. Only steamy fumes of $HCl$ or $HF$ are observed.
2. Bromide ($Br^-$): Bromide is a strong enough reducing agent to reduce the $H_2SO_4$ to $SO_2$ gas.

   $$2HBr + H_2SO_4 \rightarrow Br_2 + SO_2 + 2H_2O$$ You observe steamy fumes ($HBr$) and red/brown fumes ($Br_2$) and the smell of sulfur dioxide ($SO_2$).

3. Iodide ($I^-$): Iodide is the strongest reducing agent in the group. It is oxidised extremely easily, reducing $H_2SO_4$ fully, leading to multiple sulfur products (oxidation state changes from +6 down to -2).

   The overall reaction is complex, but key products are: Purple vapour ($I_2$), $SO_2$, elemental sulfur ($S$), and hydrogen sulfide ($H_2S$, smells of rotten eggs).

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3. Identifying Halide Ions: The Silver Nitrate Test

We use acidified silver nitrate solution, $AgNO_3(aq)$, to test for the presence of halide ions ($Cl^-$, $Br^-$, $I^-$) in solution. A precipitate of the silver halide ($AgX$) is formed:

$$\text{Halide test: } Ag^+(aq) + X^-(aq) \rightarrow AgX(s)$$

3.1 Key Steps and Observations

1. Acidification: The $AgNO_3$ solution must be acidified, usually with dilute nitric acid ($HNO_3$).

• Why? To prevent precipitation of other ions that might be present (like $OH^-$ or $CO_3^{2-}$), which would interfere with the result. For instance, $Ag_2CO_3$ or $AgOH$ would form white precipitates, leading to a false positive for chloride ions.

2. Precipitation Colours:

• Chloride ($Cl^-$): Forms White precipitate ($AgCl$).
• Bromide ($Br^-$): Forms Cream precipitate ($AgBr$).
• Iodide ($I^-$): Forms Yellow precipitate ($AgI$).

Don't worry if distinguishing between white and cream is tricky! The next step helps clarify.

3. Solubility in Aqueous Ammonia ($NH_3$):

The solubility of the silver halides decreases down the group, reflecting the increasing strength of the ionic bonding and lattice energy in $AgX$.

• $AgCl$: Soluble in dilute $NH_3(aq)$.
• $AgBr$: Soluble only in concentrated $NH_3(aq)$.
• $AgI$: Insoluble in $NH_3(aq)$ (even concentrated).

Memory Aid: C.C.I. (Chlorine - soluble in Concentrated, Iodine - Insoluble). This isn't quite right as Chlorine dissolves in dilute, but it reminds you that Chloride is the most soluble and Iodide is the least!

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4. Uses of Chlorine and Chlorate(I)

Chlorine is produced industrially by the electrolysis of concentrated brine (sodium chloride solution). It has vital uses, particularly in sterilisation and sanitation.

4.1 Reaction of Chlorine with Water

When chlorine is added to water, two key reactions can occur, depending on the conditions:

A) Disinfecting Water (Essential for water treatment)

In the absence of strong sunlight, chlorine undergoes a disproportionation reaction (where it is both oxidised and reduced) with water:

$$Cl_2(aq) + H_2O(l) \rightleftharpoons HClO(aq) + HCl(aq)$$

The product, Chloric(I) acid ($HClO$), is the active agent. It is a powerful oxidising agent and kills bacteria by disrupting their cell membranes.

Societal Context: The use of chlorine in water treatment has historically saved millions of lives by eliminating waterborne diseases (like cholera). Despite the small risk of chlorine forming potentially toxic chlorinated organic compounds, the benefits to public health far outweigh its toxic effects.

B) Reaction in Sunlight

If chlorine is left exposed to strong sunlight, the chloric(I) acid decomposes, releasing oxygen:

$$2Cl_2(aq) + 2H_2O(l) \rightarrow 4HCl(aq) + O_2(g)$$

This means chlorinated water must be stored carefully to maintain its effectiveness.

4.2 Reaction of Chlorine with Cold, Dilute Aqueous Sodium Hydroxide (NaOH)

This is another important disproportionation reaction used to manufacture bleach (sodium chlorate(I), $NaClO$):

$$Cl_2(aq) + 2NaOH(aq) \rightarrow NaCl(aq) + NaClO(aq) + H_2O(l)$$

The resulting solution contains $NaClO$, which is widely used as household bleach and for large-scale disinfection. It is only stable in cold conditions.