Chemistry (9701) | Group 17

⚗ Comprehensive Study Notes: Group 17 Elements (The Halogens) (AS Level Chemistry 9701) ⚗

Welcome to Group 17! This group contains the highly reactive non-metals known as the Halogens (Fluorine, Chlorine, Bromine, Iodine, and Astatine). Understanding the Halogens is essential because they show wonderfully clear and predictable trends, which are perfect for testing your knowledge of periodicity, bonding, and redox reactions. Let’s dive in!

11.1 Physical Properties of Group 17 Elements

The Halogens all exist as diatomic molecules (\(X_2\)). Down the group (F to I), we see predictable changes in physical properties due to increasing atomic size and mass.

1. Colours and Physical States

As you move down Group 17, the elements become darker and change state from gas to liquid to solid at room temperature and pressure (r.t.p.):

• Fluorine (\(F_2\)): Pale yellow gas
• Chlorine (\(Cl_2\)): Greenish-yellow gas
• Bromine (\(Br_2\)): Reddish-brown liquid (easily vaporises)
• Iodine (\(I_2\)): Grey/Black solid (sublimes to a purple vapour)

2. Trend in Volatility (Melting and Boiling Points)

Volatility (how easily a substance turns into a gas) decreases down the group, meaning the melting and boiling points increase.

Why?

Halogen molecules (\(X_2\)) are simple molecular structures. The forces holding the molecules together are weak intermolecular forces, specifically instantaneous dipole-induced dipole forces (also known as London dispersion forces or van der Waals' forces).

• Down the group, the atoms have more electrons.
• More electrons mean the electron cloud is larger and more easily distorted (more polarizable).
• This results in stronger id-id forces between the molecules.

Key Takeaway: Because stronger forces require more energy to overcome, volatility decreases and boiling/melting points increase from F₂ to I₂.

3. Bond Strength of Halogen Molecules (\(X-X\))

The energy needed to break the covalent bond in the \(X_2\) molecule generally decreases down the group.

Explanation:

• As the atoms get larger (from Cl to I), the bond length increases.
• The overlap between the valence orbitals becomes less effective.
• This results in a weaker covalent bond, requiring less energy to break (lower bond enthalpy).

! Common Mistake Alert !

While the general trend is decreasing, Fluorine (\(F_2\)) is an anomaly. The F–F bond is weaker than the Cl–Cl bond. This is due to the small size of F atoms, causing significant repulsion between the non-bonding electron pairs (lone pairs) in the molecule.

✅ Key Takeaway (11.1): Volatility decreases and boiling points increase due to stronger id-id forces (more electrons = more polarizable). Bond strength generally decreases down the group.

11.2 Chemical Properties: Halogens as Oxidising Agents

Halogens are highly reactive non-metals. They readily gain one electron to achieve a noble gas configuration, forming a halide ion (\(X^-\)). This means they are powerful oxidising agents (they accept electrons).

1. Relative Reactivity and Oxidising Power

The reactivity and the strength as an oxidising agent decreases down the group:

$$F_2 > Cl_2 > Br_2 > I_2$$

Explanation of the Trend:

Oxidising power depends on how readily the atom accepts an electron. This is influenced by the atomic radius and shielding:

• Fluorine is the smallest atom with the least shielding. The nucleus exerts a very strong pull on incoming electrons.
• Iodine is the largest atom with maximum shielding. The nucleus’s attraction to incoming electrons is much weaker.

☞ Analogy: Think of it like gravity. Fluorine is like a black hole—it grabs electrons instantly. Iodine is like the moon—it still pulls, but much less strongly because the electron is far away and shielded by many energy levels.

2. Displacement Reactions (Showing Oxidising Power)

A more reactive halogen (higher up the group) will displace a less reactive halide ion (lower down the group) from its aqueous salt solution.

(a) Chlorine displacing Bromide and Iodide:

Chlorine water is added to colourless potassium bromide solution:

$$\text{Cl}_2\text{(aq)} + 2\text{Br}^-\text{(aq)} \longrightarrow 2\text{Cl}^-\text{(aq)} + \text{Br}_2\text{(aq)}$$ (Solution turns orange/brown due to aqueous bromine.)

Chlorine water is added to colourless potassium iodide solution:

$$\text{Cl}_2\text{(aq)} + 2\text{I}^-\text{(aq)} \longrightarrow 2\text{Cl}^-\text{(aq)} + \text{I}_2\text{(aq)}$$ (Solution turns brown due to aqueous iodine, or if organic solvent is added, purple/pink.)

3. Reactions with Hydrogen (Formation of Hydrogen Halides)

Halogens react with hydrogen gas (\(H_2\)) to form hydrogen halides (HX):

$$X_2 + H_2 \longrightarrow 2HX$$

The reactivity in this reaction decreases down the group, reflecting their oxidising power:

• \(F_2\): Explosive reaction, even in the dark and cold.
• \(Cl_2\): Explosive in sunlight (requires UV light or heat to start).
• \(Br_2\): Requires heating and a catalyst.
• \(I_2\): Requires continuous heating and a catalyst, and the reaction is reversible.

✅ Key Takeaway (11.2.1-2): Halogen reactivity decreases down the group. They are oxidising agents, and a more reactive halogen displaces a less reactive halide ion.

11.2 (Continued) Thermal Stability of Hydrogen Halides (HX)

The hydrogen halides (HF, HCl, HBr, HI) are all covalent gases. Their thermal stability refers to how resistant they are to breaking down upon heating:

$$2HX \longrightarrow H_2 + X_2$$

Thermal stability decreases down the group:

$$HF > HCl > HBr > HI$$

Explanation:

This trend is explained directly by the bond strength (bond energy) of the H-X bond:

• As the halogen atom size increases (F to I), the covalent bond length in H-X increases.
• The electron pair is held further from the halogen nucleus.
• This weakens the bond, making it easier to break (decompose) upon heating.

Therefore, HI is the least stable and decomposes most easily when heated.

11.3 Some Reactions of the Halide Ions (X⁻)

Halide ions (\(Cl^-\), \(Br^-\), \(I^-\)) have already gained an electron. To react further, they must lose an electron, meaning they act as reducing agents.

1. Relative Reducing Power

The ability of the halide ion to lose an electron (reducing power) increases down the group:

$$\text{I}^- > \text{Br}^- > \text{Cl}^-$$

Explanation:

• Iodide ion (\(I^-\)) is the largest ion. Its valence electrons are held furthest from the nucleus and are most heavily shielded.
• These outer electrons are therefore easiest to lose.

2. Reaction with Concentrated Sulfuric Acid (\(H_2SO_4\))

Concentrated \(H_2SO_4\) is a powerful oxidising agent. When heated with solid halide salts, the subsequent reaction depends entirely on the reducing power of the halide ion.

(a) Chloride ions (\(Cl^-\)) – Weakest Reducer

Chloride ions are too weak to reduce \(H_2SO_4\). Only an acid-base reaction occurs, displacing the volatile hydrogen halide gas, HCl:

$$\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(conc)} \longrightarrow \text{NaHSO}_4\text{(s)} + \text{HCl(g)}$$ Observation: Steamy white fumes of HCl gas. No redox.

(b) Bromide ions (\(Br^-\)) – Moderate Reducer

Bromide ions are strong enough to reduce some of the \(H_2SO_4\) to sulfur dioxide (\(SO_2\)):

Step 1: Acid displacement (similar to chloride) $$\text{NaBr(s)} + \text{H}_2\text{SO}_4\text{(conc)} \longrightarrow \text{NaHSO}_4\text{(s)} + \text{HBr(g)}$$ Step 2: Redox reaction (Reduction of \(H_2SO_4\)) $$\text{2HBr(g)} + \text{H}_2\text{SO}_4\text{(conc)} \longrightarrow \text{Br}_2\text{(g)} + \text{SO}_2\text{(g)} + \text{2H}_2\text{O(l)}$$ Observation: Steamy white fumes (HBr) plus red-brown fumes (\(Br_2\)) and pungent gas (\(SO_2\)).

(c) Iodide ions (\(I^-\)) – Strongest Reducer

Iodide ions are strong reducing agents and reduce the \(H_2SO_4\) right down to Sulfur (\(S\)) or Hydrogen Sulfide (\(H_2S\)), passing through \(SO_2\):

Step 1: Acid displacement $$\text{NaI(s)} + \text{H}_2\text{SO}_4\text{(conc)} \longrightarrow \text{NaHSO}_4\text{(s)} + \text{HI(g)}$$ Step 2: Extensive Redox (HI is immediately oxidized) $$\text{8HI(g)} + \text{H}_2\text{SO}_4\text{(conc)} \longrightarrow 4\text{I}_2\text{(g)} + \text{H}_2\text{S(g)} + 4\text{H}_2\text{O(l)}$$ Observation: Purple vapour (\(I_2\)) and black solid residue, plus smell of rotten eggs (\(H_2S\)) and yellow precipitate (S).

3. Precipitation with Aqueous Silver Ions (\(Ag^+\))

This is the standard laboratory test for halide ions. Halides form insoluble precipitates (salts) with aqueous silver nitrate (\(AgNO_3\)).

$$\text{Ag}^+\text{(aq)} + X^-\text{(aq)} \longrightarrow \text{AgX(s)}$$ (The reagent is acidified silver nitrate, usually with dilute \(HNO_3\), to prevent precipitation of other ions like carbonates.)

The precipitates are then tested for solubility in aqueous ammonia (\(NH_3\)):

| Halide Ion | Precipitate Colour (AgX) | Solubility in NH₃(aq) | | :--- | :--- | :--- | | \(Cl^-\) | White (\(\text{AgCl}\)) | Soluble in dilute \(\text{NH}_3\text{(aq)}\) | | \(Br^-\) | Cream (\(\text{AgBr}\)) | Partially soluble in concentrated \(\text{NH}_3\text{(aq)}\) | | \(I^-\) | Pale Yellow (\(\text{AgI}\)) | Insoluble in \(\text{NH}_3\text{(aq)}\) |

☞ Did You Know? The solubility differences are explained because silver forms a complex ion, \([\text{Ag}(\text{NH}_3)_2]^+\), when reacted with ammonia. The AgI lattice energy is too high for this complexation to fully dissolve the solid.

✅ Key Takeaway (11.3): Halide reducing power increases down the group. I⁻ is strong enough to reduce conc. H₂SO₄ extensively, while Cl⁻ is not.

11.4 The Reactions of Chlorine: Disproportionation

Disproportionation is a specific type of redox reaction where the same element is simultaneously oxidized and reduced.

1. Chlorine with Cold Aqueous Sodium Hydroxide (NaOH)

Chlorine reacts with cold, dilute aqueous NaOH to form sodium chloride (NaCl) and sodium chlorate(I) (NaClO):

$$\text{Cl}_2 + 2\text{NaOH} \longrightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}$$

Interpretation by Oxidation Number:

• Cl in \(\text{Cl}_2\) starts at oxidation state 0.
• In \(\text{NaCl}\), Cl is -1 (Reduction).
• In \(\text{NaClO}\), Cl is +1 (Oxidation).

This reaction is used to produce household bleach (sodium chlorate(I)).

2. Chlorine with Hot Aqueous Sodium Hydroxide (NaOH)

If the reaction mixture is hot and concentrated, a different disproportionation occurs, forming sodium chloride (NaCl) and sodium chlorate(V) (\(\text{NaClO}_3\)):

$$3\text{Cl}_2 + 6\text{NaOH} \longrightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}$$

Interpretation by Oxidation Number:

• Cl in \(\text{Cl}_2\) starts at oxidation state 0.
• In \(\text{NaCl}\), Cl is -1 (Reduction).
• In \(\text{NaClO}_3\), Cl is +5 (Oxidation).

3. Chlorine in Water Purification

Chlorine is added to drinking water to kill bacteria, making it safe for consumption. This process relies on a disproportionation reaction with water:

$$\text{Cl}_2\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{HCl(aq)} + \text{HOCl(aq)}$$ (Chlorine is reduced from 0 to -1 in HCl, and oxidized from 0 to +1 in HOCl.)

• The active species that kills bacteria is Hypochlorous acid (\(\text{HOCl}\)), which is a powerful oxidising agent.
• In alkaline conditions, \(\text{HOCl}\) dissociates to form the Hypochlorite ion (\(\text{ClO}^-\)), which is also an active disinfectant.

HOCl and \(\text{ClO}^-\) kill bacteria by oxidising essential compounds within the microbial cells.

✅ Key Takeaway (11.4): Disproportionation is key! Chlorine is both reduced and oxidized. Temperature and concentration dictate the product formed (NaClO or \(\text{NaClO}_3\)).