Hello Future Chemists! Understanding Halide Ion Reactions (Syllabus 11.3)

Welcome to a fascinating corner of Inorganic Chemistry! This chapter focuses on the powerful yet predictable reactions of the halide ions: Chloride (\(\text{Cl}^-\)), Bromide (\(\text{Br}^-\)), and Iodide (\(\text{I}^-\)).

Why is this important? Learning these reactions is crucial because it demonstrates key periodic trends, specifically how reducing power changes down Group 17. It also covers the classic identification tests you’ll perform in the lab using silver nitrate and ammonia!

Don’t worry if redox reactions seem challenging; we will break down the chemistry into simple, predictable steps.

Part 1: Halide Ions as Reducing Agents

The Concept of Reducing Power

A reducing agent is a substance that causes reduction by itself being oxidised (losing electrons). Halide ions (\(\text{X}^-\)) are anions, meaning they have gained an electron to achieve a stable noble gas configuration. To act as a reducing agent, they must lose this extra electron, forming the neutral halogen element (\(\text{X}_2\)).

Trend in Reducing Power Down Group 17

The ability of the halide ions to act as reducing agents increases as you move down the group (from Cl to I).

  • Weakest: \(\text{Cl}^-\) (Chloride)
  • Medium: \(\text{Br}^-\) (Bromide)
  • Strongest: \(\text{I}^-\) (Iodide)
Explanation of the Trend

This trend is explained by the size of the ion:

  1. Going down the group, the halide ions get larger (\(\text{I}^-\) is the largest).
  2. The outer valence electron is therefore further away from the positively charged nucleus.
  3. The attraction between the nucleus and the outer electron is weaker (due to increased atomic radius and shielding).
  4. It is therefore easier to remove that electron (i.e., easier to oxidise the ion).

Key Takeaway: The bigger the halide ion, the weaker the hold on its outer electrons, and the better it is at donating them (acting as a reducing agent).

Testing Reducing Power with Concentrated Sulfuric Acid (\(\text{H}_2\text{SO}_4\))

Concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)) is a good acid (it donates \(\text{H}^+\)) and also a reasonably strong oxidising agent. When it reacts with a solid halide salt (e.g., NaCl, NaBr, NaI), the outcome depends entirely on the reducing power of the halide ion.

Part 2: Reactions of Halide Ions with Concentrated Sulfuric Acid (11.3.2b)

1. Chloride Ions (\(\text{Cl}^-\)) - Only Acid-Base Reaction

Chloride ions are the weakest reducing agents among the halides and cannot reduce \(\text{H}_2\text{SO}_4\). Therefore, only a simple acid-base reaction occurs where the concentrated acid displaces the more volatile hydrogen halide gas, \(\text{HCl}\).

Step 1: Reaction (Displacement)

Equation:
$$ \text{NaCl}(\text{s}) + \text{H}_2\text{SO}_4(\text{l}) \rightarrow \text{NaHSO}_4(\text{s}) + \text{HCl}(\text{g}) $$

Observation:

Steamy white fumes of hydrogen chloride gas (\(\text{HCl}\)).

Why Fumes are Steamy: \(\text{HCl}\) gas reacts instantly with water vapour in the air, forming tiny droplets of hydrochloric acid, which appear as white fumes.

2. Bromide Ions (\(\text{Br}^-\)) - Acid-Base and Reduction

Bromide ions are stronger reducing agents than chlorides. The reaction occurs in two stages: first, acid-base displacement, then subsequent oxidation of the hydrogen halide.

Step 1: Acid-Base Reaction (Produces HBr)

$$ \text{NaBr}(\text{s}) + \text{H}_2\text{SO}_4(\text{l}) \rightarrow \text{NaHSO}_4(\text{s}) + \text{HBr}(\text{g}) $$

Step 2: Reduction Reaction

The \(\text{HBr}\) produced is a strong enough reducing agent to reduce some of the hot concentrated \(\text{H}_2\text{SO}_4\). In this process, \(\text{H}_2\text{SO}_4\) (Sulfur oxidation state +6) is reduced to sulfur dioxide, \(\text{SO}_2\) (Sulfur oxidation state +4), and bromide is oxidised to bromine, \(\text{Br}_2\).

Balanced Equation (required by syllabus):
$$ \text{2HBr}(\text{g}) + \text{H}_2\text{SO}_4(\text{l}) \rightarrow \text{Br}_2(\text{g}) + \text{SO}_2(\text{g}) + \text{2H}_2\text{O}(\text{l}) $$

Observations:

Steamy white fumes (HBr) followed by dense orange/brown fumes of bromine vapor (\(\text{Br}_2\)).

3. Iodide Ions (\(\text{I}^-\)) - Acid-Base and Strong Reduction

Iodide ions are the strongest reducing agents. They reduce the \(\text{H}_2\text{SO}_4\) much more effectively, often reducing the sulfur all the way down to Sulfur dioxide (\(\text{SO}_2\)), elemental Sulfur (S), or even hydrogen sulfide (\(\text{H}_2\text{S}\)).

Step 1: Acid-Base Reaction (Produces HI)

$$ \text{NaI}(\text{s}) + \text{H}_2\text{SO}_4(\text{l}) \rightarrow \text{NaHSO}_4(\text{s}) + \text{HI}(\text{g}) $$

Step 2: Reduction Reaction

The \(\text{HI}\) reduces \(\text{H}_2\text{SO}_4\) readily. The syllabus specifically requires the overall reaction, typically focusing on the initial reduction to \(\text{SO}_2\).

Balanced Equation (Reduction to \(\text{SO}_2\)):
$$ \text{2HI}(\text{g}) + \text{H}_2\text{SO}_4(\text{l}) \rightarrow \text{I}_2(\text{s}) + \text{SO}_2(\text{g}) + \text{2H}_2\text{O}(\text{l}) $$

Observations:

Steamy white fumes (HI) followed by:

  • Purple fumes (iodine vapor, \(\text{I}_2\)) and/or black solid (sublimed solid iodine).
  • A gas with a rotten egg smell (\(\text{H}_2\text{S}\)) or suffocating smell (\(\text{SO}_2\)).

Quick Review: Halides with Conc. H₂SO₄

Cl⁻: Only \(\text{HCl}\) gas (Steamy fumes). No reduction.
Br⁻: \(\text{HBr}\) gas + \(\text{Br}_2\) liquid/gas (Orange/Brown fumes). Reduction to \(\text{SO}_2\).
I⁻: \(\text{HI}\) gas + \(\text{I}_2\) solid/gas (Purple fumes/Black solid). Reduction to \(\text{SO}_2\), S, or \(\text{H}_2\text{S}\).

Part 3: Testing for Halide Ions with Aqueous Silver Ions (11.3.2a)

This is a standard qualitative analysis technique used to identify which halide ion is present in a solution. It involves two steps: precipitation using silver nitrate, and then testing the solubility of the precipitate in aqueous ammonia.

Step 1: Precipitation Reaction with Aqueous Silver Nitrate (\(\text{AgNO}_3\))

The reagent used is aqueous silver nitrate (\(\text{AgNO}_3\)), which is often acidified with dilute nitric acid (\(\text{HNO}_3\)). This acidification prevents interference from other anions, like carbonate, which might also form precipitates with \(\text{Ag}^+\).

General Ionic Equation:
$$ \text{Ag}^+(\text{aq}) + \text{X}^-(\text{aq}) \rightarrow \text{AgX}(\text{s}) $$

The precipitates formed are silver halides, \(\text{AgX}(\text{s})\):

  1. Chloride (\(\text{Cl}^-\)): Forms Silver Chloride (\(\text{AgCl}\)) – a White precipitate.
  2. Bromide (\(\text{Br}^-\)): Forms Silver Bromide (\(\text{AgBr}\)) – a Cream precipitate.
  3. Iodide (\(\text{I}^-\)): Forms Silver Iodide (\(\text{AgI}\)) – a Pale Yellow precipitate.
Memory Aid: W-C-PY

To remember the colours down the group (Cl, Br, I): White, Cream, Pale Yellow.

Step 2: Confirmation Test using Aqueous Ammonia (\(\text{NH}_3(\text{aq})\))

Since the colours (especially white and cream) can be difficult to distinguish accurately, we use aqueous ammonia to test the solubility of the precipitates. Solubility depends on the stability of the complex ion formed when silver is dissolved by ammonia.

(Note: Although the reaction mechanism involves the formation of the diamminesilver(I) complex, \([\text{Ag}(\text{NH}_3)_2]^+\), the syllabus states that the formula and its formation are not required. You only need to state the solubility observed.)

Halide Ion Precipitate (\(\text{AgX}\)) Solubility in Aqueous Ammonia
Chloride (\(\text{Cl}^-\)) White Soluble in dilute \(\text{NH}_3(\text{aq})\)
Bromide (\(\text{Br}^-\)) Cream Partially Soluble (only dissolves in concentrated \(\text{NH}_3(\text{aq})\))
Iodide (\(\text{I}^-\)) Pale Yellow Insoluble in concentrated \(\text{NH}_3(\text{aq})\)

The difference in solubility helps you definitively identify the halide, even if the colours were ambiguous initially.

Key Takeaway Summary: Halide Ions

Reducing Power (with H₂SO₄):

\(\text{Cl}^-\) (Weakest) → Only \(\text{HCl}\) formed (No reduction).
\(\text{Br}^-\) → \(\text{HBr}\) formed and reduced to \(\text{Br}_2\) (Orange/Brown).
\(\text{I}^-\) (Strongest) → \(\text{HI}\) formed and heavily reduced to \(\text{I}_2\) and sulfur compounds (Purple/Black).

Identification (with \(\text{AgNO}_3\) then \(\text{NH}_3\)):

\(\text{AgCl}\) (White) → Soluble in dilute \(\text{NH}_3\).
\(\text{AgBr}\) (Cream) → Soluble in concentrated \(\text{NH}_3\).
\(\text{AgI}\) (Pale Yellow) → Insoluble in concentrated \(\text{NH}_3\).