Halide ions form molecules by losing electrons

Halide ions are the negatively charged ionic forms of the halogens fluorine, chlorine, bromine and iodine. They have a 1- charge and are named by adding ‘-ide’ to the halogen name (e.g. chloride, bromide, iodide).

Halide ions react by losing an electron to form neutral halogen molecules. For example:

X^-^ ➔ 1⁄2X_2_ + e^-^

Where X represents a halogen atom.

This electron loss results in the halide being oxidised, as its oxidation number increases from -1 to 0.

As the halide ion loses an electron, it causes another substance to be reduced - so halide ions act as reducing agents.

Reducing power of halide ions increases down group 7

The reducing power of halide ions refers to their ability to donate electrons to other substances.

This reducing power of halide ions increases down group 7 because:

• Ionic radius increases down group 7 as more electron shells are added.
• The increasing ionic radius leads to the outer electrons being farther from the positive nucleus.
• The outer electrons experience more shieldingfrom inner electron shells.
• The electrostatic attraction between the outer electrons and nucleus gets progressively weaker.
• The increase in ionic radius and shielding outweigh the increase in nuclear charge so it becomes easier for larger halide ions to lose electrons and becomes oxidised.

Therefore, fluoride is the weakest reducing agent and iodide is the strongest.

Reactions of halide ions with sulfuric acid

When halide ions react with concentrated sulfuric acid, hydrogen halide gases are initially produced. Subsequent reactions depend on the relative reducing powers of the hydrogen halides formed.

Fluoride and chloride

Misty white fumes of HF or HCl gas are seen:

• NaF_(s) + H₂SO_4(aq) ➔ NaHSO_4(aq) + HF_(g)

• NaCl_(s) + H₂SO_4(aq) ➔ NaHSO_4(aq) + HCl_(g) These halides have low reducing power, so no further redox reactions occur.

• The oxidation number of sulfur remains at +6.

• The oxidation numbers of F- and Cl- remain at -1.

Bromide

Misty fumes of HBr gas are produced initially:

• NaBr_(s) + H₂SO_4(aq) ➔ NaHSO_4(aq) + HBr_(g) Bromide ions then reduce H₂SO₄ to SO₂. Orange bromine vapour and choking SO_2_ gas are observed:

• 2HBr_(g) + H₂SO_4(aq) ➔ Br_2(g) + SO_2(g) + 2H₂O_(l) Sulfur is reduced to as its oxidation number decreases from +6 in H₂SO₄ to +4 in SO₂.

Bromide is oxidised as its oxidation number increases from -1 in HBr to 0 in Br₂.

Iodide

Misty fumes of HI gas are produced initially:

• NaI_(s) + H₂SO_4(aq) ➔ NaHSO_4(aq) + HI_(g) Iodide ions then reduce H₂SO₄ to SO₂. Violet iodine vapour and choking SO₂ gas are observed:

• 2HI_(g) + H₂SO_4(aq) ➔ I_2(g) + SO_2(g) + 2H₂O_(l) Iodide ions then reduce SO₂ further to produce H₂S gas. Violet iodine vapour and the rotten egg smell of H₂S gas are observed:

• 6HI_(g) + SO_2(g) ➔ H₂S_(g) + 3I_2(s) + 2H₂O_(l) Sulfur is reduced to as its oxidation number decreases from +4 in SO₂ to -2 in H₂S.

Iodide is oxidised as its oxidation number increases from -1 in HI to 0 in I₂.

During the reduction, the oxidation number of sulfur passes through 0 and some yellow, solid sulfur may be seen.

The reactions of chloride, bromide and iodide ions with sulfuric acid is summarised in the table below.

Testing for halide ions

Aqueous silver nitrate (AgNO₃) solution is used to test for the presence of halide ions (Cl^-, Br^-, I^-) in a sample.

The method is:

1. Add dilute nitric acid to the sample to remove any ions like carbonates that may interfere by also forming precipitates with silver nitrate, confounding the results.
2. Add silver nitrate solution; silver ions react with halide ions to form silver halide precipitates: Ag⁺_(aq) + X^-_(aq) ➔ AgX_(s)

Where X = Cl, Br or I

1. Observe precipitate colour to identify which halide ion is present:

Some silver halides have similar colours so to confirm the identity of the halide present, add excess ammonia solution. Silver halides have different solubilities in dilute and concentrated ammonia solution: