trends and reactions of halogens

Group 7 elements (halogens): Fluorine, chlorine, bromine, iodine, and astatine - non-metals with similar chemical properties due to having seven outer electrons. What is a displacement reaction?

When a more reactive halogen displaces a less reactive halide (negatively charged ions formed when halogens gain an electron) from its salt.

Explain the trend of boiling points down the group.

Boiling points increase down the group due to greater van der waals forces (larger molecules with more electrons).

Meaning that more energy is required to separate molecules.

Name the halogen, its standard state, its colour and its colour in aqueous solution.

Fluorine - gas, pale yellow, n/a

Chlorine - gas, pale green, pale green

Bromine - liquid, red-brown, orange

Iodine - solid, grey, brown

Solid iodine easily vaporises to iodine gas which is a purple vapour.

Explain the electronegativity of halogens.

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

Electronegativity decreases down group 7.

As atomic radius increases and shielding increases, the attraction between the nucleus and the shared electrons decreases.

Fluorine is the most electronegative element on the period table.

Explain the oxidising power of halogens.

Oxidising ability decreases down the group as the halogens are less likely to be reduced (gain electrons) themselves.

Fluorine is the strongest oxidising agent.

Iodine is the weakest.

Atomic radius and shielding increases down the group meaning that it is harder to attract electrons.

Give examples of displacement reactions.

The oxidising ability of halogens can be demonstrated through displacement reactions. A mor reactive (more strongly oxidising) halogen displace a less reactive halide from its compound.

Cl2 + 2NaBr — 2NaCl + Br2

A colour change is observed if a reaction takes place as the halogen in solution determines its colour - give the examples of the colour changes and equations that occur.

Reaction:

Cl2 + 2Br- — 2Cl- + Br2

Observation:

Pale green (cl2) to orange solution (Br2)

Reaction:

Cl2 + 2I- — 2Cl- + I2

Observation:

Pale green to brown solution

Reaction:

Br2 + 2I- — 2Br- + I2

Observation:

Orange solution to brown solution

I2 + Cl- — no reaction

Observation:

Solution stays brown (I2 is less oxidising than Cl2)

In these examples, the more reactive halogen displaces the less reactive halide and is reduced itself (gains electrons, and forms a - ion)

What are halide ions and what is a reducing agent?

Halide ions - anions formed when halogens gain an electron (Cl-, Br-)

Reducing agent - a species that donates electrons and is itself oxidised.

Explain the reducing power of the halides.

The reducing power of the halides increases down the group.

The ionic radius and shielding increases, meaning that there is a weaker attraction to the outer electron and it is therefore easier to donate electrons.

The trend in reducing ability can be demonstrated experimentally through action of the solid sodium halides with concentrated sulfuric acid. How. Can we tell that a halide ion has a stronger reducing ability in these experiments.

The more the oxidation state of the S in H2SO4 is decreased, the stronger the reducing ability of the halide ion must be.

Explain the reaction of solid sodium chloride with H2SO4.

H2SO4 + NaCl - NaHSO4 + HCl

The chloride ion is a very weak reducing agent. The oxidising state of the sulfur has not changed. This is an acid based reaction where the H2SO4 acts as an acid (H+ donor) and the Cl- acts as a base (H+ acceptor).

Steamy fumes of HCl are observed.

Explain the reaction between solid sodium bromide and H2SO4.

H2SO4 + NaBr — NaHSO4 + HBr

The first reaction is a simple acid-base reaction.

As Br- is a stronger reducing agent, the HBr reacts with H2SO4 further to produce sulfur dioxide SO2.

2Br- — Br2 + 2e-

H2SO4 + 2H+ + 2e- — SO2 + 2H2O

Overall:

2H+ + 2Br- + H2SO4 — Br2 + SO2 + 2H2O

The S has been reduced from +6 in H2SO4 to +4 in SO2.

Steamy fumes of HBr, the orange/brown vapour Br2, and a chocking gas SO2 are all observed.

Explain the reactions between solid sodium iodide and H2SO4.

H2SO4 + NaI — NaHSO4 + HI

The first reaction is a simple acid-base reaction.

As I- is a stronger reducing agent, the HI reacts with H2SO4 further to produce sulfur dioxide, SO2 and elemental sulfur, S, and hydrogen sulfide, H2S.

H2SO4 + 6H+ + 6I- — 3I2 + S + 4H2O - yellow solid formed

H2SO4 +8H+ + 8I- — 4I2 + H2S + 4H2O - rotten egg smelling gas

Iodide ions can reduce the S in H2SO4 from +6 down to -2 in H2S demonstrating its position as the strongest reducing agent.

State the sulfur containing product and its observation.

SO2 - choking gas

S - yellow solid sodium halides

H2S - rotten egg smelling gas

State the halide, the sulfur containing product it can form and then the oxidation state of S in the product.

Cl- = NaH2SO4, +6

Br- = NaH2SO4, +6, SO2, +4

I- = NaH2SO4, +6, SO2, +4, S, 0, H2S, -2

What observations are seen with HI, HBr, Br2, HCl, I2

HI - steamy fumes

HBr - steamy fumes

Br2 - brown gas

HCl - steamy fumes

I2 - black solid/purple vapour