Trends in properties (3.2.3.1)

What are the colours and states at room temperature of the group 7 elements?

F_2(g)- pale yellow

Cl_2(g) - yellow-green

Br_2(l)- dark red

I_2(s)- shiny grey-black - purple gas - brown when dissolved in H₂O - purple in organic solvents

What is the trend in melting point in group 7?

As you go down the group the melting point increases as the molecules have more electrons so they have stronger London forces which require more energy to break.

What is the trend in electronegativity in group 7?

The electronegativity decreases down the group as the atomic radius and shielding increase so there is a weaker attraction between the molecules and bonding pair.

What is the trend in oxidising ability of the halogens?

The halogens are good oxidising agents as they readily accept electrons

The oxidising power of the halogens decreases down the group as their attraction to electrons decreases due to the increase in atomic radius and shielding

This means that the halogens can displace the halide ion of a halogen below it in the group

What are the observations in the displacement reactions

Cl₂ + 2Br^- → 2Cl^- + Br₂

A orange solution of Br₂ is formed

Cl₂ + 2I^- → I₂ + 2Cl^-

A brown solution of I₂ is formed

Br₂ + 2I^- → I₂ + 2Br^-

An orange solution of I₂ is formed

What is the trend in reducing ability of the halides?

The reducing ability of the halides increases down the group as electrons are easier to lose from larger ions due to larger atomic radius and shielding

What are the reactions and observations of NaF, NaCl, NaBr, HBr, NaI and HI with H₂SO₄?

NaF_(s) + H₂SO_4(aq) → NaHSO_4(aq) + HF_(g) - not a redox reaction - white fumes of HF observed

NaCl_(s) + H₂SO_4(aq) → NaHSO_4(aq) + HCl_(g) - not a redox reaction - white fumes of HCl observed

NaBr_(s) + H₂SO_4(aq) → NaHSO_4(aq) + HBr_(g)

2HBr_(g) + H₂SO_4(aq) →Br_2(g) + 2H₂O_(l) + SO_2(g) - red-brown gas of Br₂

Iodine:

NaI_(s) + H₂SO_4(aq) → NaHSO_4(aq) + HI_(g) - Misty fumes of HI

2HI_(g) + H₂SO_4(aq) → I_2(s) + 2H₂O_(l) + SO_2(g) - purple vapour of I₂

6HI_(g) + H₂SO_4(aq) → S_(s) + 3I_2(s) + 4H₂O_(l) - yellow solid from S

H_2​SO₄​_(aq) + 8HI_(g) → 4I_2(s) + H₂S_(g) + 4H₂O_(l) - strong smell of bad eggs from H₂S

To remember the iodine reactions just memorise what sulfur turns into as half-equations:

SO₄^2- → SO₂

SO₄^2- → S
SO₄^2- → H₂S

I^- → I₂

The greater the reducing power, the longer the reaction as the halide is powerful enough to reduce more species

How do you test for halide ions?

• Add acidified silver nitrate.

  • It is acidified to prevent any false positives from CO₃^- ions. The acid can’t be HCl as it contains Cl^- ions

AgCl - white precipitate

AgBr - cream precipitate

AgI - yellow precipitate

As the colours are quite similar, to distinguish further dilute ammonia is added

If the precipitate dissolves the halide ion was Cl^-

No change is seen with the other two ions

To distinguish between Br^- and I^- concentrated ammonia is added

If the precipitate dissolves the halide ion was Br^-

No change is seen in I^-