Inorganic (Year 1)

The group 7 elements.Note: this is because electronegativity increases going across a period, but the noble gases are not electronegative as they don’t form covalent bonds.

2Mg + TiCl₄ -> 2MgCl₂ + Ti.

It acts as a reducing agent as its oxidation state increases from 0 to +2, meaning electrons are lost.Note: use this explanation for why anything is a reducing agent (reverse for oxidising agents too of course).

It’s called milk of magnesia and treats indigestion.Note: it doesn’t treat constipation, as it neutralises stomach acid.

9.

Note: this is because it is only sparingly soluble, so few OH⁻ ions dissociate:

8-12.Note: I’d say 10. Remember it is more alkaline than Mg(OH)₂ which has a pH of 9 but is still not very alkaline as it is only partially soluble.

Increasing soil pH in agriculture.

It is used for X-rays for medical diagnosis as barium meal.

It is safe to use for this purpose because it is insoluble, meaning it won't release harmful barium ions when ingested.

Note: one mark scheme I saw only accepted barium meal and X-rays, but others have had medical diagnosis I think. That’s why you should just say all 3.

Cl₂ + H₂O <-> HCl + HClONote: remember it is reversible. This is also an example of a disproportionation reaction.

2Cl₂ + 2H₂O -> 4HCl + O₂Note: this is because the HClO decomposes into HCl and O₂ in sunlight.

Cl₂ + 2NaOH -> NaCl + NaOCl + H₂O.

The NaOH is cold, dilute and aqueous:

Sodium chlorate (1).Note: always remember the '(1)'.

Bleach.

ClO⁻ + 2I⁻ + 2H⁺ -> I₂ + H₂O + Cl⁻

This forms a brown solution due to the I⁻ getting oxidised to I₂.

A black precipitate can also be formed if they are oxidised all the way to forming iodate ions IO₃ -, which form the black precipitate NaIO₃.

It is potentially toxic if overdosed.Note: only use this if you specify that it is 'only potentially toxic if overdosed', otherwise they are unlikely to accept it.

It is wasteful as most treated water isn’t used for drinking so doesn’t need to be sterile.

It causes eye irritation for some people.

Some people find the taste unpleasant.

2X⁻ → X₂ + 2e⁻.

Cl⁻ + H₂SO₄ -> HCl (g) + HSO₄⁻

Acid-base only.

Steamy fumes (HCl).

It shows it has little reducing power as the reaction is acid-base, not redox, and it doesn’t react vigorously or produce many products.

Br⁻ + H₂SO₄ -> HBr (g) + HSO₄⁻

2Br⁻ + H₂SO₄ + 2H⁺ -> Br₂ (g) + SO₂ (g) + 2H₂O

The formation of HBr is acid-base.

Br⁻ is oxidised and H₂SO₄ is reduced, so it is also a redox reaction.

Steamy fumes (HBr).

Brown fumes (Br₂).

Orange solution (Br₂).

Colourless gas (SO₂).

It shows it has medium reducing power, as one redox reaction occurs and it reacts more vigorously than chloride ions but less than iodide ions.

I⁻ + H₂SO₄ -> HI (g) + HSO₄⁻

2I⁻ + H₂SO₄ + 2H⁺ -> I₂ (g+s) + SO₂ (g) + 2H₂O

6I⁻ + H₂SO₄ + 6H⁺ -> 3I₂ + S (s) + 4H₂O

8I⁻ + H₂SO₄ + 8H⁺ -> 4I₂ + H₂S (g) + 4H₂O

The formation of HI is acid-base.

The rest of the ionic equations are redox.

Steamy fumes (HI).

Purple fumes (I₂).

Black solid (I₂).

Colourless gas (SO₂).

Yellow solid (S).

Foul/egg smelling gas (H₂S).

It shows it has a great reducing power, as three redox reactions occur and it reacts vigorously to form lots of products.

They act as bases/ proton acceptors.Note: therefore the role of H₂SO₄ is also just an ‘acid’ or ‘proton donor’.

H₂SO₄ + 6H⁺ + 6e⁻ → S + 4H₂ONote: Write H₂SO₄ as a single compound - don’t separate it into its ions (actually this doesn’t matter much). Then, just treat it like any other half-equation question (balance Os, then Hs, then e⁻) - don’t try to form any other products.

Yellow gas.

Green gas.

Orange liquid.

Grey solid that easily forms purple vapour upon heating.

Pale green (basically colourless)

Orange.

Brown.

Colourless.

It is an antiseptic.

Effervescence. This is because H₂ gas is produced.

Say ‘add water’ first (assuming it’s not already in solution).

Add acidified silver nitrate to a solution containing the metal halide. This will form a silver halide precipitate.

The precipitate will be white for Cl⁻, cream for Br⁻, and yellow for I⁻.Note: you can also identify the halide ion present in the precipitate using a different test (details in another note)

Add excess dilute or concentrated ammonia solution.Note: In some mark schemes you have to specify it is an ammonia solution, just ‘ammonia’ is insufficient.

Silver chloride will dissolve when dilute (or concentrated) ammonia is added to form a colourless solution.

Silver bromide will only dissolve when concentrated ammonia is added.

Silver iodide will not dissolve.

An acid is needed to remove other anions so they don’t give a false positive by forming a precipitate.

Nitric acid is used as it contains nitrate ions like the silver nitrate.

The solution remains colourless.Note: this is because silver fluoride is soluble in water, so it doesn’t form the white/ cream/ yellow precipitate found in the rest of group 7.

AgCl + 2NH₃ -> [Ag(NH₃)₂]⁺+ Cl⁻.

By adding an acidified soluble chloride (e.g. sodium chloride).

It will form a white precipitate.Note: this is basically the test for halides but the other way round.

You must acidify it with hydrochloric acid to remove any carbonate ions present, which is important as barium carbonate is a white insoluble solid that would be mistaken for barium sulfate.Note: hydrochloric acid is used as it contains chloride ions like the barium chloride.

A white precipitate forming.Note: always mention this when describing the test for CO₂ just in case.

You should always add NaOH as the metal must be group 1, as group 2 metals are likely to form precipitates with other anions in the solution.Note: you could also use KOH.

A colourless solution is produced (as Sr(OH)₂ and Ba(OH)₂ are soluble) and effervescences occur (because H₂ is produced).

Titration:

1. Take a known volume of the solution.

2. Titrate with an acid/ base of known concentration.

3. Calculate moles of reactant in solution.

Evaporation:

1. Take a known volume of the solution.

2. Evaporate the filtrate to dryness.

3. Weigh the residue.

1. It can cause some of the solid to be poured off and lost.

2. It remains contaminated with soluble impurities.

Instead filter, rinse with cold distilled water, then dry.

To get rid of soluble impurities.

To get rid of the sulfuric acid used to form it.Note: this is an applied version of the general ‘why are crystals washed?’ question.

You should dry it by pressing it between filter paper or in warm air.Note: if heating is used, it must be specified that the heat is gentle.

No - always list the test for every substance.

However, if you don't know the test for the remaining substance but know what it is (via process of elimination), say this because it might get you an extra mark.

It shows a carbonate ion was present in the ionic compound as the gas being formed is CO₂.

A white precipitate will form with MgCl₂ as Mg(OH)₂ is insoluble.

There will be no visible change with BaCl₂ as Ba(OH)₂ is soluble, so forms a colourless solution.