Inorganic Chemistry

• Oxidation is losing
• Reduction is gaining

Oxidation

Reduction

Redox reaction

Accepts electrons and gets reduced

Donates electrons and gets oxidised.

Ions

0

0

The same as it's charge. (+1 for for Na example)

the ion charge.

0

• -2
• -1
• +2
• +1

• +1
• -1

Oxidation number

Increase

Decrease

Disproportionation.

• 1 - Each element down group 2 has an extra electron shell compared to the one above.
• 2 - The extra inner shells shield the electrons from the attraction of the nucleus.
• 3 - Also, the extra shell means that the other electrons are further away from the nucleus, which greatly reduces the nucleus's attraction.

• 0
• +2
• M²⁺
• 2 electrons in their outer shell.

• Metal hydride
• Hydrogen
• Decrease

Characteristic flame colours.

White solid chlorides.

Bases.

• Water
• Metal hydroxides
• Strongly alkaline

MgO is an exception - it only reacts slowly and the hydroxide isn't very soluble.

• Strongly
• Soluble.

Neutralise.

MO _(s) + H₂0 _(l) → M(OH)_{2 (aq)}

MO _(s) + 2HCl _(aq) → MCl_{2 (aq)} + H₂0 _(l)

• _+H20
• M(OH)_{2 (s)} → M²⁺ _(aq) + 2OH^- _(aq)

M(OH)_{2 (aq)} + HCl _(aq) → MCl_{2 (aq)} + 2H₂0 _(l)

Compound Anion.

Increase

Doubly

• Group 2 element Hydroxide (OH^-) Sulfate (SO₄^2-)
• Mg ↓ ↑
• Ca ↓ ↑
• Sr ↓ ↑
• Ba ↓ ↑

Barium

Sparingly

Changes.

• Broken down
• Heated

Increases.

Positively charged ion.

• Less
• Less
• More

Less

• Charge
• Distortion
• Less

• 2+
• 1+

Thermally stable

Except Li₂CO₃ which decomposes to Li₂O and CO₂

Oxide and Carbon Dioxide

MCO_{3 (s)} → MO _(s) + CO_{2 (g)}

Nitrate and oxygen.

2MNO_{3 (s)} → 2MNO_{2 (s)} + O_{2 (g)}

Except LiNO₃ which decomposes to form LiNO, NO₂ and O₂

• Oxide
• Nitrogen dioxide
• Oxygen

2M(NO₃)_{2 (s)} → 2MO _(S) + 4NO_{2 (g)} + O_{2 (g)}

Hw long it takes until oxygen is produced. (i.e to relight a glowing split)

OR

How long it takes until a brown gas (NO₂) is produced. this needs to be done in a fume cupboard because NO₂ is toxic.

How long it takes for carbon dioxides to be produce. You test for carbon dioxide using lime water - which is a saturated solution of calcium hydroxide. This turns cloudy with carbon dioxide.

Red

Orange/Yellow

Lilac

Red

Blue

Brick red

Crimson

Green

Mix a small amount of the compound you are testing with a few drops of hydrochloric acid

Heat a piece of platinum or nichrome wire in a hot Bunsen flame to clean it

Dop the wire into the compound/acid mixture. Hold it in a very hot flame and note the colour produced.

The energy absorbed from the flame causes electrons to move to higher energy levels. The colours are seen as the electrons gall back down to lower energy levels, releasing energy in the form of light, The difference in energy between the higher and lower levels determines the wavelength of the light released which determines the colour of the light.

Electron Transition.

Group 7

• X
• X₂
• Negative Ion (X^-)

• F₂
• Pale yellow
• Gas
• 1s² 2s² 2p⁵

• Cl₂
• Green
• Gas
• 1s² 2s² 2p⁶ 3s² 3p⁵

• Br₂
• Red-Brown
• Liquid
• 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁵

• I₂
• Grey
• Solid
• 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁵

Low

Organic.

• In water - Virtually Colourless
• Hexane - Virtually Colourless

• In water - Yellow/Orange
• In hexane - Orange/Red

• In water - Brown
• In hexane - Pink/Violent

Less

• Gaining
• Reduced
• Oxidise
• Oxidising

• Larger
• Shielded
• Reactive

Fluorine

• Increase
• Gas

Disproportionation.

Equation : X₂ + 2NaOH → NaXO + NaX +H₂0

Ionic Equation : X₂ + 2OH^- → XO^- + X^- + H₂0

Equation : 3X₂ + 6NaOH → NaXO₃ + 5NaX + 3H₂O

Ionic Equation : 3X₂ + 6OH^- → XO₃^- + 5X^- + 3H₂O

-1

Oxidise.

Weaker

2Fe(s) + 3Cl2 (g) → 2FeCl3 (s)

• Fe is oxidised: 2Fe → 2Fe³⁺ + 6e^-
• Cl is reduced: 3Cl₂ + 6e^- → 6Cl^-

S_8(s) + 4Cl_2(g) → 4S₂Cl_2(l)

• S is oxidised to +1
• Cl is reduced to -1

• Iron (III)
• Green
• Orange

• Reducing
• Nucleus
• Electrons
• Weaker
• 1) The ions get bigger, so the electrons are further away from the positive nucleus
• 2) There are extra inner electron shells, so there's a greater shielding effect

Hydrogen

KF _(s) + H₂SO_{4 (l)} → KHSO_{4 (s)} + HF _(g)

KCl _(s) + H2_SO_{4 (l)} → KHSO_{4 (s)} + HCl _(g)

• 1) HF or HCl is formed. You'll see misty fumes as the gas comes in contact with moisture in the air.
• 2) But HF and HCl aren't strong enough reducing agents to reduce the sulphuric acid so the reaction stops there.
• 3) It is not a redox reaction - the oxidation states of the halide and sulphur stay the same (-1 and +6)

• KBr _(s) + H₂SO_{4 (l)} → KHSO_{4 (s)} → + HBr _(g)
• 2HBr _(aq) + H₂SO_{4 (l)} → Br_{2 (g)} + SO_{2 (g)} + 2H₂O _(l)

Ox state of S : +6 → +4 (Reduction)

Ox state of Br : -1 → 0 (Oxidation)

• 1) The first reaction gives misty fumes of HBr
• 2) But the HBr is a stronger reducing agent than HCl and reacts with the H₂SO₄ in a redox reaction.
• 3) The reaction produces choking fumes of SO₂ and orange fumes of Br₂

• KBr _(s) + H₂SO_{4 (l)} → KHSO_{4 (s)} → + HBr _(g)
• 2HBr _(aq) + H₂SO_{4 (l)} → Br_{2 (g)} + SO_{2 (g)} + 2H₂O _(l)

Ox state of S : +6 → +4 (Reduction)

Ox state of Br : -1 → 0 (Oxidation)

• 1) The first reaction gives misty fumes of HBr
• 2) But the HBr is a stronger reducing agent than HCl and reacts with the H₂SO₄ in a redox reaction.
• 3) The reaction produces choking fumes of SO₂ and orange fumes of Br₂

KI _(s) + H₂SO_{4 (l)} → KHSO_{4 (s)} + HI _(g)

2HI _(g) + H₂SO_{4 (l)} → I_{2 (s)} + SO_{2 (g)} + 2H₂0 _(l)

• Ox state of S : +6 → +4 (Reduction)
• Ox state of I : -1 → 0 (Oxidation)

6HI _(g) + SO_{2 (g)} → H₂S _(g) +3I_{2 (s)} + 2H₂0 _(l)

• Ox state of S : +4 → -2 (Reduction)
• Ox state of I : -1 → 0 (Oxidation)

• 1) Same initial reaction giving HI gas
• 2) The HI then reduces H₂SO₄ as before.
• 3) But HI keeps going and reduces the SO₂ to H₂S. The H₂S gas is toxic and smells of bad eggs.

Colourless

• Soluble
• Acids
• Blue
• Red
• HCl _(g) → H⁺ _(aq) + Cl^- _(aq)

• Hydrochloric
• Hydrobromic
• Hydroiodic

White fumes.

• For example:
• Hydrogen chloride gives ammonium chloride.

NH_{3 (g)} + HCl _(g) → NH₄Cl _(s)

• Displaced
• Reactive

• Oxidising
• Color changes

KCl _(aq) - No reaction

KBr _(aq) - Orange solution (Br₂) formed.

KI _(aq) - Brown solution (I₂) formed.

KCl _(aq) - No reaction

KBr _(aq) - No reaction

KI _(aq) - Brown solution (I₂) formed.

KCl _(aq) - No reaction

KBr _(aq) - No reaction.

KI _(aq) - No reaction.

• Organic solvent
• Organic solvent
• Aqueous

Cl_{2 (aq)} + 2Br^- _(aq) → 2Cl^- (aq) + Br_{2 (aq)}

Cl_{2 (aq)} + 2I^- _(aq) → 2Cl^- _(aq) + I_{2 (aq)}

Br_{2 (aq)} + 2I^- _(aq) → 2Br^- _(aq) + I_{2 (aq)}

No reaction

Below

Cl_{2 (aq)} + 2Br^- _(aq) → 2Cl^- _(aq) + Br_{2 (aq)}

• 0 → -1
• -1 → 0

Silver Nitrate

• 1) First you add dilute nitric acid to remove ions that might interfere with the test.
• 2) Then you add silver nitrate solution (AgNO_{3 (aq)})
• A precipitate is formed (of the silver halide)

Ag⁺ _(aq) + X^- _(aq) → AgX_(s)

Where X is F, Cl, Br or I

Each silver halide has a different solubility in ammonia.

Fluoride F^- : No precipitate

Chloride Cl^- : White precipitate, dissolves in dilute NH_{3 (aq)}

Bromide Br^- : Cream precipitate, dissolves in conc. NH_{3 (aq)}

Iodide I^- : Yellow precipitate, insoluble in conc. NH_{3 (aq)}

Sunlight

Decomposes

2AgBr → 2Ag + Br₂

Concentration.

Neutralise.

1) You measure out some alkali using a pipette and put it in a flask, along with some indicator (such as phenolphthalein)

2) Then you do a rough titration to get an idea where the end point is (the point where the alkali is exactly neutralised and the indicator changes colour). Add the acid to the alkali using a burette - giving the flask a regular swirl.

3) Now you would carry out a accurate titration. Tun the acid into within 2cm³ of the end point, then add the acid dropwise. If you don't notice exactly when the solution changed colour then you've overshot and your result won't be accurate.

4) Record the amount of acid used to neutralise the alkali. It's best the repeat this process a few times making sure you get the same answer each time.

A burette

Finished

• Colour
• Small

Methyl orange - Turns yellow to red when adding acid to alkali

Phenolphthalein - Turns red to colourless when adding acid to alkali

First write a balanced equation and decide what you know and what you need to know:

HCl (25cm³ 0.5M) + NaOH (35cm³ xM) → NaCl + H₂0

• No of moles HCl → (Conc. x Volume) / 1000
• → (0.5 x 25) / 1000 = 0.0125

• From the equation you know 1 mole of HCl neutralises 1 mole of NaOH.
• So 0.0125 HCl must neutralise 0.0125 moles of NaOH

• Conc. NaOH → (Moles NaOH x 1000) / Volume
• → (0.0125 x 1000) / 35 = 0.36 mol dm^-3

Error

Maximum

Equipment.

• 0.05
• 0.05

• Exact
• 0.005

• Reduce Errors.
• Precise Equipment.

• Initial
• Final
• 0.05
• 0.1

Percentage Uncertainty = (Uncertainty / Reading) x 100

(0.1 / 10) x 100 = 1%

(0.1 / 20) x 100 = 0.5%

• This shows that the larger the volume used, the uncertainty percentage decreases.
• The same principle can be applied to other measurements such as weighing solids.

• Systematic
• Random

They are the same every time you repeat an experiment. They may be caused by the set-up or equiptment you're using. If the 10.00 cm³ pipette you're using to measure out a sample for titration actually only measures 9.95 cm³, your sample will be about 0.05 cm³ too small every time you repeat the experiment.

Random errors vary. They're what make the results a big different each time you repeat an experiment. The errors when you make a reading from a burette are random. You have to estimate or round the level when it's between two marks - so sometimes your figure will be above the real one, and sometimes it will be below.

• Random
• Reliable
• Systematic
• Accurate.

1) Find the percentage uncertainty for each bit of equiptment

2) Add the individual percentage uncertainty together. This gives the percentage uncertainty in the final result.

3) Use this to work out the actual total uncertainty in the final result.

First work out the percentage uncertainty for each volume measurement.

The KOH volume of 10.00 cm³ has an uncertainty of 0.06 cm³ :

(0.06 / 10.00) x 100 = 0.60 %

The HCl volume of 27.3 cm³ has an uncertainty of 0.1 cm³

(0.1 / 27.3) x 100 = 0.37%

Find the percentage uncertainty in the final result:

0.6 + 0.37 = 0.97%

Uncertainty of the final answer:

0.97% of 1.365 mol dm^-3 = 0.013 mol dm^-3

• Oxidising
• Concentrated
• Ions

Stage 1 - Use a sample of oxidising agent to oxidise as much iodide as possible.

Measure out a certain volume of potassium iodide (V) (the oxidising agent) - say 25 cm3

Add this to an excess of acidic potassium iodide solution. The iodate (V) ions in the potassium iodide (V) solution oxidise some of the iodide ions to iodide.

Stage 2 - Find out how many moles of iodine have been produced.

You do this by titrating the resulting solution with sodium thiosulfate. (You need to know the conc. of the sodium thiosulfate solution)

The iodine in the solution tracts with the thiosulfate ions like this:

I₂ + 2S₂O₃^2- → SI^- + S₄O₆^2-

Titration of Iodine with Sodium Thiosulfate.

• 1) Put all the solution from stage 1 in a flask.
• 2) From the burette, add sodium thiosulfate solution to the solution in the flask
• 3) It's had to see the end point, so when the iodine colour fades to pale yellow, add 2 cm3 of starch solution (to detect the presence of iodine). The solution in the conical flask will go dark blue showing theres still some iodine there.
• 4) Add sodium thiosulfate one drop at a time until the blue colour disappears
• 5) When this happens, it means that all the iodine has just been reacted
• 6) Now you can calculate the number of moles of iodine in the solution.
• You would then calculate the number of moles of iodine produced in stage one. In this case it equals 6.66 x 10^-4 moles.

Calculate the concentration of the oxidising agent.

You look back at your original equation:

IO₃^- _(aq) + 5I^- _(aq) + 6H⁺ _(aq) → 3I_{2 (aq)} + 3H₂O _(l)

The equation shows that 1 mole of iodate (V) ions produces 3 moles of iodine. 25 cm³ of potassium iodate (V) solution produced 6.66 x 10^-4 moles of iodine. So there must have been 6.66 x 10^-4 / 3 = 2.22 x 10^-4 moles of iodate (V) ions.

There would be the same number of moles of potassium iodate (V) in the solution. So now it's straightforward to find the conc. of potassium iodate (V) solution:

N = (CV/1000)

2.22 x 10^-4 = (Conc. x 25) / 1000

Conc. of potassium iodate (V) solution = 0.00888 mole dm^-3

• Contaminated
• Clean
• Rinse

• Read
• Bottom

• Repeat
• Average

• Wash
• New
• Clean