Topic 8

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Everything from this topic in one place, for the last read before an exam. If anything here is unfamiliar, go back to the full page.

Definitions to learn word for word

Oxidation number rules (in priority order)

Rule Oxidation number
uncombined element 0
total in a species the overall charge (0 for a compound)
simple ion the charge on the ion
Group 1, Group 2, aluminium +1, +2, +3
fluorine −1
hydrogen +1 (metal hydrides −1)
oxygen −2 (peroxides −1; with fluorine +2)
Cl, Br, I −1 (positive when bonded to O or F)

Write the sign first: +7. A charge has the sign after: 1−. Roman numerals in names are oxidation numbers: manganate(VII) has Mn at +7, iron(III) has Fe at +3.

Equations from half-equations

  1. Balance the atom being changed, then O with H₂O, then H with H⁺, then charge with e⁻.
  2. Multiply so both half-equations have the same number of electrons.
  3. Add and cancel the electrons, and any water or H⁺ that appears on both sides.
MnOX4X−+8 HX++5 FeX2+→MnX2++4 HX2O+5 FeX3+ \ce{MnO4- + 8H+ + 5Fe^2+ -> Mn^2+ + 4H2O + 5Fe^3+}

Check atoms and charge: left −1+8+10=+17-1 + 8 + 10 = +17; right +2+15=+17+2 + 15 = +17.

Groups 1 and 2

Property Trend down the group Reason
First ionisation energy falls outer electron further out and better shielded
Reactivity of the metal rises easier to lose outer electrons
Solubility of Group 2 hydroxides rises a trend to learn; the reason is beyond the spec (page 12)
Solubility of Group 2 sulfates falls a trend to learn; BaSO₄ is the least soluble
Thermal stability of carbonates and nitrates rises falling charge density, less polarisation of the anion

Group 1 lithium is out of line: its carbonate and nitrate decompose on heating, because Li⁺ has a high charge density. Charge density = charge ÷ radius: Mg²⁺ 27.8, Ca²⁺ 20.0, Ba²⁺ 14.7, Li⁺ 13.2, Na⁺ 9.8, K⁺ 7.2 (charge per nm).

MCOX3(s)→MO(s)+COX2(g)2 M(NOX3)X2(s)→2 MO(s)+4 NOX2(g)+OX2(g) \ce{MCO3(s) -> MO(s) + CO2(g)} \qquad \ce{2M(NO3)2(s) -> 2MO(s) + 4NO2(g) + O2(g)}

Group 1 nitrates (not lithium) give the nitrite and oxygen: 2 NaNOX3→2 NaNOX2+OX2\ce{2NaNO3 -> 2NaNO2 + O2}.

Flame colours and tests

Test Result
Li⁺ red
Na⁺ yellow
K⁺ lilac
Ca²⁺ orange-red
Sr²⁺ red
Ba²⁺ green
Mg²⁺ no colour
Carbonate or hydrogencarbonate dilute acid: fizzing; gas turns limewater milky
Sulfate dilute HCl, then BaCl₂(aq): white precipitate
Ammonium NaOH(aq) and warm: gas turns damp red litmus blue
Halide AgNO₃(aq) after dilute HNO₃: AgCl white, AgBr cream, AgI yellow

Silver halides in ammonia: AgCl dissolves in dilute NH₃; AgBr dissolves only in concentrated NH₃; AgI does not dissolve.

Group 7

Cl₂ Br₂ I₂
State and colour pale green gas red-brown liquid grey-black solid
In water very pale green orange brown
In a non-polar solvent pale green orange-red violet
Boiling temperature / °C −34 59 184
Oxidising power strongest middle weakest
ClX2+2 NaOH→NaCl+NaClO+HX2O3 ClX2+6 NaOH→5 NaCl+NaClOX3+3 HX2O \ce{Cl2 + 2NaOH -> NaCl + NaClO + H2O} \qquad \ce{3Cl2 + 6NaOH -> 5NaCl + NaClO3 + 3H2O}

Titration in six steps

  1. Rinse the burette with the solution it will hold, fill it, and remove the air from the tip.
  2. Pipette 25.0 cm³ into a clean conical flask, add a few drops of indicator.
  3. Do a rough titration, then repeat, adding drop by drop near the end point, until two concordant titres (within 0.10 cm³).
  4. Mean of the concordant titres only. Do not include the rough one.
  5. Moles of the known solution: n=cVn = cV with VV in dm³.
  6. Use the equation ratio to get moles of the unknown, then c=n÷Vc = n \div V. To get g dm⁻³ multiply by MrM_\text{r}.

Uncertainty: percentage uncertainty = (absolute uncertainty ÷ measured value) × 100. Burette titre: two readings, so double the single-reading uncertainty (±0.05 each gives ±0.10).

The ten traps

  1. Writing the oxidation number with the sign after (2+ for +2), or the ion charge with the sign first.
  2. Forgetting that the subscript multiplies (O₂ in SO₂ gives −4 in total).
  3. Forgetting that oxygen in peroxides and hydrogen in metal hydrides is −1.
  4. Adding half-equations without balancing the electrons first.
  5. Leaving H⁺ or H₂O on both sides of a final equation.
  6. Calling a reaction a disproportionation without naming the one element that goes both up and down.
  7. Explaining thermal stability with "bond strength" instead of charge density and polarisation.
  8. Giving the flame colour as "white" for magnesium, or the reds from memory without the mark-scheme wording.
  9. Doing the sulfate test without acid first, so carbonate also gives a white precipitate.
  10. Including the rough titre in the mean, or forgetting the mole ratio.

Mixed test

  1. Work out the oxidation number of Cr in CrX2OX7X2−\ce{Cr2O7^2-} and of Mn in MnOX4X−\ce{MnO4-}.
  2. Write the half-equation for the reduction of MnOX4X−\ce{MnO4-} to MnX2+\ce{Mn^2+} in acid, and for the oxidation of FeX2+\ce{Fe^2+} to FeX3+\ce{Fe^3+}, and combine them.
  3. Chlorine reacts with cold dilute sodium hydroxide. Write the equation and show that it is a disproportionation.
  4. 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide needs 21.40 cm³ of hydrochloric acid for neutralisation. Calculate the concentration of the acid, and the percentage uncertainty in the titre from the burette alone (±0.05 cm³ per reading).
  5. Explain why magnesium carbonate decomposes at a lower temperature than barium carbonate.
  6. A white solid gives a lilac flame, fizzes with dilute acid and gives a gas that turns limewater milky. Name the solid and justify each observation.
  7. Describe what you would see when concentrated sulfuric acid is added to solid sodium iodide, and name the sulfur-containing products with their oxidation numbers.
  8. Describe how you would use silver nitrate and ammonia to distinguish chloride, bromide and iodide ions.
  9. Explain, using oxidation numbers, why bromine water turns brown when added to potassium iodide solution.
  10. Predict the state and colour of astatine at room temperature, and say whether chlorine would displace astatide ions.

Answers cover these until you have tried the questions

  1. Cr₂O₇²⁻: the total is −2 and O is −2, so 2Cr+7(−2)=−22\text{Cr} + 7(-2) = -2, 2Cr=+122\text{Cr} = +12, Cr is +6. MnO₄⁻: Mn+4(−2)=−1\text{Mn} + 4(-2) = -1, Mn is +7.

  2. Reduction: MnOX4X−+8 HX++5 eX−→MnX2++4 HX2O\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}. Oxidation: FeX2+→FeX3++eX−\ce{Fe^2+ -> Fe^3+ + e-}. Multiply the iron half-equation by 5 and add: MnOX4X−+8 HX++5 FeX2+→MnX2++4 HX2O+5 FeX3+\ce{MnO4- + 8H+ + 5Fe^2+ -> Mn^2+ + 4H2O + 5Fe^3+}

  3. ClX2+2 NaOH→NaCl+NaClO+HX2O\ce{Cl2 + 2NaOH -> NaCl + NaClO + H2O}. Chlorine goes from 0 to −1 in NaCl (reduced) and from 0 to +1 in NaClO (oxidised). One element, chlorine, is both oxidised and reduced, so this is disproportionation.

  4. n(NaOH)=0.100×25.0÷1000=0.00250n(\text{NaOH}) = 0.100 \times 25.0 \div 1000 = 0.00250 mol. The ratio is 1 : 1, so n(HCl)=0.00250n(\text{HCl}) = 0.00250 mol. c=0.00250÷0.02140c = 0.00250 \div 0.02140 = 0.117 mol dm⁻³. Titre uncertainty is 2×0.05=±0.102 \times 0.05 = \pm0.10 cm³, so the percentage uncertainty is 0.10÷21.40×1000.10 \div 21.40 \times 100 = 0.47 %.

  5. Mg²⁺ is smaller with the same 2+ charge, so it has a higher charge density (27.8 against 14.7). It polarises the carbonate ion more, weakening a C–O bond, so less energy is needed to decompose MgCO₃, which decomposes at a lower temperature.

  6. Potassium carbonate (or hydrogencarbonate). The lilac flame shows K⁺. Fizzing with acid shows carbonate (or hydrogencarbonate) giving carbon dioxide, and limewater turns milky as CO₂ forms white calcium carbonate.

  7. Steamy fumes of HI, purple fumes and a black solid (I₂), a yellow solid (S), and a rotten-egg smell (H₂S), plus colourless SO₂. Sulfur is +6 in H₂SO₄ and goes to +4 in SO₂, 0 in S and −2 in H₂S.

  8. Acidify with dilute nitric acid, add silver nitrate solution: white AgCl, cream AgBr, yellow AgI. Add dilute ammonia: only AgCl dissolves. Add concentrated ammonia to the others: AgBr dissolves, AgI stays.

  9. Bromine is a stronger oxidising agent than iodine, so Br₂ oxidises I⁻ to brown I₂: BrX2+2 IX−→2 BrX−+IX2\ce{Br2 + 2I- -> 2Br- + I2}. Bromine goes from 0 to −1 (reduced) and iodine from −1 to 0 (oxidised).

  10. Astatine is a black solid (following the trend of grey-black iodine). Chlorine is a stronger oxidising agent than astatine, so it would displace astatide ions: ClX2+2 AtX−→2 ClX−+AtX2\ce{Cl2 + 2At- -> 2Cl- + At2}.