Transition Elements Worked ExamplesIGCSE Chemistry 0620

(a) Transition elements have high melting points, high densities and form coloured compounds. Q and S fit all three; P and R have low melting points and densities and form white chlorides.

So the transition elements are Q and S. Evidence: melting points above 1500 °C; densities above 7 g per cm³; coloured chlorides.

(b) Two different chlorides mean Q forms ions with different oxidation numbers (for example 2+ and 3+) — the transition elements have variable oxidation numbers.

(a) The Contact process catalyst is vanadium(V) oxide: V.

(b) A 2+ ion has two fewer electrons than protons, so the atom has $23 + 2 = 25$ protons: Mn.

(c) Light blue precipitate, dissolving in excess ammonia to a dark blue solution, is the test result for copper(II): Cu.

(d) A red-brown precipitate with sodium hydroxide is iron(III): Fe.

(e) White precipitate soluble in excess of both reagents is zinc. (Aluminium ions also dissolve in excess sodium hydroxide but not in excess ammonia, and aluminium is not on the diagram anyway.) Answer: Zn.

(a) The Roman numeral is the oxidation number of the metal. In iron(III) sulfate iron has oxidation number +3, so it is present as $\text{Fe}^{3+}$ ions.

(b) Balance the charges so the total is zero. Oxide is $\text{O}^{2-}$ and sulfate is $\text{SO}_4^{\,2-}$.

(i) $\text{Cr}^{3+}$ and $\text{O}^{2-}$: two $\text{Cr}^{3+}$ (+6) balance three $\text{O}^{2-}$ (−6).

(ii) $\text{Mn}^{4+}$ needs two $\text{O}^{2-}$: $\text{MnO}_2$. (iii) two $\text{Fe}^{3+}$ (+6) balance three $\text{SO}_4^{\,2-}$ (−6): $\text{Fe}_2(\text{SO}_4)_3$. (iv) two V at +5 (+10) balance five $\text{O}^{2-}$ (−10): $\text{V}_2\text{O}_5$.

(c) In $\text{FeCl}_2$, two $\text{Cl}^-$ ions (−2) mean iron is +2: iron(II) chloride. In $\text{CuO}$, one $\text{O}^{2-}$ means copper is +2: copper(II) oxide.

(a) Green precipitate → iron(II), $\text{Fe}^{2+}$, in J. Red-brown precipitate → iron(III), $\text{Fe}^{3+}$, in K.

(b) Each $\text{Fe}^{2+}$ needs two $\text{OH}^-$ to form the neutral hydroxide:

(c) Oxygen in the air reacts with the precipitate at the surface, converting iron(II) to iron(III), which is red-brown. The oxidation number increases from +2 to +3; each iron(II) ion loses one electron, so iron is oxidised.

(d) Yes — aqueous ammonia gives the same results: green precipitate with iron(II), red-brown with iron(III), both insoluble in excess.

(a) Manganese(IV) oxide — manganese is a transition element, and transition elements and their compounds often act as catalysts. The other three are compounds of Group I or Group II metals.

(b) Read up from 40 s to the catalysed curve, then across: about 107 $\text{cm}^3$ (a reading between 105 and 110 $\text{cm}^3$ is acceptable).

(c) Moles of hydrogen peroxide (convert $\text{cm}^3$ to $\text{dm}^3$):

The ratio $\text{H}_2\text{O}_2 : \text{O}_2$ is $2 : 1$.

The catalysed curve levels off at $120 \text{ cm}^3$, so the graph agrees.

(d) Weigh the solid before adding it. At the end, filter off the solid, wash and dry it, and weigh it again — the mass is unchanged.

A, C and D are true of all metals, so potassium has them too. Simple potassium compounds such as potassium chloride are white and dissolve to give colourless solutions, whereas chromium compounds are coloured. Answer: B.

C is tempting because transition elements are known for forming ions with different charges — but potassium also forms positive ions ($\text{K}^+$), so this does not distinguish them.

(a) Transition elements have high densities: chromium, iron and copper (all above 7 g per cm³). Lithium, sodium and potassium are all below 1 g per cm³.

(b) Any two: transition elements have higher melting points, are harder / stronger, and are less soft than the Group I metals (which can be cut with a knife).

(c) The general trend is that density increases down Group I, so potassium should be denser than sodium. The chart shows potassium ($0.86$) is less dense than sodium ($0.97$), so it breaks the trend.

(a) A green precipitate that is soluble in excess sodium hydroxide but insoluble in excess ammonia is chromium(III), $\text{Cr}^{3+}$. A white precipitate with acidified silver nitrate is chloride. W is chromium(III) chloride, $\text{CrCl}_3$.

(b) Iron(II) also gives a green precipitate, but it is insoluble in excess sodium hydroxide. Here the precipitate dissolved, so the cation cannot be iron(II).

(c) Three hydroxide ions balance the 3+ charge:

(d) Transition elements form coloured compounds.

Rules: oxygen is −2 in these compounds; the oxidation numbers in a compound add up to zero; in an ion they add up to the charge on the ion.

(a) $2x + 5(-2) = 0$

(b) $x + 4(-2) = -1$

(c) $2x + 7(-2) = -2$

(d) $2x + (-2) = 0$, so $x = +1$ — copper here is in oxidation state +1, not its more familiar +2.

(e) Potassium is +1, so the $\text{MnO}_4^{\,-}$ ion contains manganese at +7: potassium manganate(VII).

(a) Copper: +2 in $\text{Cu}^{2+}$ → +1 in CuI (iodide is −1). The oxidation number decreases, so copper is reduced. Iodine: −1 in $\text{I}^-$ → 0 in $\text{I}_2$. It increases, so iodide is oxidised. Both happen together, so it is redox.

(b) The oxidising agent is reduced itself: copper(II) ions. The reducing agent is oxidised itself: iodide ions.

(c) Aqueous potassium iodide is colourless; an oxidising agent turns it brown because it oxidises iodide ions to iodine.

(d) Moles of $\text{Cu}^{2+}$, then use the $2 : 1$ ratio of $\text{Cu}^{2+}$ to $\text{I}_2$:

$M_r(\text{I}_2) = 2 \times 127 = 254$.

(e) Copper appears in two oxidation states, +2 and +1 — transition elements have ions with variable oxidation numbers. (The blue colour of copper(II) sulfate also shows that they form coloured compounds.)

(a) Mass of chlorine combined = mass of product − mass of iron.

Divide each mass by $A_r$ (Fe = 56, Cl = 35.5):

Simplest whole-number ratio Fe : Cl = 0.0200 : 0.0600 = 1 : 3. Empirical formula: $\text{FeCl}_3$.

(b) Chloride is −1, so iron is +3: iron(III) chloride. Chlorine is diatomic.

(c) With the acid, iron forms iron(II) and hydrogen:

(d) Add aqueous sodium hydroxide to each: iron(II) chloride gives a green precipitate; iron(III) chloride gives a red-brown precipitate (both insoluble in excess).

(a) Anhydrous copper(II) sulfate is white. (The blue colour of the hydrated crystals is an example of a coloured transition-metal compound.)

(b) Heating until the mass no longer changes makes sure all the water of crystallisation has been removed.

(c) Find the masses from the readings:

Convert to moles: $M_r(\text{CuSO}_4) = 64 + 32 + 4(16) = 160$; $M_r(\text{H}_2\text{O}) = 18$.

Ratio of water to copper(II) sulfate:

(d) The white solid turns blue (and becomes warm). The reaction is reversible: $\text{CuSO}_4\cdot 5\text{H}_2\text{O} \rightleftharpoons \text{CuSO}_4 + 5\text{H}_2\text{O}$.