Group Properties Worked ExamplesIGCSE Chemistry 0620

(a) Melting point decreases down Group I, so the element lowest in the list has the lowest melting point: caesium.

(b) Density generally increases down Group I, so the element at the top has the lowest density: lithium.

(c) Reactivity increases down Group I, so the most vigorous reaction with water is given by caesium.

(d) From the flame-test table: $\text{Li}^+$ red, $\text{Na}^+$ yellow, $\text{K}^+$ lilac. The answer is lithium.

(e) Argon has 18 electrons (2,8,8). A Group I atom loses one electron to form a 1+ ion, so the atom needs 19 electrons: proton number 19 is potassium ($\text{K}^+$ is 2,8,8).

(a) These three descriptions must be learned exactly as the syllabus words them: chlorine is a gas, pale yellow-green; bromine is a liquid, red-brown; iodine is a solid, grey-black.

The state changes gas → liquid → solid down the group because the melting and boiling points increase down the group.

(b) Bromine is $\text{Br}_2$. Diatomic means each molecule is made of two atoms joined by a covalent bond.

(c) The density increases down Group VII (from fluorine to iodine).

(a) Observations are things you can see or hear. Any two of: the potassium floats; it moves around on the surface; it fizzes (bubbles); it melts into a ball; it burns with a lilac flame; it disappears.

(b) Group I metals react with water to form the metal hydroxide and hydrogen. Balance the hydrogen atoms by using $2\text{H}_2\text{O}$.

Check: left side 2 K, 4 H, 2 O; right side 2 K, 2 + 2 = 4 H, 2 O. Balanced.

(c) Potassium hydroxide is an alkali, so the solution becomes alkaline and thymolphthalein turns blue (it is colourless in acidic and neutral solutions).

(d) Moles of potassium ($A_r = 39$):

From the equation, 2 mol K give 1 mol $\text{H}_2$. Then use $24\,000 \text{ cm}^3$ per mole at r.t.p.

(a) A colour change to the colour of a different halogen means a displacement happened. Chlorine turns the bromide solution orange (bromine formed) and the iodide solution brown (iodine formed), so chlorine displaces both.

Bromine turns the iodide solution brown (iodine formed) but does not react with chloride. Iodine displaces neither. A more reactive halogen displaces a less reactive one from its halide, so the order is chlorine > bromine > iodine.

(b) Bromine is less reactive than chlorine, so it cannot displace chlorine from chloride ions. No reaction occurs and the only colour is that of the bromine that was added.

(c) The potassium ions are spectator ions and are left out. Bromine gains electrons to form bromide ions; iodide ions lose electrons to form iodine.

Check: 2 Br and 2 I on each side; total charge $2-$ on each side.

(d) In those dimples a halogen would be added to its own halide ion (e.g. chlorine to chloride), so no displacement is possible and the test gives no information.

(a) Bromine is released, so the solution turns from colourless to orange (yellow-orange is also accepted).

(b) Chlorine is more reactive than bromine, so it displaces it:

(c) The $\text{Mg}^{2+}$ ions are unchanged (spectator ions), so remove them. Two bromide ions are needed for each chlorine molecule.

(d) Oxidation is loss of electrons and reduction is gain of electrons. Each bromide ion loses an electron (it is oxidised to bromine) and each chlorine atom gains an electron (chlorine is reduced to chloride ions). Both happen at the same time, so the reaction is redox.

(e) The oxidising agent is the substance that oxidises another and is itself reduced: chlorine.

Down Group I: melting point decreases, density increases and reactivity increases. Caesium is lower than sodium, so it has a lower melting point, a higher density and reacts faster.

That is row B.

C is tempting because two of its three entries are right, but density increases down the group. A and D have the reactivity trend the wrong way round — the halogen trend (reactivity decreasing down the group) has been applied to Group I by mistake.

(a) The points rise by roughly 95–120 °C per period. Continuing the line (or adding the average rise, about 110 °C, to iodine's 114 °C) gives an estimate of about 220 °C. Any value clearly above iodine's melting point that follows the trend (roughly 200–300 °C) is acceptable.

(b) Room temperature (about 25 °C) is far below the predicted melting point, so astatine is a solid. The colours darken down the group (pale yellow-green → red-brown → grey-black), so astatine is predicted to be black.

(c) Density increases down Group VII, so astatine is more dense than iodine.

(d) Reactivity decreases down Group VII, so chlorine is more reactive than astatine and displaces it. Sodium ions are spectators.

(a) Sodium compounds and burning sodium give a yellow flame.

(b) A Group I metal and a halogen form an ionic halide: sodium bromide. Bromine is diatomic, so two sodium atoms are needed for each $\text{Br}_2$ molecule.

(c)(i) Moles of sodium, then use the 1 : 1 ratio of Na to NaBr. $M_r(\text{NaBr}) = 23 + 80 = 103$.

(ii) 2 mol Na react with 1 mol $\text{Br}_2$, and $M_r(\text{Br}_2) = 2 \times 80 = 160$.

Check by conservation of mass: $0.69 + 2.40 = 3.09$ g, the same as the mass of sodium bromide.

(a) Yellow flame → $\text{Na}^+$. Cream precipitate with acidified silver nitrate → $\text{Br}^-$ (chloride would give white, iodide yellow). X is sodium bromide.

(b) Silver ions and bromide ions combine to form insoluble silver bromide:

(c) Hydrochloric acid contains chloride ions, which would give a white precipitate of silver chloride with silver nitrate, so the result would not show which halide was in X.

(d) Test 3: chlorine is more reactive than bromine, so it displaces bromine from the bromide ions, turning the solution orange. Test 4: iodine is less reactive than bromine, so no displacement occurs — the brown colour is just the iodine that was added. This negative result supports the identification.

(a) Convert each volume to moles using $24\,000 \text{ cm}^3$ per mole at r.t.p.

Relative molecular mass = mass ÷ moles.

Halogens are diatomic, so divide by 2 to find $A_r$: A has $A_r = 35.5$, which is chlorine ($\text{Cl}_2$); B has $A_r = 19$, which is fluorine ($\text{F}_2$).

(b) Reactivity decreases down Group VII, so fluorine (B) is more reactive. Chlorine (A) is less reactive than fluorine, so it would not displace fluorine from fluoride ions — no reaction.

(a) Each metal M reacts as $2\text{M(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{MOH(aq)} + \text{H}_2\text{(g)}$, so moles of $\text{H}_2$ = moles of metal ÷ 2, and $V = n \times 24\,000 \text{ cm}^3$.

Lithium ($A_r = 7$):

Sodium ($A_r = 23$) and potassium ($A_r = 39$):

(b) Match the final volumes: X = lithium (60), Z = sodium (about 18), Y = potassium (about 11). This agrees with the gradients: Y is steepest at the start because potassium is the most reactive, X is the least steep.

(c) The initial gradient shows the rate, which depends on reactivity — potassium is the most reactive, so it is fastest. The final volume depends only on the number of moles of metal. Potassium has the largest $A_r$, so the same mass contains the fewest moles of atoms and gives the least hydrogen.

(a) The colourless solution turns brown as iodine forms; with this much iodine a grey-black solid may also appear. Bubbles are seen as the gas enters.

(b) Chlorine is more reactive than iodine; sodium ions are spectators.

(c)(i) Moles of iodide ions (convert $\text{cm}^3$ to $\text{dm}^3$ first):

The ratio $\text{Cl}_2 : \text{I}^-$ is $1 : 2$.

(ii) One mole of $\text{I}_2$ forms per mole of $\text{Cl}_2$; $M_r(\text{I}_2) = 2 \times 127 = 254$.

(d) Chlorine is a toxic gas, so any excess must be removed safely by the fume cupboard.