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A-Level Chemistry – Inorganic

Periodicity, groups 2 and 7, transition metals, aqueous ions.

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12 questions  ·  44 marks  ·  answers included  ·  A-Level Chemistry (Edexcel/AQA/OCR style)
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This worksheet covers:

  • Periodicity
  • Groups 2 and 7
  • Transition metals
  • Aqueous ion chemistry

A calculator and periodic table/data booklet may be used.

Questions

Section A – Periodicity and Group 2

1

Describe and explain the general trend in first ionisation energy across Period 3, from sodium to argon, referring to any exceptions to the general trend.

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First ionisation energy generally increases across Period 3 because nuclear charge increases while shielding stays roughly constant (same outer shell), so the outer electron is held more strongly and more energy is needed to remove it. There is a drop from magnesium to aluminium because the outer electron in Al is in a 3p sub-shell, which is higher in energy and slightly shielded by the 3s electrons, so it is easier to remove. There is a further drop from phosphorus to sulfur because in S the fourth 3p electron pairs up in an orbital, and the extra electron-electron repulsion makes it easier to remove than the singly occupied 3p electron in P.
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2

(a) Write a balanced equation for the reaction of magnesium with steam.(b) Write a balanced equation for the reaction of calcium with cold water.(c) Explain why calcium reacts more vigorously with water than magnesium does.

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(a) $\text{Mg}(s) + \text{H}_2\text{O}(g) \rightarrow \text{MgO}(s) + \text{H}_2(g)$.(b) $\text{Ca}(s) + 2\text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(aq) + \text{H}_2(g)$.(c) Down Group 2, atomic radius increases and the outer electrons are further from the nucleus and more shielded, so the first two ionisation energies decrease. This means the outer electrons are lost more easily, so calcium is a stronger reducing agent and reacts more vigorously than magnesium.
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3

Group 2 carbonates decompose on heating: $\text{MCO}_3 \rightarrow \text{MO} + \text{CO}_2$. Explain why magnesium carbonate decomposes at a lower temperature than barium carbonate.

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The smaller $\text{Mg}^{2+}$ ion has a higher charge density than the larger $\text{Ba}^{2+}$ ion, so it polarises (distorts the electron cloud of) the carbonate ion more strongly. This weakens the carbon–oxygen bonds in the carbonate ion more for $\text{MgCO}_3$, so less thermal energy is needed to break it down, meaning $\text{MgCO}_3$ decomposes at a lower temperature than $\text{BaCO}_3$.
[3]
4

A $0.481\,\text{g}$ sample of a Group 2 metal $\text{M}$ reacts completely with excess dilute hydrochloric acid, producing $288\,\text{cm}^3$ of hydrogen gas measured at room temperature and pressure (molar volume $=24000\,\text{cm}^3\,\text{mol}^{-1}$): $\text{M} + 2\text{HCl} \rightarrow \text{MCl}_2 + \text{H}_2$.Calculate the molar mass of $\text{M}$ and identify the metal.

Show answer
$n(\text{H}_2)=\dfrac{288}{24000}=0.0120\,\text{mol}=n(\text{M})$ (1:1 ratio). $M_r=\dfrac{0.481}{0.0120}=40.1$. This matches calcium ($A_r=40.08$), so $\text{M}$ is calcium.
[4]

Section B – Group 7 (the halogens)

5

Explain, in terms of atomic structure, why the oxidising power of the halogens decreases down Group 7.

Show answer
Down the group, atomic radius increases and the incoming electron is added to a shell further from the nucleus with more shielding from inner shells. The nuclear attraction for an incoming electron is therefore weaker, so the halogens become less able to gain an electron (accept electrons less readily), and oxidising power decreases.
[3]
6

Chlorine water is added to a colourless solution of potassium bromide.(a) State the observation.(b) Write the ionic equation for the reaction that occurs.

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(a) The solution turns orange (as bromine is formed).(b) $\text{Cl}_2(aq) + 2\text{Br}^-(aq) \rightarrow 2\text{Cl}^-(aq) + \text{Br}_2(aq)$.
[3]
7

Solid sodium halides are separately treated with concentrated sulfuric acid.Describe and explain the difference in observations between sodium chloride and sodium iodide, in terms of the reducing power of the halide ions.

Show answer
With $\text{NaCl}$, only steamy (white) fumes of $\text{HCl}$ are seen, because $\text{Cl}^-$ is not a strong enough reducing agent to reduce the sulfur in $\text{H}_2\text{SO}_4$. With $\text{NaI}$, as well as $\text{HI}$ fumes, a black solid (iodine) and a gas with a rotten-egg smell ($\text{H}_2\text{S}$) are formed, because $\text{I}^-$ is a much stronger reducing agent than $\text{Cl}^-$ (reducing power increases down the group) and can reduce $\text{H}_2\text{SO}_4$ (sulfur, $+6$) all the way down to $\text{H}_2\text{S}$ (sulfur, $-2$).
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8

Chlorine reacts with cold, dilute sodium hydroxide solution: $\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}$.Calculate the maximum mass of $\text{NaClO}$ ($M_r=74.5$) produced when $3.55\,\text{g}$ of chlorine gas ($M_r=71.0$) reacts completely.

Show answer
$n(\text{Cl}_2)=\dfrac{3.55}{71.0}=0.0500\,\text{mol}$. Since the ratio $\text{Cl}_2:\text{NaClO}$ is $1:1$, $n(\text{NaClO})=0.0500\,\text{mol}$. Mass $=0.0500\times74.5=3.73\,\text{g}$.
[3]

Section C – Transition metals and aqueous ions

9

(a) Define the term transition metal.(b) Explain why scandium and zinc are not classified as transition metals, using electron configurations.

Show answer
(a) A transition metal is a d-block element that forms at least one stable ion with a partially filled (incomplete) d sub-shell.(b) Scandium only forms the $\text{Sc}^{3+}$ ion, which has the configuration $[\text{Ar}]$ with an empty 3d sub-shell; zinc only forms $\text{Zn}^{2+}$, which has the configuration $[\text{Ar}]3d^{10}$, a full 3d sub-shell. Neither ion has a partially filled d sub-shell, so neither element is classified as a transition metal.
[4]
10

Aqueous ammonia is added dropwise, then in excess, to a pale blue solution of aqueous copper(II) sulfate.(a) State and explain the observations.(b) Write an equation for the reaction occurring in excess ammonia.

Show answer
(a) With a small amount of ammonia, a pale blue precipitate of $\text{Cu(OH)}_2$ forms (ammonia acts as a base, providing $\text{OH}^-$). In excess ammonia, the precipitate dissolves to give a deep/royal blue solution, as ammonia acts as a ligand and substitutes for water molecules to form the complex ion $[\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$.(b) $\text{Cu(OH)}_2(s) + 4\text{NH}_3(aq) \rightarrow [\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+}(aq) + 2\text{OH}^-(aq)$.
[4]
11

Calculate the oxidation state of chromium in the dichromate ion, $\text{Cr}_2\text{O}_7^{2-}$, showing your working.

Show answer
Let the oxidation state of Cr be $x$. Oxygen is $-2$: $2x + 7(-2) = -2 \Rightarrow 2x – 14 = -2 \Rightarrow 2x = 12 \Rightarrow x = +6$. Chromium is in the $+6$ oxidation state.
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12

In an acidified redox titration, $\text{MnO}_4^-$ reacts with $\text{Fe}^{2+}$: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$.A $25.0\,\text{cm}^3$ sample of $\text{Fe}^{2+}$ solution required $22.40\,\text{cm}^3$ of $0.0200\,\text{mol dm}^{-3}$ acidified $\text{KMnO}_4$ for complete reaction.Calculate the concentration of the $\text{Fe}^{2+}$ solution, in $\text{mol dm}^{-3}$.

Show answer
$n(\text{MnO}_4^-)=0.0200\times\dfrac{22.40}{1000}=4.48\times10^{-4}\,\text{mol}$. From the equation, $n(\text{Fe}^{2+})=5\times n(\text{MnO}_4^-)=2.24\times10^{-3}\,\text{mol}$. Concentration $=\dfrac{2.24\times10^{-3}}{25.0/1000}=0.0896\,\text{mol dm}^{-3}$.
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