IB DP Chemistry Structure 1.4 Counting particles by mass: The mole HL Paper 2- Exam Style Questions - New Syllabus

IBDP Chemistry HL Paper 2 - All Chapters

Question

Thallium is a heavy metallic element found in group \(13\) of the periodic table.
(a) \(30\%\) of thallium atoms have \(122\) neutrons, while the remaining atoms have \(124\) neutrons.
(i) Determine the nuclear symbol of the thallium isotope that contains \(122\) neutrons. Use section \(7\) of the data booklet (atomic number of thallium, \(\mathrm{Tl}=81\)). [1]
(ii) Calculate the relative atomic mass of thallium, giving your answer to two decimal places. [2]
(b) Thallium(I) sulfate has the chemical formula \(\mathrm{Tl_2SO_4}\).
(i) This compound contains both ionic and covalent bonds. Identify which particles are connected by covalent bonds and which are connected by ionic bonds.
(ii) Distinguish between covalent bonding and ionic bonding in terms of valence electron behaviour.
(iii) Name the enthalpy change that represents the strength of attraction between ions in an ionic lattice.
(iv) Write a chemical equation for the formation of aqueous thallium(I) sulfate by reacting solid thallium(I) hydroxide with sulfuric acid.
(v) Calculate the volume of sulfuric acid with concentration \(2.00\,\text{mol dm}^{-3}\) needed to completely react with \(10.0\,\text{g}\) of thallium(I) hydroxide. Use section \(1\) of the data booklet (molar masses: \(\mathrm{Tl}=204.38\), \(\mathrm{O}=16.00\), \(\mathrm{H}=1.01\)).
(c) The atomic emission spectrum of thallium includes an intense green line.
(i) State which feature of an atomic emission spectrum can be used to determine the ionization energy of an element.
(ii) Calculate the wavelength corresponding to the first ionization energy of thallium. Use sections \(1\), \(2\) and \(9\) of the data booklet and the values \[ h = 6.63 \times 10^{-34}\,\text{J s}, \quad c = 3.00 \times 10^{8}\,\text{m s}^{-1}, \quad N_A = 6.02 \times 10^{23}\,\text{mol}^{-1} \] First ionization energy of thallium \(= 589\,\text{kJ mol}^{-1}\).
(iii) Explain why thallium has a lower first ionization energy than lead, referring to nuclear charge and electron shielding.

Most-appropriate topic codes (IB Chemistry 2025):

• Structure 1.1: The nuclear atom — parts (a)(i), (a)(ii)
• Structure 2.1: Ionic bonding and lattice enthalpy — parts (b)(i), (b)(iii)
• Structure 2.2: Covalent bonding — parts (b)(i), (b)(ii)
• Reactivity 1.1: Stoichiometric relationships — part (b)(v)
• Reactivity 1.2: Energy cycles and enthalpy changes — part (b)(iii)
• Structure 1.3: Electron configurations — parts (c)(i), (c)(iii)
• Structure 1.4: Atomic spectra — part (c)(ii)
▶️ Answer/Explanation

(a)(i)
Thallium has atomic number \(81\). The mass number of the isotope is therefore: \[ 81 + 122 = 203 \] Nuclear symbol: \[ {}^{203}_{81}\mathrm{Tl} \]

(a)(ii)
The second isotope has a mass number of: \[ 81 + 124 = 205 \] Relative atomic mass: \[ A_r = (0.30 \times 203) + (0.70 \times 205) = 204.40 \]

(b)(i)
Covalent bonds are present between sulfur and oxygen atoms within the \(\mathrm{SO_4^{2-}}\) ion. Ionic bonding occurs between \(\mathrm{Tl^+}\) ions and \(\mathrm{SO_4^{2-}}\) ions.

(b)(ii)
Covalent bonding involves the sharing of valence electrons, whereas ionic bonding involves the transfer of electrons.

(b)(iii)
Lattice enthalpy.

(b)(iv)
\[ 2\mathrm{TlOH}(s) + \mathrm{H_2SO_4}(aq) \rightarrow \mathrm{Tl_2SO_4}(aq) + 2\mathrm{H_2O}(l) \]

(b)(v)
Molar mass of \(\mathrm{TlOH} = 204.38 + 16.00 + 1.01 = 221.39\,\text{g mol}^{-1}\).
Amount of \(\mathrm{TlOH} = \dfrac{10.0}{221.39} = 0.0452\,\text{mol}\).
Required amount of \(\mathrm{H_2SO_4} = 0.0226\,\text{mol}\).
Volume required: \[ \dfrac{0.0226}{2.00} = 0.0113\,\text{dm}^3 = 11.3\,\text{cm}^3 \]

(c)(i)
The series limit (continuum) at the high-frequency end of the emission spectrum.

(c)(ii)
Energy per atom: \[ E = \dfrac{589 \times 10^3}{6.02 \times 10^{23}} = 9.78 \times 10^{-19}\,\text{J} \] Wavelength: \[ \lambda = \dfrac{hc}{E} = 2.03 \times 10^{-7}\,\text{m} \]

(c)(iii)
Thallium has a lower effective nuclear charge than lead and experiences similar shielding, so its outer electron is less strongly attracted to the nucleus.

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