A Level H2 Chemistry Periodic Table Quiz

Answer Key - A-Level Chemistry H2 Quiz: Periodic Table

1. Mg has a stable $3\text{s}^2$ configuration. Al has a $3\text{p}^1$ electron which is further from the nucleus and more shielded by the $3\text{s}^2$ electrons, making it easier to remove. [2]

2. Na is larger than Cl. Across the period, nuclear charge increases while shielding remains relatively constant; the stronger nuclear attraction pulls electrons closer to the nucleus. [2]

3. Removing the second electron from Na requires breaking a stable noble gas configuration ($2\text{p}^6$), which involves a much larger energy jump compared to removing an electron from Mg's $3\text{s}^1$ orbital. [2]

4. Electronegativity increases. Nuclear charge increases and atomic radius decreases, increasing the attraction for a shared pair of electrons. [2]

5. Oxygen has a paired electron in the $2\text{p}$ orbital. Inter-electronic repulsion between the paired electrons makes the first electron easier to remove than the unpaired electron in nitrogen's $2\text{p}$ orbital. [2]

6. $\text{Mg}^{2+}$ is smaller than $\text{Na}^+$. Both have the same electron configuration ($2\text{p}^6$), but $\text{Mg}^{2+}$ has a higher nuclear charge (+12 vs +11), pulling the electrons more strongly. [2]

7. Electronegativity: The ability of an atom in a molecule to attract the shared pair of electrons. Electron affinity: The energy change when an electron is added to a gaseous atom. [2]

8. $\text{Ca(s)} + 2\text{H}_2\text{O(l)} \rightarrow \text{Ca(OH)}_2\text{(aq/s)} + \text{H}_2\text{(g)}$ [2]

9. Down the group, ionic radius increases. Lattice energy decreases more significantly than hydration energy. The overall enthalpy of solution becomes more exothermic/less endothermic. [3]

10. $\text{BaCO}_3$ is more stable. $\text{Ba}^{2+}$ has a larger radius and lower charge density than $\text{Mg}^{2+}$, so it polarises the carbonate ion less, making it harder to decompose. [3]

11. $\text{MgO(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Mg}^{2+}\text{(aq)} + \text{H}_2\text{O(l)}$ [2]

12. Down the group, the size of the sulfate ion is large and constant. The hydration energy of the cation decreases more rapidly than the lattice energy as the cation size increases. Thus, $\text{BaSO}_4$ is less soluble. [3]

13. Observation: Magnesium burns with a bright white light to form a white powder. Equation: $2\text{Mg(s)} + \text{O}_2\text{(g)} \rightarrow 2\text{MgO(s)}$. [2]

14. $\text{Be}$ has a very high charge density, giving $\text{BeO}$ significant covalent character and the ability to react with both acids and bases. $\text{Ba}$ has low charge density, making $\text{BaO}$ purely ionic and basic. [2]

15. $\text{I}_2 < \text{Br}_2 < \text{Cl}_2 < \text{F}_2$. Oxidizing power increases up the group as atomic radius decreases and electronegativity increases, making it easier to attract/gain an electron. [3]

16. $\text{Cl}_2\text{(aq)} + 2\text{KBr(aq)} \rightarrow 2\text{KCl(aq)} + \text{Br}_2\text{(aq)}$. Observation: Colourless solution turns orange/yellow. [2]

17. Transition metals have partially filled d-orbitals. Ligands cause d-orbital splitting. Electrons absorb visible light to transition between these levels; the complementary colour is observed. Main group metals have empty or full d-shells, so no such transitions occur. [3]

18. Colour change: Blue to yellow-green. Equation: $[\text{Cu(H}_2\text{O)}_6]^{2+}\text{(aq)} + 4\text{Cl}^-\text{(aq)} \rightleftharpoons [\text{CuCl}_4]^{2-}\text{(aq)} + 6\text{H}_2\text{O(l)}$. [3]

19. The energy levels of the $4\text{s}$ and $3\text{d}$ orbitals are very close. Therefore, varying numbers of electrons can be lost from both orbitals without a huge energy penalty. [2]

20. $\text{I}_2$ has a higher boiling point. Both are non-polar, but $\text{I}_2$ has more electrons, leading to stronger London dispersion forces (instantaneous dipole-induced dipole). [2]