A Level H2 Chemistry Periodic Table Quiz

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

1. Mg has a higher nuclear charge (more protons) than Na, resulting in a stronger electrostatic attraction between the nucleus and the outer electrons, requiring more energy to remove the first electron. [2]

2. Atomic radius decreases across Period 3. This is because the nuclear charge increases while the shielding effect remains relatively constant, pulling the outer electrons closer to the nucleus. [2]

3. Fluorine is more electronegative than chlorine. F has a smaller atomic radius and fewer shielding electrons, allowing the nucleus to attract bonding electrons more strongly. [2]

4. The outer electron of Al is in the 3p subshell, which is higher in energy (less stable) and further from the nucleus than the 3s electron of Mg, making it easier to remove. [2]

5. $\text{Na}_2\text{O}$ is strongly basic; $\text{Al}_2\text{O}_3$ is amphoteric. This is because Na is more electropositive, making its oxide more ionic and basic, whereas Al has a higher charge density and more covalent character. [2]

6. Phosphorus exists as $\text{P}_4$ molecules, while sulfur exists as $\text{S}_8$ molecules. (Wait, corrected: Sulfur $\text{S}_8$ actually has a higher boiling point due to larger molecular size/stronger London forces). Correction for key: Sulfur has a higher boiling point than phosphorus due to larger molecular size ($\text{S}_8$ vs $\text{P}_4$) leading to stronger London dispersion forces. [2]

7. Oxides change from basic ($\text{Na}_2\text{O}$, $\text{MgO}$) to amphoteric ($\text{Al}_2\text{O}_3$) to acidic ($\text{SiO}_2$, $\text{P}_4\text{O}_{10}$, $\text{SO}_3$, $\text{Cl}_2\text{O}_7$). [2]

8. The second electron is removed from a full 2p shell (core electron), which is much closer to the nucleus and experiences significantly less shielding than the 3s electron. [2]

9. $\text{Mg}(\text{s}) + 3\text{H}_2\text{O}(\text{g}) \rightarrow \text{MgO}(\text{s}) + 3\text{H}_2(\text{g})$ [2]

10. The hydration energy of the cation decreases less rapidly than the lattice energy of the hydroxide as the cation size increases down the group. [2]

11. $\text{Mg}^{2+}(\text{aq}) + \text{CO}_3^{2-}(\text{s}) + 2\text{H}^+(\text{aq}) \rightarrow \text{Mg}^{2+}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})$ (Accept: $\text{MgCO}_3(\text{s}) + 2\text{H}^+(\text{aq}) \rightarrow \text{Mg}^{2+}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})$) [2]

12. $\text{Cl}_2$ is a stronger oxidizing agent than $\text{Br}_2$. Chlorine has a smaller atomic radius and higher electronegativity, making it more effective at attracting electrons. [2]

13. The colorless solution turns brown (due to $\text{I}_2$ formation) and a dark precipitate may appear. [2]

14. The $\text{H}-\text{I}$ bond is longer and weaker than the $\text{H}-\text{F}$ bond due to the larger size of the iodide ion, making it easier for the bond to break and release $\text{H}^+$. [2]

15. $\text{Ca}(\text{OH})_2(\text{aq}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{CaSO}_4(\text{s}) + 2\text{H}_2\text{O}(\text{l})$ [2]

16. Transition metals have partially filled d-orbitals. Ligands cause these d-orbitals to split into different energy levels. Electrons absorb visible light to jump between these levels; the remaining light is transmitted as the complementary color. [3]

17. The energy difference between the 3d and 4s orbitals is very small, allowing different numbers of d-electrons to be lost/shared. [2]

18. They provide an alternative reaction pathway with a lower activation energy, often by adsorbing reactants onto their surface or forming intermediate complexes. [2]

19. The color changes (typically to yellow/green-yellow). This is because the ligand field changes, altering the energy gap between d-orbitals and thus changing the wavelength of light absorbed. [2]

20. $[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$ or $[\text{Cu}(\text{NH}_3)_4]^{2+}$ [2]