2024 H2 Chem P1 Q6 — Mole Solution

The answer is B.
Using \(A_r\): Si = 28.1, P = 31.0, S = 32.1, O = 16.0, Cl = 35.5. Work out the formula from each percentage and compare with the element's highest oxidation state.
• B: \(\text{SiO}_2\) has \(\%\text{O} = \frac{32.0}{60.1} \times 100 = 53.2\%\). This matches, and Si is +4 in \(\text{SiO}_2\), which is its highest oxidation state (Group 14). Correct.
• A is wrong: \(\text{PCl}_3\) has \(\%\text{Cl} = \frac{106.5}{137.5} \times 100 = 77.5\%\), so P is only +3. With P in its highest state, +5, \(\text{PCl}_5\) would contain \(\frac{177.5}{208.5} \times 100 = 85.1\%\) Cl.
• C is wrong: \(\text{P}_4\text{O}_6\) has \(\%\text{O} = \frac{96.0}{220.0} \times 100 = 43.6\%\), so P is only +3. \(\text{P}_4\text{O}_{10}\) (P +5) would contain \(\frac{160.0}{284.0} \times 100 = 56.3\%\) O.
• D is wrong: \(\text{SO}_2\) has \(\%\text{O} = \frac{32.0}{64.1} \times 100 = 49.9\%\), so S is only +4. \(\text{SO}_3\) (S +6) would contain \(\frac{48.0}{80.1} \times 100 = 59.9\%\) O.