Assuming that only particles emitted during natural radioactive decay are α and β particles, which o — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Natural Radioactive Decay Rules In natural radioactive decay, only **α and β particles** are emitted: - α decay: $A \to A-4$, $Z \to Z-2$ - β decay: $A \to A$, $Z \to Z\pm1$ Starting from $\ce{^{235}_{92}U}$: any product must satisfy $A = 235 - 4n_\alpha$ for some integer $n_\alpha \geq 0$ (since β-decay does not change A). Step 2 - Check Each Option **(A) $\ce{^{231}_{90}Th}$**: $\Delta A = 235 - 231 = 4$ → one α decay needed. $\Delta Z = 92 - 90 = 2$ → one α decay gives $Z = 90$. Consistent (no β needed). ✓ Possible. **(B) $\ce{^{227}_{89}Ac}$**: $\Delta A = 235 - 227 = 8$ → two α decays. After two α: $Z = 92 - 4 = 88$. From 88 to 89: one β⁻ decay. Sequence: $2\alpha + 1\beta^-$. ✓ Possible. **(C) $\ce{^{235}_{89}Ac}$**: $A = 235$ → zero α decays (since A unchanged). But starting from $Z = 92$: with only β decays (no α), $A$ stays 235 and $Z$ can only change by ±1. To reach $Z = 89$ from $Z = 92$, we need $\Delta Z = -3$. With $\beta^-$, $Z$ increases; with $\beta^+$, $Z$ decreases. Three $\beta^+$ decays would give $Z = 89$. But $\beta^+$ requires extra mass that must come from somewhere — and more importantly, $\ce{^{235}_{92}U}$ decay chain does not include $A=235$ products with $Z=89$. In the standard $(4n+3)$ Actinium series, all nuclides satisfy $A = 4n+3$: $235 = 4(58)+3$, so Ac-235 must satisfy $235 \mod 4 = 3$ ✓, but we need to check if such a decay path exists. From U-235 to Ac-235: $\Delta Z = 3$ decrease with no change in $A$ requires 3 β⁺ or EC decays — these are **not natural** decay types in the heavy-element series. U-235 undergoes α then β⁻, not β⁺. ✗ **Not possible.** **(D) $\ce{^{207}_{82}Pb}$**: This is the terminal stable nuclide of the Actinium series. ✓ Possible. $$\boxed{\text{Answer: C — }\ \ce{^{235}_{89}Ac}\ \text{ cannot result from natural decay of U-235}}$$