Halide reduction of H2SO4 and halogen volatility trend

Steps:

• Concentrated H2SO4 with NaCl yields HCl and NaHSO4 (S at +6, no reduction).
• With NaBr, it yields Br2 and SO2 (lowest S at +4).
• With NaI, HI reduces H2SO4 to H2S (lowest S at -2; -4 likely denotes strong reduction unique to iodide).
• Thus, NaX is NaI (X = iodine); volatility decreases down group (Cl > Br > I), so iodine (X) is less volatile than bromine (Y).

Why B is correct:

• NaI reaction produces H2S (S = -2 via H2SO4 + 8HI → 4I2 + H2S + 4H2O), and iodine (bp 184°C) is less volatile than bromine (bp 59°C), matching conditions.

Why the others are wrong:

• A: NaBr gives lowest S = +4 (SO2), not -2; bromine more volatile than chlorine.
• C: NaBr gives lowest S = +4; iodine less volatile than bromine (reverses trend).
• D: NaI fits reduction, but astatine less volatile than iodine (reverses trend).

Final answer: B