A sodium salt X with dilute H2SO4 evolves a gas that blackens lead acetate paper. The gas, passed into aqueous SO2, gives a yellow precipitate: 2H2S + SO2 -> 3S + 2H2O. X and the moles of electrons transferred per mole of sulphur formed are respectively

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JEE Advanced levelSingle correct optionDifficulty 3 of 3qa034

A sodium salt X with dilute $H X_{2} SO X_{4}$ evolves a gas that blackens lead acetate paper. The gas, passed into aqueous $SO X_{2}$ , gives a yellow precipitate: $2 H X_{2} S + SO X_{2} 3 S + 2 H X_{2} O$ . X and the moles of electrons transferred per mole of sulphur formed are respectively

A $Na X_{2} S X_{2} O X_{3}$ ; $\frac{4}{3}$
B $Na X_{2} S$ ; $\frac{4}{3}$
C $Na X_{2} S$ ; 4
D $Na X_{2} S$ ; 2

Answer: B

A gas blackening lead acetate is

H X_{2} S

, so X is

Na X_{2} S

. In the reaction, 2 S go from -2 to 0 (4 electrons lost) and one S goes from +4 to 0 (4 electrons gained), so 4 electrons are transferred for 3 S. This is 4/3 per mole of S. Thiosulphate does not blacken lead acetate paper.

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A sodium salt X with dilute HX2​SOX4​ evolves a gas that blackens lead acetate paper. The gas, passed into aqueous SOX2​, gives a yellow precipitate: 2HX2​S+SOX2​​3S+2HX2​O. X and the moles of electrons transferred per mole of sulphur formed are respectively

Answer: B

More questions on Dilute acid group anions: carbonate, sulphite, sulphide, nitrite

A sodium salt X with dilute $H X_{2} SO X_{4}$ evolves a gas that blackens lead acetate paper. The gas, passed into aqueous $SO X_{2}$ , gives a yellow precipitate: $2 H X_{2} S + SO X_{2} 3 S + 2 H X_{2} O$ . X and the moles of electrons transferred per mole of sulphur formed are respectively