Question

What are the products formed when $NaOH$ reacts with excess $S$ ?
(A) $N{a_2}{S_5},N{a_2}S{O_4}$
(B) $N{a_2}S,N{a_2}{S_2}{O_3}$
(C) $N{a_2}{S_5},N{a_2}{S_2}{O_3}$
(D) $N{a_2}{S_2}{O_3}$

Answer 1

We know that sodium hydroxide when boiled with an excess of sulfur, a disproportionate reaction occurs. In a disproportionate reaction, the same element in a compound is getting oxidized and reduced simultaneously. Here $S$ is undergoing a disproportionation reaction.

Complete answer: Sodium Hydroxide is an inorganic compound. It is also known as caustic soda. It is ionic and is a white crystalline solid. It has a formula $NaOH$ with cation $N{a^ + }$ and anion $O{H^ - }$ . Sulfur is a group $16$ element in the periodic table. It is non-metallic in nature and has a symbol $S$ . Sulfur is bright yellow, solid at room temperature. When sodium hydroxide is boiled with an excess of sulfur, the following reaction occurs:

It is a disproportionation reaction. A disproportionation reaction is a redox reaction in which the same element in a compound is getting oxidized and reduced simultaneously. So here sulfur $\left( S \right)$ is undergoing a disproportionation reaction. Its oxidation state changes from zero to $- 2$ in sodium sulfide to $+ 4$ in sodium thiosulfate.
$N{a_2}S$ is sodium sulfide. It is a colorless solid and water-soluble. And it gives a strong alkaline solution when dissolved in water.
$N{a_2}{S_2}{O_3}$ is called sodium thiosulfate. It is an inorganic compound that is used as an antidote in cyanide poisoning and also an antifungal drug.

Thus, the correct option is $A$.

Note: It is important to write the correct stoichiometric coefficients in the equation. Different reactions occur if we change the stoichiometric coefficients. Look at the following equations:

So if we compare these three equations, we can see that different products are formed if we change the stoichiometric coefficients.