This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. An organic compound was fused with sodium and the water extract was treated with silver nitrate solution when a black precipitate was obtained. The test suggests (a) absence of halogen (b) presence of sulphur (c) presence of nitrogen (d) presence of halogen

An organic compound was fused with sodium, and the resulting solution was treated with silver nitrate. The test resulted in the formation of a black precipitate.

We will now analyze each of the given choices in the multiple-choice question to determine which one is correct. (a) Absence of halogen: The formation of a black precipitate with silver nitrate suggests a reaction, which indicates the presence of a certain chemical element, not the absence of one. (b) Presence of sulphur: When sulphur is mixed with silver nitrate, Silver sulphide (Ag\(_2\)S) is formed which is black precipitate. However, we'll only be able to confirm this option if no other option produces a black precipitate as well. (c) Presence of nitrogen: Nitrogen does not react with silver nitrate to produce a black precipitate; thus, this option can be excluded. (d) Presence of halogen: When a halogen (such as chloride, bromide or iodide) reacts with silver nitrate, a precipitate of silver halides (such as AgCl, AgBr, or AgI) is formed. Among these, AgBr (Silver Bromide) is a cream-colored precipitate and AgI (Silver Iodide) forms as yellow precipitate whereas AgCl (Silver Chloride) forms a white precipitate. None of these are black precipitates, and thus this option can be excluded.

Based on the analysis in Step 2, the only choice which results in the formation of a black precipitate when treated with silver nitrate is (b) presence of sulphur. So, the correct answer is: (b) presence of sulphur