Tests for the Halogens

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Imagine an organic compound with nitrogen in it, but no halogen at all. We make its sodium fusion extract, add silver nitrate straight away, and see a white precipitate. White precipitate with silver nitrate, so chlorine is present, right? But the compound has no chlorine! So where did that white precipitate come from? We will find out today, with the tests for the halogens. And these tests come up often in JEE questions on practical chemistry. So, let's begin.

0:37Silver nitrate test: boil with nitric acid first

So, we start from the sodium fusion extract. First, we boil it with dilute nitric acid, HNO three; the small three below is a subscript, counting the oxygen atoms, and the plus sign means we add the acid. Then we add silver nitrate solution, A g N O three on screen. Okay? Now, why this boiling step? Remember our puzzle: the compound had nitrogen in it. So during the fusion with sodium, what did that nitrogen become? Yes, exactly, it became sodium cyanide! And any sulphur becomes sodium sulphide. Boiling with nitric acid drives them out: the cyanide as HCN, hydrogen cyanide, and the sulphide as H two S, hydrogen sulphide. With me? And if we skip this step? Silver ions precipitate cyanide and sulphide too: white silver cyanide, A g C N, and black silver sulphide, A g two S on screen. So that solves our puzzle! The white precipitate was silver cyanide, and it was mistaken for silver chloride. Clear? Now a trap question for you. Why nitric acid? Why not hydrochloric acid, or sulphuric acid? Hydrochloric acid, H C l, brings its own chloride ions, so it would itself give silver chloride, a false result! And sulphuric acid, H two S O four, could precipitate silver sulphate. Silver nitrate is not affected by nitric acid, so nitric acid is the safe choice. Got it?

2:10Reading the silver halide precipitate

Now the precipitate tells us which halogen it is, from two things: its colour, and what ammonia solution does to it. Chloride gives silver chloride, A g C l, a white precipitate, and it dissolves in ammonia solution. Okay? Bromide gives silver bromide, A g B r, pale yellow, and only sparingly soluble in ammonia, so just a little of it dissolves. Right? And iodide gives silver iodide, A g I on screen, yellow, and it does not dissolve in ammonia at all. So read the order: soluble, sparingly soluble, insoluble. With me? Why does silver chloride dissolve? Because it forms a complex with ammonia. Silver chloride plus two N H three gives the complex in square brackets: silver with two ammonia molecules, the N H three in round brackets taken twice, and then C l outside the brackets. The arrow means gives. Clear? Now, a question for you. Fluorobenzene has a halogen too, fluorine. So what will its extract show with silver nitrate? It shows no precipitate at all! The fusion gives sodium fluoride, but silver fluoride, A g F, is soluble in water, so nothing comes down. So this test cannot detect fluorine. Careful, this one is easy to miss. Got it?

3:27The layer test for bromine and iodine

Now a second test, only for bromine and iodine: the layer test. We acidify the extract and add a little carbon tetrachloride, C C l four. Chloroform, CHC l three, or carbon disulphide, C S two, shown in the brackets, also work. Okay? Then we add chlorine water drop by drop, shaking all the time. Now tell me, what does the chlorine do to the bromide and iodide ions? Yes, exactly, it oxidises them and sets the halogen free! Two bromide ions, B r with a small minus up top for the negative charge, plus C l two, give B r two, which is bromine, plus two chloride ions; the arrow means gives. Right? The same way, two iodide ions with chlorine give iodine, the I two on screen, plus two chloride ions. With me? The free halogen collects in the organic layer and colours it. Orange brown means bromine, and violet means iodine. Clear?

4:27Example 1: pale yellow precipitate, partly soluble in NH3

Now, example one. In the silver nitrate test, we get a pale yellow precipitate, and it dissolves only partly in concentrated ammonia. Identify the halogen, and confirm it with a second test. Pause the video and try it first. Done? Okay, let's check. The precipitate is yellowish, so it must be iodine, right? No, no, careful, that is the trap, so look at both clues together. The colour is pale yellow, and the precipitate dissolves partly, which means sparingly soluble. That is silver bromide, so the halogen is bromine. Okay? Silver iodide would not dissolve in ammonia at all, so iodine is ruled out. Right? Now confirm it with the layer test. With carbon tetrachloride and chlorine water, the carbon tetrachloride layer turns orange brown. That is bromine, confirmed by two tests. Got it?

5:27Example 2: layer test with both bromide and iodide

Now example two, a harder one. In the layer test, the extract contains both bromide and iodide. Which colour appears first, and why? Pause the video and try it first. Done? Okay, let's check. Chlorine water is the oxidising agent here. So which ion does it oxidise first, bromide or iodide? In other words, which one is the stronger reducing agent? Yes, exactly, it is iodide. Iodide is the stronger reducing agent, so chlorine oxidises it first. With me? So iodine is set free first, and the organic layer turns violet first. Okay? Now we keep adding chlorine water, more and more. So the layer simply stays violet, and the bromine stays hidden, right? No, no, wait, the colour changes! With excess chlorine water, the iodine itself is oxidised further, to iodate, and iodate is colourless. So the violet goes away. Clear? And now the orange brown of bromine shows. So violet first, then orange brown, and that tells us both iodine and bromine are present. Got it?

6:42Quick revision: nitric acid step, silver halides, layer test

So, quick revision. Boil the extract with dilute nitric acid first, then add silver nitrate; and remember, fluoride gives no precipitate. Silver chloride is white and soluble in ammonia, silver bromide is pale yellow and sparingly soluble, and silver iodide is yellow and insoluble. In the layer test, violet is iodine and orange brown is bromine. Now go and try the practice problems on this page. You will find them easy now. All the best!