Tests for Halogens and Phosphorus

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Someone heats an organic compound on a clean copper wire in a flame, and the flame turns green. Green flame on copper means a halogen is present, right? But then the silver nitrate test on the same compound shows no halogen at all. So which test is telling the truth: the flame, or the silver nitrate? Keep your answer in mind; by the end of this video, you will know exactly which one to trust. Today we learn the tests for halogens and for phosphorus, and questions on these do come up in JEE Main. So, let's begin.

0:37Halogens: boil with nitric acid first

So, first the halogens: chlorine, bromine and iodine. After sodium fusion, the halogen sits in the sodium fusion extract as halide ions, and we test them with silver nitrate, the formula on screen. The small number below the symbols is a subscript, okay? But suppose the compound also had nitrogen or sulphur. Then the extract also has cyanide and sulphide ions. What will they do with silver nitrate? They form precipitates too, yes, exactly! On screen, C N with a small minus sign up top is the cyanide ion, and S with a two minus is the sulphide ion; those raised signs are the charges. Silver cyanide is white, and silver sulphide is black, and that would confuse the result. Right? So the fix is simple. Before adding silver nitrate, boil the extract with dilute nitric acid. This decomposes the sodium cyanide and the sodium sulphide, and expels them as HCN and H two S, hydrogen cyanide and hydrogen sulphide. Remember this step; questions often ask why it is done. Clear?

1:49Silver nitrate: read the colour, then the ammonia

Now add silver nitrate. Chlorine gives silver chloride, a white precipitate. And it dissolves readily in ammonium hydroxide, the N H four O H on screen, which is ammonia solution. Okay? Bromine gives silver bromide: pale yellow, and only sparingly soluble in ammonia, so just a little of it dissolves. Got it? And iodine gives silver iodide: yellow, and insoluble. Now, pale yellow and yellow are very close colours, aren't they? So how can you be sure which one you have? Yes, use the ammonia! Readily soluble means chlorine, sparingly soluble means bromine, and insoluble means iodine. So always read two things: the colour, and the solubility in ammonia. With me?

2:37The layer test: bromine or iodine?

Now, how do we confirm bromine or iodine? We use the layer test. Take the acidified extract, add carbon tetrachloride, C C l four, or chloroform, CHC l three, which is in brackets because either solvent works, then add chlorine water, and shake; the plus signs just mean we add them together. Now tell me: the chlorine in chlorine water meets bromide ions. What happens? Chlorine liberates bromine, yes, exactly! Chlorine plus two bromide ions gives two chloride ions plus bromine; the arrow means gives. The free bromine dissolves in the carbon tetrachloride layer, the organic layer, and colours it orange-brown. Okay? And if the halogen is iodine, the organic layer turns violet instead. So orange-brown means bromine, and violet means iodine. Now here is a tricky one. Can this layer test detect chlorine? It cannot, and here is why. The test works only because chlorine water liberates the halogen. But chlorine water cannot liberate chlorine from chloride, so the layer test tells us nothing about chlorine. For chlorine, we depend on silver nitrate and ammonia.

3:53Beilstein's test: quick, but not conclusive

Clear? Next, Beilstein's test, a really quick one. Put a little of the compound on a copper wire and hold it in a flame. With a halogen, the flame turns green or blue-green. Right? And why does the flame turn green? Because volatile copper halides form, and they colour the flame. So, the flame is green. Can you tell me which halogen it is? That is right, you cannot tell. Chlorine, bromine and iodine all form copper halides, so the test does not say which halogen. With me? And there is a bigger problem. Some compounds with no halogen at all also give a green flame: urea, thiourea, and some pyridine derivatives. So Beilstein's test is not conclusive. Remember our puzzle from the start? Hold that thought; we finish it in Example two. Got it?

4:44Phosphorus: the yellow ammonium phosphomolybdate

Finally, let us test for phosphorus. For this test, the phosphorus in the compound must end up as phosphate. Phosphate is phosphorus joined to oxygen, so the phosphorus has to be oxidised. So what kind of reagent do we heat the compound with? We need an oxidising agent, very good! We use sodium peroxide, the formula on screen, and the capital P stands for phosphorus. The phosphorus becomes sodium phosphate. Okay? Next, the solution is boiled with nitric acid. Sodium phosphate, plus three nitric acid, gives phosphoric acid, plus three sodium nitrate. Clear? Then we add ammonium molybdate. Phosphoric acid, plus twelve ammonium molybdate, plus twenty one nitric acid, gives ammonium phosphomolybdate, plus twenty one ammonium nitrate, plus twelve water. The brackets around N H four hold the ammonium groups, and the dot in the middle joins the two parts of one formula. Right? And what do we see? We see a yellow precipitate of ammonium phosphomolybdate, and that means phosphorus is present. So for phosphorus, think yellow. Got it?

5:56Example 1: violet chloroform layer, which halogen and why

Now, let us try Example one. The extract, shaken with chloroform and chlorine water, gives a violet organic layer. Which halogen is present, and what is the chemistry? Pause the video and try it first. Done? Okay, let's check. Violet organic layer: that is iodine, straight from our layer test. Now for the chemistry part. The extract had iodide ions, not free iodine. So where did the free iodine come from? From the chlorine water, correct! Just as with bromine, chlorine liberates iodine: chlorine plus two iodide ions gives two chloride ions plus iodine. Right? And the iodine dissolves in the chloroform with a violet colour. So the halogen is iodine, and the chemistry is chlorine liberating iodine from iodide, and that is the complete answer.

6:50Example 2: green flame, Prussian blue, no AgNO3 precipitate

Now Example two, a harder one, and it is our puzzle from the start. One compound gives three results. First, Beilstein's test gives a green flame. Second, the extract gives Prussian blue. And third, silver nitrate gives no precipitate, after the extract is boiled with nitric acid. What can you conclude? Pause the video and try it first. Done? Let me test you first. The flame is green, so a halogen is present, right? No, no, careful, that is exactly the trap! Let us go clue by clue. Prussian blue is the test for nitrogen, from the earlier lesson in this chapter. So nitrogen is present. Okay? Next, the extract was boiled with nitric acid, and then silver nitrate gave no precipitate. This is the reliable test for halogens, so there is no halogen. With me? So the green flame came from the nitrogen compound, through volatile copper cyanide. It did not come from a halogen at all. Clear? So the conclusion is nitrogen present, and no halogen. And that solves our puzzle from the start. When Beilstein's flame and the silver nitrate test disagree, trust the silver nitrate test, done after boiling the extract with nitric acid. See how the clues fit together?

8:13Quick revision: silver nitrate, layer test, phosphorus test

So, quick revision. For halogens, boil the extract with nitric acid, then add silver nitrate; the colour and the solubility in ammonia tell you which halogen. In the layer test, orange-brown is bromine and violet is iodine, and Beilstein's test is not conclusive. For phosphorus, use sodium peroxide, then nitric acid and ammonium molybdate, and look for the yellow precipitate. Now go and try the practice problems on this page. You will do well. All the best!