Mixed Problems: Set 2 (Detection of Elements)

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a compound with nitrogen right there in its formula. Fuse it with sodium, and run the nitrogen test. The Prussian blue never comes, and the test says there is no nitrogen! So how can a nitrogen compound fail the nitrogen test? Keep that question in mind. This is Set two of the mixed problems on detecting the elements, in the JEE Main style, and questions like these come up often. First we sharpen three tools with quick questions from the set, and then we solve two problems, our puzzle included. So, let's begin.

0:39Tool one: why fuse with sodium?

So, tool one, and the first quick question from the set is an easy one. Which metal is used to prepare the Lassaigne extract? Yes, it is sodium! In an organic compound, nitrogen, sulphur and the halogens are held by covalent bonds, so they give no ionic tests. Sodium turns them into ionic sodium salts; the arrow means becomes, and X stands for any halogen. The chart on the right shows all the tests we then do on this sodium fusion extract. With me? Here is how. Sodium, carbon and nitrogen give N a C N, sodium cyanide. And two sodium plus sulphur give N a two S, sodium sulphide; the small two written low is a subscript, and it counts the sodium atoms. Okay? And sodium plus a halogen gives the sodium halide, where X can be chlorine, bromine or iodine. Right? Now look at the first equation again. Nitrogen comes out only as cyanide: C N with a small minus sign written high, which is its charge. Or, when sulphur is also there, as thiocyanate, SCN minus, which we meet next. Both of them need carbon. Hold on to this line, because it will solve our puzzle. Clear?

1:55Tool two: the colours for nitrogen and sulphur

Now tool two, the colours. Next question from the set: in the nitrogen test, the Prussian blue colour is due to which compound? Here it is: F e four, then F e C N six in square brackets, taken three times. The square brackets hold one complex ion, and the round brackets group the cyanide. It is the first box in the bottom row of the chart. Okay? Its name is iron three hexacyanoferrate two. The Roman numerals in brackets are oxidation states: the iron outside the square bracket is plus three, and the iron inside is plus two. Now another one from the set. The extract gives a black precipitate with lead acetate. Which element does it show? It is sulphur, very good! Sulphide ions meet lead ions and give P b S, lead sulphide, which is black. Right? Now a statement question. Statement one says that a blood red colour with F e C l three, iron three chloride, shows nitrogen and sulphur together. Statement two says the colour is due to the F e SCN ion, with a charge of two plus. Are both of them true? Statement one is true. When nitrogen and sulphur are both present, they can stay together as thiocyanate, SCN with a minus charge. And iron three chloride turns that extract blood red. With me? And here is the reaction. F e three plus, plus SCN minus, gives F e SCN in square brackets, with a charge of two plus. That complex ion is the blood red colour, so statement two is true as well. Clear?

3:37Tool three: the halogens

Now tool three, the halogens, which is the fourth branch of the chart. Another statement question from the set. Statement one: the extract is boiled with dilute HNO three, nitric acid, before adding A g N O three, silver nitrate. Statement two: this removes cyanide and sulphide ions. Are both true, and is the second the reason for the first? Yes, both are true, and statement two is the reason! Boiling with the acid drives out HCN and H two S, hydrogen cyanide and hydrogen sulphide. Otherwise cyanide and sulphide would also give precipitates with silver nitrate, and fool us. Okay? Now add silver nitrate. Chloride gives A g C l, silver chloride: white, and readily soluble in ammonium hydroxide, the N H four O H on screen, which is ammonia solution. Right? Bromide gives A g B r, silver bromide: pale yellow, and only sparingly soluble. Now a quick one from the set. Silver nitrate gives a yellow precipitate, insoluble in ammonia. Which halogen is it? That is iodine, correct! Silver iodide, A g I on screen, is yellow and insoluble. So read two things every time: the colour, and the solubility in ammonia. With me? Last one in this tool, the layer test. The acidified extract is shaken with C C l four, carbon tetrachloride, and chlorine water. The carbon tetrachloride layer turns orange brown. Which halogen is it, and what does the chlorine water do? It is bromine, yes, exactly! Chlorine water oxidises bromide to bromine: C l two, plus two bromide ions, gives two chloride ions, plus B r two. Clear? The free bromine dissolves in the carbon tetrachloride layer and colours it orange brown. With iodide, you get iodine instead, and the layer turns violet. Got it?

5:42Example 1: why hydroxylamine fails the nitrogen test

Now, Example one from Set two, and it is our puzzle! Which of these does not give Lassaigne's test for nitrogen: urea, hydroxylamine, aniline or acetamide? Pause the video and try it first. Done? Okay, let's check. All four compounds contain nitrogen. So all four give the test, right? No, no, careful, look at the carbon first! Urea is N H two CONH two, aniline is C six H five N H two, and acetamide is C H three CONH two. All three have carbon as well as nitrogen, so all three give sodium cyanide. Okay? But hydroxylamine is N H two O H: nitrogen, hydrogen and oxygen, and no carbon at all. Right? And remember tool one: nitrogen shows up only as cyanide or thiocyanate, and both need carbon. The double arrow means so. No carbon, so no cyanide, so no Prussian blue. Got it? So the answer is hydroxylamine. And that solves our puzzle: a compound can contain nitrogen and still fail the test, simply because it has no carbon. Clear?

6:55Example 2: thiourea fused with excess sodium

Now, Example two, a medium one. Thiourea is fused with excess sodium. Which tests are positive? Pause the video and try it first. Done? Okay, let's check. Thiourea is N H two CSNH two, so it has carbon, nitrogen and sulphur, all three. Nitrogen and sulphur in one compound, so the answer is the blood red colour, isn't it? Oh no, careful, because that is exactly the trap! Blood red comes with a little sodium, when sodium, carbon, nitrogen and sulphur stay together as sodium thiocyanate. But read the question again: excess sodium. Okay? With excess sodium, the thiocyanate breaks up. Sodium thiocyanate, plus two sodium, gives sodium cyanide, plus sodium sulphide. With me? Now cyanide and sulphide are separate and free. Sodium cyanide gives Prussian blue, and sodium sulphide gives the sulphide tests. Clear? So the answer is Prussian blue and the sulphide tests, but not blood red. Always check whether the question says a little sodium or excess sodium. Got it?

8:07Quick revision: N, S and halogen tests on the extract

So, quick revision. First, nitrogen shows up only as cyanide or thiocyanate, so with no carbon, there is no Prussian blue. Second, nitrogen and sulphur together give blood red with a little sodium, but excess sodium separates them, and then the separate tests work. And third, for halogens, boil with nitric acid first, then add silver nitrate, and check both the colour and the solubility in ammonia. Now go and try the three practice problems on this page. You can do it. All the best!