Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Caro's acid is H two S O five. Let us find the oxidation state of sulphur the usual way, with hydrogen at plus one and every oxygen at minus two. Two hydrogens give plus two, five oxygens give minus ten, so sulphur must be plus eight. Can sulphur really be plus eight? The highest state the course gives for sulphur is plus six, so something is wrong here, right? Today we find out what. This is Part A of the unit problem bank on the p block, at JEE Main level, and the course suggests trying it as a thirty minute test after chapters one to six. We will build two tools, and then solve two of its questions together, with a quick one on the way. So, let's begin.

0:49Tool 1: oxygen is not always -2

So, the first tool. Oxygen is usually minus two. But the course lists the exceptions: in peroxides it is minus one, and in superoxides it is minus one half, the fraction on screen. And with fluorine, O two F two has plus one, and O F two has plus two; the small numbers after each symbol count the atoms. Okay? In this video, the peroxides matter most. Every peroxide has an O O link; the dash between the two oxygens on screen is a single bond, oxygen to oxygen. In ionic peroxides this is the peroxide ion, O two with a raised two minus for its charge. Two oxygens share that minus two, so each oxygen is minus one. Right? Now a quick one, question eight of Part A, the Q eight on screen. Here are N a two O two, P b O two, M n O two and B a O two. Look, P b O two and B a O two even look alike, with two oxygens on one metal. So which of these are true peroxides? Sodium peroxide and barium peroxide, yes, exactly! N a two O two and B a O two are the only two that contain the peroxide ion. Clear? And here is the trap. P b O two looks just like B a O two, but it is lead four oxide; the Roman four in brackets means lead is in the plus four state. It has ordinary oxide ions, O with a raised two minus, and no O O bond, so with acids it gives no hydrogen peroxide, the H two O two on screen. And M n O two has no peroxide ion either. Got it?

2:25Tool 2: boiling points of hydrides

Now the second tool, the boiling points of hydrides. First, as molecules get heavier, the dispersion forces between them get stronger, so the boiling point rises. Okay? Second, there is hydrogen bonding. Ammonia, N H three, water, H two O, and H F are hydrogen bonded, because nitrogen, oxygen and fluorine are small and very electronegative. These bonds lift the boiling point. Right? But phosphine, P H three, and H two S are not hydrogen bonded, because phosphorus and sulphur are not electronegative enough. Clear? So, tell me this: in group sixteen, which hydride do you think has the highest boiling point? Water, yes, exactly! From H two S to H two S e to H two T e, the molar mass rises, and so does the boiling point; each pointed sign means less than. But hydrogen bonding lifts water above all of them. Got it? Group seventeen tells the same story: H C l, less than H B r, less than H I, less than H F, again because H F is hydrogen bonded. In both groups, hydrogen bonding puts the first hydride on top. Keep this pattern in mind, because example two will test it. With me?

3:41Example 1: S is +6 in Caro's acid H2SO5, one O-O bond

Now, example one, which is question five of Part A, and it is our puzzle from the start. What is the oxidation state of sulphur in Caro's acid, and how many O O bonds does it have? Pause the video and try it first. Done? Okay, let's check. Caro's acid is peroxomonosulphuric acid, H two S O five. The word peroxo is the clue: there is an O O link inside. In the course figure on the right, the bottom one is Marshall's acid, H two S two O eight, and its S, O, O, S bridge is a peroxo link; the double lines in the figure are double bonds. The four acids in the top row have no O O link at all. Okay? Now the oxidation state of sulphur; let us call it x. Two hydrogens give plus two, and five oxygens at minus two give minus ten. So x equals plus eight, right? No, no, wait, that is exactly the trap from our puzzle! The course lists the states of sulphur as minus two, plus two, plus four and plus six, so plus eight cannot be right. Draw the structure first: H O, then O, then S O two, then O H, with a dash for each single bond. There is one O O link, between the first two oxygens. Clear? Those two peroxo oxygens are minus one each, so two times minus one gives minus two. Right? The other three oxygens are ordinary, at minus two each, so three times minus two gives minus six. With me? So two, plus x, minus two, minus six, equals zero, and the double arrow means this gives x equals plus six. That is the highest state of sulphur, just as the course says. Got it? So the answer is plus six, and one O O bond. And our puzzle is solved: plus eight came from treating the two peroxo oxygens as ordinary oxygens at minus two. Very good!

5:38Example 2: boiling point order of NH3, PH3, AsH3, SbH3

Now example two, which is question seven of Part A, a medium one. Arrange N H three, P H three, A s H three and S b H three by boiling point; these are the hydrides of nitrogen, phosphorus, arsenic and antimony. Pause the video and try it first. Done? Okay, let's check. Water was on top in group sixteen, and H F in group seventeen. Ammonia is hydrogen bonded too, so ammonia boils the highest here, right? Take the three without hydrogen bonds first. From P H three to A s H three to S b H three, the molar mass rises, so the dispersion forces get stronger and the boiling point rises. Okay? Now bring in ammonia: its hydrogen bonding lifts it above P H three and A s H three. Right? But not above S b H three, and that is the trap! S b H three has a large molar mass, and its strong dispersion forces outweigh the hydrogen bonding in ammonia. Clear? So the increasing order is P H three, less than A s H three, less than N H three, less than S b H three. If you put ammonia on top, you followed the pattern from group sixteen too far. Got it? One more warning: the melting point is a different story. There, ammonia does come out on top: P H three, less than A s H three, less than S b H three, less than N H three. So read the question carefully, boiling or melting. With me?

7:12Quick revision: peroxo oxygen, true peroxides, SbH3 vs NH3

So, quick revision. Oxygen is minus two, but minus one in a peroxo O O link, so draw the structure first. True peroxides contain the peroxide ion, and P b O two is not one of them. Boiling points rise with molar mass, and hydrogen bonding lifts H F, water and ammonia. But in group fifteen, S b H three still beats ammonia. Now go and try the other questions of Part A, and the practice problems on this page. All the best!