Overview

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Think of diborane, B two H six. Two boron atoms, side by side in the formula. So the two borons must be bonded to each other, right? Hmm, are you sure about that? That exact question sits in this chapter, the unit problem bank for the p block elements. Questions like this do come up in JEE, and they catch people who trust the formula. Today: how to use this bank, two of its problems, and this puzzle, solved. So, let's begin.

0:37How to use this chapter

So, this chapter has three parts. Part A has eight solved problems at JEE Main level, then two more for practice. Okay? The instruction is simple: attempt Part A as a thirty minute test, after Chapters one to six. So sit with a clock, and open the solutions only at the end. Right? Then Part B: seven longer problems, at JEE Advanced level, and two for practice. With me? And what kind of problems are they? Mostly counting problems, like how many of these molecules have a certain bond, and reaction-sequence problems, where one product leads to the next. They combine several groups, so you need the whole unit at once. Got it? And Part C: six concept checks, the ideas that are often tested in statement form, like true or false. Our diborane puzzle lives there. Clear?

1:32Part C: read the structure, not the formula

Now, Part C, and here is our puzzle, as a statement. True or false: B two H six, diborane, has a B B bond, a bond between the two borons; the short dash means a bond. The answer is right there in the figure, on the left. Look at the two borons, the circles with B inside them. Is there a line joining them directly? There is no line between them, so the statement is false. The two borons are held together only through the two BHB bridges, the curved bonds through the top and the bottom hydrogens. Four more hydrogens sit at the ends, the terminal ones. Okay? And each bridge is a three centre, two electron bond: three atoms share just one pair of electrons. On screen, three c dash two e. These bridges are also called banana bonds. The figure on the right is the aluminium chloride dimer, from Chapter one; today we only need the left side. Right?

2:28Part C: is it really a peroxide?

Here is one more Part C statement. True or false: P b O two is a peroxide. There are two oxygen atoms in the formula, after all. So, true or false? If you said true, careful, the statement is false. P b O two is lead four oxide; the Roman numeral four in brackets is the oxidation state of lead. Its oxygens are oxide ions, O with a two minus charge, written up top. With me? And the proof: with acids, it gives oxygen, O two, or, with hydrochloric acid, chlorine, C l two, but not hydrogen peroxide, H two O two. Clear? So remember the rule: a true peroxide, like sodium peroxide or barium peroxide, contains the peroxide ion, O two with a two minus charge, and that ion has an O O bond. Keep that O O bond in mind, because our first example is built on it. Let's go.

3:23Example 1: Caro's acid, oxidation state of S and O-O bonds

Now, Example one, from Part A, okay? What is the oxidation state of sulphur in Caro's acid, and how many O O bonds, oxygen to oxygen bonds, does it have? Pause the video, and try it first. Done with it? Okay, let's check. Caro's acid is H two S O five, peroxomonosulphuric acid. Its structure: H O, dash, O, dash, S O two, dash, O H, where each dash is a single bond. Look at the middle: one oxygen bonded to another oxygen, the peroxo link. Clear? Now the oxidation state of sulphur; call it x. Each hydrogen is plus one, and every oxygen is minus two, as usual. Two hydrogens give plus two, five oxygens give minus ten, so the sulphur comes out at plus eight, right? No, no, wait, that cannot be right! Sulphur has only six valence electrons, so it can never go above plus six. Here is the mistake: the two oxygens of the O O link are peroxo oxygens, and peroxo oxygen is minus one, not minus two. The other three oxygens stay at minus two, and each hydrogen is plus one. Got it? So, two from the hydrogens, plus x, minus two from the peroxo pair, minus six from the other three, equals zero. The double arrow on screen means therefore, and the oxidation state of sulphur, x, is plus six. Okay? And the O O bonds? Just one, the peroxo link. So the answer is plus six, and one bond. See how the O O bond changed the whole calculation? With me?

5:04Example 2: count same-element bonds, B2H6 to H2S2O8

Now a harder one, from Part B, JEE Advanced level, a counting problem. Among B two H six, B three N three H six, N two O, N two O four, H two S two O three, and H two S two O eight, how many contain a covalent bond between two atoms of the same element? Pause the video and try it. Hint: draw each structure, and do not trust the formula. Done with it? Okay, let's check, one by one. First, diborane, our puzzle from the start! Two borons in the formula, but no B B bond, because the borons are joined through the hydrogen bridges. So diborane does not count. Right? Next, B three N three H six, borazine. Three borons and three nitrogens in one ring. Is there a B B bond, or an N N bond, anywhere in it? No, there is not. The ring alternates, boron, nitrogen, boron, nitrogen, so borazine has only B N bonds. It does not count either. Clear? Now the two oxides of nitrogen. N two O, laughing gas, is built as nitrogen, nitrogen, oxygen, so the two nitrogens are bonded to each other. And N two O four also has an N N bond. So that is two that count. Okay? Next, H two S two O three, thiosulphuric acid. It has a sulphur to sulphur bond, so it counts. That makes three. With me? And the last one, H two S two O eight, peroxodisulphuric acid, also called Marshall's acid. Its two sulphurs are not bonded to each other. So it does not count, right? Oh no, careful, read the question again! It says two atoms of the same element, any element. Between the two sulphurs sits a peroxo bridge: sulphur, oxygen, oxygen, sulphur. That is an O O bond, so Marshall's acid does count. Got it? So the count is N two O, N two O four, thiosulphuric acid and Marshall's acid. The answer is four. And our puzzle? Diborane has two borons, but no bond between them, so our puzzle is solved. Okay?

7:18Quick revision: how to use Parts A, B and C

So, quick revision. Part A: a thirty minute test after Chapters one to six. Part B: JEE Advanced level, mostly counting, so draw every structure before you count. Part C: statements; read the structure, not the formula, just like diborane. Now open Part A, start your clock, and begin. All the best!