Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Picture two bottles with the same formula on the label: cobalt chloride with four ammonia molecules. One compound is violet, and the other is green. Yet one mole of each, with excess silver nitrate, gives just one mole of silver chloride. Same formula, same test, two colours. So how can that be? The answer is in this chapter, the unit problem bank for coordination compounds, and puzzles like this do come up in JEE exams. So, let's begin.

0:37How to use this chapter

So, this chapter has three parts. Part A has sixteen solved problems at JEE Main level. Okay? Attempt Part A as a thirty minute test, after Chapters one to seven of this unit, with a clock running, just like the real exam. Right? Then comes Part B, with twelve solved problems at JEE Advanced level: identification problems, where you work out a formula or structure, counting problems, like the number of isomers, and calculations. Our violet and green puzzle is one of them. With me? And Part C has twelve concept checks: the ideas most often tested in statement form, true or false. Clear? Each part ends with one practice problem for you to try.

1:24Part B: one formula, two colours

Got it? Now, let us solve our puzzle. On screen: C o C l three, dot, four N H three. The small numbers written low count atoms, and the centred dot joins the ammonia molecules to cobalt chloride. One form is violet, the other green. Okay? And each one gives one mole of A g C l, silver chloride, per mole. So tell me, how many chloride ions sit outside the square bracket? Just one, yes, exactly! Only chloride outside the bracket precipitates: the C l minus on screen, with a raised minus for its charge. So one chloride is outside, and the other two chlorides sit inside the bracket, together with the four ammonias. Both forms are open square bracket, C o, N H three four times, C l two, close square bracket, C l. Right? But wait, same formula, and the same ions in water. So what can still be different? The arrangement in space, yes! In the figure, the black dot is the metal, M, grey balls are ligand A, here ammonia, and orange balls are ligand B, here chloride. So our complex is of type M A four B two, the two middle drawings: the chlorides are cis, next to each other, or trans, opposite. The other drawings, for M A two B two, and fac and mer for M A three B three, we do not need today. Clear? So the two compounds are geometrical isomers. The violet form is cis, and the green form is trans. Same formula, but a different arrangement in space, and our puzzle is solved.

3:08Part C: true or false?

Got it? Now, Part C gives you short statements, and you decide: true or false. Statement one: the primary valency is ionisable. What do you say? This one is true. The primary valency is the oxidation state of the metal, and it is satisfied by counter ions, the ions outside the bracket. Our puzzle, and both examples coming up, are built on this. Okay? Statement two: all octahedral cobalt three complexes are diamagnetic. On screen, C o with three in Roman numerals, in brackets, means cobalt three. True or false? This one is false. Look at C o F six in square brackets, with a raised three minus for its charge. It is high spin, with four unpaired electrons, so it is not diamagnetic. Right? And statement three: a complex shows the colour of the light it absorbs. Is that true? No, this is false too. We see the complementary colour, not the colour that is absorbed. So in statement questions, read every word slowly. With me?

4:20Example 1: Co(III) complex formula from the AgCl test

Now, Example one, from Part A, based on a JEE Main question from twenty seventeen. One hundred millilitres of zero point one molar C o C l three, dot, six H two O, with excess silver nitrate, A g N O three, precipitates one point two times ten to the power twenty two chloride ions. The capital M means molar, moles per litre. Identify the complex. Pause the video and try it first. Done? Okay, let's check. First, the moles of the complex: molarity times volume in litres. One hundred millilitres is zero point one zero zero litre, so zero point one times zero point one zero zero gives zero point zero one mole. Okay? Now, the formula has three chlorides. So the precipitate must be three times zero point zero one, which is zero point zero three mole, right? No, no, wait, the question gives us the number of chloride ions precipitated. So one point two times ten to the power twenty two, over Avogadro's number, six times ten to the power twenty three, gives zero point zero two mole of silver chloride. With me? And zero point zero two over zero point zero one gives two. So there are two ionisable chlorides per formula unit, not three. Clear? The metal is cobalt three, and its coordination number is six. So the third chloride is inside the bracket, and five water molecules fill the other five places. Right? So the answer is open square bracket, C o, H two O five times, C l, close square bracket, then C l two, dot, H two O, and the sixth water sits outside the bracket. Got it?

6:12Example 2: Pt(IV) complex formula from freezing point, i = 3

Now a harder one, Example two, from Part B, at JEE Advanced level. An aqueous solution of P t C l four, dot, four N H three, of molality zero point zero one zero mole per kilogram, freezes zero point zero five five eight kelvin below pure water. K with a small f, the cryoscopic constant of water, is one point eight six kelvin kilogram per mole. Write the formula of the complex. Pause the video and try it. Here is a hint: count the ions. Done? Okay, let's check. The depression in freezing point, delta T f, with a small f for freezing, equals i times K f times m. Delta is the Greek capital letter, the little triangle, and m is the molality. And i is the van't Hoff factor: moles of particles in solution for each mole of solute taken. Okay? So i equals zero point zero five five eight, over one point eight six times zero point zero one zero. The bottom is zero point zero one eight six, so i is exactly three. Clear? Now, i is three. So there are three chloride ions outside the bracket, right? Oh no, careful, that is the trap. The factor i counts every ion in solution, and the complex ion is one of them. So three ions means one complex cation and just two chloride ions. Got it? Ammonia is neutral and there are four chlorides, so platinum is plus four, written P t with four in Roman numerals, and its coordination number is six. So four ammonias and two chlorides fill the six places inside. Clear? So the answer is open square bracket, P t, N H three four times, C l two, close square bracket, then C l two. See, Example one used silver chloride and this one a freezing point, but both count what is outside the bracket. With me?

8:17Quick revision: AgCl only from outside Cl, i counts all ions

So, quick revision. First, attempt Part A as a thirty minute test after Chapters one to seven, then go on to Part B and Part C, okay? Second, only the chloride outside the square bracket gives silver chloride. And third, the van't Hoff factor, i, counts every ion, the complex ion too. Now open Part A, start your clock, and do try the practice problem at the end of each part. All the best!