Overview

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Potassium permanganate is purple. And colour in transition metal ions usually comes from d to d transitions, right? But in permanganate, the manganese has no d electrons at all. So where does the purple come from? Hmm, strange, isn't it? That exact question sits in this chapter, the unit problem bank for the d and f block elements. Questions like this do come up in JEE, often as true or false statements. In this video, we will see how to use this bank, solve this puzzle, and work through two of its permanganate problems. So, let's begin.

0:41How to use this chapter

So, this chapter has three parts. Part A has eight solved problems at JEE Main level. Okay? And the instruction is clear: attempt Part A as a thirty minute test, after Chapters one to six of this unit. So, keep a clock running, just like a real test. Right? Then comes Part B: six solved problems at JEE Advanced level. With me? These are identification problems, where you name a compound or a metal from its colours and its tests, and calculation problems, like titrations. Clear? And Part C has six concept checks, the ideas that are often tested in statement form, true or false. Our purple puzzle lives right there. Got it? And each part ends with one practice problem for you to try on your own.

1:33Part C: purple, with no d electrons?

Okay? Now, Part C, and here is our puzzle, as a statement. K M n O four, potassium permanganate, gets its colour from a d to d transition; the small four written low counts the oxygen atoms, and the dash reads as d to d. Look at the figure. The third ion is permanganate, M n O four minus, where the raised minus sign is its charge, and it is labelled purple, d zero, where the raised zero counts the d electrons. So, true or false? The statement is false. In permanganate, manganese, M n, is plus seven, so all its d electrons are gone, and it is d zero. And no d electrons means no d to d transition at all. Right? So its purple comes from charge transfer, from the ligand to the metal; here an electron moves from an oxygen to an empty manganese orbital. This gives an intense colour, even with d zero. So, having no d electrons does not always mean colourless. Clear? And the same is true for the first two ions in the figure: chromate, C r O four two minus, which is yellow, and dichromate, C r two O seven two minus, which is orange. Chromium is plus six in both, so again d zero, and again charge transfer. The angle marked on dichromate, about one hundred and twenty six degrees, is its chromium, oxygen, chromium bridge. And in the other three ions, the dashed bond points behind the page. The last ion, manganate, is green, with one d electron, d one. With me?

3:10Part C: the same product in every medium?

Here is one more Part C statement, and this one feeds straight into our examples. Permanganate, M n O four minus, is reduced to the same product in all media. True or false? The statement is false, because the medium decides the product. In acid, permanganate gives M n two plus, with two plus as the raised charge. Manganese goes from plus seven to plus two; the arrow means goes to. So it gains five electrons, and we write n equals five, where n is the number of electrons gained. Okay? In neutral or faintly alkaline solution, the product is M n O two, manganese dioxide. Plus seven to plus four, so n is three. Right? And in strong alkali, it stops at manganate, M n O four two minus. Plus seven to plus six, so n is just one. With me? Now the acid one in full. M n O four minus, plus eight H plus, plus five e minus, give M n two plus, plus four H two O, water; e minus is an electron. Keep that five in mind, because both our examples are built on it. Clear?

4:27Example 1: moles of KMnO4 per mole of FeSO4 in acid

Now, Example one, from Part A, at JEE Main level. How many moles of K M n O four oxidise one mole of F e S O four, ferrous sulphate, in acid? Pause the video and try it first. Done? Okay, let's check. Ferrous sulphate gives iron two plus, F e two plus, and permanganate oxidises it to iron three plus. Here is the balanced equation: M n O four minus, plus eight H plus, plus five F e two plus, give M n two plus, plus five F e three plus, plus four H two O, water. Okay? Notice the five irons. Each iron two plus gives just one electron, going from plus two to plus three. And each permanganate takes five, as we just saw. Got it? So, one permanganate for five irons. There is a five in the ratio, so the answer is five moles, right? No, no, wait, read the question again! It asks for moles of K M n O four per mole of F e S O four. One mole of K M n O four oxidises five moles of F e S O four. With me? So for one mole of F e S O four, we need one over five, which is zero point two moles of K M n O four. The answer is zero point two. Five was the number of irons, not the number of permanganates. Clear?

5:59Example 2: percentage of iron from a KMnO4 titration

Now a harder one, from Part B, at JEE Advanced level, a calculation. Here it is: a zero point five zero gram sample containing iron is dissolved, and all the iron, F e, is reduced to iron two plus. It is titrated with twenty five point zero millilitres of zero point zero two zero molar K M n O four; the capital M means molar, moles per litre. Find the percentage of iron. Pause the video and try it. Here is a hint: use the one to five ratio from Example one. Done? Okay, let's check. First, the moles of permanganate: molarity times volume in litres. Twenty five millilitres is zero point zero two five litre. Zero point zero two zero times zero point zero two five gives five point zero times ten to the power minus four mole. Okay? Now the iron. Many students write the same number of moles for the iron, five point zero times ten to the power minus four, right? Oh no, careful, remember Example one! Each permanganate oxidises five iron two plus ions. So multiply by five: five times zero point zero two zero times zero point zero two five gives two point five times ten to the power minus three mole of iron. Got it? Now the mass. Two point five times ten to the power minus three mole, times fifty six, the molar mass of iron, gives zero point one four zero gram. Right? And the percentage: zero point one four zero over zero point five zero, times one hundred, gives twenty eight percent. The answer is twenty eight percent. See, that same factor of five from Example one decided the whole answer. With me?

7:52Quick revision: Part A test, KMnO4 colour, 1 to 5 with Fe2+

So, quick revision. First, attempt Part A as a thirty minute test after Chapters one to six, then go on to Parts B and C, okay? Second, K M n O four is d zero, so its purple comes from charge transfer, not from d to d. And third, in acid, one permanganate oxidises five iron two plus. Now open Part A, start your clock, and begin, and do try the practice problem at the end of each part. All the best!