Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a one-minute question, in the JEE Main style. Iron two plus has six d electrons, and manganese three plus has only four. So their magnetic moments must be different, right? Well, hold that thought; example one will settle it. Five chapters on the d and f blocks are done, and now it is time to test yourself. This chapter gives you three sets of mixed problems in the JEE Main style, so let us see how to use them. Let's begin.

0:31How this chapter works

So, first, why this chapter matters. The d and f block unit gives about two questions in a JEE Main shift. And those questions come in a few fixed types. Okay? First, counting questions: a magnetic moment, or the colour of an ion. Second, reactions of K M n O four, potassium permanganate, or K two C r two O seven, potassium dichromate; the small numbers written low count the atoms. Right? Third, statements about the lanthanoids, which you judge true or false. And, up to twenty twenty three, a match of ores or refining methods. Clear? Now let me ask you this. Our iron and manganese question from the start: which of these types is it? It is a counting question, a magnetic moment count, yes, exactly! Now the three sets. Set one covers properties and magnetism, Set two covers dichromate and permanganate, and Set three covers the metals and the f block. Got it? Each set has eight solved problems, and each one should take you about a minute. Then one practice problem closes the set, for you to try on your own. With me? And the real JEE Main questions, from twenty twenty one to twenty twenty six, are at the end of Chapters one to five. The problems in this chapter are original, written in the same style. Right?

1:56Tool for Set 1: count the unpaired electrons

Now, the one tool you need for Set one. Mu, the Greek letter on screen, is the spin-only magnetic moment. Mu equals the square root of n, times n plus two in brackets, in B M, which stands for Bohr magnetons; the tick shaped sign is the square root. And n is the number of unpaired electrons, not the total. Okay? So how do we count n? For free or high spin three d ions, from d one to d five, n goes from one to five; the small raised number counts the d electrons. From d six to d nine, the new electrons pair up, so n comes down as four, three, two and then one. Right? Now colour. For most transition metal ions, the colour comes from an electron jumping from one d level to a higher one, a d to d transition; the short dash joins the two d levels. And for that jump, the d set must be partly filled. The deep colours of permanganate and dichromate come another way, by charge transfer, so this rule is only for d to d colour. Clear? Here is a quick one from Set one. Titanium three plus, zinc two plus, cobalt two plus and vanadium three plus; the raised plus signs show the charges. Which one is colourless in water? Zinc two plus, yes, exactly! It is three d ten, so all five d orbitals are full, and there is no place for an electron to jump to. No d to d transition, so it is colourless. Got it?

3:29Tool for Set 2: count the electrons transferred

Now Set two, the reactions of permanganate and dichromate. In a redox reaction, every electron lost by the reductant is gained by the oxidant. So equivalents of oxidant equal equivalents of reductant. With me? In numbers: n one times M one times V one, equals n two times M two times V two. The small one and two written low label the oxidant and the reductant; M is the molarity, and V is the volume. And n is the n factor, the electrons gained or lost by one formula unit. Okay? First, K M n O four in acid. Manganese goes from plus seven to plus two; the arrow means goes to. Five electrons are gained, so the n factor is five. Right? Now a quick one from Set two. What is the n factor of K M n O four in neutral solution? Is it five again? No, it is three, because the medium changes the product! In neutral solution, manganese goes from plus seven only to plus four, as M n O two, manganese dioxide. And in strong alkali, the n factor is just one. So always read the medium first. Clear? Next, K two C r two O seven has two chromium atoms, and each goes from plus six to plus three. Two times three is six, so its n factor is six. And iron two plus, going to iron three plus, has an n factor of one. Got it?

4:56Example 1: which ion has the same spin-only moment as Fe2+

Now, Example one, from Set one, and it settles our opening question. Which ion has the same spin-only moment as iron two plus: manganese three plus, cobalt two plus, or nickel two plus? Pause the video and try it first. Done? Okay, let's check. Iron is three d six, four s two, and the two four s electrons leave first, so iron two plus is three d six. Six d electrons leave four unpaired, so n is four. Mu is root twenty four, about four point nine zero Bohr magnetons; the wavy equals sign means approximately. Okay? Now manganese three plus, which, as the badge says, is d four. Only four d electrons against six, so it cannot match, right? Oh no, careful, that is exactly the trap! Manganese is three d five, four s two; remove the two four s electrons and one d electron, and manganese three plus is three d four. The formula needs unpaired electrons, not the total number of d electrons. The four d electrons of manganese three plus all go in singly, one per orbital, so n is four, the same as iron two plus. Right? Now the other two. Cobalt two plus is three d seven, high spin, so n is three. Nickel two plus is three d eight, so n is two. Neither of them matches. Clear? So the answer is manganese three plus. And our opening puzzle is solved: d four and d six both have four unpaired electrons, so they have the same moment. With me?

6:35Example 2: volume of 0.1 M K2Cr2O7 to oxidise 30 mL of Fe2+

Now Example two, a step harder: a medium problem from Set two. What volume of zero point one molar K two C r two O seven oxidises thirty millilitres of zero point two molar iron two plus, in acid? Capital M means molar, moles per litre, and m L means millilitres. Pause the video and try it first. Done? Okay, let's check. Many students write zero point one times V equals zero point two times thirty, and get sixty millilitres. Is that right? No, no, wait, they forgot the n factors! Dichromate takes six electrons, and iron two plus gives only one. Okay? So, six times zero point one times V equals one times zero point two times thirty. Both volumes are in millilitres, so no conversion is needed. Right? The left side is zero point six times V, and the right side is six. So V is ten millilitres; the double arrow means it follows that. Each dichromate ion takes the electrons of six iron two plus ions, so a small volume is enough. Got it?

7:44Quick revision: spot the type, spin-only formula, n-factor

So, quick revision. First, spot the type: a count, a reaction or a statement. Second, for counts, mu is the square root of n times n plus two Bohr magnetons, with n the unpaired electrons, and a d to d colour needs a partly filled d set. Third, for reactions, always multiply by the n factor, and for permanganate, read the medium. Now open Set one, give each problem about a minute, and then try the practice problem at the end of each set. All the best!