Overview

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Oxygen does not really want a second electron: putting two electrons on an oxygen atom costs plus seven hundred and three kilojoules per mole. And calcium must lose two electrons, which costs another seventeen hundred and thirty five. Yet calcium oxide forms happily, and even releases energy. So who pays the bill? Hold on to that question. Today: the unit problem bank for chemical bonding, how to use it, and this exact puzzle, solved. So, let's begin.

0:36How to use this chapter

So, let's open this chapter. Part A has eighteen solved problems, at JEE Main level. Okay? And the instruction: attempt them as a forty minute test, after Chapters one to seven. Sit with a clock, and open the solutions only at the end. Right? Now, a quick one for you. Eighteen problems, forty minutes. How much time can you give each problem? Forty over eighteen; that is the division sign, and the wavy equals sign means approximately. About two point two, so a little over two minutes a problem, that's it. Read, spot the idea, answer. Clear? Then Part B: longer problems, at JEE Advanced level. We will open that up in a minute. Got it? Part C checks the ideas that are often tested as statements, true or false. Short, but every word matters. With me? And at the very end, the unit test: fourteen short questions, a quick recall drill for the whole unit.

1:38Part B: two or three ideas in one problem

Okay? Now for Part B, tell me: what makes a problem JEE Advanced level? Bigger numbers? Longer calculations? Not really! Each problem combines two or three ideas of the unit. That is the secret. Each idea alone, you already know. The real skill is joining them. Right? For example, dipole moment plus bond angle, in water: H two O, where the small two counts the hydrogen atoms. The angle decides how the bond moments add. That is our Example two. Okay? Next, lattice enthalpy plus ionic charge. The question mark is a choice: N a O, or N a two O, okay? Only one of them actually forms, and the charges on the ions, through the lattice enthalpy, decide which. Two ideas, one question. Clear?

2:25Part C: one word can flip a statement

Now for Part C, here is one statement from it. Every pi bond is weaker than every sigma bond. Pi and sigma are the Greek letters on screen, the two kinds of covalent bond. True or false? If you said true, careful! It is false. The pi bond in C double bond C is about two hundred and seventy, while the sigma bond in F F, fluorine to fluorine, is only one hundred and fifty eight. Both in kilojoules per mole, the k J mol to the minus one on screen. Okay? Sigma stronger than pi is a rule for bonds between the same pair of atoms. Change the atoms, and the rule can break. With me? So, a topper's trick for statements: when one says all, always or every, hunt for one counterexample. One is enough to make it false. Right?

3:16Example 1: lattice enthalpy of CaO by Born-Haber cycle

Now, our puzzle from the start, as a medium problem from Part A, okay? Find the lattice enthalpy of calcium oxide, C a O, from this data. Delta, the little triangle, means an enthalpy change, and the small letters below say which one. Sublimation of calcium, delta sub H: one hundred and seventy eight. Delta i one plus delta i two, the first two ionisation enthalpies: seventeen hundred and thirty five. Half of delta diss H of O two, half the dissociation enthalpy: two hundred and forty nine. Delta e g one plus delta e g two, the two electron gain enthalpies of oxygen: plus seven hundred and three. And delta f H, the enthalpy of formation: minus six hundred and thirty five, all in kilojoules per mole. Pause the video, and try it first. Done? Okay, let's check. We use the Born Haber cycle from Chapter one: delta f H equals delta sub H, plus delta i H, plus half delta diss H, plus delta e g H, minus capital U, the lattice enthalpy we want. Here the ionisation and electron gain terms are each a sum of two. Right? So, put in the numbers: minus six hundred and thirty five equals one hundred and seventy eight, plus seventeen hundred and thirty five, plus two hundred and forty nine, plus seven hundred and three, minus U, okay? That plus seven hundred and three is the bill from our puzzle: the second electron gain is strongly endothermic. Clear? Add the four terms on the right: two thousand eight hundred and sixty five. Got it? Now bring U to the other side. So U is two thousand eight hundred and sixty five, minus six hundred and thirty five, right? No, no, careful! This is the trap. We have minus six hundred and thirty five equals two thousand eight hundred and sixty five, minus U, yes? Move U to the left and the minus six hundred and thirty five to the right, and its sign flips. The double arrow means therefore: U equals two thousand eight hundred and sixty five, plus six hundred and thirty five. Okay? Three thousand five hundred kilojoules per mole! So who pays the bill? The lattice! The O two minus ion, where the small two minus on top is its charge, costs energy, and the huge lattice enthalpy pays for it. Puzzle solved. With me?

5:40Example 2: O-H bond moment from the dipole of water

Now a hard one, from Part B, okay? The dipole moment of water is one point eight four debye; the capital D on screen stands for debye. The HOH angle is one hundred and four point five degrees, the small circle being the degree sign. Find the O H bond moment. Pause the video, and try it first. Done? Okay, let's check. Mu sub O H, the Greek letter mu with O H written small, is the moment of one bond. Two bonds give one point eight four, so each is one point eight four over two, zero point nine two debye, right? No, no, careful! Dipole moments are vectors, each along its own bond. Two equal moments, mu sub O H, sit at one hundred and four point five degrees, and mu, the dipole moment of water, is their vector sum. You cannot add them like plain numbers. Clear? For two equal moments, the vector sum from Chapter two becomes simple: mu equals two times mu sub O H, times cos of one hundred and four point five degrees over two. Cos is the cosine. Half the angle, because the sum points along the line that cuts the angle in half. Okay? Half of one hundred and four point five is fifty two point two five degrees. Its cos is about zero point six one two, and two times that is one point two two four. So one point eight four equals one point two two four, times mu sub O H, with me? Now divide: mu sub O H equals one point eight four over one point two two four, about one point five zero debye. That wavy equals sign again: approximately. Right? So, two ideas joined, exactly as Part B promised: the shape gives the angle, and the angle decides how the moments add. Got it?

7:30Quick revision: Born-Haber equation and bond moments

So, quick revision. Part A is a forty minute test, Part B joins ideas, and in Part C, watch every word. Born Haber: delta f H equals delta sub H, plus delta i H, plus half delta diss H, plus delta e g H, minus U, and watch the signs. Two equal O H moments at an angle theta, the Greek letter on screen, give mu equals two mu sub O H, times cos of theta over two. Now open Part A, start your clock, and begin. All the best!