Overview

Transcript

0:00Introduction

Hello everyone! Okay, one quick question before we start. Take ammonium bromide, N H four B r. Nitrogen, hydrogen, bromine: every atom in it is a non-metal. So it must be a purely covalent compound, right? Hmm, hold on to your answer. Today is a special lesson. You have finished the six chapters on chemical bonding, and now it is time to test yourself. This chapter gives you three timed sets of problems in the JEE Main style. I will show you how to use them, and we will solve this exact question together. So, let's begin.

0:40How to use this chapter

So, what is inside this chapter? Three timed sets, with fifteen problems in each. They are original problems, written in the style of recent JEE Main papers. Okay? Now, one question for you. In the real JEE paper, does a question tell you which chapter it comes from? No, it never does! Spotting the idea is your job. So each set mixes the ideas of two chapters. Set one covers ionic bonding, Lewis structures and polarity, from Chapters one and two. Clear? Set two covers shapes, lone pairs and hybridisation, from Chapters three and four. Got it? And Set three covers molecular orbitals and intermolecular forces, from Chapters five and six. With me? Now the most important instruction. Allow twenty minutes per set. So sit with a clock, exactly as if it is the real paper. Right? And the real past questions? They are at the end of Chapters one to six, okay? Here you practise the same style, against the clock.

1:46Topper's trick: eighty seconds a problem

Now let me share a small trick that toppers use. Before you start a set, count. Fifteen problems, and twenty minutes on the clock. So how much time can you give each problem? Twenty minutes is twelve hundred seconds. Now divide; the sign you see on screen is the division sign. Twelve hundred over fifteen: eighty seconds, that's it. So, about one minute and twenty seconds for each problem. Not much, right? Read it, spot the idea, write the answer. Now suppose one problem takes you four minutes, but you get it right. Right answer, so all good? No, no! Those extra minutes were stolen from the other problems. So if you are still stuck at eighty seconds, mark that problem, move on, and come back to it at the end. Collect the easy marks first. Clear?

2:38Example 1: which salts have both ionic and covalent bonds

So, remember ammonium bromide from the start? Here it is, in the very first problem of Set one. How many of these contain both ionic and covalent bonds: sodium nitrate, N a N O three; potassium cyanide, KCN; calcium chloride, C a C l two; ammonium bromide, N H four B r; and magnesium oxide, M g O? The small numbers at the bottom count the atoms. Pause the video, and try it first. Done? Okay, let's check. First, the key idea. The ionic bond is the attraction between the ions. Covalent bonds hold the atoms together inside a polyatomic ion, an ion made of more than one atom. So in each compound, just look for a polyatomic ion. Okay? Sodium nitrate is N a plus and the nitrate ion, N O three minus. The small plus and minus at the top right are the charges. Inside the nitrate ion, nitrogen and oxygen share electrons, so those bonds are covalent. The double arrow on screen means therefore: both kinds of bond. Same with potassium cyanide: K plus, and the cyanide ion, C N minus. Both again. Right? Calcium chloride: C a two plus, and two chloride ions. The two chlorines are two separate C l minus ions, with no bond between them. So, ionic only. Got it? Now our question from the start. Ammonium bromide has only non-metals. So it is purely covalent, right? No, no, careful! This is the trap. Ammonium bromide is made of two ions: the ammonium ion, N H four plus, and the bromide ion, B r minus. The ionic bond holds these two ions together, and inside the ammonium ion, the nitrogen hydrogen bonds are covalent. So, both! Only non-metals, and still ionic. Clear? Last, magnesium oxide: M g two plus, and O two minus. Two simple ions, no polyatomic ion, so ionic only. With me? So the answer is three: sodium nitrate, potassium cyanide and ammonium bromide, thanks to their polyatomic ions, N O three minus, C N minus and N H four plus. If you said purely covalent at the start, no problem; now you know the trap. One quick look at each compound, well inside eighty seconds. Okay?

5:13Example 2: total lone pairs on Xe in four compounds

Now a harder one, from Set two. Find the number of lone pairs on xenon, X e, in four compounds: X e O four, X e O three, X e F two, and X e O F two. Then add them. Pause the video, and try it first. Done? Okay, let's check. We use the quick count from Chapter three. Pairs equal one half of the whole bracket: V plus M, minus small c, plus small a, okay? Capital V is the central atom's valence electrons, and capital M the number of monovalent atoms attached, like hydrogen and the halogens. Small c is the positive charge, and small a the negative charge. Then, lone pairs equal pairs minus attached atoms. Clear? So, first we need V for xenon. How many valence electrons does xenon have? Eight, very good! Xenon is a noble gas. And all four molecules are neutral, so small c and small a are zero. X e O four: oxygens attached by double bonds add nothing, so half of eight, four pairs. Four atoms attached, so zero lone pairs. Right? X e O three: again half of eight, four pairs. But only three atoms attached, so one lone pair. With me? X e F two: two fluorines, each monovalent, so M is two. Half of eight plus two, half of ten: five pairs. Two atoms attached, so three lone pairs. Clear? Now the last one, X e O F two. Three atoms attached: one oxygen and two fluorines. So M is three, right? No, no, careful! M counts only monovalent atoms. The oxygen, attached by a double bond, adds nothing, so M is just two. Half of eight plus two: five pairs. But when you subtract, count every attached atom, all three. So five minus three: two lone pairs. Got it? Now add them all: zero plus one plus three plus two. Six lone pairs in total. Four molecules, one formula, and the oxygen trap avoided. That is the speed each set needs. Okay?

7:37Quick revision: timing, ionic plus covalent, pair count

So, quick revision. Three timed sets of fifteen problems: twenty minutes a set, about eighty seconds a problem. An ionic compound with a polyatomic ion has both: covalent bonds inside that ion, and ionic bonds between the ions. And for lone pairs, pairs equal one half of the whole bracket, V plus M minus small c plus small a, and oxygen attached by a double bond adds nothing. Now open Set one, start your clock, and begin. You did the hard work in six chapters, and this is where it pays off. All the best!