Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a one-minute question, in the JEE Main style. How many lone pairs are there in xenon tetrafluoride, counting every atom? Xenon has two lone pairs, so the answer is two, right? Hmm, hold that thought; example one will settle it. Five chapters on the p block 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.

0:32How this chapter works

So, first, why this chapter matters. The p block often gives two or three questions in a JEE Main shift. And these questions come in three types. Okay? First, counting questions. You count the lone pairs, the pi bonds, with the Greek letter pi on screen, the bridges, or the oxidation states. Right? Second, statement pairs, where you decide which of the statements are true. And third, the reactions of the important compounds. Clear? Now let me ask you this. Our xenon question from the start: which of these three types is it? It is a counting question, yes, exactly! Now look at the three sets. Set one covers groups thirteen and fourteen, Set two covers groups fifteen and sixteen, and Set three covers groups seventeen and eighteen. Got it? Each set has eight solved problems, and each one should take you about a minute. Then two practice problems close 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?

1:50A quick statement pair from Set 1

Now, a quick statement pair, from Set one. The Roman numerals one and two number the statements, and the small numbers below the symbols count the atoms. Statement one says P b O two gives hydrogen peroxide with dilute acids. Statement two says S n C l two reduces mercury two chloride; the Roman two in brackets is the oxidation state of mercury. So, is each statement true or false? Let us take statement one first. P b O two has two oxygens on one metal atom, just like a peroxide. So with an acid it should give hydrogen peroxide, and statement one is true, right? No, we have to be careful here, because that is the trap! P b O two is lead four oxide; the Roman four in brackets means lead is in the plus four state. In brackets it says no O dash O bond, meaning no oxygen atom is bonded to another oxygen atom, and a peroxide needs that bond. So it is not a peroxide at all. Okay? So it gives no H two O two, hydrogen peroxide, with acids, and statement one is false. Now judge statement two completely on its own. Got it? S n C l two plus two H g C l two give H g two C l two, plus S n C l four; the arrow means gives. The down arrow means the H g two C l two comes out as a precipitate, and the word in brackets tells you it is white. Clear? So S n two plus, with the small raised two plus for its charge, is a good reducing agent, and statement two is true. The answer is: statement one false, statement two true. Judge each statement on its own, every time. With me?

3:38Example 1: total lone pairs in XeF4, on every atom

Now, example one, from Set three, and it settles our opening question. Find the total number of lone pairs in X e F four, xenon tetrafluoride, on all atoms. Pause the video and try it first. Done? Okay, let's check. This figure from Chapter five shows four xenon compounds, and the second one is X e F four, square planar. Xenon has eight valence electrons, and four of them go into the four bonds to fluorine. Okay? So the other four electrons make two lone pairs on xenon; see the two grey lobes, one above and one below. Four bond pairs and two lone pairs make it square planar, with xenon s p three d two; the small raised numbers sit on the p and the d. With me? So the answer is two, just as we guessed at the start, right? Oh no, careful, because the question says on all atoms! Each fluorine has seven valence electrons and uses one in its bond, so seven minus one leaves six, which makes three lone pairs. The figure does not draw them, and that is exactly why students miss them. Four fluorines times three gives twelve. Clear? So the total is two plus twelve, which is fourteen, not two. In counting questions, read exactly what to count, and then count on every atom. Got it?

5:03Example 2: volume of NH3 at STP from 10.1 g of Mg3N2

Now example two, a step harder: a medium problem from Set two. What volume of ammonia at STP is formed when ten point one grams of M g three N two, magnesium nitride, reacts with excess water? STP is standard temperature and pressure, where one mole of gas takes twenty two point four litres; the small raised minus one after mol means per mole. Pause the video and try it first. Done? Okay, let's check. This is a reaction question, so we balance the equation first. M g three N two plus six H two O give three M g O H two, magnesium hydroxide, plus two N H three, ammonia. In magnesium hydroxide, the O H in brackets is taken twice. Okay? Now the amount, n, in moles. Capital M, the molar mass of magnesium nitride, is one hundred point nine grams per mole. Ten point one over one hundred point nine gives zero point one zero zero moles. Got it? So zero point one zero zero moles of magnesium nitride gives zero point one zero zero moles of ammonia, right? No, that is exactly the trap, so be careful! Look at the equation: one mole of M g three N two gives two moles of N H three, since it holds two nitrogen atoms. So the ammonia is zero point two zero zero moles. With me? Finally, V, the volume, is zero point two zero zero times twenty two point four. That gives four point four eight litres; the capital L means litres. Balance first, then use the mole ratio, and the question is done in a minute. Clear?

6:53Quick revision: counting, statement pairs, reactions

So, quick revision. Counting questions: read what to count, and count on every atom. Statement pairs: judge each statement on its own. Reactions: balance the equation first, then use the mole ratio. Now open Set one and begin, about a minute for each problem, and then try the practice problems. All the best!