Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Here are two pairs. First, fluorine minus and aluminium three plus, and both of them carry exactly ten electrons. Second, nitrogen and oxygen, side by side in period two. In each pair, the second one has more protons. More protons means a stronger pull, so it should hold its electrons tighter, right? Well, in the first pair this works perfectly. In the second pair, it fails! Today we find out why. This is Part A of the unit problem bank, at JEE Main level, and the course suggests trying it as a thirty minute test after chapters one to six. We will solve two of its questions together, so let's begin.

0:48Tool 1: same electrons? Compare the protons

So, the first tool. Atoms or ions with the same number of electrons are called isoelectronic. Okay? Take nitrogen three minus; the raised three minus is its charge. Nitrogen has seven electrons and gains three, so seven plus three gives ten. Oxygen two minus has eight plus two, again ten. So both have the same ten electrons. Which ion do you think is larger? Nitrogen three minus, yes, exactly! Oxygen has eight protons pulling on those ten electrons, and nitrogen has only seven. So among isoelectronic species, the radius falls as the nuclear charge, capital Z, rises. Right? And the course data agree. Nitrogen three minus is one hundred and seventy one picometres, and oxygen two minus is one hundred and forty; the p m in brackets means picometres, and the greater than sign means larger. Clear? Now a quick trap. Compare oxygen two minus with sulphur two minus. Sulphur has sixteen protons, and oxygen only eight. More protons, so sulphur two minus must be smaller, right? Oh no, careful, count the electrons first! Oxygen two minus has ten electrons, but sulphur two minus has sixteen plus two, which is eighteen. So these two are not isoelectronic, and the nuclear charge rule does not apply. With me? Instead, we use the group trend. Oxygen and sulphur are in the same group, and going down a group adds a new shell. So sulphur two minus is larger than oxygen two minus. Got it?

2:28Tool 2: ionisation enthalpy across a period

Now the second tool. Take an atom, X, as a gas; the small g in brackets means gas. The arrow means it becomes the gaseous ion, X plus, after losing one electron, the e with a minus sign. The energy needed is the first ionisation enthalpy, delta i H one: the Greek capital delta, a small i for ionisation, H for enthalpy, and a small one for the first electron. The graph on the right plots it, in kilojoules per mole, against the atomic number for the first twenty elements. Okay? Across a period, Z effective rises; that is the capital Z with the small e f f below it, the effective nuclear charge. So the ionisation enthalpy generally rises. On the graph, the line climbs from lithium up to neon, and again from sodium up to argon. Right? So here is question four of Part A, a quick one. Which group contains the elements with the lowest first ionisation enthalpy in each period? Look at the graph and tell me. Group one, the alkali metals, very good! Lithium, sodium and potassium sit at the bottom of the troughs, right after each noble gas peak. One line, and that question is done. Clear? But look closely, because the climb is not smooth. Beryllium is above boron, and magnesium above aluminium; here the greater than sign means a higher ionisation enthalpy. The s arrow p in brackets marks where the filling moves from an s subshell to a p subshell. Boron loses a two p electron, shielded by the two s pair, and it comes off more easily than the penetrating two s electron of beryllium. Got it? And the second kind of dip comes where p three becomes p four; the raised numbers count the p electrons. Nitrogen is above oxygen, and phosphorus is above sulphur. This is the second pair from our puzzle, and we will crack it in example two. Order questions often test these dips. With me?

4:34Example 1: order Al3+, Mg2+, Na+, F- by increasing radius

Now, example one, which is question two in Part A of the course. Arrange aluminium three plus, magnesium two plus, sodium plus and fluorine minus in increasing order of radius, which means the smallest first. Pause the video and try it first. Done? Okay, let's check. The very first thing we check is the number of electrons. Fluorine has nine and gains one, sodium has eleven and loses one, magnesium has twelve and loses two, and aluminium has thirteen and loses three. Okay? Every one of them gives ten, so all four are isoelectronic, and the nuclear charge decides. Here Z is nine, eleven, twelve and thirteen. The question says increasing, so we write them in this same order, with fluorine minus first, right? No, no, wait, that is the trap! That is the order of increasing nuclear charge. But a higher nuclear charge pulls the same ten electrons closer and gives a smaller ion, so the order runs backwards. Right? So the smallest is aluminium three plus, with thirteen protons on ten electrons. The order is aluminium three plus, less than magnesium two plus, less than sodium plus, less than fluorine minus; each less than sign means a smaller radius. Clear? And the course radii in picometres agree. Aluminium three plus is fifty four, magnesium two plus is seventy two, sodium plus is one hundred and two, and fluorine minus is one hundred and thirty six. So in the first pair of our puzzle, more protons really does mean a stronger pull. Very good!

6:15Example 2: N, O, F, Ne by first IE, why O falls below N

Now example two, which is question one of Part A, and this one hides our puzzle. Arrange nitrogen, oxygen, fluorine and neon in increasing order of first ionisation enthalpy. Pause the video and try it first. Done? Okay, let's check. All four are in period two, and ionisation enthalpy rises across a period. So the answer is just nitrogen, oxygen, fluorine, neon, in order of atomic number, right? Oh no, careful, look at the graph first! At atomic number seven, nitrogen, there is a small peak, and at eight, oxygen, the line dips. Nitrogen's outer subshell is two p three, which means three electrons, one in each of the three p orbitals. That is exactly half-filled, and a half-filled subshell is extra stable. Okay? Oxygen is two p four. Its fourth p electron has to pair up in one orbital, and the paired electrons repel each other. So one of them comes off more easily. Right? So oxygen, at one thousand three hundred and fourteen, is below nitrogen, at one thousand four hundred and two kilojoules per mole; the raised minus one on mol means per mole. And our puzzle is solved: oxygen has more protons, yet it gives up an electron more easily. With me? After that, the normal trend takes over. Fluorine is at one thousand six hundred and eighty one, and neon, a noble gas at the top of the peak, is at two thousand and eighty. Got it? So the increasing order is oxygen, less than nitrogen, less than fluorine, less than neon. If you wrote nitrogen before oxygen, that was the trap. Clear?

8:05Quick revision: isoelectronic rule, IE trend and its dips

So, quick revision. Count the electrons first, and if the species are isoelectronic, the higher nuclear charge gives the smaller radius. Ionisation enthalpy rises across a period, and group one is the lowest in each period. And watch the dips: boron and aluminium, where s changes to p, and oxygen and sulphur, where p three becomes p four. Now go and try the other questions of Part A and the practice problem on this page. All the best!