Mixed Problems: Set 3

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Nitrogen has seven protons, and oxygen, next to it, has eight. In both atoms, the electron we remove sits in the two p subshell. So oxygen, with one extra proton, should hold its electrons more tightly, right? And yet, it is easier to pull an electron out of oxygen than out of nitrogen! How can that be? Hold on to that question, because we will answer it today. This is Set three of the mixed problems, the hard set, in the JEE Main style. We will look at the dips in ionisation enthalpy, sharpen two tools, then solve two problems, our puzzle included. So, let's begin.

0:44First ionisation enthalpy: the trend and its dips

So, first, this graph from Chapter three shows the first ionisation enthalpy, in kilojoules per mole, against Z, the atomic number. We write it delta i H one: capital Greek delta, a subscript i for ionisation, capital H for enthalpy, and a subscript one for the first electron. Across a period, it generally increases. Okay? But look closely, the climb is not smooth. Beryllium is above boron, and magnesium is above aluminium, as the greater than signs show. Boron has one more proton than beryllium, so why is it lower? Yes, it depends on which electron we remove! In boron and aluminium, the outer electron is a p electron, and a p electron is easier to remove than a penetrating s electron. Right? And the second pair of dips: nitrogen is above oxygen, and phosphorus is above sulphur. On the graph, oxygen, the point right after nitrogen, sits lower. That is our puzzle, and Example two will explain it. With me?

1:50Hydrogen-like ions: one electron only

Now, the first tool. Some species have just one electron: hydrogen itself, the H on screen, then helium plus and lithium two plus, where the raised plus signs show the charge. We call them hydrogen-like. Okay? For these, the ionisation energy, capital E, equals thirteen point six times Z squared over n squared, in electron volts, the e V on screen. Here Z is the atomic number, the charge on the nucleus, and n is the shell the electron is in. Clear? When the electron is in the first shell, n equals one. Then E is simply thirteen point six times Z squared. Got it? Here is a quick one from Chapter three. Helium plus has Z equal to two. So its ionisation energy is how many times hydrogen's thirteen point six? Yes, exactly, it is four times, because two squared is four. Thirteen point six times four gives fifty four point four electron volts. Right? And lithium two plus, with Z equal to three? Three squared is nine, and thirteen point six times nine gives one hundred and twenty two point four electron volts. Keep this number in mind. With me?

3:06Slater's rules: the effective nuclear charge

Alright, now the second tool. An outer electron does not feel the full pull of the nucleus, because the other electrons screen it. So Z effective, a capital Z with a small e f f below, equals Z minus capital S, the screening by the other electrons. Okay? Slater's rules tell us how big S is. Rule one says that each other electron in the same n s and n p group adds zero point three five. Right? Rule two says that each electron in the shell just below, shell n minus one, adds zero point eight five. Clear? And rule three says that each electron in the deeper shells adds a full one point zero zero. Got it? Let us try sodium, from Chapter two. Sodium has eleven electrons: two in the first shell, eight in the second, and one alone in the third, the three s electron. Which electrons screen it, and by how much? Nothing else shares its group, so rule one gives nothing. Shell two gives eight times zero point eight five, and shell one gives two times one; the small raised numbers count the electrons in each subshell. So Z effective is eleven minus eight point eight, which gives two point two. The outer electron feels only about two of the eleven protons! Okay?

4:22Example 1: ionisation energy of Be3+ (217.6 eV)

Now, Example one, from Set three. Find the ionisation energy of beryllium three plus, B e with a raised three plus. Pause the video, and try it first. Done? Okay, let's check. Beryllium has four electrons, and three plus means three are gone. Four minus three leaves one electron, and the double arrow says it is hydrogen-like, so our formula works. With me? Now, which Z do we put in? The ion has a charge of three plus, so Z equals three, right? Oh no, careful here, because Z is the charge on the nucleus, and beryllium has four protons. Three plus is only the charge on the whole ion. Z equal to three gives one hundred and twenty two point four, the value for lithium two plus, which is the wrong ion. Clear? So, with n equal to one, E equals thirteen point six times four squared over one squared, which is thirteen point six times sixteen. Right? And thirteen point six times sixteen gives two hundred and seventeen point six electron volts. That is sixteen times the value for hydrogen. Got it?

5:32Example 2: Slater Z eff of N and O, and why N beats O

Now remember our puzzle? Here it is, as Example two from Set three. With Slater's rules, find Z effective for a two p electron of nitrogen, and of oxygen. Does this alone explain why nitrogen's delta i H is greater than oxygen's? Pause the video, and try it first. Done? Okay, let's check. Nitrogen is one s two, two s two, two p three. For one two p electron, four others share its group in shell two, and shell one has two. So Z effective is seven minus the bracket, four times zero point three five plus two times zero point eight five, and that gives three point nine. Clear? Now oxygen, with two electrons in the first shell and six in the second. For one two p electron, how many others are in its own group? Yes, exactly, there are five, because we leave our own electron out of the six. So for oxygen, one s two, two s two, two p four, Z effective is eight minus the bracket, five times zero point three five plus two times zero point eight five, and that gives four point five five. Got it? Now compare the two numbers. Oxygen's two p electron feels four point five five, and nitrogen's only three point nine. So oxygen holds its electron more tightly, and its ionisation enthalpy should be higher, right? No, no, wait, remember the graph, where nitrogen is the higher one. Z effective is larger for oxygen, but delta i H is larger for nitrogen. So the double arrow says it: Z effective alone cannot explain it. With me? So what is the real reason? In oxygen, two p four, one two p orbital holds a pair of electrons, and the two repel each other. That makes one of them easier to remove. Okay? And nitrogen, two p three, has a half-filled two p subshell, which is stable. Right? So the answer is three point nine for nitrogen, four point five five for oxygen, and no: pairing repulsion explains it, not Z effective. That solves our puzzle!

7:39Quick revision: IE dips, 13.6 Z2/n2 and Slater's rules

Clear? So, quick revision. Across periods two and three, the first ionisation enthalpy shows four small dips: beryllium above boron, magnesium above aluminium, nitrogen above oxygen, and phosphorus above sulphur. For one-electron ions, E equals thirteen point six times Z squared over n squared electron volts, with Z the nuclear charge. Slater's rules: Z effective is Z minus S, with zero point three five, zero point eight five and one point zero zero. And nitrogen beats oxygen because of pairing repulsion in two p four, not Z effective. Now try the practice problem on this page. You can do it. All the best!