First Ionisation Enthalpy and Its Trends

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Nitrogen has seven protons, and oxygen, right next to it, has eight. More protons means a stronger pull, so oxygen should hold its electrons harder, right? Well, the real numbers say no. Pulling an electron out of nitrogen needs one thousand four hundred and two kilojoules per mole, and out of oxygen only one thousand three hundred and fourteen. Today we learn ionisation enthalpy, its trends, and where those trends break. Order questions on this come up often in JEE, so let's begin.

0:38First ionisation enthalpy

So, take an isolated atom, X, as a gas; the small g in brackets means gas. Remove one electron, the e with a minus sign, and the arrow takes us to the gaseous ion, X plus; the raised plus is its charge. The enthalpy change 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. Okay? Now tell me: to pull an electron away from its atom, is energy given out, or taken in? It is taken in, yes, exactly! The nucleus holds the electron, so energy must be supplied, and delta i H one is always positive. Clear? We usually give it in kilojoules per mole; the raised minus one on mol, and on atom, simply means per. And to convert, one electron volt per atom, with e V for electron volt, equals ninety six point five kilojoules per mole. Got it?

1:39What makes ionisation enthalpy large

Alright, so what makes this energy large? There are four things to watch. First, the size: in a smaller atom, the outer electron sits closer to the nucleus, so it is held tighter. Okay? Second, Z effective, the capital Z with the small e f f below it. It is the effective nuclear charge, the pull an electron really feels after screening. Higher Z effective means a stronger pull. Right? Before the third point, tell me this: in the same shell, which gets closer to the nucleus, an s electron or a p electron? The s electron, very good! It penetrates closer to the nucleus. So penetration goes s, then p, then d, then f, as the greater-than signs on screen show, and a more penetrating electron is harder to remove. With me? And fourth, half-filled and completely filled subshells are extra stable, so removing an electron from them costs more. This one hides the answer to our puzzle. Okay?

2:44The trends on the graph

Now look at this graph: delta i H one, in kilojoules per mole, for the first twenty elements. Across a period, atoms get smaller and Z effective rises, so the ionisation enthalpy generally increases, from lithium to neon, and again from sodium to argon. Right? Down a group, a new shell is added, so the outer electron sits farther from the nucleus, and the ionisation enthalpy generally decreases. Look at lithium, sodium and potassium: each is lower than the one before. Clear? Now tell me, which family of elements sits at the top of each peak? The noble gases, exactly, helium, neon and argon, with their full shells! And at the bottom, right after each peak, sit the alkali metals, with one outer s electron that comes off easily. Got it?

3:38Exceptions across a period

But look closely: the climb has small dips, at boron, oxygen, aluminium and sulfur. Take boron: it comes right after beryllium, with one more proton, so it should be higher than beryllium, right? But no, boron is actually lower! Beryllium, B e, is eight hundred and ninety nine kilojoules per mole, and boron, B, only eight hundred and one. Period three repeats it: magnesium, M g, is above aluminium, A l. Okay? So why does this happen? Remember penetration. Beryllium loses a two s electron, which penetrates close to the nucleus. Boron loses a two p electron, shielded by the two s pair and higher in energy, so it comes off more easily. The same goes for the three p electron of aluminium. Clear? Now, our puzzle. Nitrogen, at one thousand four hundred and two, is above oxygen, at one thousand three hundred and fourteen, and phosphorus, P, is above sulfur, S, the same way. Nitrogen's outer subshell is two p three; the small raised three counts its electrons. So what is special about that? It is exactly half-filled, very good, with one electron in each of the three p orbitals! And a half-filled subshell is extra stable, so removing an electron from it costs more. Okay? And oxygen is two p four: one p orbital now holds a pair, and the paired electrons repel, so one of them comes off more easily. That is why oxygen is lower, and our puzzle is solved! Now the trap: the trend breaks exactly where s changes to p, and where p three becomes p four, and many order questions test these spots. Got it?

5:27Groups 13 and 14: poor d and f screening

Now some surprises down a group. In group thirteen, gallium, G a, almost equals aluminium; the wavy equals sign means almost equal. And thallium, T l, is above indium, I n, though it is lower down. In group fourteen, lead, P b, is above tin, S n, okay? The reason is that the d and f electrons added before these elements screen the nucleus poorly, so the outer electron feels a bigger pull than expected. Clear?

5:59Example 1: order B, C, N, O by first ionisation enthalpy

Now, example one, an easy one to warm up. Arrange boron, carbon, nitrogen and oxygen 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. So the answer is simply boron, carbon, nitrogen, oxygen, along the period, right? Oh no, careful, that is not the full story! The trend gives the easy parts: the less-than signs show boron below carbon, and carbon below oxygen. Okay? But nitrogen has the half-filled two p three, which is extra stable. So nitrogen is above oxygen, exactly like our puzzle from the start. Right? So the order is boron, less than carbon, less than oxygen, less than nitrogen. Nitrogen is the highest; if you put oxygen last, that was the trap. Got it?

6:57Example 2: group 13 order, why Tl comes second after B

Now example two, a little harder. Arrange the group thirteen elements in decreasing order of first ionisation enthalpy. The values in kilojoules per mole are: boron, eight hundred and one; aluminium, five hundred and seventy seven; gallium, five hundred and seventy nine; indium, five hundred and fifty eight; and thallium, five hundred and eighty nine. Pause the video and try it first. Done? Okay, let's check. Down a group, ionisation enthalpy decreases. So the order is just boron, aluminium, gallium, indium, thallium, from top to bottom, right? No, no, wait, that is the trap! The trend fails in group thirteen, so read the numbers. Boron, at eight hundred and one, is far above the rest, so boron comes first. Okay? The biggest of the rest is thallium, at five hundred and eighty nine. Then gallium, at five hundred and seventy nine, just above aluminium, at five hundred and seventy seven. That is the poor d and f screening at work. Clear? And indium, at five hundred and fifty eight, is the lowest of all. With me? So the final order is boron, greater than thallium, greater than gallium, greater than aluminium, greater than indium. Thallium is second, right after boron! Order questions like this come up often in JEE, so read the values carefully.

8:30Quick revision: always positive, trends, key exceptions

So, quick revision. First, delta i H one is always positive, and larger for small atoms with a high Z effective. Second, it generally rises across a period and falls down a group. Third, watch the exceptions: beryllium above boron, nitrogen above oxygen, gallium almost equal to aluminium, and thallium above indium. Now go and try the practice problems on this page. All the best!