Overview

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Three elements sit side by side in period two. Their first ionisation enthalpies are one thousand four hundred and two, then one thousand three hundred and fourteen, and then one thousand six hundred and eighty one kilojoules per mole. But across a period, ionisation enthalpy should rise, right? So why does the middle one drop? Is the data wrong? Hold that thought. This is the unit problem bank for Classification of Elements and Periodicity. Today: how to use this bank, a few of its problems, and this puzzle, solved. So, let's begin.

0:41How to use this chapter

So, this chapter has three parts. Part A has five solved problems at JEE Main level, and then one more for practice. Okay? And the instruction is clear: attempt Part A as a thirty minute test, after Chapters one to six. So, sit with a clock, and check the answers only at the end. Right? Then Part B: four longer problems at JEE Advanced level, and one for practice. With me? And Part C: four concept checks, and one for practice. These are the ideas that are often tested in statement form, where a single word can change the answer. Clear?

1:21Part C: one word decides the answer

Now, on to Part C, and these look easy, but they test whether your ideas are exact. Here is one. Ionisation enthalpy is the enthalpy change when an atom in the gas state loses an electron. Is it ever negative? Yes or no? No, it is never negative, because removing an electron always needs energy. So ionisation enthalpy is always positive. Got it? Now here is the next one. Electron gain enthalpy is the enthalpy change when an atom in the gas state gains an electron. For most elements it is negative, which means energy is released. So, can it ever be positive? Yes, it can, and the course names the cases: the noble gases, the group two elements, and nitrogen have positive, or near zero, values. The noble gases have full shells, group two has a full s subshell, and nitrogen has a half-filled two p subshell, so an extra electron is not welcome. Okay? And one more case: every second electron gain. On screen, O minus, in the gas state, shown by the small g in brackets, plus an electron, e minus, gives O two minus; the arrow means gives. This costs plus eight hundred and forty four kilojoules per mole; the minus one written small after mol means per mole. Every second electron gain is endothermic: energy has to be put in. So if a statement says electron gain enthalpy is always negative, it is false. In Part C, read every word. Right?

2:58Example 1: order Al3+, Mg2+, Na+, F- by increasing radius

Now, Example one, from Part A, JEE Main level. Arrange aluminium three plus, magnesium two plus, sodium plus, and F minus, the fluoride ion, in increasing radius. The small numbers and signs written above are the charges on the ions. Pause the video, and try it first. Done? Okay, a quick first guess. Of these four elements, fluorine has the smallest atom. So F minus should be the smallest ion, right? Oh no, careful, ions are not atoms! First, count the electrons. The aluminium atom has thirteen electrons, and three plus means it lost three, so thirteen minus three gives ten. Magnesium: twelve minus two gives ten. Sodium: eleven minus one gives ten. And fluorine gained one: nine plus one gives ten. So all four have ten electrons! Clear? Same number of electrons: we call these isoelectronic species. And for isoelectronic species, the radius falls as the nuclear charge, capital Z, rises. Same ten electrons, but more protons pulling them in. With me? Z is thirteen for aluminium, twelve for magnesium, eleven for sodium, and nine for fluorine; the greater than signs put them in order. So aluminium three plus is pulled in the most. Right? So, in increasing radius: aluminium three plus, less than magnesium two plus, less than sodium plus, less than F minus. F minus is the largest, the exact opposite of our first guess! Got it? And the course's radii agree. In picometres, written p m on screen, they are fifty four, then seventy two, then one hundred and two, then one hundred and thirty six. But remember, this rule needs equal electron numbers, so always count the electrons first. Okay?

4:57Example 2: IE 1402, 1314, 1681 gives N, O, F; the oxygen dip

Now, Example two, from Part B, JEE Advanced level, and this is our puzzle! The first ionisation enthalpies of three consecutive elements X, Y and Z of period two are one thousand four hundred and two, then one thousand three hundred and fourteen, and then one thousand six hundred and eighty one kilojoules per mole. Identify them, and explain the dip. The graph on the right is from Chapter three of this unit; use it if you like. Pause the video and try it. Done? Okay, let's check. First, the pattern: X, then Y is lower, then Z is higher again. So the middle one, Y, dips. Now the graph. Up the side is delta i H one, the first ionisation enthalpy, in kilojoules per mole; the triangle is the Greek letter capital delta, and the small i and one written below mean ionisation, first. Along the bottom is the atomic number, written capital Z on the graph; careful, that is not our element Z, okay? Now, across a period, the first ionisation enthalpy generally rises. But look at period two on the graph, from lithium to neon. Where does the curve dip? At two places, yes! At boron, where s changes to p, because a two p electron is easier to remove than a penetrating s electron. And at oxygen, where p three becomes p four. The arrow means changes to, and the small number above p counts the p electrons. Clear? So which dip is ours? Both have the same shape: up, down, up. That is the trap! Around boron, beryllium is eight hundred and ninety nine, boron is eight hundred and one, and carbon is one thousand and eighty six. Those are far too low, because our numbers are all above thirteen hundred. So it is not boron, okay? Around oxygen, nitrogen is one thousand four hundred and two, oxygen is one thousand three hundred and fourteen, and fluorine is one thousand six hundred and eighty one. These match our numbers exactly! So X is nitrogen, two p three, Y is oxygen, two p four, and Z is fluorine. The brackets show the two p subshell of each. Right? Now, explain the dip. Nitrogen's half-filled two p three is stable. In oxygen's two p four, two electrons must share one orbital, and the paired electron is repelled by its partner, so it is easier to remove. This is pairing repulsion. Puzzle solved: the data was right all along; the dip is an exception, not a mistake. Got it?

7:46Quick revision: IE sign, isoelectronic radius, B and O dips

So, quick revision. Part A as a thirty minute test, and Part B at JEE Advanced level. Part C: ionisation enthalpy is never negative, but electron gain enthalpy can be positive. For isoelectronic species, count the electrons, then a larger Z means a smaller radius. And for ionisation enthalpy, trend first, then the exceptions: in period two, the dips at boron and oxygen, and the numbers decide which. Now open Part A, start your clock, and begin. Then try the practice problems on this page. All the best!