Group 18: Noble Gases and Xenon Compounds

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. The noble gases have completely filled shells, and we call them inert. And yet, in nineteen sixty two, a chemist named Bartlett made a compound of xenon. And his clue came from oxygen! How can oxygen tell you anything about xenon? That sounds strange, right? Xenon compounds come up often in JEE, and by the end of this lesson, you will know the answer. So, let's begin.

0:31Group 18: full shells, very high ionisation enthalpy

So, group eighteen: helium, neon, argon, krypton, xenon and radon. Each has a closed shell, n s two n p six; the small raised numbers count the electrons. Helium has just one s two. Okay? Their ionisation enthalpies are very high. For xenon, it is one thousand one hundred and seventy kilojoules per mole; the raised minus one on mol means per mole. And their electron gain enthalpies are positive. So tell me, will a noble gas atom lose an electron, or gain one? Neither, that is exactly right! They neither lose nor gain electrons easily, so they are chemically inert. Clear? Now, boiling points. They rise down the group, from four point two kelvin for helium, up to two hundred and eleven kelvin for radon; the capital K means kelvin. The reason is that dispersion forces increase down the group. With me?

1:28Uses: why helium, why argon?

Now, a few uses. Balloons are filled with helium. But hydrogen, H two, is even lighter. So why not use hydrogen? Safety, that is correct! Hydrogen is explosive, while helium is light and non-inflammable. Okay? Divers breathe a helium and oxygen mixture, joined by the dash, because helium is only a little soluble in blood. And argon gives an inert atmosphere in arc welding. Right?

1:57Bartlett's clue: oxygen and xenon

Now, back to our puzzle. In nineteen sixty two, Bartlett made O two plus, P t F six minus. The square brackets hold the P t F six minus ion, and the raised plus and minus are charges. So the oxygen molecule has lost an electron. Okay? Now compare the first ionisation enthalpies. For O two, it is one thousand one hundred and seventy five, and for xenon, one thousand one hundred and seventy kilojoules per mole. They are almost the same! So, if you were Bartlett, what would you react with P t F six next? Xenon, yes, exactly! He reacted xenon with P t F six and got a red solid, X e plus, P t F six minus, the first xenon compound. That is how oxygen gave the clue. Got it?

2:45Xenon fluorides: the conditions decide

So, the fluorides. Xenon in excess, the word in brackets, with fluorine at six hundred and seventy three kelvin and one bar, gives X e F two. The arrow means gives, the conditions sit over it, and the small two is a subscript. Okay? With more fluorine, xenon to fluorine in the ratio one to five, at eight hundred and seventy three kelvin and seven bar, we get X e F four. Now push further: a ratio of one to twenty, and sixty to seventy bar. Which fluoride do you expect? X e F six, very good! Xenon plus three F two, at five hundred and seventy three kelvin and sixty to seventy bar, gives X e F six. Right? So remember the rule. Excess xenon gives X e F two, and more fluorine with higher pressure gives the higher fluorides. Clear?

3:35Hydrolysis of xenon fluorides

Next, hydrolysis, the reaction with water. Two X e F two plus two H two O gives two X e, plus four H F, plus O two. The xenon is set free. Okay? X e F six plus three H two O gives X e O three plus six H F, and xenon is plus six on both sides. So there is no change in oxidation state. Right? Now X e F four, the tricky one. Six X e F four plus twelve H two O gives four X e, plus two X e O three, plus twenty four H F, plus three O two. Xenon starts at plus four. Where does it end up? Both down and up, that is correct! Some xenon goes down to zero, as free xenon, and some goes up to plus six, in X e O three. One element, reduced and oxidised in the same reaction: that is disproportionation. The oxygen of water is also oxidised, to O two. With me? And X e O three is a colourless, explosive solid. Got it?

4:41Shapes: count the pairs on xenon

Now, shapes. Take X e F two. Xenon has eight valence electrons, and the two fluorines add one each. So we take half of the bracket, eight plus two, and that gives five pairs: two bond pairs and three lone pairs. Three lone pairs, only two atoms attached, so X e F two must be bent, right? No, no, careful, it is linear! Five pairs means s p three d hybridisation; the three is a superscript. In the first drawing, the grey lobes are the lone pairs. All three sit around the middle, in the equatorial positions, so the fluorines point straight up and down. Okay? X e F four has six pairs, two of them lone, so s p three d two. The two lone pairs sit above and below, opposite each other, so the four fluorines form a flat square: square planar. Right? In X e O three, xenon has three bonds to oxygen, the double lines, and one lone pair. So it is s p three, and pyramidal. Clear? And X e O F four, the last drawing, also has one lone pair: s p three d two, square pyramidal. With me?

5:52Example 1: oxidation state of Xe in XeOF4, XeO2F2, XeO3F2

Now example one, an easy one. What is the oxidation state of xenon in X e O F four, X e O two F two, and X e O three F two? Pause the video and try it first. Done? Okay, let's check. Let x be the oxidation state of xenon. Oxygen counts minus two, fluorine minus one, and the total is zero. In X e O F four, x minus two minus four equals zero, so x equals plus six. Okay? In X e O two F two, minus four from oxygen and minus two from fluorine, so x equals plus six again. Now the last one, X e O three F two. What do you get? Plus eight, very good! Here, x minus six minus two equals zero, so x equals plus eight. Got it?

6:41Example 2: XeF6 partial hydrolysis, total lone pairs

Now example two, a little harder. Which xenon compound, on partial hydrolysis, gives X e O two F two? And how many lone pairs does it contain, in all? Pause the video and try it first. Done? Okay, let's check. X e F six plus two H two O gives X e O two F two plus four H F, so the compound is X e F six. Now, how many lone pairs does X e F six have, in all? One lone pair, you say? Yes, the xenon carries one lone pair. But oh no, careful, the question says in all! What about the six fluorines? Each fluorine has three lone pairs, that is correct. Six times three gives eighteen. Okay? So the total is one plus eighteen, which is nineteen lone pairs. Counting only the xenon gives one, and that is the trap. Clear?

7:41Quick revision: Bartlett's clue, xenon fluorides, shapes

So, quick revision. Xenon and O two have almost equal ionisation enthalpies: that was Bartlett's clue. More fluorine and pressure take you from X e F two, to X e F four, to X e F six. And the shapes: X e F two linear, X e F four square planar, X e O three pyramidal. Now go and try the practice problems on this page. All the best!