General Trends in Group 16

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Look at two fluorides of sulphur, S F six and S F four. S F six is so unreactive that even steam does not hydrolyse it; it is used as an insulating gas. But water hydrolyses S F four rapidly. Two fewer fluorine atoms, and a completely different story! By the end, you will know exactly why. Group sixteen trends come up often in JEE, so let's begin.

0:30Group 16 at a glance

Group sixteen is the oxygen family, the chalcogens. Oxygen and sulphur are non-metals, selenium and tellurium are metalloids, and polonium is a radioactive metal. The valence shell is n s two, n p four, with n the outer shell and the raised numbers counting electrons. Okay? Now size, the covalent radius, in picometres, the p m after the slash. Sixty six for oxygen, then one hundred and four, one hundred and seventeen, one hundred and thirty seven, and one hundred and forty six for polonium. The biggest jump is from oxygen to sulphur. Right? Next, the first ionisation enthalpy: the Greek capital delta with a small i, then H with a small one. It is lower than in group fifteen, because removing one paired p electron relieves repulsion. Now, electron gain enthalpy, the delta with a small e g on the badge. Oxygen is the most electronegative, so its value is the most negative, right? Oh no, careful, that is the trap! In kilojoules per mole, where mol to the minus one means per mole, oxygen is minus one hundred and forty one, but sulphur is minus two hundred. So sulphur, not oxygen, is the most negative. With me? The reason is size: the incoming electron enters the small, compact two p shell of oxygen, and the electrons already there repel it strongly. Clear?

2:01Oxidation states and the inert pair effect

Now, oxidation states. Oxygen is almost always minus two. The exceptions are peroxides, at minus one, and superoxides, at minus one half, the fraction on screen. Okay? And in its two fluorides, O two F two and O F two, oxygen even becomes positive; the small low numbers count atoms. We work out these values in example one. Right? The other members, sulphur, selenium, tellurium and polonium, show minus two, plus two, plus four and plus six. Now tell me, as we go down the group, which becomes more stable, plus four or plus six? Yes, exactly, plus four becomes more stable than plus six as we go down! This is the inert pair effect: the n s two pair of the heavier elements is held more tightly, and forms weaker bonds. Got it?

2:55Why oxygen is different

Now, oxygen, the first member, is different from the rest. Three things make it so: its small size, its very high electronegativity, and no d orbitals at all. Clear? So its covalency is normally just two. It rises to three in H three O plus, the raised plus showing a positive charge, and four is the maximum. Okay? Oxygen forms a strong p pi, p pi double bond, a pi bond between two p orbitals, pi being the Greek letter. So dioxygen, O two, is diatomic. Sulphur has the same valence shell. So is sulphur diatomic too? No, not at all, sulphur forms S eight rings instead! For sulphur, two S S single bonds, the dash, beat one S S double bond, the two lines. So sulphur shows strong catenation, chains and rings, while oxygen shows very little. With me?

3:54Hydrides H2O to H2Te: bond angle, acidity, boiling point

Next, the hydrides, H two E, with E any element of the group; the small circle means degrees. The bond angle is one hundred and four point five in H two O, ninety two in H two S, ninety one in H two S e, and ninety in H two T e. Okay? The E H bond enthalpy and the thermal stability both fall down the group; the pointed sign means greater than. So the E H bond gets weaker. Then is the hydride a stronger acid, or a weaker one? Yes, exactly, a weaker bond means a stronger acid! So acidity and reducing character rise from H two O to H two T e. Now, the boiling point, b p on the badge, rises as we go down, from H two S to H two T e. So water, at the top of the group, should boil the lowest, right? Oh no, careful, water boils the highest of all! Here m p means melting point, and this pointed sign means less than. Both the boiling and the melting points go H two S, H two S e, H two T e, then H two O, because of hydrogen bonding in water. Got it?

5:12Halides and oxides

Now the halides. The hexafluorides, S F six, S e F six and T e F six, are octahedral, and S F six, our insulating gas, is exceptionally inert. Okay? S F four is see-saw, S C l two is bent, and S two C l two is a dimer: C l, S, S, C l, joined by single bonds. Right? The oxides E O two and E O three are acidic. And the reducing power of E O two falls from S O two to T e O two. Clear? But note this one: S e O two is an oxidising agent. It is easy to miss, so mark it, okay?

5:53Example 1: oxidation state of O in OF2, O2F2, H2O2, KO2

Now example one, an easy one. Give the oxidation state of oxygen in O F two, O two F two, H two O two, and K O two. Pause the video and try it first. Done? Okay, let's check. Fluorine is always minus one, hydrogen and potassium are plus one, and let x stand for oxygen. Now, O F two. Oxygen is usually minus two, so the answer is minus two, right? No, no, wait, that is the trap! Fluorine is more electronegative than oxygen, so fluorine takes minus one, and in a neutral molecule the oxidation states add up to zero. So x plus two times minus one equals zero, and the double arrow means this gives x equals plus two. Okay? Next, O two F two. Two times x, plus two times minus one, equals zero, so x equals plus one. Right? For H two O two, two times plus one, plus two times x, equals zero, so x equals minus one. That is the peroxide value. Got it? And for K O two, plus one, plus two times x, equals zero, so x equals minus one half, the superoxide value. So, plus two, plus one, minus one and minus one half. Clear?

7:18Example 2: why SF6 resists hydrolysis but SF4 does not

Now example two, which settles our puzzle. S F six is not hydrolysed even by steam, but S F four is rapidly hydrolysed. Explain why. Pause the video and try it first. Done? Okay, let's check. In S F six, the sulphur atom is completely shielded by six fluorine atoms. It is coordinatively saturated, with no room for another group. Okay? So water cannot reach the sulphur, and sulphur has no low-lying vacant orbital to accept it. Now, S F four has only four fluorine atoms. So what does water find on the sulphur there? Yes, exactly, it finds room to attack! S F four has a lone pair and an open site on sulphur, so water attacks it. Right? The reaction, with the arrow meaning gives: S F four plus two H two O gives S O two plus four H F, and it is balanced. So our puzzle is solved: six fluorine atoms guard the sulphur completely, four leave it open. With me?

8:25Quick revision: inert pair, anomalous oxygen, hydride trends

So, quick revision. Down the group, plus four gets more stable than plus six: the inert pair effect. Oxygen is different: no d orbitals, a p pi, p pi double bond, and hydrogen bonding. And from H two O to H two T e, the bond angle and stability fall, while acidity rises. Now go and try the practice problems on this page. All the best!