Interhalogens, Polyhalides and Pseudohalides

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle first. Chlorine and fluorine join to make C l F, C l F three and C l F five. But you will not find C l F two or C l F four in the list. The even numbers are missing, which is strange, right? And a second puzzle: liquid B r F three, made of molecules, still conducts electricity. Today we solve both. Interhalogens come up often in JEE, so let's begin.

0:30Interhalogens: two different halogens

So, an interhalogen is a compound of two different halogens, written X X prime sub n: X, then X with a small dash called prime, and a small n below, the subscript. Here n is one, three, five or seven, and X, written first, is the larger, less electronegative halogen. So in B r C l, which halogen plays the role of X here? Bromine, yes, exactly! It is larger and less electronegative than chlorine. The first family is X X prime: C l F, B r F, I F, B r C l, I C l and I B r. Right? Next, X X prime three, the small three meaning three atoms: C l F three, B r F three, I F three, and I C l three, which is dimeric, I two C l six in brackets. Clear? Then X X prime five: C l F five, B r F five and I F five. And X X prime seven has only one member, I F seven. With me? We make them from the halogens. C l two plus F two gives two C l F, in the first equation. The arrow means gives, and the number above it is the temperature, four hundred and thirty seven kelvin, with capital K for kelvin, okay? And C l two plus three F two, with fluorine in excess, the word in brackets, at five hundred and seventy three kelvin, gives two C l F three.

1:49Shapes by VSEPR: count the lone pairs

Now for the shapes, using VSEPR theory. C l F three has three bonds to fluorine. So it must be trigonal planar, a flat triangle, right? Oh no, careful, that is the trap! Chlorine uses three of its seven valence electrons in bonds, and the four left over make two lone pairs. So it has three bond pairs and two lone pairs. So C l F three and B r F three are bent T-shaped. Okay? Next, C l F five, B r F five and I F five have five bond pairs and one lone pair, so they are square pyramidal. Right? And I F seven has seven bond pairs and no lone pair, so it is pentagonal bipyramidal. Count the lone pairs first. Got it?

2:32Reactions of interhalogens

Now let us look at reactivity. The X X prime bond, the dash between them, is polar, because the two halogens differ in electronegativity. It breaks more easily in reactions, so interhalogens are more reactive than the halogens, except fluorine, and I C l is more reactive than I two. Clear? Next, hydrolysis, the reaction with water. Many students write I C l plus water gives H I and HOC l. Is that correct? No, no, wait, it is the other way round! I C l plus H two O gives H C l plus HOI, okay? The rule: the smaller halogen goes to the halide, and the larger one to an oxoacid, an acid containing oxygen. Let us check two more. C l F plus water gives H F plus HOC l. And B r F five plus three H two O gives H B r O three, plus five H F, fluorine again as the halide. With me? And C l F three and B r F three are strong fluorinating agents. Uranium plus three C l F three gives U F six, uranium hexafluoride, plus three molecules of C l F, okay?

3:43Polyhalides

Next come the polyhalides. I two plus I minus, where the raised minus is the negative charge, forms I three minus, the linear triiodide ion. The two half arrows mean an equilibrium. That is why iodine dissolves much better in aqueous K I, potassium iodide, than in water. Right? Next, C s I three is not caesium three plus with three iodides. Caesium has one valence electron, so it forms only C s plus. So C s I three is C s plus with I three minus, and C s B r three is C s plus with B r three minus. Clear? Two more ions: I C l two minus is linear, and I C l four minus is square planar. Got it? And KHF two contains K plus and the FHF minus ion, in square brackets, held together by hydrogen bonding. Okay?

4:34Pseudohalides

Finally, pseudohalide ions, ions that behave like halide ions. C N minus, SCN minus, OCN minus, CNO minus, N three minus, and S e C N minus. With me? Just like halide ions, they form insoluble silver salts. And they dimerise to give pseudohalogens, such as C N in brackets sub two, and SCN in brackets sub two. Right? But be careful here. Carboxylate ions, such as C H three COO minus, are not pseudohalides. Clear?

5:05Example 1: why n is always odd in interhalogens, no ClF2

Now example one, which solves our first puzzle. Why do interhalogens always contain an odd number of halogen atoms around the central one? Pause the video and try it first. Done? Okay, let's check. The central halogen has seven valence electrons, an odd number. If it uses n of them in bonds, how many are left over? Seven minus n, yes, exactly! For n equal to one, three, five and seven, that leaves six, four, two and zero. All even, so every leftover electron stays paired, as lone pairs. Okay? But if n were even, say two, seven minus two leaves five, an odd number, so one electron stays unpaired. That is why C l F two is not in our list. Clear? So the answer: all the electrons remain paired only when n is odd. And that solves our first puzzle. Got it?

6:05Example 2: why liquid BrF3 conducts, shapes of its ions

Now example two, a little harder, and our second puzzle. Liquid B r F three conducts electricity. Explain this, and give the shapes of the ions formed. Pause the video and try it first. Done? Okay, let's check. To carry a current, what must a liquid contain? Ions, yes, exactly! Liquid B r F three makes its own ions; this is self-ionisation. Two B r F three, in equilibrium, give B r F two plus and B r F four minus. With me? Now for the shapes. In B r F two plus, bromine has seven valence electrons, minus one for the plus charge, so six. Two go into the two bonds, and four are left, which is two lone pairs. So four pairs: two bond pairs, the b p, and two lone pairs, the l p, so the ion is bent. Clear? Now, B r F four minus has four bonds. So it must be tetrahedral, right? Let us count the pairs. Seven plus one for the minus charge gives eight electrons. Four go into the four bonds, and four are left, two lone pairs. So six pairs, four bond and two lone, and the ion is square planar, not tetrahedral! These two ions carry the current, and that solves our second puzzle. Okay?

7:26Quick revision: odd n, shapes, hydrolysis, triiodide ion

So, quick revision. Interhalogens are X X prime sub n, with n odd and X the larger halogen. Count lone pairs for shapes: C l F three is bent T, B r F five is square pyramidal. In hydrolysis, the smaller halogen goes to the halide, the larger to an oxoacid. Iodine with iodide gives the triiodide ion, and pseudohalides behave like halides. Now try the practice problems on this page. All the best!