Set 2 (Shapes, Lone Pairs and Hybridisation)

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Boron trifluoride, B F three, and chlorine trifluoride, C l F three. Both have three fluorines round one central atom. So both are flat triangles, with one hundred and twenty degree angles, right? B F three, yes. But C l F three is T-shaped, and its smallest angle between fluorines is only about eighty seven and a half degrees! Same three fluorines, so why different shapes? In this second problem set, one short method will settle shapes, lone pairs and hybridisation. Questions like these come up often in JEE Main. So, let's begin.

0:43Step 1: count the pairs, then the lone pairs

So, most problems in this set start with one quick count from the VSEPR chapter. Electron pairs round the central atom equal one half of, V plus M, minus c, plus a. The small fraction is one half. Okay? Capital V: the valence electrons of the central atom. Capital M: the monovalent atoms attached to it, hydrogen and the halogens, each adding one. Right? Now the charge. Suppose the species is a negative ion. Do we add that charge, or subtract it? Add it, exactly! Each negative charge is one extra electron. So small c, the positive charge, is subtracted; small a, the negative charge, is added. And oxygen or sulphur on a double bond adds nothing. Clear? Then, lone pairs equal pairs, minus the number of attached atoms. Every attached atom uses up one pair, even with a double bond. With me? Now back to our puzzle. Use the count on B F three and on C l F three. The small numbers count the atoms. How many lone pairs on each central atom? Done? Boron has three valence electrons, and the three fluorines make M three. Half of six: three pairs. Three atoms attached, so zero lone pairs. Okay? Now chlorine: seven valence electrons, plus three fluorines. Half of ten: five pairs! But only three atoms are attached, so two lone pairs. See? Chlorine trifluoride hides two lone pairs. That is the whole difference. Got it?

2:22Step 2: pairs fix the arrangement and hybridisation

Step two. The pairs decide the arrangement; the figure shows all five, each with its angle and an example. The thick line comes towards you, the dashed line goes away, and the small circle means degrees. Under the tetrahedron, the small dash after twenty eight means minutes: one hundred and nine degrees, twenty eight minutes. Two pairs, linear; three, trigonal planar; four, tetrahedral. The hybridisation follows the same count: s p, s p two, s p three, with the small raised numbers as superscripts. Okay? So tell me. C l F three has five pairs. Which arrangement, and which hybridisation? Trigonal bipyramidal, s p three d, yes! That is the fourth drawing: ninety and one hundred and twenty degrees. Notice: the lone pairs count in the hybridisation too, not only the bonds. Right? And six pairs: octahedral, every angle ninety degrees, s p three d two. So S F six, X e F four and B r F five, six pairs each, are all s p three d two, though their shapes differ. Clear?

3:37Step 3: lone pairs pick their seats

Step three, and this is the step that needs care. When we name a shape, we look at the atoms only, not the lone pairs. The lone pairs push, but they never appear in the name. Okay? Now, C l F three again: five pairs, two of them lone pairs. In the fourth drawing, top and bottom are axial; the three round the middle are equatorial. So, where do the two lone pairs sit? Equatorial, correct! An equatorial seat has only two neighbours at ninety degrees; an axial seat has three. So the lone pairs take the equatorial seats, where they are pushed the least. Right? So with five pairs: one lone pair gives a see-saw, two a T shape, three a straight line. Here, l p means lone pair. And that solves our puzzle! In C l F three, the two lone pairs sit in the middle belt, and the three fluorines form a T shape. With me? Now six pairs. X e F four has six pairs, and two of them are lone pairs. So the two lone pairs sit side by side, ninety degrees apart, right? No, no, careful! This is a classic trap. Lone pair to lone pair repulsion is the strongest, so the two lone pairs go opposite each other, one hundred and eighty degrees apart. We call that trans. Clear? So with six pairs: one lone pair gives a square pyramid, and two opposite lone pairs give a square plane. That is X e F four: four fluorines, but square planar, not tetrahedral! Got it? Last idea: lone pairs squeeze bond angles. The wavy equals sign means about; the pointed sign, less than. So C l F three, about eighty seven and a half degrees, is less than P F three, about ninety eight, less than B F three, one hundred and twenty. C l F three starts from ninety, and its lone pairs squeeze even that!

5:33Example 1: shapes of PCl4+ and ICl4+

Okay? Example one, an easy one from this set. Give the shape of P C l four plus, and of I C l four plus. The small plus at the top right is the charge. Pause the video, and try it first. Done? Okay, let's check. P C l four plus: phosphorus has five valence electrons, four chlorines make M four, and we subtract one for the charge. Half of eight: four pairs, all attached, so zero lone pairs: tetrahedral. The double arrow means, so. Right? Now I C l four plus. Iodine is a halogen, with seven valence electrons. Same four chlorines, same plus charge. So, how many pairs, and what shape? Half of seven plus four minus one: half of ten, five pairs. Very good! Four atoms attached, so one lone pair, in an equatorial seat: see-saw. And if you added the charge instead? Six pairs, and a completely wrong shape. So watch the sign of the charge, okay?

6:40Example 2: lone pairs on Xe in XeO4, XeO3, XeF2, XeOF2

Now a harder one, JEE Main level. Find the lone pairs on xenon in X e O four, X e O three, X e F two and X e O F two, and add them up. Pause and try. Hint: watch the oxygens! Done? Okay, let's check. X e O four first. Xenon has eight valence electrons. And there are four oxygens, so M is four, right? No, no, careful! Oxygen is not monovalent. Each oxygen here is on a double bond, so it adds nothing, and M is zero. Half of eight: four pairs. Four oxygens attached, so zero lone pairs. Okay? X e O three: again half of eight, four pairs. Three atoms attached, so one lone pair. Right? X e F two: now the fluorines do count. Half of eight plus two: five pairs. Two atoms attached, so three lone pairs. Clear? Now the tricky one: X e O F two, with one oxygen and two fluorines. How many lone pairs on xenon? Two, correct! Only the two fluorines go into M: half of eight plus two, five pairs. But three atoms are attached, the oxygen as well, so five minus three: two lone pairs. With me? So the total is zero plus one plus three plus two: six. See the trap? Oxygen adds nothing to the pairs, but it still uses up one pair as an attached atom. Well done!

8:18Quick revision: pair count, hybridisation, lone pair seats

So, quick revision. One: pairs equal one half of V plus M minus c plus a; lone pairs equal pairs minus attached atoms. Two: the pairs give the hybridisation: four, s p three; five, s p three d; six, s p three d two. Three: lone pairs go equatorial with five pairs, opposite with six, and the shape names the atoms only. Now work through the rest of this set on this page; give all fifteen problems twenty minutes, and try each one before you open its solution. All the best!