Hybridisation Involving d Orbitals

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Sulfur, in its ground state, has only two unpaired electrons. So, two bonds, right? But sulfur hexafluoride exists: one sulfur atom holding six fluorine atoms. Six bonds! Six bonds from two unpaired electrons? Strange, isn't it? And oxygen, just above sulfur in the same group, can never do this. Today we find out why. This idea comes up often in JEE, so let's begin.

0:31Where do the extra unpaired electrons come from?

So, sulfur, the S on screen. Its ground state is three s two, three p four. The big three is the shell number, and the small number on top counts the electrons, okay? Four electrons in three p orbitals: one pair and two single electrons. So, only two unpaired. Right? Now, the third shell also has d orbitals, five of them, called the three d orbitals, and in sulfur they are all empty. So tell me, how can sulfur get six unpaired electrons? Yes, exactly! Promote them. One three s and one three p electron move up into two vacant three d orbitals: three s one, three p three, three d two. One plus three plus two: six unpaired electrons! And these d orbitals are in the same shell, the third shell. Clear? These six orbitals, one s, three p and two d, mix into six hybrid orbitals, called s p three d two. Here the small three and two on top count the p and d orbitals used, okay? Each hybrid makes one bond with a fluorine: six bonds in S F six. The small six below the F counts the fluorine atoms. Puzzle solved! With me? Now, oxygen is in the same group. So why can't oxygen form O F six? Because oxygen's valence shell is the second shell, n equals two; n is the shell number. The second shell has only s and p orbitals. No two d orbitals at all! Nothing to promote into, so oxygen holds at most four electron pairs. Got it?

2:07Five pairs: sp3d hybridisation

Alright, a quick recall. Hybridisation comes from the electron pairs on the central atom: sigma bonds plus lone pairs. Sigma is the Greek letter on screen. Okay? Five pairs need five hybrids, but one s and three p give only four. So the fifth comes from a d orbital: s p three d, that is one s, three p and one d, as the brackets show. The five hybrids point to the corners of a trigonal bipyramid: three round the middle, and two straight up and down, along one axis. Right? But there are five d orbitals. Two pairs lie along one axis; call it the z axis. Which d orbital joins in? d z squared, correct! That is d with a small z squared written below. Its big lobes lie along the z axis, just where we need them. This comes up often as a one-line question. Clear? So, which molecules use s p three d? P C l five, where P is phosphorus and C l is chlorine, then S F four, and C l F three. Okay? But wait. S F four has only four bonds. So s p three, right? Then why is it in this list? Oh no, careful! Count the pairs, not just the bonds. Sulfur has six valence electrons; four go into the bonds, and two are left: one lone pair. Four plus one: five pairs, s p three d. This is the classic trap. Got it?

3:35Six and seven pairs: sp3d2 and sp3d3

Next, six pairs: one s, three p, two d, that is s p three d two. Which two d orbitals? d x squared minus y squared, written as d with x squared minus y squared below, and d z squared. Okay? These six hybrids point to the corners of an octahedron. S F six is the example, and B r F five; B r is bromine. Five bonds, but bromine keeps one lone pair, so six pairs. So the pairs are octahedral, though the B r F five molecule itself is square pyramidal, since the lone pair is not counted in the shape. Right? And seven pairs? One s, three p, three d: s p three d three. The shape is a pentagonal bipyramid: five round the middle, and two along the axis. I F seven is the classic case. With me? Now look at all three. Five pairs, one d. Six pairs, two d. Seven pairs, three d. See a pattern? Yes! Here is a topper's trick: the number of d orbitals used equals pairs minus four. Because one s and three p are always used first. But only for five to seven pairs, okay?

4:44dsp2: square planar complexes

Now one more, a little different: d s p two. The d comes first. It mixes one inner d orbital, d x squared minus y squared, with one s and two p orbitals; why inner, we will see in a moment. So, how many hybrids? Four, correct! And all four lie in one plane, pointing to the corners of a square: square planar. Why d x squared minus y squared? Its lobes lie along x and y, exactly in that square. Clear? We see this in these complexes. The square brackets hold the complex ion, its charge sits at the top right, and the round brackets group a ligand. N i is nickel, C N is cyanide, P t is platinum, C u is copper, and N H three is ammonia. Okay? And why inner d? Nickel two plus uses a three d orbital, one shell below its four s and four p orbitals. And notice, our pairs minus four trick does not apply here: four pairs, yet one d. Got it?

5:46Example 1: hybridisation of PCl5, SF6 and IF7

Let's use it. Example one. Give the hybridisation of phosphorus in P C l five, sulfur in S F six, and iodine in I F seven. Pause the video, and try it first. Done? Okay, let's check. Phosphorus has five valence electrons, all used in five bonds. Five sigma bonds, no lone pair. The double arrow means, so: s p three d. Right? Sulfur: six valence electrons, six bonds, no lone pair. Six pairs, so s p three d two. Our puzzle molecule! Okay? Iodine: seven valence electrons, seven bonds, no lone pair. Seven pairs, so s p three d three. The trick agrees: one, two and three d orbitals. Easy, right?

6:35Example 2: hybridisation of XeF2, XeF4 and XeF6

Now a harder one. Example two. Xenon, X e, is a noble gas, yet it forms X e F two, X e F four, and X e F six. Give the hybridisation of xenon in each. Pause and try. Hint: count the lone pairs. Alright, X e F four first. Four bonds, so s p three, right? No, no, careful! That is the trap. Xenon has eight valence electrons, and each xenon fluorine bond, the dash on screen, uses one. The leftover electrons sit as lone pairs. Got it? X e F two: two electrons used, six left, three lone pairs. Two plus three: five pairs, so s p three d. Okay? X e F four: four used, four left, two lone pairs. Four plus two: six pairs, so s p three d two, not s p three! Right? And X e F six: six used, two left, one lone pair. Six plus one: seven pairs, so s p three d three. Same answers as Example one, but only because we counted the lone pairs. Clear?

7:48Quick revision: sp3d, sp3d2, sp3d3 and dsp2

So, quick revision. Pairs are sigma bonds plus lone pairs: five, six and seven pairs give s p three d, s p three d two, and s p three d three. Then, s p three d uses d z squared, and d s p two uses the inner d x squared minus y squared, for square planar complexes. And extra unpaired electrons come from promotion into vacant d orbitals of the same shell. Now go and try the practice problems on this page. You will do great. All the best!