Hybridisation and Geometry

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Take two complexes of cobalt three plus, each with six ligands around cobalt in an octahedron. With ammonia, the complex has no unpaired electron at all. But with fluoride, it has four unpaired electrons! Same metal ion, same shape, so why the difference? Valence bond theory answers this. Questions on it come up often in JEE, so let's begin.

0:31Valence bond theory: lone pairs into hybrid orbitals

So, what does valence bond theory say? The metal ion offers a set of empty orbitals, and these mix, or hybridise, okay? Each ligand donates a lone pair into one hybrid orbital. The ligand gives both electrons, so this is a coordinate bond, and a sigma bond, the Greek letter sigma on screen. Right? Now let me ask you this. If six ligands attach, how many hybrid orbitals does the metal need? Six, yes, exactly, one for each lone pair! So the number of hybrid orbitals equals the coordination number, C N for short, and their type fixes the geometry. Clear?

1:11The coordination number fixes the shape

Now the table you must know. Coordination number two gives s p, and a linear shape. For example, open square bracket, A g, then N H three in round brackets, two times, close bracket, and a plus. The small numbers below count groups; the plus up high is the charge. Okay? Coordination number four with s p three, the raised three on the p, is tetrahedral, like N i, C O four times, or Z n, N H three four times, two plus, both in square brackets. Clear? Now careful here. Coordination number four, so the complex must be tetrahedral, right? No, no, wait, there is a second shape! With d s p two, it is square planar, like N i, C N four times, two minus. So for coordination number four, find the hybridisation before you name the shape. Got it? Coordination number five gives d s p three, trigonal bipyramidal, like F e, C O five times. With me? And coordination number six is octahedral, in two ways: d two s p three, or s p three d two. Notice where the d sits. More on this soon, okay? Which d orbitals are used? In d s p two, it is d x squared minus y squared, with s, p x and p y. In both octahedral types, it is d x squared minus y squared, and d z squared. The small letters below say which way each orbital points. Clear?

2:48Inner-orbital or outer-orbital?

Now, the octahedral case. The n minus one d, with n minus one in round brackets, means the d orbitals of the shell below the outer one. If two of them are free, after any pairing the ligands force, we get d two s p three, an inner-orbital or low-spin complex. Okay? But tell me this. What if the ligand is weak and cannot force pairing? Then, for most d counts, two inner d orbitals are not free. So where does the metal find two d orbitals? Yes, exactly, it goes up, to the outer n d orbitals! That gives s p three d two, an outer-orbital complex, and since the electrons stay unpaired, it is high-spin. To remember it: in d two s p three the d comes first, inner, and in s p three d two it comes last, outer. Right?

3:41Which ligands force pairing?

So who decides, inner or outer? It is the ligand. Strong ligands force pairing: cyanide, C N minus, carbon monoxide, C O, nitrite, N O two minus, ethylenediamine, written e n, and bipyridyl, written b i p y. Okay? With M three plus, a three d metal ion with charge plus three, ammonia and oxalate, C two O four two minus, also force pairing. And with cobalt three plus, even water, H two O, does it! Right? Weak ligands do not force pairing: iodide, bromide, chloride, thiocyanate, SCN minus, fluoride and hydroxide, O H minus. With me? And here is the catch. With M two plus, a plus two ion such as manganese, iron or cobalt two plus, water and ammonia are weak. So always check the charge on the metal, okay? And four d and five d metals almost always give low-spin complexes; palladium two and platinum two, in Roman numerals, are square planar. Clear?

4:49Shortcuts from the d-electron count

Now three shortcuts from the d electron count. With d one, d two or d three, the raised number counting the d electrons, two three d orbitals are always free, so the octahedral complex is d two s p three, inner. The double arrow means this gives. Okay? Now a classic trap. Nickel two plus is d eight, with six ammonias around it. Even if every electron pairs up, can two three d orbitals be free? No, they cannot, and that is the trap! Eight paired electrons fill four three d orbitals, so at most one is freed. So octahedral d eight, like nickel two plus, is s p three d two, outer, even with a fairly strong ligand. Got it? And with ten d electrons, as in zinc two plus, copper one plus or nickel zero, with the raised zero, no three d orbital is free. So coordination number four means s p three, tetrahedral. Clear?

5:47Example 1: hybridisation and shape of [Zn(NH3)4]2+

Let's try the first example. Give the hybridisation, shape and magnetic nature of Z n, N H three four times, in square brackets, two plus. Pause the video and try it first. Done? Okay, let's check. Zinc two plus is three d ten, the raised ten counting the electrons. All five three d orbitals are full, so none is free. Right? So the four ammonia lone pairs enter the four s and the three four p orbitals, giving s p three, tetrahedral. Every electron is paired, so it is diamagnetic. Our d ten shortcut said exactly this. See how quick that was?

6:30Example 2: [Co(NH3)6]3+ vs [CoF6]3-, inner or outer orbital

Now, remember our puzzle from the start? Let's finish it. Compare C o, N H three six times, in square brackets, three plus, with C o F six, in square brackets, three minus, in hybridisation and magnetic behaviour. Here is a hint: find the d count first. Pause the video and try it first. Done? Okay, let's check. In both, cobalt is three plus, and cobalt three plus is three d six. Okay? Now the key question. With cobalt three plus, is ammonia a strong ligand or a weak one? Strong, yes, very good! With cobalt three plus, ammonia forces pairing. The six electrons pair up in three of the three d orbitals, so two three d orbitals are free. Right? That is d two s p three, inner orbital. Every electron is paired, so it is diamagnetic. Clear? Now fluoride, a weak ligand, so no pairing. Six electrons in five three d orbitals give one pair and four single electrons, so four unpaired. So the metal uses the four s, three four p and two four d orbitals. Got it? That is s p three d two, outer orbital. Its moment uses the spin-only formula from the next lesson: mu, the Greek letter with a small s below, equals the square root of n times n plus two, in B M, Bohr magnetons, with n the number of unpaired electrons. Four gives root twenty four, about four point nine zero Bohr magnetons; the wavy equals sign means about. So the puzzle is solved: ammonia pairs the electrons, and fluoride does not! With me?

8:18Quick revision: sp3 vs dsp2, d2sp3 vs sp3d2

So, quick revision. Hybrid orbitals equal the coordination number, and the hybridisation fixes the shape. Coordination number four is s p three, tetrahedral, or d s p two, square planar. For six, if two three d orbitals are free, it is d two s p three, inner; if not, s p three d two, outer. Now try the practice problems on this page. All the best!