Effective Atomic Number and the 18-Electron Rule

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Iron and carbon monoxide make a neat molecule: one iron atom holding five carbon monoxide groups. Nickel holds four of them. So manganese should simply make a molecule with one manganese atom and five carbon monoxide groups, right? But that is not what happens! Manganese carbonyl has two manganese atoms bonded to each other, with ten carbon monoxide groups. Why would a metal pair up like that? One number explains it, and that number is thirty six. Questions on this do come up in JEE, and they take just a few seconds once you know the idea. So, let's begin.

0:44EAN: counting the electrons round the metal

So, here is the whole idea in one line. EAN stands for effective atomic number. It counts all the electrons round the metal in a complex. EAN equals Z, minus the oxidation state of the metal, plus two times C N, where C N, in round brackets, is the coordination number, and the cross on screen just means times. Let us read it piece by piece, okay? Capital Z is the atomic number of the metal. So the neutral metal atom starts with that many electrons, Z of them. Right? But in a complex, the metal often carries a positive oxidation state. So let me ask you this: if the metal is in the plus three state, how many electrons has it lost? Three electrons, yes exactly! So we subtract the oxidation state. Plus three means three electrons gone, and plus two means two electrons gone. With me? And now the ligands give electrons to the metal. Each donor atom gives one pair, that is two electrons. The coordination number, C N, counts the donor atoms, so we add two times the coordination number. Two electrons per donor atom. Clear?

1:58The target: the next noble gas

Now, why do we count at all? Iron has atomic number twenty six. Moving forward along the periodic table, which noble gas comes after iron? Krypton, with thirty six electrons, very good! On screen, K r is krypton, with thirty six. Then X e, xenon, with fifty four, and R n, radon, with eighty six. These three numbers are our targets. Okay? So here is the rule. Many stable complexes reach the next noble-gas number. The electrons of the metal, plus the electrons from the ligands, add up to thirty six, fifty four or eighty six. Got it? And here is the same thing said another way: the metal ends up with eighteen valence electrons. That is why you will also hear it called the eighteen-electron rule. Same idea, two names. Right?

2:52Metal carbonyls: almost all obey the rule

Now, almost all metal carbonyls obey this rule. Carbonyls are complexes with carbon monoxide, C O, as the ligand. Take N i, C O four, in square brackets: nickel with four C O groups, and the small four below tells us how many. C O is neutral, and the whole molecule has no charge, so nickel is in the zero oxidation state. Okay? Nickel has atomic number twenty eight. Minus zero, plus two times four. Twenty eight plus eight gives thirty six. That is krypton again. Clear? And look, the same with F e, C O five, iron with five C O groups: twenty six plus ten gives thirty six. And with C r, C O six, chromium with six C O groups: twenty four plus twelve gives thirty six. Thirty six for all three. In these neutral carbonyls there is nothing to subtract, okay?

3:44Careful: many complexes, not all

Now, a small trap. Here is N i, N H three six, in square brackets, with two plus written up high, the charge on the ion. That is nickel with six ammonia molecules. Ammonia is neutral, so nickel is plus two. Work it out and tell me, does this complex reach thirty six? Nickel has atomic number twenty eight, and we subtract two. Six ammonia molecules, each with one donor atom, give two times six, that is twelve. Twenty eight minus two plus twelve gives thirty eight, not thirty six! So this complex does not obey the EAN rule. Right? So remember, the rule talks about many stable complexes, not all of them. It is a guide for many complexes, not a law. In the exam, never assume the answer is thirty six; always calculate it. Got it?

4:39Example 1: EAN of Fe in K4[Fe(CN)6] and Co in [Co(NH3)6]3+

Alright, let's solve a question. Find the EAN of iron in K four, F e, C N six. The square brackets hold the complex ion, iron with six cyanide groups, C N, in round brackets, and the small four below K means four potassium ions. Careful, here C N means cyanide, not the coordination number! And find the EAN of cobalt in C o, N H three six, in square brackets, three plus: cobalt with six ammonia molecules, and the three plus up high is the charge on the ion. Pause the video, and try it first. Done? Okay, let's check. Iron first. Four potassium ions sit outside, each plus one, so the complex ion carries minus four. So iron is minus four, right? No, no, wait, that is exactly the trap! Minus four is the charge on the whole ion, not on iron. Each cyanide carries minus one. So, let x be the oxidation state of iron. Then x plus six times minus one equals minus four. The double arrow means therefore: x equals plus two. Okay? Now we use the formula. Iron has atomic number twenty six, and we subtract two for the oxidation state. Six cyanides, each with one donor atom, so the coordination number is six, and two times six is twelve. Twenty six minus two plus twelve gives thirty six. With me? Now cobalt. Ammonia, N H three, is neutral, so the whole three plus charge belongs to cobalt. So cobalt is plus three. Right? Cobalt has atomic number twenty seven. Twenty seven minus three plus twelve gives thirty six. So both complexes come to thirty six, the electron count of krypton. Clear?

6:29Example 2: why manganese carbonyl is Mn2(CO)10, not Mn(CO)5

Now, remember our puzzle from the start? Let's finish it. Explain why manganese carbonyl is M n two, C O ten, and not M n, C O five. The small two below M n means two manganese atoms, and the small ten means ten C O groups in all. Pause the video, and try it first. Done? Okay, let's check. Suppose M n, C O five existed. Manganese has atomic number twenty five, and it is in the zero state, because C O is neutral. Five C O groups give two times five, that is ten. Twenty five plus ten gives thirty five. That is one short of thirty six. Okay? So each manganese needs just one more electron. Look at the real formula: it has two manganese atoms. So where can each manganese get one more electron? Yes, exactly, from the other manganese! Two units join through a manganese to manganese bond, shown by the dash between the two M n symbols. This bond is a shared pair, one electron from each manganese, so it adds one electron to each manganese. Right? So thirty five plus one gives thirty six, krypton again, that is eighteen valence electrons. Two units joining into one molecule is called dimerisation, and dimerisation completes the eighteen-electron count. That is why manganese carbonyl is M n two, C O ten, and our puzzle is solved. See how one number explained it?

8:01Quick revision: EAN formula, noble-gas targets, charge trap

So, quick revision. EAN equals Z, minus the oxidation state, plus two times the coordination number. The target is the next noble gas, thirty six, fifty four or eighty six, that is eighteen valence electrons. And always use the oxidation state of the metal, not the charge on the whole ion. Now go and try the practice problem on this page. All the best!