Mixed Problems: Set 3 (Colour, Carbonyls, Stability and Uses)

Transcript

0:00Introduction

Hello everyone! Let us start with something from the school lab. Drive off all the water from blue copper sulphate crystals, and the powder left behind is white. Same copper, same sulphate. So where did the blue colour go? Hold that thought; we answer it soon. This is Set three of the mixed problems: colour, carbonyls, stability and uses. These ideas come up often in JEE Main, and with the right tool, each problem takes about a minute. So, let's begin.

0:33Colour: an electron jumps the gap

So, first, let us look at colour. On the left of the figure is the octahedral case: the d orbitals split into the dark blue t two g trio below and the orange e g pair above. The smaller tetrahedral splitting on the right is not needed today. The gap is delta naught, the Greek capital delta with a small o below. Light of exactly this energy makes one electron jump from t two g to e g; the arrow means goes to. This is the d to d transition. Okay? Now, suppose a complex absorbs orange light. What colour do you see? Many students say orange, but no, we see blue! The orange is taken out, and our eyes see the complementary colour. So, absorbs orange, looks blue. This is a very common trap. Right? Now let us look at the size of the gap. Delta naught equals h c over lambda, where h is the Planck constant, c the speed of light, and lambda, the Greek letter, the wavelength absorbed. Lambda is in the denominator, so a larger delta naught means a shorter wavelength. Clear? And here is the answer to our opening puzzle. In the blue crystals, water is the ligand that splits the d orbitals of copper. Without ligands, there is no splitting, so there is no gap to jump. That is why anhydrous C u S O four, with the small four below, is white. Got it? There is one more condition. The electron needs a place to jump from, and an empty place to jump to. So d zero and d ten ions, the raised number counting the d electrons, give no d to d colour: d zero has no electron to jump, and d ten has no empty d orbital to jump into. With me?

2:16Carbonyls: give and take, then count to 18

Now, let us move to carbonyls. In the figure, the grey circle M is the metal, and carbon monoxide, C O, is on the right. The orange arrow is sigma donation, sigma being the Greek letter: the carbon lone pair goes into an empty metal orbital. Okay? Then the metal gives back, and this is pi back donation. The green arrow shows a filled metal d orbital donating into the empty pi star orbital of carbon monoxide, pi with a small star. Each effect strengthens the other, and this give and take is called synergic bonding. Is that clear? Now, pi star is an antibonding orbital. So when electrons enter it, what happens to the bond between carbon and oxygen? It gets weaker, yes, exactly! More back donation means a weaker C O bond and a lower stretching frequency, nu C O, the Greek letter nu with C O written small below. Right? Now comes the counting rule. Add the valence electrons of the metal, e with a raised minus, and two from each carbon monoxide. Most carbonyls reach eighteen; this is the eighteen electron rule. Got it? And when the metal has an odd number of electrons, like manganese or cobalt, the count is odd. Then two metal atoms join by an M M bond, the dash between the two M's, and each metal counts that bond as one electron, reaching eighteen. With me?

3:41Stability and uses

Now, let us talk about stability. Ligands join a metal ion one at a time, and each step has a stepwise constant: K one, K two, and so on, as the dots show, the small numbers below counting the steps. The overall constant is beta n, the Greek letter beta with a small n below; a large beta means a stable complex. So, is beta n the sum of the K's, or their product? It is the product, yes! Beta n equals K one times K two, and so on, up to K n. Adding them is a common trap. Take logs, and the product becomes a sum: log beta n equals log K one, plus log K two, up to log K n. Right? Next comes the chelate effect. Three e n ligands, ethane one two diamine, each holding with two arms, fill the same six places as six ammonia molecules. Yet N i, e n three, is much more stable than N i, N H three six, each in square brackets with a raised two plus. And why is that? Mainly because replacing the ammonia molecules sets more particles free, so the entropy increases, as the brackets say. Clear? Finally, the uses: short facts that come up often. EDTA treats lead poisoning, and cis-platin inhibits tumour growth. Got it? Cobalt is the metal in vitamin B twelve, with the twelve written small below, and rhodium is the metal in Wilkinson's catalyst. And hypo fixes a photographic film: it dissolves the unexposed silver bromide, A g B r, as A g, S two O three, two, in square brackets with a raised three minus. With me?

5:19Example 1: which complex is colourless? Count d electrons

Now, Example one, an easy one. Which of these is colourless? C u, H two O four, two plus; C u, C N four, three minus; N i, H two O six, two plus; and T i, H two O six, three plus, each in square brackets with the charge raised. Pause the video and try it first. Done? Okay, let's check. Water is neutral, so in the first one copper is two plus, which is d nine. Nickel is two plus, d eight, and titanium is three plus, d one. All partly filled, so all coloured: blue, green and purple. Okay? Now, the second complex is also copper. So it must be blue as well, right? No, no, wait, find the oxidation state first! Let x be the oxidation state of copper. Four cyanides give four times minus one, and the total is minus three. So x equals plus one; the double arrow means it follows. Clear? Copper plus one is three d ten, completely filled. There is no empty d orbital to jump into, so no d to d transition, and it is colourless. The answer is C u, C N four, three minus. Just like our white copper sulphate: no jump, no colour. Got it?

6:41Example 2: which carbonyl breaks the 18-electron rule

Now Example two, a step harder, on carbonyls. Which carbonyl does not obey the eighteen electron rule? C r, C O six; V, C O six; F e, C O five; or N i, C O four, each in square brackets. Pause the video and try it first. Done? Okay, let's check. Carbon monoxide is neutral, so each metal is in the zero oxidation state. Chromium is three d five, four s one: six valence electrons. Six carbon monoxides give six times two, which is twelve, and six plus twelve is eighteen. Right? Iron has eight, plus five times two, so eighteen. Nickel has ten, plus four times two, eighteen again. With me? Now look at vanadium. It has six carbon monoxides, just like chromium. So eighteen again, right? Careful here, because vanadium is three d three, four s two, only five valence electrons, one less than chromium. Five plus twelve gives seventeen. So V, C O six does not obey the rule. Clear? And seventeen is an odd count, so the complex is paramagnetic. And being just one short of eighteen, it easily takes one more electron to become V, C O six, minus, with eighteen. Got it?

8:02Quick revision: colour, carbonyls, stability constants

So, quick revision. Colour: we see the complement, a larger delta naught means a shorter wavelength, and d zero and d ten give no d to d colour. Carbonyls: back donation weakens the bond between carbon and oxygen, and metal electrons plus two per carbon monoxide usually make eighteen. Stability: beta n is the product of the K's, so the logs add, and chelates are more stable. Now try the practice problem on this page, then the other solved problems in this set. You can do this. All the best!