Colour of Complexes

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Copper sulphate crystals are bright blue. Heat them strongly, the water is driven off, and the powder turns white! The copper is still the same, so where did the blue colour go? Strange, right? Today we find where the colour of a complex comes from, and how to turn a colour into an energy. This comes up often in JEE, so let's begin.

0:26Where the colour comes from

So, where does the colour come from? Take a complex with a partly filled d subshell. The ligands split its d orbitals, with a gap called delta, the Greek capital delta on screen. It absorbs light whose energy is exactly delta. Okay? The complex has taken in this energy. What does it use it for? Which electron moves, and where to? Yes, exactly, an electron jumps up, from the lower set, t two g, to the higher set, e g; the small letters sit below, and the arrow means goes to. Both are d orbitals, so this is a d d transition, with a dash between the d's. Clear? Now, here is a trap. Suppose a complex absorbs blue green light. Then its solution looks blue green, right? No, no, wait, that is the trap! The blue green light was absorbed, so it never reaches your eye. You see the rest: the complementary colour of the light absorbed. With me?

1:25Complementary colours: absorbed and seen

So which colour is the complement of which? Here is the table, in pairs. Absorbs violet, looks yellow, or yellow green; absorbs yellow, looks violet. Okay? Now you try. Suppose a complex absorbs orange light. What colour does it look? It looks blue, very good! Absorbs blue, looks orange yellow; absorbs orange, looks blue. Right? And the last pair: absorbs green, or blue green, looks red purple; absorbs red, looks green, or blue green. Got it? Now a topper's trick: there are only three pairs, violet with yellow, blue with orange, and green with red, and each pair works both ways. Remember three pairs, not six rows. Clear? Take the titanium complex: open square bracket, T i, H two O in round brackets with a small six below, close bracket, three plus, the raised three plus being the charge. It absorbs near four hundred and ninety eight nanometres, n m on screen, and looks purple. Right?

2:29From wavelength to Delta o

Now let's put numbers on colour. The octahedral gap, delta naught, the delta with a small o below, equals h nu, which is h c over lambda. Here h is Planck's constant, nu, the Greek letter like a v, is the frequency, c is the speed of light, and lambda is the wavelength. Okay? That is for one ion. For one mole, multiply by N A, Avogadro's constant, the capital N with a small A below. So per mole, delta naught equals N A times h c, over lambda. Right? Often we get a wavenumber, nu bar, the nu with a bar on top, which is one over lambda, in per centimetre, the c m with a raised minus one. And a shortcut: a wavenumber of one per centimetre is about eleven point nine six joules per mole; the wavy equals sign means about. Clear? Now let me ask you this. If delta naught becomes bigger, does the absorbed wavelength get longer, or shorter? It gets shorter, yes, exactly! Lambda is in the denominator, so a bigger gap means a shorter wavelength and a higher wavenumber. Got it? See this in two nickel complexes. Open square bracket, N i, H two O six, close bracket, two plus, is green; with three e n ligands, ethylenediamine, it is violet. And why does the colour change? Because e n is a stronger ligand, so delta naught is larger and the absorption moves to a shorter wavelength, changing the colour we see. With me?

3:56No d–d colour: when and why

Now, which ions give no d d colour? Think about d zero and d ten; the raised number counts the d electrons. Can either have a d d transition? No, neither of them can! In d zero there is no d electron to jump; in d ten the e g set is full, with nowhere to jump. So S c three plus, T i four plus, Z n two plus and C u plus show no d d colour. Okay? But here is a twist. K M n O four, potassium permanganate, is intensely purple, yet its manganese is plus seven, d zero! Its colour is ligand to metal charge transfer: an electron moves from an oxygen orbital to the metal, an allowed and intense transition. The same goes for K two C r two O seven, potassium dichromate. Right? Now, our puzzle from the start. Without ligands there is no splitting, so no d d transition. So anhydrous C u S O four is white, while C u S O four dot five H two O, the dot joining five waters, is blue. Drive off the water, and the blue goes too. Puzzle solved, clear?

5:04Example 1: 20300 cm-1 to 493 nm, so Ti(H2O)6 3+ looks purple

Let's try the first example, our titanium complex. Its delta naught is twenty thousand three hundred per centimetre. Find the wavelength absorbed and the colour of the solution. Pause the video and try it first. Done? Okay, let's check. Lambda is one over the wavenumber, so one over twenty thousand three hundred, in centimetres, gives four point nine three times ten to the power minus five centimetre; the cross means times. Okay? Now change to nanometres; there are ten to the power seven nanometres in a centimetre. So lambda is about four hundred and ninety three nanometres, and that is blue green light. Right? Now the question asks for the colour of the solution. What do you say? If you said blue green, that is our trap again! Blue green is absorbed. We see its complement, red purple, so the solution looks purple, like the titanium complex earlier. See how it fits?

5:59Example 2: Cr(H2O)6 3+ absorbs 575 nm, Delta o is 208 kJ/mol

Now a slightly harder one. The chromium complex, C r, H two O six, in square brackets, three plus, absorbs at five hundred and seventy five nanometres. Find delta naught in kilojoules per mole. Hint: work out one ion first, then a mole. Pause the video and try it first. Done? Okay, let's check. The energy E of one ion, in joules, is h c over lambda: six point six two six times ten to the power minus thirty four, times three times ten to the power eight, over five hundred and seventy five times ten to the power minus nine. Be careful here: if you forget that ten to the power minus nine, you are off by a billion! Okay? Multiply out the top, divide by the bottom, and keep track of the powers of ten. The energy comes to three point four six times ten to the power minus nineteen joule. Right? So, is this our final answer? No, not yet! That is for one ion, and the question wants kilojoules per mole. So multiply by Avogadro's constant, six point zero two two times ten to the power twenty three. Got it? Three point four six times six point zero two two is about twenty point eight, giving two point zero eight times ten to the power five joules per mole. Divide by one thousand, and delta naught is about two hundred and eight kilojoules per mole, and that is our answer!

7:21Quick revision: complementary colour, Delta o, d0 and d10

So, quick revision. In a d d transition we see the complement of the colour absorbed. Delta naught is h c over lambda for one ion, times N A per mole; a larger delta naught means a shorter wavelength. The d zero and d ten ions show no d d colour, and potassium permanganate gets its colour from charge transfer. Now try the practice problems on this page. You can do it. All the best!