Catalysis, Complexes, Interstitial Compounds and Alloys

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. There is an alloy called German silver. So, how much silver do you think is in German silver? Is it half? Is it a quarter? Hold on to your guess; we will settle it in Example one. Today we study four more properties of transition metals: catalysis, complexes, interstitial compounds and alloys. Short questions on these do come up in JEE, and they are quick marks. So, let's begin.

0:30Why transition metals are good catalysts

So, let us begin with catalysis. Transition metals and their compounds are very good catalysts, for two reasons. Here is the first question for you. Which property of transition metals lets them take up electrons, and then give them back? Yes, exactly, variable oxidation states! The metal changes its oxidation state easily, so it can take up electrons from the reactants and then give them back. Okay? And the second reason is the surface. The metal surface adsorbs the reactant molecules, which means it holds them on its surface. This raises their concentration at the surface and weakens their bonds, so they react faster. Clear?

1:12Catalysts to remember

Now let us learn the catalysts you should know by name. In the Haber process, the catalyst is iron, written F e, with molybdenum, M o, as the promoter. Right? In the contact process, it is vanadium pentoxide, V two O five; the small numbers written below count the atoms. And for the hydrogenation of oils, it is nickel, written N i, okay? Next comes polythene, made from ethene by Ziegler–Natta polymerisation; this route gives high density polythene. The catalyst is titanium tetrachloride, T i C l four, plus triethyl aluminium. In that second formula, the brackets hold C two H five, an ethyl group, and the small three outside means three of them. Careful here, the catalyst is the pair together, so write both compounds. With me? In the Wacker oxidation of ethene, the catalyst is palladium chloride, P d C l two. And in the Deacon process, it is copper two chloride, C u C l two. Got it? And last, manganese dioxide, M n O two, catalyses the decomposition of potassium chlorate, K C l O three. Learn this list once, and these are easy marks. Okay?

2:25Why so many complexes?

Now let us look at complexes. Transition metal ions form a very large number of them. So tell me, what kind of ion would attract lone pairs of electrons strongly? Think about its size and its charge. Yes, exactly, a small ion with a high charge! So much positive charge, packed into a small ion, pulls hard on electron pairs. Right? But pulling is not enough; the ion also needs somewhere to keep those electrons. Transition metal ions have vacant d orbitals of suitable energy, and these accept the lone pairs. So small size, high charge and vacant d orbitals: write all three reasons in the exam. Clear?

3:07Interstitial compounds

Next, let us look at interstitial compounds. In a metal lattice, there are small holes between the metal atoms. Small atoms, hydrogen, boron, carbon and nitrogen, shown in brackets as H, B, C and N, sit in these holes. That is why these compounds are called interstitial. Okay? Here are some examples: titanium carbide, T i C, then M n four N, F e three H, and T i H one point seven. Wait, look at the last one. One point seven hydrogen atoms for each titanium atom? Is that a misprint? No, it is not a misprint! Interstitial compounds are non-stoichiometric, so the ratio of atoms need not be a simple whole number. Now here is the next question. With small atoms sitting in the holes, will the compound be harder or softer than the pure metal? Yes, it will be harder, because the small atoms in the holes lock the lattice. So interstitial compounds are very hard, and their melting points are higher than that of the pure metal. Right? And two more properties. They keep their metallic conductivity, and they are chemically inert. So the full list is hard, high melting, conducting, inert and non-stoichiometric. Got it?

4:23Why alloys form so easily

And now the last property, alloys. Transition metals form alloys with each other very easily. Why is that? Here is a hint: think about the sizes of their atoms. Yes, exactly, their atomic radii are similar, within about fifteen percent of each other; the sign after fifteen means percent. So one metal atom can simply take the place of another in the lattice. Okay? Here are some alloys to remember. Gun metal is copper, tin and zinc, written C u, S n and Z n. That is eighty seven percent copper, ten percent tin and three percent zinc. Clear? Stainless steel is iron, chromium and nickel, F e, C r and N i; the dashes simply join the metals of the alloy. And solder is lead and tin, written P b and S n. With me?

5:14Example 1: constituents of brass, bronze and German silver

Now, Example one, and it is our puzzle from the start. Give the main constituents of brass, bronze and German silver. Pause the video and try it first. Done? Okay, let's check. Brass is copper and zinc, written C u and Z n. Clear? Bronze is copper and tin, written C u and S n. Brass has zinc and bronze has tin, so do not swap them. Now, German silver. With a name like that, it must be mostly silver, right? Oh no, careful, German silver contains no silver at all! It is copper, zinc and nickel, written C u, Z n and N i. So the answer to our puzzle is zero silver. Do not let the name fool you. Got it?

6:00Example 2: how Fe3+ catalyses iodide with peroxodisulphate

Now, Example two, a little harder. Show how iron three plus catalyses the reaction of iodide with peroxodisulphate. On screen, two iodide ions, I minus, plus S two O eight with charge two minus, give I two, plus two S O four with charge two minus. The small raised signs are the charges, and the arrow means gives. Pause the video and try it first. Done? Okay, let's work it out together. Iodide is a negative ion, and peroxodisulphate is a negative ion too. So why is their direct reaction slow? Yes, exactly, two negative ions repel each other! So without a catalyst, the reaction is one slow step between two anions. Right? Now add iron three plus. In step one, two iron three plus ions take electrons from two iodide ions. They become two iron two plus ions, and iodine, I two, forms. But a catalyst must be given back at the end. So what must happen to iron two plus next? That is correct, it must go back to plus three! In step two, two iron two plus ions give electrons to one peroxodisulphate ion. We get back two iron three plus ions, and two sulphate ions, S O four with charge two minus. Clear? Now look at the charges. In step one, positive iron meets negative iodide. In step two, positive iron meets negative peroxodisulphate. Oppositely charged ions attract, so both steps are fast. With me? So here is the answer. Two fast steps replace one slow step, and iron shuttles between plus three and plus two. Iron three plus is given back at the end, unchanged. And see, that is variable oxidation states at work, our very first reason. Got it?

7:51Quick revision: catalysts, complexes, interstitials, alloys

So, quick revision. First, catalysts work through variable oxidation states and surface adsorption. Second, complexes form because the ions are small, highly charged, and have vacant d orbitals. Third, interstitial compounds are very hard, high melting and inert. And fourth, alloys form easily because the atomic radii are similar, and German silver has no silver! Now go and try the two practice problems on this page. You will find them easy. All the best!