Copper and Its Compounds

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Drop a piece of copper into dilute hydrochloric acid, and nothing happens, not a single bubble of hydrogen. Now drop it into nitric acid, and the copper dissolves! Both are acids. So why does one attack copper and the other leave it alone? Keep your answer ready; we will settle it in Example one. Today we see copper with acids, blue vitriol, and copper sulphate with iodide, cyanide and ammonia. These reactions come up often in JEE, so let's begin.

0:34Copper and acids

So, let us start with copper and acids. Copper does not displace hydrogen, so dilute hydrochloric acid, H C l, gives no hydrogen gas, H two. The small two written low counts the atoms, so H two means two hydrogen atoms joined together. Okay? Now take nitric acid, HNO three. This one does dissolve copper. And you know the school rule: metal plus acid gives salt plus hydrogen. So the gas from copper and dilute nitric acid must be hydrogen, right? No, and this is the first trap: the gas is nitric oxide, N O, and not hydrogen at all. Three copper, plus eight HNO three, give three copper nitrate, plus two N O, plus four water. The arrow means gives, and in copper nitrate, the brackets with a small two outside mean two nitrate groups. With me? With concentrated nitric acid, the gas changes. Copper, plus four HNO three, give copper nitrate, plus two N O two, nitrogen dioxide, plus two water. So dilute acid gives N O, and concentrated acid gives N O two. Clear? And hot, concentrated sulphuric acid dissolves copper too. Copper, plus two H two S O four, give copper sulphate, plus S O two, sulphur dioxide, plus two water. Again, no hydrogen at all; the acid itself is reduced. Got it?

2:05Blue vitriol, CuSO₄·5H₂O

Next, blue vitriol, written C u S O four, dot, five H two O, in the heading. The raised dot joins the water of crystallisation to the salt. So there are five water molecules. Are all five bonded to the copper? No, only four of them are bonded to copper! So the formula is better written like this: copper with four waters inside square brackets, then sulphate, dot, one more water. The square brackets hold the complex, with the four waters bonded directly to copper. Okay? Around the copper, two oxygen atoms of the sulphate join those four waters. Four plus two makes six, and these six complete an octahedron. Right? So the fifth water is not on copper at all. It is held by hydrogen bonds, to the sulphate and to the coordinated waters. Clear? On heating, blue vitriol loses its water in stages, not all at once. Anhydrous copper sulphate, with no water at all, is white, and it turns blue again when water is added. With me?

3:13Copper sulphate with iodide

Now, copper sulphate with iodide. With chloride, the complex ion C u C l four, in square brackets, two minus, exists; the raised two minus is its charge. Iodide is also a halide. So C u I four, two minus, should exist too, right? That is the trap, because iodide reacts with copper two plus instead. Two copper two plus, plus four I minus, give two copper iodide, C u I, plus iodine, I two. Okay? Look at the oxidation states. Copper goes from plus two to plus one, and iodide is oxidised to iodine. So copper two plus and iodide react with each other; they do not stay together as a complex. Right? That is why C u I four, two minus, does not exist, while C u C l four, two minus, does. This pair comes up often, so remember it. Clear? And the copper iodide comes out white. This reaction is the basis of the iodometric estimation of copper. Got it?

4:14Ammonia, moist air and Fehling's solution

Now add ammonia solution to copper sulphate, drop by drop. First, a pale blue precipitate of copper hydroxide, C u, O H in brackets, two, appears. Okay? Keep adding, and with excess ammonia it dissolves to give a deep blue solution. That deep blue is the complex ion with four ammonia molecules on copper, in square brackets, with a charge of two plus. Right? Next, copper in moist air. Two copper, plus water, plus carbon dioxide, plus oxygen, give copper carbonate, dot, copper hydroxide. The dot joins the two parts into one compound. With me? This is basic copper carbonate, and it forms a green layer on the surface. That is exactly why a copper roof turns green. Clear? And last, Fehling's solution. It contains a copper two tartrate complex; the Roman numeral two in brackets means copper is in the plus two state. Reducing sugars and aliphatic aldehydes give red C u two O, copper one oxide. Aromatic aldehydes, like benzaldehyde, do not. Got it?

5:20Example 1: no H2 from dilute HCl, yet Cu dissolves in HNO3

Now, Example one, and it is our puzzle from the start. Why does copper not liberate hydrogen with dilute H C l, yet dissolve in nitric acid? Pause the video and try it first. Done? Okay, let's check. Look at E naught for copper two plus over copper; the small circle on E means standard, and the slash separates the ion from the metal. This value is positive. So, can H plus ions oxidise copper? Correct, they cannot oxidise copper. So no hydrogen gas is formed, and that is why dilute H C l does nothing to copper. Okay? But nitric acid brings nitrate ions, and nitrate is a strong enough oxidant. It is reduced to N O or N O two, the very gases from our equations earlier. Right? So here is the answer: nitrate, not H plus, is the oxidant. And that solves our puzzle. Of these two acids, only nitric acid brings an oxidant strong enough for copper. Clear?

6:25Example 2: KCN added to CuSO4 solution until in excess

Now, Example two, a little harder. What happens when KCN, potassium cyanide, is added gradually to copper sulphate solution, until it is in excess? Pause the video and try it first. Done? Okay, let's check. With the first additions, copper two plus, plus two C N minus, give copper cyanide with two cyanide groups, written in brackets with a two outside. Okay? But this compound does not last. Two of these give two C u C N, plus cyanogen, written C N in brackets, two. Cyanogen is a gas. Right? Look at the copper here. It went from plus two to plus one, just as it did with iodide, and the cyanide was oxidised to cyanogen. With me? Now keep adding cyanide until it is in excess. Each C u C N picks up three more cyanide ions, so the copper ends up with four. Copper is plus one, and each cyanide is minus one. So what is the charge on the complex? Yes, exactly, plus one and minus four make three minus! C u C N, plus three C N minus, give the complex ion: C u, with C N in brackets, four, all inside square brackets, with a charge of three minus. Clear? So the answer is cyanogen gas first, and then the colourless complex. Colourless makes sense, because copper one is d ten; the raised ten counts its d electrons. With the d orbitals completely filled, there is no d to d transition. Got it?

7:56Quick revision: Cu with acids, blue vitriol, KI, KCN, NH3

So, quick revision. First, copper gives no hydrogen with these acids; nitric acid gives N O when dilute, and N O two when concentrated. Second, in blue vitriol, four waters sit on copper, and the fifth is held by hydrogen bonds. And third, iodide and cyanide take copper two down to copper one, while excess ammonia gives the deep blue complex. Now go and try the two practice problems on this page. You will find them easy. All the best!