Zinc and Nickel

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle from the salt analysis lab. Nickel sulphide is a black solid, and once it has formed, dilute hydrochloric acid cannot dissolve it. Now, in group two we pass hydrogen sulphide through a solution in dilute hydrochloric acid. So nickel sulphide should come down right there, shouldn't it? But it does not. It waits for group four. So how can a solid that will not dissolve in the acid also refuse to form in the acid? By the end of this lesson you will know, and you will meet two lovely tests on the way: zinc's white precipitate that vanishes in excess alkali, and nickel's rosy-red precipitate with dimethylglyoxime. These tests come up often in JEE practical chemistry. So, let's begin.

0:47Group IV: white ZnS and black NiS

Group four also has other cations, such as manganese, but in this lesson we focus on two of them, zinc and nickel; the Roman four is in the heading. The group reagent is hydrogen sulphide, H two S, passed into a solution made alkaline with ammonium hydroxide, N H four O H on screen; the small numbers are subscripts. In this alkaline solution, hydrogen sulphide ionises freely, so we get plenty of sulphide ion, S two minus; the raised two minus is its charge. Okay? First, zinc two plus; the raised two plus is its charge. It comes down as zinc sulphide, Z n S, and zinc sulphide is white. And it is soluble in dilute hydrochloric acid, written H C l on screen. Right? Next, nickel two plus. Its sulphide, N i S, is black, and it is insoluble in dilute hydrochloric acid. Clear? So how do we bring nickel sulphide back into solution? We use aqua regia, because nickel sulphide dissolves in it. Now let me ask you this. Suppose your group four precipitate has both, white zinc sulphide and black nickel sulphide. How would you separate them? Yes, exactly, we use cold dilute hydrochloric acid. Zinc sulphide dissolves and goes into solution, while nickel sulphide remains behind. With me?

2:15The nickel sulphide puzzle

Now, back to our opening puzzle. In group two, the Roman two on screen, hydrogen sulphide is passed in dilute hydrochloric acid, so the concentration of sulphide ion is low; the square brackets mean concentration. So why does nickel sulphide not form here? Think about its solubility product. The K s p of nickel sulphide, its solubility product, is too high; the small s p below the K is just a label. At that low sulphide concentration, the ionic product of nickel and sulphide ions never exceeds it, so nickel sulphide does not precipitate in group two. Right? But in group four the solution is alkaline, and there nickel sulphide does form. Once formed, it changes to a less soluble form, so dilute hydrochloric acid cannot dissolve it afterwards. So that solves our puzzle. Clear?

3:09Zinc: sodium hydroxide and ferrocyanide

Now the zinc tests. Dissolve the white zinc sulphide in dilute hydrochloric acid, and then add sodium hydroxide slowly. Sodium hydroxide is written N a O H on screen. We get a white precipitate of zinc hydroxide; the hydroxide group, O H, sits in round brackets, taken twice. Okay? Now, what happens if we keep adding sodium hydroxide, in excess? Does the precipitate just grow? The precipitate does not grow at all, it dissolves, forming sodium zincate. In its formula, zinc with four hydroxide groups sits inside square brackets, that is the zincate ion, with two sodium ions outside. Right? The second zinc test uses potassium ferrocyanide. It gives a white or bluish-white precipitate: two zinc ions with one ferrocyanide group, which is iron with six cyanide groups, C N in round brackets, all inside square brackets. Got it? Now here is a trap. Your salt gives a white hydroxide that dissolves in excess sodium hydroxide. So the cation must be zinc, right? That is exactly the trap, because aluminium hydroxide also dissolves in excess sodium hydroxide. So this test alone cannot tell zinc from aluminium. With me? So use ammonia instead. Zinc hydroxide dissolves in excess ammonia as this complex ion: zinc with four ammonia molecules, N H three, inside square brackets, and the raised two plus is the charge of the whole ion. Aluminium hydroxide does not dissolve in ammonia. Clear?

4:46Nickel: the rosy-red dimethylglyoxime test

Now nickel's confirmatory test. Take the aqua regia solution of nickel sulphide, make it ammoniacal, and add dimethylglyoxime, written DMG for short, in round brackets. Out comes a rosy-red precipitate. Okay? Here is its formula: nickel with two d m g units, in round brackets with a small two, all inside square brackets. Each DMG molecule loses a proton when it binds, and the ammonia removes these protons. So what do you think happens if we forget the ammonia, and the solution stays acidic? In acid, the complex does not form, so no rosy-red precipitate. So never skip the ammonia. Right? This complex is square planar and diamagnetic. And it is a chelate, which means each ligand holds the nickel at two points, like a claw. Clear? Those two points are two nitrogen atoms, so each ligand closes a five-membered ring with the nickel, two rings in all. And the two ligands are also linked by two hydrogen bonds, shown as O H dot dot dot O on screen; the dotted part is the hydrogen bond. With me? One last point, about colour. Zinc two plus is three d ten; the raised ten counts the electrons in the three d orbitals. With no d to d transitions, zinc salts are colourless. Nickel two plus is three d eight, a partly filled set, so d to d transitions can absorb visible light, and nickel salts are green. And three d eight has two unpaired electrons in the free ion; keep those in mind for example two. Got it?

6:23Example 1: white sulphide from H2S in ammoniacal solution

Alright, example one, an easy one. When hydrogen sulphide is passed into an ammoniacal solution, a white precipitate forms. Which cation is present? Pause the video and try it first. Done? Okay, let's check. Hydrogen sulphide in an ammoniacal solution is the group four reagent, so we are in group four. And in group four, zinc sulphide is white, while nickel sulphide is black. Okay? So the cation present is zinc two plus. Very good if you said zinc! Clear?

6:59Example 2: oxidation state, dsp2 and magnetism of Ni(dmg)2

Now example two, a little harder. Find the oxidation state, the hybridisation and the magnetic behaviour of nickel in its dimethylglyoxime complex. Pause the video and try it first. Done? Okay, let's check. Each DMG loses one proton, H plus, when it binds. So each ligand carries a charge of minus one, and the two of them together give minus two. Okay? The complex is neutral; there is no charge outside its square brackets. So x, the oxidation state of nickel, plus two times minus one, must equal zero. That gives x equals plus two. Right? Now the magnetic part. We just saw that nickel two plus is three d eight, with two unpaired electrons. So the complex must be paramagnetic, right? That is exactly the trap, because this complex is square planar, and square planar nickel here uses d s p two hybridisation; the raised two means two p orbitals are used. Clear? To make d s p two hybrids, nickel needs one empty three d orbital, so its eight three d electrons pair up in the other four. No unpaired electrons are left, so the complex is diamagnetic. With me? So the answer is plus two, d s p two, square planar, and diamagnetic. Very good if you avoided the paramagnetic trap! See how the lesson's clues fit together?

8:31Quick revision: ZnS, NiS, sodium zincate and the DMG test

So, quick revision. First, in group four, white zinc sulphide dissolves in dilute hydrochloric acid, while black nickel sulphide needs aqua regia. Second, zinc gives a white hydroxide that dissolves in excess sodium hydroxide, and a white or bluish-white precipitate with potassium ferrocyanide. Third, nickel in ammoniacal solution gives a rosy-red, square planar, diamagnetic complex with dimethylglyoxime. Now go and try the four practice problems on this page. All the best!