Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Let me start with a question. Iron, copper, silver and gold are all solid metals. Mercury is a metal too, and yet it is a liquid at room temperature. So why is mercury a liquid? Hold that thought; it is one of today's checks. This is Part C of the unit problem bank: six quick concept checks on the d and f blocks. Each one looks like a one-word answer, and each one hides a trap. Ideas like these come up often in JEE, inside statement questions. So, let's begin.

0:37Is zinc a transition element?

So, check number one: is zinc a transition element? The d block is groups three to twelve, and the three d series runs from scandium, S c, all the way to zinc, Z n. So zinc sits right inside the d block. That makes it a transition element, right? Hmm, not so fast, it depends on the definition. Here it is: a transition element has an incompletely filled d subshell, either in the atom, or in one of its common oxidation states. Clear? Now, zinc is three d ten, four s two; the raised numbers count the electrons in each subshell. Zinc two plus, with the raised two plus as its charge, is three d ten. Full in the atom, and full in the ion. So the answer is no, zinc is not a transition element. Right? And the same goes for cadmium and mercury, C d and H g. But copper two plus is three d nine, incompletely filled, so copper is a transition element. With me?

1:41Does every lanthanoid show +4?

Next, the lanthanoids, from cerium to lutetium. For all of them, plus three is the characteristic oxidation state. Okay? Plus two or plus four appear mainly when the ion reaches an empty, a half-filled or a completely filled f set. That is f zero, f seven or f fourteen; the raised number counts the f electrons. Right? So cerium four plus is four f zero, and terbium four plus is four f seven. Both are strong oxidising agents. Clear? So, does every lanthanoid form a plus four compound? The answer is no, only a few do, such as cerium, praseodymium, neodymium, terbium and dysprosium; that is C e, P r, N d, T b and D y on screen. So if a statement says all the lanthanoids show plus four, that statement is false. Got it?

2:40Why is mercury a liquid?

Now our opening question, mercury. What holds a transition metal together? It is metallic bonding. Besides the outer s electrons, written n s, the unpaired d electrons of the shell just below also take part; n minus one, in brackets, is that shell below. That is why transition metals have high melting points. Okay? So now tell me. Zinc, cadmium and mercury are all d ten. How many unpaired d electrons can they give to the bonding? None at all, yes, exactly! So zinc, cadmium and mercury have low melting points. Right? And mercury goes one step further. Its outer electrons are five d ten, six s two. That six s pair is held very tightly, and the five d ten shell is full. With me? So its metallic bonding is very weak, and mercury is a liquid at room temperature. That is our opening puzzle, solved. Clear?

3:39Is the colour of KMnO₄ a d–d transition?

Now let us talk about colour. First, the d to d transition; the dash between the two d letters is read as to. Here a d electron absorbs visible light and jumps to a higher d level. Okay? So, simple d zero and d ten ions, like scandium three plus and zinc two plus, have no d to d transition, and they are colourless. Right? Now take K M n O four, potassium permanganate. Manganese is three d five, four s two. Here it is plus seven, written M n with the Roman numeral seven in brackets, so all seven outer electrons are gone: three d zero. And yet the salt is deep purple. Clear? So let me ask you this: is the colour of potassium permanganate due to a d to d transition? The answer is no, because with zero d electrons there is nothing to jump. Instead, an electron moves from an oxygen, the ligand, to an empty manganese orbital; the arrow points from O to M n. This is called ligand to metal charge transfer. Got it? The same is true for yellow chromate, C r O four two minus, and orange dichromate, C r two O seven two minus; the raised two minus is the charge. Chromium is plus six and d zero in both. Keep these two in mind for Example one. With me?

5:03Example 1: is chromate to dichromate a redox reaction?

Now, Example one, an easy one. Is the change from chromate to dichromate a redox reaction? Hint: find the oxidation state of chromium in each ion. Pause the video and try it first. Done? Okay, let's check. Two chromate ions and two H plus give one dichromate ion and water. The small numbers written low count the atoms, and the two half arrows mean the reaction goes both ways. Okay? And see the colours: yellow chromate in base, orange dichromate in acid. The colour changes, so something must be oxidised or reduced, right? Oh no, careful, let us actually calculate. Let x be the oxidation state of chromium. In chromate, x plus four times minus two equals minus two, the charge on the ion. So x is plus six; the double arrow means so. Okay? In dichromate, two x plus seven times minus two equals minus two. So two x is twelve, and x is plus six again. Clear? So chromium stays plus six. Nothing is oxidised or reduced, and the answer is no, it is not a redox reaction. Both colours come from charge transfer, so a colour change alone does not mean redox. Got it?

6:24Example 2: KMnO4 product in acid, neutral and alkaline media

Now, Example two, a step harder. Is permanganate reduced to the same product in all media? Hint: think about acid, neutral and strongly alkaline solution. Pause the video and try it first. Done? Okay, let's check. Most of us first meet permanganate in acid, where it gives M n two plus. So it gives M n two plus every time, right? No, wait, that is the trap. It is true in acid: M n O four minus, plus eight H plus, plus five e minus, give M n two plus, plus four molecules of water; e minus is the electron. Manganese goes from plus seven to plus two, so it gains five electrons. Okay? But in neutral or faintly alkaline solution, M n O four minus, plus two molecules of water, plus three e minus, give M n O two, plus four O H minus; O H minus is the hydroxide ion. Right? So here manganese goes from plus seven to plus four, taking just three electrons, and the product is manganese dioxide. Clear? And in strongly alkaline solution, M n O four minus takes only one electron and becomes M n O four two minus, the manganate ion. Manganese goes from plus seven to plus six. With me? So the answer is no. We get M n two plus in acid, M n O two in neutral or faintly alkaline solution, and M n O four two minus in strong alkali. The medium decides the product. Got it?

8:02Quick revision: Zn and Hg, lanthanoid +4, colour, KMnO4

So, let us do a quick revision. First, zinc, cadmium and mercury are d ten, so they are not transition elements, and weak metallic bonding makes mercury a liquid. Second, plus three is the characteristic state of the lanthanoids, and only a few of them, like cerium and terbium, show plus four. Third, d zero oxoanions get their colour from charge transfer, and chromate to dichromate is not redox. And fourth, permanganate gives M n two plus in acid, M n O two in neutral solution, and manganate in strong alkali. Now go and try the practice problem on this page. All the best!