Tin and Lead
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Tin and lead for JEE Main and Advanced: tin from cassiterite, tin with dilute and concentrated nitric acid, SnCl2 as a reducing agent with HgCl2 (white, then grey), why gaseous SnCl2 is bent, tin plague, self-reduction of galena, PbO and PbO2, why PbO2 is not a peroxide, PbCl2, PbI2, the lead storage battery, and two solved problems.
Chapters
- 0:00Introduction
- 0:24Tin: from cassiterite, and with nitric acid
- 1:27SnCl₂ is a reducing agent
- 2:32Shape of gaseous SnCl₂, and tin plague
- 3:25Lead from galena: self-reduction
- 4:07Oxides of lead: PbO and PbO₂
- 5:26Lead salts and the lead storage battery
- 5:58Example 1: oxidation state of lead in litharge and red lead
- 7:04Example 2: mass of tin from 1.00 kg of cassiterite
- 8:15Quick revision: SnCl2, galena, PbO2, tin with nitric acid
Transcript
0:00Introduction
Hello everyone! Here is something strange. Cool a shiny piece of white tin below two hundred and eighty six kelvin, about thirteen degrees Celsius, and over time it slowly turns into a grey powder. Is it rusting? Is something eating it? We will find out today. Tin and lead come up often in JEE, so let's begin.
0:24Tin: from cassiterite, and with nitric acid
So, let's start with tin. Its ore is cassiterite, S n O two; the small lowered two counts the oxygen atoms. Carbon reduces it: S n O two plus two C gives S n plus two C O, carbon monoxide, and the arrow means gives. Now, with nitric acid, HNO three, will dilute and concentrated acid give the same products? No, they do not. Dilute acid: four S n plus ten HNO three gives four tin two nitrate, plus N H four N O three, ammonium nitrate, plus three H two O, water. The brackets hold one N O three group, and the two outside means two of them. And nitrogen falls from plus five all the way to minus three. With me? With concentrated acid: S n plus four HNO three gives H two S n O three, metastannic acid, plus four N O two, nitrogen dioxide, plus water. Okay?
1:27SnCl₂ is a reducing agent
Next, S n C l two, tin two chloride. It is a reducing agent: S n two plus, the raised two plus being its charge, readily changes to S n four plus. Right? The classic reaction: S n C l two plus two H g C l two, mercury two chloride, gives H g two C l two, plus S n C l four. The down arrow means a precipitate, and this one is white. Now, what if we keep adding S n C l two? What happens to that white solid? Yes, it turns grey! The excess S n C l two reduces H g two C l two further, to two H g, mercury metal, which is grey, plus S n C l four. Clear? White, then grey: this is a test for mercury two plus, and for tin two plus. Got it? S n C l two also reduces F e three plus to F e two plus, iron three to iron two, and iodine to iodide. Okay?
2:32Shape of gaseous SnCl₂, and tin plague
Now a shape question, and a common trap. Gaseous S n C l two, and M g C l two, magnesium chloride. Both have one metal atom and two chlorine atoms, so both are linear, right? No, no, careful, look at tin! It has four valence electrons: two go into the tin chlorine bonds, and two stay as a lone pair. Two bond pairs and one lone pair, three pairs, so the molecule is bent. With me? Magnesium has only two valence electrons, both used in bonding. Two pairs, so M g C l two is linear. Right? Now, our puzzle from the start. Below two hundred and eighty six kelvin, the K on screen, white tin changes into powdery grey tin. It is not rusting; it is still tin, in a different form. This is tin plague. Clear?
3:25Lead from galena: self-reduction
Now, let's turn to lead. Its ore is galena, P b S, lead sulphide. Part of it is roasted: two P b S plus three O two, oxygen, gives two P b O, lead oxide, plus two S O two, sulphur dioxide. The remaining P b S now meets this P b O, and no carbon is added. So what reduces the lead oxide? Yes, the sulphide itself! P b S plus two P b O gives three P b, plus S O two. No outside reducing agent is needed, so we call it self reduction. And as the bracket says, P b O can also be reduced by carbon. Got it?
4:07Oxides of lead: PbO and PbO₂
Lead has two important oxides. P b O, called litharge, has lead in plus two, and it is amphoteric: it reacts with both acids and bases. Okay? Then P b O two, lead dioxide: brown, and a powerful oxidising agent. And why is that? The inert pair effect: lead prefers plus two, so P b four plus is eager to become P b two plus. So, add concentrated hydrochloric acid, H C l. Which gas comes off? Yes, exactly, it is chlorine! P b O two plus four H C l gives P b C l two, plus C l two, chlorine gas, plus two water. Right? With concentrated nitric acid, it releases oxygen instead: two P b O two plus four HNO three gives two lead nitrate, plus two water, plus O two. Now, P b O two has two oxygen atoms. So it is a peroxide, and acids should give H two O two, hydrogen peroxide, right? Oh no, careful, P b O two is not a peroxide. It is lead four oxide, with lead in plus four and no oxygen oxygen bond, so acids give no hydrogen peroxide from it. This is a common trap. Clear?
5:26Lead salts and the lead storage battery
Now, some lead salts. P b C l two, lead chloride, is sparingly soluble in cold water, but soluble in hot water. With me? P b I two, lead iodide, is yellow. And lead acetate paper turns black with H two S, hydrogen sulphide, as P b S, lead sulphide, forms on it. Okay? And the lead storage battery uses lead and P b O two in thirty eight percent sulphuric acid, H two S O four. Right?
5:58Example 1: oxidation state of lead in litharge and red lead
Now, example one, an easy one. What is the oxidation state of lead in litharge, and in red lead, on average? Pause the video and try it first. Done? Okay, let's check. Litharge is P b O, lead oxide. Let x be the oxidation state of lead. Oxygen is minus two, so x minus two equals zero, which makes x equal to plus two. Got it? Red lead is P b three O four: three lead atoms and four oxygen atoms. So three x minus eight equals zero, and three x equals eight. That makes x equal to eight over three, a fraction. Can an oxidation state be a fraction? Yes, as an average, it can. Red lead has two lead atoms in plus two and one in plus four; the small raised Roman numerals next to P b on screen show these states. Two plus two plus four is eight, and eight over three is the average. Clear?
7:04Example 2: mass of tin from 1.00 kg of cassiterite
Now example two, a numerical one. What mass of tin can be obtained from one point zero zero kilograms, the k g on screen, of pure cassiterite? Pause the video and try it first. Done? Okay, let's check. One S n O two gives one S n, so one mole gives one mole. So one kilogram of ore gives one kilogram of tin, right? No, no, wait, equal moles are not equal masses; oxygen carries part of the mass. M, the molar mass of S n O two, is one hundred and eighteen point seven, plus two times sixteen point zero, which gives one hundred and fifty point seven grams per mole; mol to the minus one means per mole. With me? So every one hundred and fifty point seven grams of cassiterite holds one hundred and eighteen point seven grams of tin. Okay? Now scale it up. Mass of tin equals one point zero zero kilograms, times one hundred and eighteen point seven over one hundred and fifty point seven. That gives zero point seven eight eight kilograms of tin. Got it?
8:15Quick revision: SnCl2, galena, PbO2, tin with nitric acid
So, quick revision. Tin: S n C l two reduces mercury two chloride, white then grey, and gaseous S n C l two is bent. Lead: galena by self reduction, and P b O two oxidises H C l to chlorine, but it is not a peroxide. And tin with nitric acid: dilute gives ammonium nitrate, concentrated gives metastannic acid. Now try the practice problems on this page. All the best!
More videos in Unit 10 · p-Block Elements
- General Trends in Group 13
- Boron: Borax, Boric Acid and the Trihalides
- Diborane and Borazine
- Aluminium and Its Compounds
- General Trends in Group 14
- Carbon: Allotropes and Oxides
- Silicon: Silica, Silicates, Zeolites and Silicones
- General Trends in Group 15
- Dinitrogen and Ammonia
- Oxides of Nitrogen and Nitric Acid
- Phosphorus: Allotropes, Phosphine and Halides
- Oxides and Oxoacids of Phosphorus
- General Trends in Group 16
- Dioxygen, Oxides and Ozone
- Sulphur and Sulphur Dioxide
- Sulphuric Acid and the Oxoacids of Sulphur
- General Trends in Group 17
- Chlorine, Bleaching Powder and Hydrogen Chloride
- Oxoacids of Halogens
- Interhalogens, Polyhalides and Pseudohalides
- Group 18: Noble Gases and Xenon Compounds
- Overview
- Mixed Problems: Set 1 (Groups 13 and 14)
- Mixed Problems: Set 2 (Groups 15 and 16)
- Mixed Problems: Set 3 (Groups 17 and 18)
- Overview
- Part A: JEE Main Level
- Part C: Concept Checks