Zinc, Mercury, Silver and Gold

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Drop zinc into dilute sulphuric acid, and you see bubbles of hydrogen. Nothing strange there, right? Now drop zinc into sodium hydroxide, an alkali. And bubbles of hydrogen come out again! So, for equal masses of zinc, which one gives more hydrogen, the acid or the alkali? Keep your guess in mind. Today we study zinc, mercury, silver and gold, whose reactions come up often in JEE, so let's begin.

0:35Zinc reacts with acid and with alkali

So, first zinc, written Z n. With dilute sulphuric acid, Z n plus H two S O four gives Z n S O four, zinc sulphate, plus H two, hydrogen gas. The arrow means gives, and the small numbers written low count the atoms. Okay? Now tell me, will zinc also react with sodium hydroxide, N a O H, an alkali? Yes, it does, and it gives hydrogen again! Z n plus two N a O H gives N a two Z n O two, sodium zincate, plus H two. Reacting with both acid and alkali makes zinc amphoteric. Right? Its oxide behaves the same way. Z n O, zinc oxide, dissolves in sodium hydroxide as sodium zincate. And Z n, O H in brackets, two, zinc hydroxide, is a weak, amphoteric base. With me? Now two colour facts to remember. Zinc oxide is white when cold, and yellow when hot. And Z n S, zinc sulphide, is the only white sulphide of its group. Clear?

1:43Zinc with nitric acid

Now zinc with nitric acid, HNO three. Does it give the same product whether the acid is dilute or concentrated? No, it changes with the concentration, a common trap. Dilute nitric acid gives N two O, nitrous oxide. Four Z n plus ten HNO three gives four Z n, N O three in brackets, two, zinc nitrate, plus N two O, plus five H two O, water. Got it? With very dilute acid, we get N H four N O three, ammonium nitrate. And with concentrated acid, we get N O two, nitrogen dioxide. So always read the concentration first. Okay?

2:27Mercury: calomel and HgI₂

Next, mercury, H g, and its compound calomel, H g two C l two. Each chloride is minus one. So what is the oxidation state of mercury here? If you said plus one, very good! Two chlorides give minus two, shared by two mercury atoms, so each is plus one; that is the Roman one in brackets. It exists as H g two, two plus, where the raised two plus is the charge, and the little dash on screen is the bond joining the two mercury atoms. Right? Now add ammonia, N H three. H g two C l two plus two N H three gives H g, plus H g, N H two in brackets, C l, plus N H four C l. Mercury plus one goes down to the metal and up to plus two together: a disproportionation. The fine mercury is black, so calomel turns black. Clear? Next, H g I two, mercury two iodide, which is red. In excess K I, potassium iodide, it dissolves as K two, H g I four in square brackets; the square brackets hold the complex ion. With me? Add KOH, potassium hydroxide, and we get Nessler's reagent. It gives a brown precipitate with ammonia. Okay?

3:40Silver halides: why AgF dissolves

Now silver, written A g. Silver fluoride dissolves in water, but silver chloride does not. So what makes the fluoride different? It is the fluoride ion, F minus. F minus is very small, so its hydration energy, the energy released when water surrounds the ion, is very large. For silver fluoride, the hydration energy is greater than the lattice energy; that is the greater than sign. So it dissolves. Got it? For A g C l, A g B r and A g I, it is the other way round. Now the lattice energy is greater than the hydration energy. So these stay insoluble. Right? Silver chloride, though, dissolves in ammonia, in KCN, potassium cyanide, and in hypo. And concentrated sodium chloride gives A g C l two minus in square brackets. Clear?

4:31Silver: photography and tarnishing

In photography, light decomposes A g B r, silver bromide, on the film. Then the developer, hydroquinone, reduces the exposed grains to silver metal. With me? Then hypo, N a two S two O three, fixes the film by dissolving the unexposed silver bromide. In symbols, A g B r plus two N a two S two O three gives N a three, A g, S two O three in brackets, two, all inside square brackets, plus N a B r. Okay? Silver articles tarnish in air because traces of hydrogen sulphide, with oxygen, form black silver sulphide. Four A g plus two H two S plus O two gives two A g two S plus two H two O, water. Right?

5:14Gold: aqua regia

Finally, gold, written A u. Among the acids, only aqua regia dissolves gold, and it also dissolves H g S, mercury sulphide. Clear? Gold plus four H plus, plus N O three minus, plus four C l minus, gives A u C l four minus in square brackets, plus N O, plus two H two O, water. So aqua regia supplies hydrogen, nitrate and chloride ions. With me? Now let me ask you this: why does gold need both the nitrate and the chloride, and not just one of them? Yes, exactly! Nitrate oxidises gold to A u three plus, and chloride holds it as the complex ion, A u C l four minus. Remember, both ions are needed. Okay?

5:59Example 1: H2 volumes from zinc with acid vs with NaOH

Now Example one, and it is our puzzle from the start. Equal masses of zinc react with excess dilute H two S O four, and with excess sodium hydroxide. Compare the volumes of hydrogen. Pause the video and try it first. Done? Okay, let's check. With the acid, Z n plus H two S O four gives Z n S O four plus H two. Got it? And with the alkali, the equation has two N a O H on the left side. Many students see that two and say the alkali gives twice the hydrogen. Is that right? No, and that is the trap: that two belongs to sodium hydroxide, not to hydrogen. In both reactions, one mole of zinc gives one mole of H two; mol is short for mole. Right? Equal masses of zinc means equal moles of zinc, so equal moles of hydrogen. Measured under the same conditions, the volumes are in the ratio one to one; the double arrow means so. So our puzzle is solved: neither gives more hydrogen. Clear?

7:06Example 2: identify a red solid that dissolves in KI

Now Example two, a harder one, like a detective case. We have a red solid, insoluble in water, which dissolves when K I is added. On heating, it gives violet vapour and droplets of metal. Identify the solid. Pause the video and try it first. Done? Okay, let's check. Take the first clue: which red solid from today's lesson dissolves in potassium iodide? Yes, exactly, it is H g I two, mercury two iodide! It dissolves in K I as K two H g I four, the complex we met earlier. Okay? Now check the heating clue. H g I two breaks up into H g and I two, so the violet vapour is iodine, and the droplets are mercury. Right? So the answer is H g I two, mercury two iodide; the Roman two in brackets is the oxidation state of mercury. Got it?

8:01Quick revision: amphoteric Zn, calomel, AgF, aqua regia

So, quick revision. Zinc is amphoteric, its oxide is white when cold and yellow when hot, and with nitric acid the product depends on the concentration. Mercury: calomel turns black with ammonia, and red H g I two dissolves in excess potassium iodide. Silver and gold: silver fluoride is soluble, hypo fixes silver bromide, and among the acids, only aqua regia dissolves gold. Now try the two practice problems on this page. All the best!