Mixed Problems: Set 3 (Groups 17 and 18)

Transcript

0:00Introduction

Hello everyone! Let us start with a one-minute question. Three moles of chlorine gas bubble through hot, concentrated sodium hydroxide. How many moles of sodium chlorate come out? Three moles of chlorine, so three moles of chlorate, right? Hmm, hold that thought; Example one will settle it. This is Set three of the mixed problems, on groups seventeen and eighteen, the halogens and the noble gases. Questions like these come up often in JEE Main. So, let's begin.

0:34Chlorine in alkali: down and up at once

First, the reaction behind our question. Pass chlorine into cold, dilute sodium hydroxide. C l two plus two N a O H gives N a C l, plus N a O C l, sodium hypochlorite, plus water. The small numbers below count the atoms, and the arrow means gives. Okay? Now let me ask you this. What is the oxidation state of chlorine in sodium chloride, and in sodium hypochlorite? Let us see. In C l two, chlorine is zero. In sodium chloride it is minus one, and in sodium hypochlorite it is plus one. Right? So chlorine is reduced and oxidised at the same time. We call this disproportionation. Clear? Now make the alkali hot and concentrated. We get chloride again, at minus one, and chlorate, C l O three minus, with the raised minus for its charge, where chlorine is plus five. We balance this one in Example one. Got it?

1:34Halogens: fluorine stands apart

Next, the halogen trends. Oxidising power falls down the group: F two, then C l two, then B r two, then I two, as the greater than signs show. Fluorine is the strongest, and the brackets say why: the F F bond is weak, so it has a low bond dissociation enthalpy, and the small fluoride ion, F minus, has a very high hydration enthalpy. With me? So each halogen displaces the heavier halide ions from solution. C l two plus two B r minus gives two C l minus plus B r two; the raised minus is the charge on each halide ion. Okay? Now a quick one from this set. Which halogen can displace all the other halogens from solutions of their halides? Fluorine, yes, exactly, the strongest oxidant of all! Just one note, in the brackets: in water, fluorine also oxidises the water itself. Right? Now a statement pair from this set. Statement one: hydrogen fluoride is a weaker acid than hydrogen chloride in water. Statement two: hydrogen fluoride has the higher boiling point. Can both be true? Yes, both are true! In water, acid strength rises from H F to H I, as the less than signs show, because the bond enthalpy falls. The strong bond in hydrogen fluoride makes it the weakest acid. Clear? But boiling point is a different property. H C l boils lowest, and hydrogen fluoride highest, because its molecules are held together by hydrogen bonding. So judge each statement on its own.

3:14Hybridisation: count the electron pairs

Got it? Now hybridisation. The trick is to count the electron pairs on the central atom: the bond pairs plus the lone pairs. The figure is from Chapter five; it shows four xenon compounds, with the lone pairs on xenon drawn as grey lobes. Okay? Look at X e F two first, the first one in the figure. Xenon makes two bonds, and the three grey lobes are its three lone pairs. Two plus three gives five pairs, so xenon is s p three d, with a small raised three on the p, and the molecule is linear. Right? In X e F four, four bonds and two lone pairs, one above and one below, give six pairs. So xenon is s p three d two, with a raised two on the d as well, and the molecule is square planar. With me? Now you try. In B r F three, bromine trifluoride, bromine has seven valence electrons and makes three bonds. How many electron pairs does bromine have? Five pairs, yes, exactly! Three electrons go into the bonds, and the other four make two lone pairs. So bromine is s p three d, just like xenon in X e F two, and the molecule is bent T-shaped. Clear? And in I F seven, iodine uses all seven valence electrons in seven bonds, with no lone pair left. Seven pairs means s p three d three, and the molecule is pentagonal bipyramidal. Got it?

4:46Example 1: moles of NaClO3 from 3 mol Cl2 in hot NaOH

Now Example one, from Set three, and it settles our opening question. How many moles of N a C l O three, sodium chlorate, are formed when three moles of C l two react with hot, concentrated N a O H? Pause the video and try it first. Done? Okay, let's check. Hot and concentrated, so we get chlorate. The balanced equation: three C l two plus six N a O H gives five N a C l, plus N a C l O three, plus three molecules of water. Okay? So three moles of chlorine give three moles of chlorate, just as we guessed at the start, right? Oh no, careful, because that is the trap! Count the chlorine atoms: three C l two hold six of them. Five go down to minus one, as sodium chloride, and only one goes up to plus five, as chlorate. Clear? Let us check the electrons. Five atoms gain one electron each, five times one, which is five. And the single atom going up to plus five loses five electrons, one times five. Gained equals lost, so the equation balances. With me? So the answer: three moles of C l two give just one mole of sodium chlorate. Not three, and that settles our opening question! Got it?

6:11Example 2: available chlorine in bleaching powder

Now Example two, a step harder. Five point zero zero grams of bleaching powder is treated with excess K I, potassium iodide, and acetic acid. The iodine needs forty point zero millilitres of zero point five zero zero molar thiosulphate; the capital M means molar. Find the percentage of available chlorine, the chlorine the sample can give, as a percentage of its mass. Pause the video and try it first. Done? Okay, let's check. Millilitres times molarity gives millimoles, the m mol on screen. Forty point zero times zero point five zero zero gives twenty point zero millimoles of thiosulphate. Okay? Thiosulphate reacts with iodine; this is iodometry. Two S two O three two minus, the raised two minus being the charge, plus I two gives S four O six two minus, tetrathionate, plus two iodide ions. Got it? So twenty point zero millimoles of thiosulphate means twenty point zero millimoles of iodine, right? No, no, wait, look at the equation again! Each I two needs two thiosulphate ions. So the iodine is twenty point zero over two, which gives ten point zero millimoles. Clear? Now, this iodine was set free from potassium iodide by the available chlorine, one C l two for each I two, just like the displacement we saw earlier. So the chlorine is also ten point zero millimoles. With me? Now the mass. The molar mass of C l two is seventy point nine grams per mole; the raised minus one means per mole. Ten point zero millimoles times seventy point nine gives seven hundred and nine milligrams, which is zero point seven zero nine grams. Got it? Finally, zero point seven zero nine over five point zero zero, times one hundred, gives fourteen point two percent of available chlorine. Two thiosulphate ions for each iodine, and you are done. Okay?

8:20Quick revision: Cl2 in alkali, HF, hybridisation, iodometry

So, quick revision. Chlorine in alkali disproportionates: cold gives hypochlorite, hot gives chlorate. Fluorine is the strongest oxidant, and hydrogen fluoride is the weakest of the four acids, yet boils highest. For hybridisation, add bond pairs and lone pairs: five, six and seven pairs give s p three d, s p three d two and s p three d three. And in iodometry, one C l two frees one I two, which takes two thiosulphate ions. Now go and try the practice problems on this page. All the best!