Mixed Problems: Set 1 (Werner's Theory, Names and Isomers)

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. Take one cobalt complex with one e n ligand, two ammonias, one chloride and one bromide. How many stereoisomers can it have? Four feels like a safe answer, right? Hmm, hold on, the true number is bigger, and by the end you will see why. This is Set one of the mixed problems on coordination compounds: the theory of Werner, names and isomers. Questions like these come up often in JEE Main, and each one should take about a minute. We will build three quick tools, then solve two problems together. So, let's begin.

0:43Tool 1: inside or outside the square bracket?

So, tool one, from Werner's theory. In a complex, the metal shows two kinds of valency. The primary valency equals the oxidation state, and it is ionisable. Okay? The secondary valency equals the coordination number. It is satisfied by the ligands, and it is non-ionisable. Right? Now the key rule. Only the ions outside the square bracket ionise in water. So only a chloride outside the bracket, C l minus, where the small raised minus sign shows its charge, gives a precipitate of silver chloride, written A g C l. The chloride inside the bracket stays put. Clear? Now a quick one, question one of this set. One mole of C o C l three, dot, five N H three, is treated with excess silver nitrate, A g N O three. The small numbers below the letters count atoms, and the dot joins the two parts of the formula. It has three chlorines, so it gives three moles of silver chloride, right? No, no, careful, first ask where each chloride sits! Cobalt holds six donor atoms here: five ammonias and one chloride, all inside the bracket. So the formula is open square bracket, C o, N H three in round brackets five times, C l, close bracket, then C l two. Only the two chlorides outside ionise, so the answer is two moles of silver chloride. Got it?

2:11Tool 2: naming a complex, step by step

Now tool two, names. The cation is named first, then the anion, as in simple salts. Okay? Inside the complex, the ligands come first, in alphabetical order, and then the metal. And the multiplying prefixes listed in the round brackets on screen, di, tri and tetra, are ignored while sorting. Right? After the metal comes its oxidation state, in Roman numerals inside round brackets. So the three capital I's on screen are the Roman numeral three, and we read it as cobalt three. With me? Anionic ligands, the negative ones, end in the letter o, so chloride becomes chlorido, and hydroxide becomes hydroxido. Neutral ligands keep their names, except a few: water, H two O, is aqua, and ammonia, N H three, is ammine, spelt with a double m in the middle. Got it? Now tell me this. Suppose the whole complex carries a negative charge. Does the name of the metal change? Yes, it does change: for an anionic complex, the metal's name ends in ate. Take question seven of this set: K two, open square bracket, Z n, O H in round brackets four times, close bracket. Potassium is the cation, so it comes first, and zinc is plus two here. So the name is potassium tetrahydroxidozincate two. Clear?

3:34Tool 3: counting stereoisomers

Now tool three, isomers. In an octahedral complex, the metal has six positions, in three opposite pairs. Two ligands opposite each other are trans, and two side by side are cis. Okay? Next, e n, short for ethane one two diamine. It is bidentate: it binds through two donor atoms, its two nitrogens. So its two ends can only reach two cis positions, never trans. Right? Now, a complex is chiral when it cannot be superimposed on its mirror image. So let me ask you this: what must a complex lack to be chiral? Symmetry, yes, exactly! When a complex is chiral, it has no plane of symmetry, no centre of symmetry, and no other improper axis either. Then it and its mirror image are two different isomers, a pair of enantiomers. With me? So, total stereoisomers equals the number of geometrical isomers, plus one extra for each chiral one. One extra, not two, because the chiral form is already counted once. Got it?

4:40Example 1: IUPAC name of [Co(NH3)4(H2O)Cl]Cl2

Alright, example one, an easy one, question six of this set. Name this complex: open square bracket, C o, N H three in round brackets four times, H two O in round brackets, C l, close bracket, then C l two. Pause the video and try it first. Done? Okay, let's check. Two chlorides sit outside, so the complex is a cation, with the raised two plus charge on screen. Right? Let x be the oxidation state of cobalt. Ammonia and water are neutral, and the chloride inside is minus one. So x, plus four times zero, plus zero, plus, in brackets, minus one for the chloride, equals plus two, which gives x equals plus three. The double arrow means therefore, so we write cobalt three, in Roman numerals. Okay? Now the order: ammine, aqua and chlorido. With four ammines, the name says tetraammine, and tetra starts with t. So aqua comes first, right? No, no, wait, that is exactly the trap in this question! We ignore the multiplying prefixes while sorting. Ammine and aqua both start with the letter a, and in the second letter, m comes before q. So the order is ammine, aqua, chlorido. Clear? Now put it together, as one word, then chloride for the anion outside. Tetraammineaquachloridocobalt three chloride is the full name. Got it?

6:12Example 2: stereoisomers of [Co(en)(NH3)2ClBr]+

Now, remember our puzzle from the start? Let's solve it, with question thirteen of this set. Open square bracket, C o, e n in round brackets, N H three in round brackets twice, C l, B r, close bracket, with a raised plus charge. How many geometrical isomers, and how many stereoisomers? Pause the video and try it first. Done? Okay, let's check. The e n ligand takes two cis places. Of the four places left, two are trans to the e n nitrogens, and the other two are trans to each other, one above and one below. The two ammonias, the chloride and the bromide fill these four places. With me? So, where can the two ammonias go? List the cases first. Case one: the two ammonias trans to each other, one above and one below. Then the e n, the chloride and the bromide lie in one plane with the cobalt, and that plane is a mirror plane. So case one gives one isomer, and it is achiral. Okay? Case two: both ammonias trans to e n. Now the chloride and the bromide go above and below, and a mirror plane passes through them and the cobalt, cutting the e n down the middle. So case two also gives one isomer, and it is achiral too. Right? Case three: one ammonia trans to e n, and the other one above or below. Opposite that second ammonia, we can put the chloride or the bromide, so this case gives two isomers. Neither of them has a mirror plane, so each one is chiral. Got it? So, one plus one plus two gives four geometrical isomers. Clear? Then four geometrical isomers means four stereoisomers, right? Oh no, careful, that is exactly the trap from our puzzle! The two chiral isomers of case three each come with a mirror image partner, so we add one for each. Four plus two gives six stereoisomers. Puzzle solved: the two extra forms are those mirror images. With me?

8:23Quick revision: AgCl test, naming rules, isomer count

So, quick revision. First, only the ions outside the square bracket ionise, as the silver chloride test shows. Second, for names: ligands alphabetically, prefixes ignored, then the metal and its Roman numeral, with the ending ate for an anionic complex. Third, stereoisomers equal the geometrical isomers, plus one for each chiral one. Now go and try the practice problem on this page. All the best!