Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. The crystals C o C l three, dot, six H two O have three chlorine atoms in every formula unit. Add excess silver nitrate, and all three chlorides should come down as silver chloride, right? Hmm, not so fast, because in a JEE Main question only two of every three come down. So where is the third chloride hiding? This is Part A of the unit problem bank on coordination compounds. We will sharpen two tools, then solve two of its questions. So, let's begin.

0:38Tool 1: which chlorides precipitate?

So, the first tool is Werner's theory: the metal in a complex shows two kinds of valency. The primary valency is ionisable, and it equals the oxidation state; negative ions satisfy it. Okay? The secondary valency is non-ionisable, and it equals the coordination number, the number of donor atoms on the metal. Ligands satisfy it, neutral ones like ammonia and water, or negative ones like chloride. Right? So let me ask you this: when you add silver nitrate, which chloride ions can it pull out as silver chloride? Only the ones outside the square brackets, yes, exactly! The brackets hold the coordination entity, the metal with its ligands, which stays intact in water. Ions outside are counter ions; they ionise, and chloride counter ions give A g C l, silver chloride, with A g N O three, silver nitrate; the small numbers written low count the atoms. Clear? Werner saw this. One mole of C o C l three, dot, five N H three gives two moles of silver chloride; the raised dot joins the parts of the formula, and mol means moles. Cobalt has coordination number six, so five ammonias, the N H three in round brackets with a five, and one chloride sit inside the square bracket. The double arrow means therefore: two chlorides stay outside, and the formula is open square bracket, C o, N H three five, C l, close bracket, C l two. With me? Here is a quick one on the way, from Part A, question ten. How many ions per formula unit does K four, open square bracket, F e, C N six, close bracket, potassium ferrocyanide, give in water? Four K plus ions, and one complex anion, F e C N six, with the raised four minus as its charge, which stays in one piece. So, five ions; the cyanides never come off separately. Got it?

2:40Tool 2: the spin-only magnetic moment

Now let us build the second tool. The Greek letter mu, with a small s below it, is the spin-only magnetic moment. It equals the square root of n times the quantity n plus two, in B M, which stands for Bohr magnetons. Here n is the number of unpaired electrons. Okay? Put in n from one to five, and you get one point seven three, two point eight three, three point eight seven, four point nine zero, and five point nine two Bohr magnetons. Right? Now, to find n, look at the d electrons. An octahedral d six ion has two choices. Here t two g and e g name the lower and upper sets of d orbitals, and the small raised numbers count their electrons. High spin is t two g four, e g two, with four unpaired electrons; low spin is t two g six, all paired, so n is zero. Clear? So let me ask you this: for the same d six ion, what decides whether we get high spin or low spin? Yes, exactly, it is the ligand! Strong ligands force the electrons to pair: cyanide, C O, and en, which is ethane one two diamine. And for C o three plus, cobalt in the plus three state, ammonia, oxalate, C two O four two minus, and even water also pair the electrons. Keep this rule for cobalt three plus; do not stretch it to every metal ion. With me? Weak ligands do not pair the electrons: iodide, bromide, chloride, thiocyanate, fluoride and hydroxide. Keep fluoride in mind; it comes up often. Got it?

4:22Example 1: spin-only moment of [CoF6]3- and [Co(C2O4)3]3-

Now, example one, which is question three of Part A, an easy one. Find the spin-only moment of two ions. The first is C o, C two O four three, in square brackets, three minus. The second is C o F six, in square brackets, three minus. Pause the video and try it first. Done? Okay, let's check. Oxalate is minus two and fluoride is minus one, so the ligands give minus six, and the ion carries minus three. So cobalt is plus three in both, and C o three plus is three d six. Okay? Let us take fluoride first. It is a weak ligand, so there is no pairing: high spin, t two g four, e g two, and n is four. Right? So mu s is the square root of four times six, that is root twenty four, which gives four point nine zero Bohr magnetons. Clear? Now let us look at oxalate. In the spectrochemical series, oxalate sits to the left of water, on the weak side. So this is high spin too, four point nine zero again, right? No, no, careful, this is exactly the trap students fall into! Tool two told us that with C o three plus, oxalate forces pairing. So C o three plus goes low spin, t two g six, n is zero, and the ion is diamagnetic. With me? So the answer is zero for the oxalate complex, and four point nine zero Bohr magnetons for fluoride. Same metal, same charge; the ligand alone made the difference. If you saw that, you are doing very well!

6:02Example 2: CoCl3.6H2O with AgNO3, JEE Main 2017

Now example two, which is question one of Part A, from JEE Main twenty seventeen, and it is our puzzle from the start. One hundred millilitres of zero point one molar C o C l three, dot, six H two O is treated with excess silver nitrate. It precipitates one point two times ten to the power twenty two chloride ions. Identify the complex. Pause the video and try it first. Done? Okay, let's check. First, the moles of complex. Zero point one molar, the capital M, times zero point one litre, the capital L, gives zero point zero one mole. Okay? Now let us count the chloride ions. Divide by Avogadro's number, taken as six times ten to the power twenty three. One point two times ten to the power twenty two, over six times ten to the power twenty three, gives zero point zero two mole of silver chloride. Right? So let me ask you this: how many chloride ions does each formula unit give? Yes, exactly, it is two! Zero point zero two over zero point zero one gives two ionisable chlorides, not three. So the third chloride hides inside the bracket, as a ligand. Clear? Now, the formula has six waters. So all six waters go inside the bracket, together with that chloride, right? No, no, wait, count the coordination number first! Cobalt has coordination number six, written C N on screen. One chloride takes one place, so only five waters fit inside the square brackets. The sixth water stays outside the bracket. Got it? So the complex is: open square bracket, C o, H two O five, C l, close bracket, C l two, dot, H two O, with that water outside. Puzzle solved: the third chlorine is bonded to cobalt, so silver nitrate cannot pull it out. If you got this one, that is really good work!

8:10Quick revision: AgCl test, spin-only formula, d6 spin states

So, quick revision. Only chloride outside the square brackets precipitates as silver chloride; the bracket stays intact. Mu s is the square root of n times the quantity n plus two, and a d six ion has four unpaired electrons in high spin, none in low spin. Fluoride is weak and does not pair, while oxalate pairs the electrons of C o three plus. Now go and try the practice problem on this page; it uses tool two. All the best!