Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a one-minute question, in the JEE Main style. Cobalt chloride with five ammonia molecules carries three chlorides. So one mole of it, with excess silver nitrate, should give three moles of silver chloride, right? Well, hold that thought; example one will settle it. This chapter gives you three sets of mixed problems in the JEE Main style, so let us see how to use them. Let's begin.

0:30How this chapter works

So, first, why this chapter matters. Coordination compounds usually give one or two questions in a JEE Main shift, and they come in a few fixed types. Okay? First, counting questions: how many species are paramagnetic, how many unpaired electrons, or how many isomers. Right? Second, matching a hybridisation or a CFSE value, the crystal field stabilisation energy. And also IUPAC names, and colour orders. Clear? Now let me ask you this. Our cobalt chloride question from the start: which of these types is it? It is a counting question, yes, exactly! We count moles of silver chloride. Now the three sets. Set one covers Werner's theory, names and isomers, and Set two covers bonding, magnetism and crystal fields. Got it? Set three covers colour, carbonyls, stability and uses. Each set has eighteen solved problems, and each one should take you about a minute. With me? And the real JEE Main questions, from twenty twenty one to twenty twenty six, are at the end of Chapters one to six. The problems here are original, in the same style. Okay?

1:44Tool for Set 1: inside or outside the square bracket

Now, the tool for Set one: Werner's theory. The metal shows two kinds of valency. The primary valency is ionisable, and it equals the oxidation state. Right? The secondary valency is non-ionisable, and it equals the coordination number. Okay? The metal and its ligands sit inside the square brackets you see on screen. Ions outside the bracket are counter ions, and they ionise in water. So only the chlorides outside precipitate as A g C l, silver chloride. Clear? For names, the ligands come in alphabetical order, then the metal, then its oxidation state in Roman numerals. Got it? And if the whole entity is an anion, the metal name ends in ate: zincate, and the Latin roots ferrate for iron and cuprate for copper. With me?

2:33Tool for Set 2: d count, ligand, then spin

Now Set two: bonding, magnetism and crystal fields. The same four steps work every time, and each arrow means then. Find the oxidation state and the d electron count, judge whether the ligand is strong or weak, and place the electrons. Okay? Strong ligands pair the electrons: C N minus, cyanide; C O, carbon monoxide; and e n, ethane one two diamine. Ammonia, N H three, is strong too with an M three plus ion, M being the metal and the raised three plus its charge. Right? For the three d metals, weak ligands do not pair: fluoride and chloride, F minus and C l minus. And water, H two O, or ammonia with an M two plus ion are weak as well. Four d and five d metals, like platinum, almost always pair, even with chloride. Clear? In an octahedral complex, only d four to d seven can be high or low spin; the raised number counts the d electrons. Got it? Then, mu with a small s, the spin-only moment, equals the square root of n, times n plus two in round brackets, in B M, Bohr magnetons; the tick shaped sign is the square root. And n is the number of unpaired electrons, not the total. With me? And the CFSE is minus zero point four for each t two g electron, plus zero point six for each e g electron, all times delta naught, the Greek capital delta with a small o. The small labels under n name the two sets. In brackets, for low spin, add P, the pairing energy, once for each extra pairing, unless the question says to ignore it.

4:10Tool for Set 3: colour and carbonyls

Okay? Now Set three. When a complex absorbs light, we see the complementary colour, so a complex that absorbs orange looks blue. Right? If delta naught gets larger, is the light absorbed of a longer wavelength or a shorter one? It is shorter, yes, exactly! Bigger gap, more energy, and more energy means a shorter wavelength. So a stronger ligand absorbs at a shorter wavelength. Clear? Finally, carbonyls. The carbon lone pair of C O goes into an empty metal orbital, along the arrow, making a sigma bond, sigma being the Greek letter. Then a filled metal d orbital gives electrons back into pi star of C O, the pi with a small star, making a pi bond. Each effect strengthens the other, and that is synergic bonding. Got it?

4:59Example 1: moles of AgCl from CoCl3.5NH3 with AgNO3

Now, Example one, an easy one from Set one, and it settles our opening question. How many moles of A g C l does one mole of C o C l three, dot, five N H three give with excess A g N O three, silver nitrate? The centred dot joins the ammonia molecules to cobalt chloride. Pause the video and try it first. Done? Okay, let's check. Ammonia is neutral and there are three chlorides, so cobalt is plus three, its primary valency. And its coordination number, the secondary valency, is six. Okay? So there are six places around cobalt, and five ammonia molecules fill only five. One chloride must go inside to fill the sixth. The formula is C o, then N H three in round brackets, five times, then C l, all inside the square brackets, and C l two outside; the small numbers written low count the groups. Right? Only the two chlorides outside the bracket are counter ions, so only they ionise. The answer is two moles of silver chloride, not three, and our opening puzzle is solved. Clear?

6:07Example 2: spin-only moment of K3[Mn(CN)6]

Now Example two, a step harder: a medium problem from Set two. Find the spin-only moment of K three, open square bracket, M n, then C N in round brackets, six times, close square bracket. Pause the video and try it first. Done? Okay, let's check. Let x be the oxidation state of manganese. Three potassium ions give three times plus one, then plus x, then six cyanides give six times minus one, and the total is zero. So x equals plus three; the double arrow means it follows that. Okay? Manganese is three d five, four s two; the raised numbers count the electrons. Manganese three plus has lost both four s electrons and one d electron, so it is three d four; the arrow means goes to. Right? Now, many students say that d four means four unpaired electrons, so the moment is four point nine zero Bohr magnetons. Is that right? No, no, wait, they forgot the ligand! Cyanide is strong, so it pairs the electrons, and this is low spin. All four go into t two g, the t with a small two g below and a raised four, and e g gets zero. Three orbitals and four electrons give one pair and two single electrons, so n is two. Got it? So mu s equals the square root of two, times two plus two in brackets, which is root eight, about two point eight three Bohr magnetons; the wavy equals sign means approximately. Four point nine zero is right only for high spin, with a weak ligand. With me?

7:45Quick revision: question types, ionisable ions, spin moment

So, quick revision. First, spot the type: a count, a match, a name or a colour order. For Set one, only the ions outside the square bracket ionise. For Set two, go from the oxidation state to the d count, judge the ligand, and then use mu s, the square root of n, times n plus two, in Bohr magnetons. Now open Set one, give each problem about a minute, and try the practice problem at the end of each set. All the best!