Mixed Problems: Set 2 (Bonding, Magnetism and Crystal Fields)

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Chromium two plus and manganese three plus both have four d electrons. With six water molecules, chromium two plus shows a spin only moment of four point nine zero Bohr magnetons. With six cyanide ions, manganese three plus shows a much smaller moment. Same four d electrons, so why a different moment? Strange, right? This is set two of our mixed problems. Questions like these come up often in JEE Main. So, let's begin.

0:37One four-step method for bonding problems

So, the bonding and magnetism problems in this set all fall to one four step method. Step one: find the oxidation state of the metal, and its number of d electrons. Step two: judge the ligand. Strong ligands pair up the d electrons, and weak ones do not. Step three: place the electrons, and count the free inner d orbitals, written n minus one in brackets, then d. With me? And step four: read off the hybridisation, geometry, unpaired electrons and mu s, the Greek letter mu with a small s below, the spin only moment. Clear?

1:16Step two: strong or weak ligand?

Step two needs a list. Strong ligands: cyanide, C N with a raised minus for its charge; carbon monoxide, C O; nitrite, N O two minus, with a small two below and a raised minus; and e n and bipy, short for ethylenediamine and bipyridine. Okay? Weak ligands: iodide, bromide, chloride, thiocyanate, fluoride and hydroxide, the six ions with a minus sign on screen. Right? Now let me ask you this. Ammonia, N H three, with a small three below: strong ligand or weak? Careful here, it depends on the metal ion, and that is a classic trap! Water, H two O, and ammonia are weak with two plus ions, written M two plus, M for the metal, such as manganese, iron and cobalt two plus. Got it? But with a three plus ion of the three d series, ammonia is strong, and with cobalt three plus, even water pairs the electrons. Clear?

2:16Step four: the spin-only moment

Next, the formula for step four. Here, mu s equals the square root of n times n plus two, with n plus two in brackets, in Bohr magnetons, written as B M on screen. And n counts only the unpaired electrons. Okay? For n from one to five, the values are one point seven three, two point eight three, three point eight seven, four point nine zero and five point nine two. Toppers learn these five numbers by heart, and so should you. Right? So, a quick check: suppose a complex shows five point nine two Bohr magnetons. How many unpaired electrons? It is five, yes, straight from the table! Now, in octahedral complexes only d four to d seven, the raised number counting d electrons, can be high or low spin. The slash separates high from low: four or two, five or one, four or zero, three or one unpaired. With me?

3:13Crystal fields: Δ_o versus P

Now the crystal field picture. On the left, the five d orbitals of the free ion all have the same energy. In the octahedral field they split about the grey dashed line, the barycentre: the dark blue trio below is t two g, with a small two g below the t; the orange pair above is e g, with a small g. The brackets beside each set name its orbitals, and the plus zero point six and minus zero point four show how far each set moves. The double arrow is delta naught, the Greek capital delta with a small o. If delta naught is greater than capital P, the pairing energy, pairing is cheaper, so low spin; with the less than sign, high spin. Okay? Now the CFSE, the crystal field stabilisation energy, with the pairing energy P ignored. Each t two g electron counts minus zero point four delta naught, and each e g electron plus zero point six; the little n with t two g or e g below is the electron count. Got it? So tell me: high spin d five is t two g three, e g two, where the raised numbers count electrons. What is its CFSE? It is zero, and here is why. Three times minus zero point four is minus one point two, and two times zero point six is plus one point two. They cancel completely, so high spin d five has zero stabilisation. Clear? Last, the tetrahedral pattern on the right, with the order reversed: t two, with a small two below, is now on top, and e is below. Delta t, with a small t below, is about four ninths of delta naught for the same metal and ligand; the wavy equals sign means approximately. So tetrahedral complexes are practically always high spin.

4:56Example 1: spin-only moment of K3[Mn(CN)6]

Okay? Now Example one, our puzzle from the start. Find the spin only moment of K three, open square bracket, M n, then C N in round brackets with a small six, close square bracket. Pause the video and try it first. Done? Okay, let's check. Call the oxidation state of manganese x. Three potassiums at plus one, six cyanides at minus one, and the total is zero. The double arrow means so: x is plus three, manganese three, with Roman three in brackets. Right? Manganese is three d five, four s two, so it loses the two four s electrons first, then one three d electron, and manganese three plus is three d four. Okay? So four d electrons, four unpaired, and four point nine zero Bohr magnetons, right? No, no, wait, that is exactly the trap! Cyanide is a strong ligand, so all four electrons go into t two g: one pair and two single electrons. Low spin, and n is two. Got it? So mu s is the square root of two times four, root eight, which gives two point eight three Bohr magnetons, not four point nine zero. Clear? And the puzzle is solved. Compare open square bracket, C r, then H two O in round brackets with a small six, close bracket, two plus. Here water is weak, because chromium is a two plus ion. So chromium stays high spin d four: four unpaired, four point nine zero. Same d four, but the ligand decides! With me?

6:29Example 2: unpaired electrons in [Co(NH3)6]3+ and [CoF6]3-

Now a harder one, Example two. For open square bracket, C o, then N H three in round brackets with a small six, close bracket, three plus, delta naught is twenty two thousand nine hundred per centimetre, and P is twenty one thousand; c m with a raised minus one is read per centimetre. For open square bracket, C o F six, close bracket, three minus, delta naught is thirteen thousand. Find the unpaired electrons in each. Pause the video and try it first. Done? Okay, let's check. Ammonia is neutral, so cobalt is plus three; six fluorides in a three minus complex also leave cobalt at plus three. Both are cobalt three plus, a d six ion. Right? Here the numbers are given, so we need no list. Each delta naught must be compared with something. With what? With P, the pairing energy, yes, exactly! It is given once, for the same cobalt three plus ion, so it serves both. For ammonia, twenty two thousand nine hundred is more than twenty one thousand, so delta naught is greater than P, okay? Pairing is cheaper, so all six electrons pair up in t two g. Low spin, t two g six, and zero unpaired electrons. Got it? For fluoride, thirteen thousand is less than twenty one thousand, so delta naught is less than P, and going up into e g is now cheaper. Clear? So it is high spin, t two g four, e g two. In t two g, one pair and two single electrons; in e g, two more single ones. Four unpaired, so the answer is zero and four. Same ion, completely different magnetism! With me?

8:17Quick revision: spin-only formula, high vs low spin

So, quick revision. Four steps: oxidation state, ligand, place the electrons, read off. The spin only moment is the square root of n times n plus two, for n unpaired electrons, from one point seven three to five point nine two Bohr magnetons. And delta naught greater than P gives low spin, less than P high spin, and tetrahedral is high spin. Now go and try the practice problem on this page. All the best!