Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Here is a statement; tell me, true or false. If an octahedral complex is paramagnetic, it must be an outer orbital complex. Many students say true straight away. Hold your answer; we settle it in the last example. This is Part C of the unit problem bank, the concept checks. Each one looks like a one-word answer, and each one hides a trap. Statements like these come up often in JEE, so let's begin.

0:30Isomers: tetrahedral and square planar

So, isomers first. The first two drawings are square planar M A two B two: a metal M with two A ligands, the grey balls, and two B ligands, the orange ones; the small twos count them. With the B ligands side by side, it is cis; opposite each other, it is trans. The four drawings after them are octahedral, cis and trans, fac and mer, as the caption says; we do not need them today. Okay? Now, check number one. Put the four ligands at the corners of a tetrahedron instead. Can a tetrahedral complex have cis and trans forms? No, it cannot, because in a tetrahedron every pair of positions is adjacent; no position lies opposite another. Right? So there is no cis and no trans, and tetrahedral complexes show no geometrical isomerism. Clear? Check number two takes us back to square planar. Can either form of M A two B two be optically active? No, and the reason is simple. The metal and all four ligands lie in one plane, and that molecular plane is itself a mirror plane. Got it? Its mirror image is the same complex. So square planar M A two B two is not optically active. With me?

1:48Carbon monoxide: donor or acceptor?

Now, carbonyls. In the figure, the grey circle M is the metal, and carbon monoxide, with its triple bond, is on the right. The orange arrow is sigma donation, sigma, the Greek letter on screen: the lone pair on carbon goes into an empty metal orbital, and the arrow means goes into. Okay? Check number three. Carbon monoxide is the strongest ligand in the spectrochemical series. So it must be a strong sigma donor, right? Oh no, careful, that is the trap! Carbon monoxide is a very weak base, so it is only a weak sigma donor. Okay? The real strength comes from the green arrow. Here a filled metal d orbital gives electrons back into the empty pi star orbital of carbon monoxide; pi is the next Greek letter, and the small star marks an antibonding orbital. This is pi back donation. With me? So the answer is no: carbon monoxide is a weak sigma donor, but a strong pi acceptor. The two effects strengthen each other: synergic bonding, as the caption says, and the note on the right says the metal to carbon bond is strengthened. Clear? Now check number four, from the same picture. The second note on the right says the carbon to oxygen bond, written C dash O, is weakened and longer. But why does back donation lengthen it? Into which orbital do those electrons go? Yes, exactly, they go into pi star, an antibonding orbital. Electrons in an antibonding orbital lower the bond order, so the C dash O bond gets longer, not shorter. Got it?

3:25Is Delta t larger than Delta o?

Next, the splitting diagram. On the far left are the five d orbitals of the free ion, all at one level. They split in the octahedral field next to them: the orange pair on top is e g, with a small g below, and it points straight at the ligands. The dark blue trio below is t two g, with a small two g below, and the double arrow between them is delta naught, the Greek capital delta with a small o below. The brackets list the orbitals in each set, and the plus and minus numbers show how far each set moves from the dashed barycentre line; we do not need those today. Okay? Now the tetrahedral field, on the right. The order is reversed, t two on top and e below, and none of the orbitals points straight at a ligand. Its gap is delta t, with a small t below. Check number five: for the same metal and the same ligands, is delta t larger than delta naught? Compare the two double arrows. No, it is smaller. For the same metal and the same ligands, delta t is about four ninths of delta naught, as the label under the tetrahedral diagram also says; the wavy equals sign means about. Right? So, with such a small gap, the electrons rarely pair, and tetrahedral complexes are practically always high spin. Clear?

4:46Example 1: are all octahedral Co(III) complexes diamagnetic?

Now, Example one. Are all octahedral cobalt three complexes diamagnetic? On screen it is C o, with the Roman numeral three in brackets. Here is a hint: compare ammonia, N H three, with fluoride, F with a raised minus. Pause the video and try it first. Done? Okay, let's check. Cobalt is three d seven, four s two; the raised numbers count the electrons. Cobalt three plus, with its charge raised, loses both four s electrons and one three d electron, so it is three d six. Okay? Take ammonia first: open square bracket, C o, N H three in round brackets, six, close bracket, three plus. With cobalt three plus, ammonia is a strong field ligand, so all six electrons pair up in t two g; the raised six counts them. In valence bond terms it is d two s p three, the raised numbers counting the orbitals used, and it is diamagnetic. Got it? Most of us meet this cobalt three complex first. So every octahedral cobalt three complex is diamagnetic, right? No, wait, the ligand decides, so take C o F six, three minus, in square brackets. Fluoride is a weak field ligand, so the electrons do not pair: high spin, t two g four, e g two. Clear? In t two g four, one orbital holds a pair and two hold single electrons, and both e g electrons are single: four unpaired electrons. In valence bond terms, no three d orbital is free, so cobalt uses outer four d orbitals: s p three d two, an outer orbital complex. With me? The spin-only moment, mu s, the Greek letter mu with a small s below, is the square root of four times the bracket four plus two, which is root twenty four. That is about four point nine zero Bohr magnetons, the B M on screen. So this fluoride complex is paramagnetic, and the answer is no. Got it?

6:57Example 2: must a paramagnetic complex be outer orbital?

Now Example two, a step harder: our opening statement. Must a paramagnetic octahedral complex be an outer orbital complex? Hint: take the chromium complex on screen, open square bracket, C r, N H three in round brackets, six, close bracket, three plus. Pause the video and try it first. Done? Okay, let's check. In Example one, the paramagnetic complex was outer orbital, and the diamagnetic one was inner. So paramagnetic always means outer orbital, right? Not so fast, let us check chromium. Ammonia is neutral, so chromium is plus three. Chromium is three d five, four s one, so chromium three plus, losing the four s electron and two three d electrons, is three d three. Okay? These three electrons go into three of the five three d orbitals, one in each. So two three d orbitals stay free, whatever the ligand. Right? So in this ammonia complex, chromium uses those two inner three d orbitals: d two s p three, an inner orbital complex. Clear? But its three electrons are still unpaired. Mu s is the square root of three times the bracket three plus two, which is root fifteen, about three point eight seven Bohr magnetons. Got it? So it is paramagnetic and inner orbital at the same time. The answer is no, and our opening statement was false. Unpaired electrons alone do not decide inner or outer; the free three d orbitals do. With me?

8:35Quick revision: isomers, CO bonding, splitting, magnetism

So, quick revision. Tetrahedral complexes show no geometrical isomerism, and square planar M A two B two is not optically active. Carbon monoxide is a weak sigma donor but a strong pi acceptor, and back donation lengthens the carbon to oxygen bond. Delta t is about four ninths of delta naught, so tetrahedral complexes are practically always high spin. And a cobalt three complex can be high spin, while a paramagnetic complex need not be outer orbital. Now go and try the practice problem on this page. All the best!