AIEEE and Earlier JEE Main Questions: The f-Block

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. The lanthanoids and the actinoids both fill f orbitals. Yet the lanthanoids mostly stay at plus three, while neptunium and plutonium can go all the way up to plus seven. So why does one family show so many more oxidation states? We will settle it with a real past-paper question in Example two. Today we go through the older entrance-exam questions on the f-block, from two thousand two to two thousand eleven. They test a small set of ideas again and again, so let's begin.

0:33Idea 1: the lanthanoid contraction

So, idea number one, and the one that shows up most in this set: the lanthanoid contraction. From L a, lanthanum, to L u, lutetium, the atomic and ionic radii fall steadily. Okay? Now look at the numbers. L a three plus is the lanthanum ion; the small three plus written up top is its charge. Its radius is about one hundred and six picometres. The wavy equals sign means about, and p m means picometres. Follow the arrow to L u three plus, at about eighty five picometres. Keep these numbers in mind for Example one. Right? Now, at each step the nuclear charge goes up by one, and one more electron comes in. More electrons, and yet a smaller atom. So what is going on? Here is the reason, and two of these papers asked exactly this. Each new electron enters the four f subshell, and one four f electron shields another very poorly. So if an option says poor shielding of one four f electron by another, that is your answer. Got it? So the effective nuclear charge, the pull the outer electrons feel, rises at every step. The electrons are pulled in, and the size shrinks. With me? And one consequence comes up often. Z r, zirconium, and H f, hafnium, sit in the same group, with hafnium one row lower. Yet zirconium is one hundred and sixty picometres, and hafnium is one hundred and fifty nine. Hafnium comes right after the fourteen lanthanoids, so the contraction almost cancels the extra size you would expect from its new shell. Clear?

2:15Idea 2: oxidation states of the lanthanoids

Idea number two: oxidation states. For the lanthanoids, plus three is the characteristic oxidation state; every one of them shows it. Okay? Plus two and plus four also occur, mainly when they give f zero, f seven or f fourteen. The small raised number counts the f electrons, so that means an empty, a half-filled, or a completely filled f subshell. Right? So here is a question from the two thousand two paper. What are the most common oxidation states of cerium? Plus three and plus four, yes, exactly! Plus three is the typical one, and C e four plus is four f zero, an empty f subshell, so cerium shows plus four as well. Got it? Now careful, because one paper offered the statement that the plus four state of cerium is not known in solution. That is wrong. Cerium four plus does exist in solution, as ceric sulphate, and it is a strong oxidant. Clear? And one more trap, from the two thousand eleven paper. All of them show plus three, but only a few show plus four. So a statement saying every member forms plus four compounds is wrong. With me?

3:28Idea 3: colour and bonding of Ln³⁺

Idea number three: colour. In the heading, L n three plus stands for the three plus ion of any lanthanoid. One paper offered this statement: lanthanoid three plus compounds are generally colourless. Is that true, or false? That statement is false, because most lanthanoid three plus ions are coloured. The colour comes from f to f transitions; the f, dash, f on screen means an electron moving between f orbitals. Okay? The exceptions are the empty and the full ones. Lanthanum three plus, with f zero, and lutetium three plus, with f fourteen, are colourless. Right? The other statements in that same question were true, and two of them are worth remembering. Bonding in these compounds is mainly ionic, and the hydroxides, L n O H three, are mainly basic; the small three after the bracket means three O H groups. Got it?

4:27Idea 4: the actinoids

Idea number four: the actinoids, from thorium, T h, to lawrencium, L r. Z is the atomic number, so they run from ninety to one hundred and three. Okay? They show many oxidation states, up to plus seven for neptunium, N p, and plutonium, P u. That is our puzzle from the start. With me? There are two reasons, and both were asked in these papers. First, the five f, six d and seven s levels are close in energy, so electrons from all of them can be used. Clear? Second, five f orbitals extend further from the nucleus than four f; we say they are less buried. So five f electrons can join in bonding; the two thousand seven paper asked exactly this. Remember the direction: it is five f that extends further, not four f. Got it?

5:21Example 1: radius of Lu3+ when La3+ is 1.06 angstrom

Now, Example one, from the two thousand two paper. The radius of L a three plus is one point zero six angstroms; the capital A with a small ring on top is the angstrom. Which is closest to the radius of L u three plus: one point four zero, one point zero six, zero point eight five, or one point six zero angstroms? Pause the video and try it first. Done? Okay, let's check. Lutetium has fourteen more electrons than lanthanum. So its ion should be bigger, one point four zero or one point six zero, shouldn't it? Oh no, careful, that is the trap! Lanthanum is fifty seven and lutetium is seventy one, and across this series the radii fall steadily. That is the lanthanoid contraction. Okay? So lutetium three plus must be smaller than one point zero six angstroms. That throws out one point four zero and one point six zero, and one point zero six as well, because the size does change. Right? Only zero point eight five angstroms is left, so that is the answer. It matches idea one too, because eighty five picometres is the same as zero point eight five angstroms.

6:36Example 2: why actinoids show more oxidation states

Clear? Now, Example two, from the two thousand eight paper, and it settles our puzzle. What is the main reason for the larger number of oxidation states of the actinoids? The options are: four f more diffuse than five f; a smaller energy gap between five f and six d than between four f and five d; a larger gap; or greater reactivity. Pause the video and try it first. Done? Okay, let's check, by throwing out the wrong options. Four f more diffuse than five f? That is reversed: five f orbitals extend further than four f, so this option is out. Okay? Next, the option greater reactivity. That is not the reason, so it is out too. Right? Two options are left, and they are opposites: a smaller gap, or a larger gap. Which one lets the actinoids use more of their electrons? The smaller gap, yes, exactly! When five f and six d are close in energy, electrons from both sets can be used, so more oxidation states become possible. Clear? So the answer is the smaller energy gap between five f and six d. And there is our puzzle solved: that is how neptunium and plutonium reach plus seven, while the lanthanoids mostly stay at plus three. Got it?

7:58Quick revision: contraction, lanthanoid and actinoid states

So, quick revision. First, the lanthanoid contraction: poor four f shielding, so the radii fall from lanthanum to lutetium, and zirconium is about the same size as hafnium. Second, the lanthanoids always show plus three, with plus two or plus four mainly for f zero, f seven or f fourteen, and most of their three plus ions are coloured. And third, the actinoids show more states, as five f extends further and lies close to six d. Now go and try the practice problem on this page. All the best!