Mixed Problems: Set 1 (Properties and Magnetism)

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Four ions are dissolved in water: titanium three plus, zinc two plus, cobalt two plus and vanadium three plus. Three of these solutions are coloured, and one is colourless. Which one is colourless? Hold that thought, we will come back to it. Today we solve a mixed set of d-block problems on unpaired electrons, magnetic moments, colour, oxidation states and ionisation enthalpies. Questions like these come up often in JEE Main. So, let's begin.

0:36Idea 1: count the d electrons of the ion

Idea number one sits behind almost every question in this set: count the d electrons of the ion. Take iron three plus, with the three plus written small and raised for the charge. How many unpaired electrons does it have? Let us work it out. When a transition metal forms an ion, the four s electrons are removed first. Okay? Iron is argon, three d six, four s two; the argon in square brackets is the argon core, and the small raised numbers count the electrons. Iron three plus loses both four s electrons and then one three d electron, so the arrow takes us to argon, three d five. Right? Five d electrons and five d orbitals, so each orbital gets one electron. The double arrow means this gives n equal to five, where n is the number of unpaired electrons. Clear? Now the pattern for free ions, or high-spin ions, as the brackets say. From d one to d five, where the dash means to, n goes from one to five. From d six to d nine, the new electrons start pairing, so n comes down to four, three, two and then one. And d zero and d ten have no unpaired electrons, so n is zero. With me?

1:50Idea 2: the spin-only magnetic moment

Idea number two. Once you know n, the moment is one formula. Mu, the Greek letter on screen, equals the square root of n times n plus two, with the n plus two in brackets, in B M, which stands for Bohr magnetons. The tick-shaped sign is the square root. This is the spin-only value: the orbital part is ignored for three d ions. Okay? Put n equal to one, two, three, four and 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. Keep these five values ready. Now find the spin-only moment of chromium three plus. Careful, chromium is one of the exceptions in its configuration. Chromium is argon, three d five, four s one, not four s two. Chromium three plus loses the four s electron and two d electrons, so it is three d three, and n is three. With me? So mu is the root of three times five, which is root fifteen. The wavy equals sign means about, so that is about three point eight seven Bohr magnetons. Clear?

2:57Idea 3: colour needs a d–d transition

Now back to our puzzle, with idea number three. In a hydrated ion, the ligands split the d orbitals, and an electron absorbs visible light. Okay? The electron jumps to a higher d level, and this jump is called a d d transition; the short dash joins the two letters d. So tell me, what must an ion have for a d d transition to happen? Yes, it needs a partly filled set of d orbitals: an electron to jump, and an empty place to jump into. So simple d zero ions, with no d electrons, and d ten ions, with every d orbital full, are colourless. The deep purple of permanganate comes from a different process, charge transfer, not from a d d transition. Right? Now our four ions. Titanium three plus is three d one, purple. Vanadium three plus is three d two, green. Cobalt two plus is three d seven, pink. All partly filled, so all coloured. And zinc two plus is three d ten, a full set, so there is no d d transition, and it is colourless. That answers our opening puzzle. Got it?

4:06Idea 4: oxidation states and electrode potentials

Idea number four. The highest oxidation state of a three d metal equals its number of three d plus four s electrons, but this holds only up to manganese, as the brackets say. Okay? So, what is the highest oxidation state of vanadium, and in which oxide do we find it? Vanadium is three d three, four s two, so three plus two gives plus five. And we find it in the oxide V two O five, where the small two and five below count the atoms. Right? Now electrode potentials. E naught, the E with a small circle, is the standard potential, here for the couple M two plus, slash, M, where M is the metal. So tell me, which of cobalt, nickel and copper has a positive E naught? Very good if you said copper! Cobalt is minus zero point two eight, nickel is minus zero point two five, and copper is plus zero point three four volt. So only copper is positive, and it is the only positive one in the three d series. Got it?

5:05Example 1: which ion has the same spin-only moment as Fe2+?

Now, example one. Which ion has the same spin-only moment as iron two plus: manganese three plus, cobalt two plus or nickel two plus? Pause the video and try it first. Done? Okay, let's check. Iron two plus is three d six. Six electrons in five orbitals means one pair, so n is four. Okay? Now manganese three plus. It has two fewer d electrons than iron two plus. So its moment must be different, and we can rule it out, right? No, no, wait, that is exactly the trap here! Manganese is three d five, four s two, so manganese three plus is three d four. Four electrons in five orbitals, all single, so n is four, the same as iron two plus. With me? Let us check the other two. Cobalt two plus is three d seven, so n is three, and nickel two plus is three d eight, so n is two. Got it? So the same n gives the same mu, root twenty four, about four point nine zero Bohr magnetons. The answer is manganese three plus. Count unpaired electrons, never total electrons. Clear?

6:09Example 2: highest second ionisation enthalpy, Cr, Mn or Fe

Now example two, a step harder. Which of chromium, manganese and iron has the highest second ionisation enthalpy? Pause the video and try it first. Done? Okay, first the tempting answer. Ionisation enthalpies rise slowly across a series, and iron comes last of these three. So iron has the highest, right? Oh no, careful, that is the trap here! The second electron is removed from the plus one ion, M plus, and it becomes M two plus. So look at the M plus ion. Okay? Chromium is three d five, four s one. Remove the four s electron, and chromium plus is three d five, a half-filled set, which is extra stable. Right? But manganese plus, three d five, four s one, and iron plus, three d six, four s one, still have a four s electron to give. With me? Now the values. Delta i H two, the triangle with a small i and then H with a small two, is the second ionisation enthalpy, in kilojoules per mole. Chromium is one thousand five hundred and ninety two. Manganese is one thousand five hundred and nine. Iron is one thousand five hundred and sixty one. Clear? So chromium is the highest, because its second electron has to break the stable d five. Before you use a trend, write the configuration of the ion. Got it?

7:25Quick revision: unpaired electrons, moment, d0 and d10 ions

So, quick revision. Ions lose their four s electrons first; count the unpaired d electrons, then use mu equals the root of n times n plus two. d zero and d ten ions have no d d transition, so simple ions like zinc two plus are colourless. And write the configuration of the ion before you trust a trend; chromium plus is three d five. Now go and try the practice problems on this page. All the best!