Mixed Problems: Set 3 (Quantum Numbers, Nodes and Configurations)

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Chromium has twenty four electrons, and each one has a spin quantum number: plus half, or minus half. Now suppose every unpaired electron in chromium has plus half. How many electrons have plus half in all? Half up, half down, so half of twenty four: twelve, right? Hmm, are you sure? Keep your answer ready; we will check it at the end of this class. This is our third mixed problem set for JEE Main: quantum numbers, nodes and configurations. And look, atomic structure is a regular in JEE Main, and these quick ones are easy marks, if you know the traps. So, let's begin.

0:49Nodes: three formulas, one trap

So, first tool: nodes. Take an orbital with principal quantum number n and azimuthal quantum number l, okay? Radial nodes equal n minus l minus one. Angular nodes equal l. And the total is n minus one. Write these three down. Got it? Now a quick one. Find the number of radial nodes of a three p orbital, plus the number of angular nodes of a four d orbital. Three p first. Here n is three, and p means l equals one. Radial: three minus one minus one, that is just one. Clear? Now four d. For angular nodes we need only l, and d means l equals two. So two angular nodes. Sum: one plus two, three. Easy, isn't it? Next, how many nodes, radial and angular together, does a five d orbital have? Here n is five, so five nodes, right? No, no, careful! The total is n minus one, not n. Five minus one: four nodes. And they split like this: radial is five minus two minus one, that is two, and angular is l, also two. Two plus two, four. It all checks out!

2:06Counting orbitals and electrons

Now, second tool: counting. Every subshell has two l plus one orbitals, and a whole shell has n squared orbitals; the small two up top means square, okay? And each orbital holds two electrons. Their spin quantum number is m sub s, m with a small s written below it. One electron has plus half, and the other has minus half. Clear? Quick question. What is the largest number of electrons with n equal to four and l equal to two? Ten, exactly! l equal to two means d, so these are the five four d orbitals. Two electrons in each; the double arrow means so: ten electrons. Now, how many electrons in an atom can have n equal to three, and m sub s equal to minus half? The shell n equals three holds two n squared, that is eighteen electrons. So eighteen, right? No, no, careful! Minus half fixes the spin. The shell has three squared, nine orbitals, and each orbital has only one electron with minus half. So nine, not eighteen. Got it?

3:12Ions: 4s leaves first, then count unpaired

Now the third tool: configurations. First, two exceptions you must know. Chromium, C r, is the argon core in square brackets, then three d five, four s one; the small number up top counts the electrons in that subshell. And copper, C u, is the argon core, then three d ten, four s one. With me? Now iron, F e, which ends in three d six, four s two. Iron three plus has lost three electrons; that is the three plus up top. Three d filled after four s, so take all three from three d: three d three, four s two, right? Wait, careful here! Cations lose their four s electrons before three d, even though four s filled first. So both four s electrons go, and then one from three d. Iron three plus is three d five: five electrons in five d orbitals, one in each, so five unpaired. The same way, cobalt two plus, C o, is three d seven: three unpaired. Nickel two plus, N i, is three d eight: two unpaired. And copper two plus is three d nine: just one. See, the five d orbitals fill singly first, and then the extra electrons pair up. And the unpaired count gives the spin-only magnetic moment. Mu, this Greek letter, equals the square root of n into n plus two, in B M, Bohr magnetons. Here n is the number of unpaired electrons, okay? So, titanium two plus, T i, is three d two: n equals two. Mu is root eight; the wavy equals sign means approximately, two point eight three B M, simple!

5:01Example 1: how many orbitals have n + l = 5

Now, example one. How many orbitals have n plus l equal to five? Pause the video, and try it first. Done? Okay, let's check. Two rules from our tools. l runs from zero up to n minus one; the dots mean every value in between. And each subshell has two l plus one orbitals. Now find the pairs that add to five; the brackets show n plus l. l zero with n five: five s, one orbital. l one with n four: four p, three orbitals. l two with n three: three d, five orbitals. Next, l three with n two: two f. Add its orbitals as well, right? Stop right there! For n equal to two, l is at most one; this sign means less than or equal to. There is no two f orbital at all. This is the JEE trap! So the total is one plus three plus five: nine orbitals. Really simple, once you skip the fake one.

6:08Example 2: electrons with ms = +1/2 in chromium

Clear? Now the harder one, JEE level, and it is our opening puzzle. In ground-state chromium, with every unpaired electron taken to have m sub s equal to plus half, how many electrons have m sub s equal to plus half? Pause and try. Hint: count the unpaired electrons first. Done? Alright, first the quick answer from the start. Half the electrons up, half down: twenty four divided by two, twelve, right? No, no, careful! Half and half works only for paired electrons. So first, the configuration. Chromium has twenty four electrons, and it is our exception: argon core, three d five, four s one. Now tell me, how many of these electrons are unpaired? Six, exactly! Five in three d, one in each orbital, and one alone in four s. Five plus one, six, and the question says all of them have plus half. The other electrons, twenty four minus six, that is eighteen, sit in pairs. Every pair has one up and one down, so nine with plus half, and nine with minus half. So six plus nine: fifteen electrons with plus half. Not twelve, but fifteen! There you go, puzzle solved: the unpaired electrons break the half and half. Got it?

7:35Quick revision: nodes, counting orbitals, magnetic moment

So, quick revision. Nodes: radial is n minus l minus one, angular is l, and the total is n minus one. Counting: two l plus one orbitals per subshell, n squared per shell, and two spins per orbital. And ions lose four s first; then count the unpaired electrons, and mu is the root of n into n plus two, in Bohr magnetons. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!