Shapes of Orbitals and Nodes

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. The nucleus is positive, the electron is negative, so the nucleus pulls the electron towards itself, right? And yet, for an electron in a p orbital, the probability density right at the nucleus is zero. Not small. Zero! How can that be? Strange, isn't it? Keep this puzzle in your mind; by the end of this lesson, you will crack it yourself. And look, in JEE, shapes and nodes often come up as quick questions: count the nodes, name the orbital. So, let's begin.

0:37Shapes: s and p

So, first the simplest one, the s orbital. In the brackets, l equals zero; l is the azimuthal quantum number from the last lesson, okay? And its shape? Spherical, like a ball. Look at the first picture on the right: the same in every direction. Clear? Next, p, where l equals one. Its shape is a dumb-bell: two lobes, lying along an axis. See the second picture, with its two coloured lobes. And notice, the two colours show the sign of the wave function in each lobe, okay? Now, there are three p orbitals: p x, p y and p z. The small letter written below the p tells you the axis it lies along. So tell me: are these three different shapes? No! Exactly the same shape. All three are the same dumb-bell; they differ only in orientation, that is, the direction they point in. So p x lies along the x axis, p y along the y axis, and p z along the z axis. Simple, right? And one more thing. Look where the two lobes of the p orbital meet: right at the nucleus. Every p orbital has a nodal plane, a flat plane passing through the nucleus. What exactly is a node? Hold that question; it is the key to our puzzle.

1:58d orbitals: five of them

Alright, now d, where l equals two. Here we get five orbitals, and their names are a little tricky. So listen carefully, okay? First, d x y. The two small letters below the d are x and y. So its lobes lie on the x axis and on the y axis, right? No, no, careful! This is the trap. Look at the third picture: the lobes of d x y sit between the x and y axes, not on them. And the same for d y z and d z x: lobes between the axes. Then which d orbital has its lobes right on the x and y axes? Yes, d x square minus y square! The small two up top means square. Its lobes lie on the x axis and the y axis. Got it? And the fifth one, d z square, is the odd one out. Look at the last picture. It has two lobes along the z axis, and a ring, like a doughnut, around the middle in the x y plane. Completely different, right? Remember this one.

3:01Nodes: where psi is zero

Now, the big idea of today: nodes. Remember psi, this Greek letter? It is the wave function. At a node, psi equals zero, so the probability of finding the electron there is zero. Not small, zero. Clear? Now think back to the shapes. The s orbital, with l equal to zero, is a plain sphere. The p orbital has l equal to one. How many nodal planes through the nucleus does each one have? Yes, exactly! The sphere has none, and the p orbital has one. Zero and one: exactly l. So, angular nodes equal l, and they are planes passing through the nucleus, okay? Just one small exception to remember. For d z square, the angular nodes are cones, not planes. The odd one out, once again! And now, our puzzle! The nodal plane of a p orbital passes right through the nucleus. On that plane, psi is zero. So a p orbital has zero probability density at the nucleus. There you go, puzzle solved: the nucleus may pull as hard as it likes, but a p electron is never found exactly there!

4:13Counting nodes

Alright, let's count nodes now. One kind we already know: angular nodes equal l. With me? Next, the radial nodes. There is a second kind of node: radial nodes, also called spherical nodes; that is the name in the brackets. Their number is n minus l minus one, where n is the principal quantum number, okay? Now, the total number of nodes. Add the two: n minus l minus one, plus l. What do you get? Yes, the l cancels! Total nodes equal n minus one. So, three formulas: angular is l, radial is n minus l minus one, and the total is n minus one. Write them down, okay? Now a topper's trick. Get the total from n, and the angular nodes from l. The radial nodes are simply what is left: the total minus the angular nodes. Got it?

5:10Example 1: radial and angular nodes of 2s, 3p and 4d

Right, let's use it. Find the radial and angular nodes of two s, three p and four d. Pause the video, and try it first. Done? Okay, let's check. Only two formulas: radial is n minus l minus one, and angular is l. Two s: n is two, and s means l equals zero. Radial: two minus zero minus one, that is one. Angular: zero. So one radial node, and no angular node. Three p: n is three, and l is one. Radial: three minus one minus one, that is one. Angular: one. Clear? Now four d. Here n is four. So the radial nodes are four minus one, that is three, right? No, no, careful! This is the JEE trap. Four minus one is the total, not the radial count. For d, l equals two. So radial: four minus two minus one, just one. And angular: two. And a quick check: radial plus angular should be n minus one. One, two and three. Matches! And notice: all three have exactly one radial node. Interesting, right?

6:29Example 2: identify the orbital from its nodes

Now a slightly harder one, JEE level. This time, the question runs backwards. An orbital has two radial nodes and one angular node. Identify it. Pause and try. Hint: start with the angular nodes. Done? Okay, first a quick guess. Two plus one makes three nodes in all. So n is three, and the orbital is three p, right? Oh no! The same trap again! Remember, the total is n minus one, not n. Let's do it properly. Angular nodes equal l, so l equals one. And l equal to one means a p orbital. Now radial: n minus l minus one equals two. Put l equal to one, so n minus one minus one equals two. The double arrow means so. So n equals four. So the orbital is four p. Not three p: four p! Topper's check: the total, n minus one, is three. And two radial plus one angular is also three. Spot on! Got it?

7:38Quick revision: shapes, radial n-l-1, angular l, total n-1

So, quick revision. Shapes: s is a sphere, p is a dumb-bell, and d has five orbitals. Nodes: radial is n minus l minus one, angular is l, and the total is n minus one. And at a node, psi is zero, so there is zero probability of finding the electron there. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!