Energies of Orbitals

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Shell three comes before shell four, right? Three is smaller than four. So every orbital of shell three should be lower in energy than every orbital of shell four, shouldn't it? And yet, in an atom with many electrons, the four s orbital sits below the three d orbital! How can a four come before a three? 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, arranging orbitals in order of energy is a common quick question. So, let's begin.

0:39Hydrogen and other one-electron species

So, let's start with the simplest atom, hydrogen, with just one electron. In hydrogen, and in any species with only one electron, the energy of an orbital depends only on n, the principal quantum number. Not on l, not on the shape. Only on n. Clear? Now tell me. In hydrogen, which one is lower in energy: two s, or two p? Neither! Yes, if you said that, exactly right. Both have n equal to two, so both have the same energy. Two s equals two p. Careful here: many students say two s is lower, because they have seen that order somewhere. In hydrogen, that is the trap. Don't fall for it! The same in shell three: three s, three p and three d, all at one energy. Orbitals with exactly the same energy are called degenerate, okay? Remember this word; it is the key word of today. So how many orbitals share one energy? All the orbitals of the level. Remember from an earlier lesson: shell n has n squared orbitals; the small two up top means squared. So in hydrogen, level n has n squared degenerate orbitals. Really simple.

1:57Many electrons: the (n+l) rule

Alright, now an atom with many electrons. Here the story changes. The subshells of a shell split apart: they no longer share one energy. So which one is lower? We need a rule. First, a quick recall from an earlier lesson. The letters s, p, d and f mean l equals zero, one, two and three, okay? Here l is the azimuthal quantum number. Now the rule. Add n and l, and write n plus l in brackets, as one number. The lower the n plus l, the lower the energy. The double arrow you see on screen means so. Got it? Right, back to our puzzle. Four s and three d. Find n plus l for each one. Yes! Four s: four plus zero, that is four. Three d: three plus two, that is five. Four is smaller, so four s is lower than three d; the less-than sign on screen means lower in energy. And there it is, puzzle solved! Three d has the smaller n, but its l of two lifts its n plus l to five. But now, a tie. Three d has n plus l equal to five, and four p also has five. So which one is lower? For a tie, the second rule: when n plus l is equal, the lower n comes first. Three d has n equal to three, and four p has four. So three d is lower than four p. Two rules, that's all: first n plus l, then n. With me?

3:38The full energy order

So, apply these two rules again and again, and you get the full order. Read it with me. One s, two s, two p, three s, three p, four s, three d, four p, five s, four d. And then: five p, six s, four f, five d, six p, seven s, five f, six d, seven p. Now a topper's trick: don't mug this up blindly. Whenever you have a doubt, just find n plus l, and the order writes itself, okay? Now, one more question. The subshells split in an atom with many electrons. But inside one subshell, do its orbitals split too? No! The orbitals of one subshell stay degenerate. Just one condition, written in the brackets: no electric field and no magnetic field acting on the atom, okay? And the last point. As the atomic number rises, the energy of a given subshell falls. So the same subshell sits lower and lower as we go to atoms of higher atomic number. Makes sense?

4:47Example 1: order 4f, 5d, 6s and 6p by the n+l rule

Now, let's use it. Arrange four f, five d, six s and six p in increasing energy, in an atom with many electrons. Pause the video, and try it first. Done? Okay, let's check. First, n plus l for each. Four f: f means l equals three, so four plus three, seven. Five d: five plus two, also seven. Six s: six plus zero, six. Six p: six plus one, seven. Now, careful. Six s has n equal to six, the biggest n here. So six s must be the highest in energy, right? No, no! This is the classic trap. Look at n plus l, not at n. Six s has the lowest n plus l, just six. So six s comes first, the lowest of all! Now three of them tie at seven: four f, five d and six p. For a tie, the lower n comes first. So four, then five, then six. So the answer: six s, then four f, then five d, then six p. And look back at our full order: exactly this piece is sitting there. Perfect match!

6:09Example 2: degenerate orbitals in H (n=3) and Fe 3d

Now a slightly harder one, JEE level. How many orbitals are degenerate in the n equals three level of hydrogen, and how many in the three d subshell of iron? Pause and try. Hint: first ask, one electron or many? Done? Okay, let's check. Hydrogen has just one electron, so its energy depends only on n. So in the n equals three level, every orbital has the same energy. How many orbitals? n squared: three squared, nine. All nine are degenerate. Now iron. Same shell three. So in iron also, all nine orbitals are degenerate, right? No, no, careful! Iron has many electrons. So three s, three p and three d split apart, just the way we saw today. Only the orbitals of one subshell stay degenerate. And every d subshell has five orbitals, remember? So in the three d subshell of iron, five are degenerate. So, nine for hydrogen, and five for iron. Same shell, different answers. One electron or many: that is the whole question. All clear?

7:22Quick revision: one electron vs many, n+l rule

So, quick revision. One electron, like hydrogen: the energy depends only on n, so two s equals two p. Many electrons: lower n plus l comes first, and on a tie, lower n. And the orbitals of one subshell stay degenerate. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!