Configurations of Ions

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. When we fill the electrons of iron, the four s orbital fills first, and the three d fills after it, right? Now iron loses two electrons and becomes iron two plus. So which two electrons leave? The last ones that came in, from three d? Last in, first out? Sounds logical, doesn't it? Hold on to that guess, because today it is going to get a big surprise. And look, the configuration of an ion is a common quick question in JEE, and you need it again for magnetic moments. So, let's begin.

0:37First, the neutral iron atom

So, first, the neutral iron atom. Its symbol is F e, and capital Z equals twenty six, in brackets, is its atomic number: twenty six electrons. The A r in square brackets is the argon core, eighteen electrons already filled. Then three d six, four s two; the small numbers up top tell us how many electrons sit in each subshell. Eighteen plus six plus two, twenty six. Okay? Now a quick recall. While filling, which subshell gets its electrons first: four s, or three d? Four s, yes! Four s has n plus l equal to four, and three d has n plus l equal to five, as written in the brackets. Lower n plus l fills first. See the diagram on the right too: follow the slanting arrows, and four s comes just before three d. Nice and clear.

1:28Cations: 4s leaves before 3d

Alright, now back to our puzzle. Iron loses two electrons. The arrow on screen means becomes, and the small two plus up top is the charge. So we get iron two plus, a positive ion, which we call a cation. Now you tell me: which two electrons leave? Most students say: three d filled last, so three d empties first. Take two electrons from three d six, and you get argon, three d four, four s two. That is the tempting answer on screen, with a question mark. Looks perfect, doesn't it? But is it right? No, no, careful! This is a classic JEE trap. When a transition metal forms a cation, electrons leave the four s orbital before the three d, even though four s filled first. So both four s electrons go, and three d six stays untouched. Iron two plus is argon, three d six, not three d four, four s two. Clear? So here is the rule for cations. First empty the four s. Only if more electrons must go, take them from three d next. Really simple, but many students lose marks on it. And a topper's trick to remember it. Four s is first in, and first out. It filled first, and it leaves first. So our opening guess, last in, first out, was wrong. Puzzle solved! Got it?

2:48Any ion in two steps

Now, let's turn this into a method for any ion. Step one: write the configuration of the neutral atom. And be careful with chromium and copper, C r and C u in the brackets, the exceptions from the last lesson. Get the atom right, and the ion is easy. Step two, for a cation: remove the four s electrons first, and only then go to three d. Exactly what we did for iron. Alright, and an anion? That is a negative ion, so it gains electrons. Does it need some special rule too? No! Just the usual order. For an anion, simply add electrons in the usual filling order, the same order we use for any neutral atom. Nothing new. The special rule is only for cations, okay?

3:33Example 1: configurations of Cu+ and Zn2+

Now an easy one. Write the configurations of copper plus, and zinc two plus. Use our two steps: neutral atom first, then remove. Pause the video, and try it. Done? Okay, let's check. Step one, neutral copper, C u, atomic number twenty nine. Copper is one of our exceptions: argon, three d ten, four s one. Now copper plus loses one electron. From where? From four s, yes! That single four s electron leaves. The double arrow means so: copper plus is argon, three d ten. See, the full three d is not touched at all. Now zinc, Z n, atomic number thirty: argon, three d ten, four s two. With me? Zinc two plus loses two electrons, and both come from four s. So zinc two plus is also argon, three d ten, the same as copper plus! And if you took electrons from three d anywhere here, you fell into the trap. Not this time, right?

4:38Example 2: Fe2+ vs Fe3+ stability, Mn2+ hard to oxidise

Now a harder one, JEE level. Which is more stable, iron two plus or iron three plus, judging by configuration? And why is manganese two plus much harder to oxidise than iron two plus? Hint: write the configurations first. Pause the video, and try it. Done? Okay, let's check. Iron two plus we already know: argon, three d six. To make iron three plus, one more electron must go, but four s is already empty. So what is iron three plus? The third electron comes from three d, exactly! Three d six becomes three d five, and three d five is a half-filled subshell. Remember the last lesson? Half-filled subshells have extra stability. So, which ion is more stable? Iron three plus, correct! Its three d five is half-filled. And iron two plus, with three d six, reaches that same three d five by losing just one electron. So, by configuration, iron three plus is more stable. Now manganese, M n, atomic number twenty five: argon, three d five, four s two. Manganese two plus loses both four s electrons, so it is argon, three d five. Oh, already half-filled! Now, oxidising it means taking away one more electron. So why does manganese two plus resist? Because oxidising it would break its half-filled three d five. Iron two plus is the opposite: losing one electron gives it a half-filled subshell. That is why manganese two plus is much harder to oxidise. See, the whole answer came from just writing configurations.

6:22Quick revision: remove 4s before 3d, half-filled 3d5

So, quick revision. Cation: write the neutral atom first, then remove four s electrons before three d. Anion: add electrons in the usual order. And a half-filled three d five gives iron three plus and manganese two plus their extra stability. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!