Electron Gain Enthalpy

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Fluorine and chlorine both sit in group seventeen. Fluorine is the smaller atom, so an incoming electron gets closer to its nucleus. So fluorine should release more energy when it gains that electron, right? Well, the real numbers say no. Chlorine releases three hundred and forty nine kilojoules per mole, and fluorine only three hundred and twenty eight. So why does the smaller atom lose? Today we find out. Order questions on this come up often in JEE, so let's begin.

0:39What is electron gain enthalpy?

So, what are we measuring? Take an isolated atom, X, as a gas; the small g in brackets means gas. Add one electron, the e with a minus sign, and the arrow gives the gaseous negative ion, X minus. The enthalpy change for this is the electron gain enthalpy, delta e g H: the Greek capital delta, a small e g below it for electron gain, and H for enthalpy. Okay? Now let me ask you this. The nucleus attracts the incoming electron, so for most atoms, is energy given out, or taken in? It is given out, yes, exactly! For most elements energy is released, so delta e g H is negative. So in this convention, energy released means a minus sign. Right? You will also hear the term electron affinity. It is the magnitude of delta e g H, its size without the sign. So when a question says magnitude, just compare the sizes. Clear?

1:39The two trends

Alright, now the trends. Across a period, atoms get smaller and the nucleus pulls harder on an incoming electron. So delta e g H becomes more negative across a period: more energy is released. With me? Now tell me. Down a group, a new shell is added each time, so the atom gets bigger. Does delta e g H become more negative, or less negative? Less negative, very good! The incoming electron sits farther from the nucleus, so less energy is released.

2:14Exceptions: the crowded small shell

Okay? So, our puzzle. By that trend, fluorine at the top should be the most negative. But fluorine, F, is minus three hundred and twenty eight, and chlorine, C l, is minus three hundred and forty nine kilojoules per mole; the minus one on mol means per mole. So fluorine is less negative, even though it sits above chlorine. Right? So what is different about fluorine? Its outer subshell, the two p, is very small. Now, if an electron comes into a tiny, crowded shell, what happens to it? It gets pushed back, yes, exactly! The small two p shell of fluorine is crowded, so the added electron is strongly repelled, and less energy is released. Chlorine's larger three p shell has more room, and that solves our puzzle. Got it? And the same thing happens in group sixteen. Oxygen, O, is minus one hundred and forty one, and sulfur, S, is minus two hundred. Again, oxygen's small two p shell is crowded. So here is the trap: in groups sixteen and seventeen, the most negative value belongs to the third period element, not to the one at the top. Clear?

3:26Positive or near-zero values

Now, does every atom release energy when it gains an electron? No, it does not, and the noble gases show it: helium, plus forty eight; neon, plus one hundred and sixteen; argon and krypton, plus ninety six each; and xenon, plus seventy seven, all in kilojoules per mole. These values are positive, so energy must be supplied. Okay? So why do the noble gases refuse an extra electron? Think about where it has to go. Their subshells are already full, very good! So the added electron must enter the next shell, far from the nucleus and well shielded, and energy has to be supplied. With me? Two more cases give positive or near-zero values. Group two elements have a full n s two subshell, where n is the outer shell and the small raised two counts its two electrons. And nitrogen, N, has a half-filled two p subshell, which is extra stable. Got it?

4:24Second electron gain, and the reverse of ionisation

Now, a second electron. Oxygen has gained one electron and become O minus, the negative ion on screen. We add one more electron to get O two minus; the raised two minus is its charge. So tell me, does a negative ion welcome another electron? No, it repels it, exactly! The O minus ion repels the incoming electron, so energy has to be supplied. This is delta e g H two, with a small two for the second electron: plus eight hundred and forty four kilojoules per mole. That is positive, and big. Okay? And this is a general rule: every second electron gain is endothermic, which means energy is taken in. Right? Finally, give a positive ion, X plus, an electron, and you get back the neutral atom, exactly the reverse of ionisation. So delta e g H of X plus equals minus delta i H of X; delta i H, with a small i, is the ionisation enthalpy, and we just flip its sign. With me?

5:29Example 1: order F, Cl, Br, I by electron gain enthalpy

Now, example one, an easy one to warm up. Arrange fluorine, chlorine, bromine and iodine by the magnitude of their electron gain enthalpies. Pause the video and try it first. Done? Okay, let's check. All four are in group seventeen, and down a group the magnitude falls. So the answer is simply fluorine, greater than chlorine, greater than bromine, greater than iodine, from top to bottom, with the greater-than sign between each pair, right? Oh no, careful, that is the trap! The trend is right for the lower three: chlorine is greater than bromine, and bromine is greater than iodine. Okay? But fluorine has the small, crowded two p shell, remember? So it is pushed below chlorine, exactly like our puzzle, but it still stays above bromine. Right? So the order is chlorine, greater than fluorine, greater than bromine, greater than iodine. If you put fluorine first, that was the trap. Got it?

6:32Example 2: delta H for O to O2- and why oxides still form

Now example two, a little harder. For oxygen, delta e g H one is minus one hundred and forty one, and delta e g H two is plus eight hundred and forty four kilojoules per mole. Find delta H, the total enthalpy change, for an oxygen atom gaining two electrons to become O two minus. And why do metal oxides still form? Pause the video and try it first. Done? Okay, let's check. Oxygen releases one hundred and forty one for one electron. So for two electrons, it is just two times minus one hundred and forty one, right? No, no, wait, that is the trap! The second electron comes into a negative ion and is repelled. So we add the two steps: delta H equals delta e g H one, plus delta e g H two. Clear? Now the numbers. Minus one hundred and forty one, plus eight hundred and forty four, gives plus seven hundred and three kilojoules per mole. Okay? So the result is positive: making a gaseous O two minus ion costs energy. But metal oxides do form. So where does that energy come from? It comes from the solid, very good! When the ions pack into the solid oxide, a large lattice enthalpy is released, and it more than pays for these seven hundred and three kilojoules. So always add the steps with their signs. Got it?

7:59Quick revision: trends, Cl vs F, S vs O, second gain

So, quick revision. First, delta e g H is negative for most elements; it becomes more negative across a period, and less negative down a group. Second, the exceptions: chlorine beats fluorine, and sulfur beats oxygen, in magnitude, and the noble gases are positive. And third, every second electron gain is endothermic. Now go and try the practice problems on this page. All the best!