- Periodic trends and Coulomb's law
- Atomic and ionic radii
- Ionization energy: group trend
- Ionization energy: period trend
- First and second ionization energy
- Worked example: Identifying an element from successive ionization energies
- Electron affinity: period trend
- Periodic trends
Worked example: Identifying an element from successive ionization energies
When electrons are removed in succession from an element, the transition from removing valence electrons to removing core electrons results in a large jump in ionization energy. By looking for this large jump in energy, we can determine how many valence electrons an element has, which in turn can help us identify the element. Created by Sal Khan.
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- I get the idea of needing more energy to remove the core electrons. But once you remove the electron from the 2nd shell, wouldn't the next electron be easier to remove. For that matter, shouldn't the first 3 removals take less and less energy, and then have a big spike to remove the 4th electron, and then less to remove the fifth?(4 votes)
- No. Each successive ionisation energy is always greater than the previous one.(2 votes)
- What about a third period element with the numbers:
IE1 1230 kJ/mol
IE2 2400 kJ/mol
IE3 3600 kJ/mol
IE4 8000 kJ/mol
IE5 10500 kJ/mol
IE6 22000 kJ/mol
IE7 27500 kJ/mol
IE8 35200 kJ/mol(1 vote)
- The biggest jump (change) is between IE5 and IE6 so I would guess that from IE6 & up we are removing core electrons.
This means that we most likely have 5 valence electrons, meaning we would be in the 15th group on the periodic table, meaning we would guess Phosphorous (P).
I'm not sure if you were challenging me or asking a question but I hope this helps & is beneficial to someone. :-)(4 votes)
- Why can't Boron or Gallium be an acceptable answer as well?(2 votes)
- "The first five ionization energies for a third-period element are shown below", key word being third-period element. Boron and Gallium have the same trend, but are in the second and fourth periods respectively. Additionally the numbers given are only true for aluminium. Boron and Gallium have different values.
Hope that helps.(2 votes)
- If the period of the element isn't given, how would we determine what element it is?(1 vote)
- If you narrow down the options to a group 13 element then you would need to know the ionization energies of that group’s elements, or their relative amounts. Ionization energy within a group generally decreases as you move from top to bottom. This is because as you move down the valence electrons are located in shells farther away from the nucleus and so feel less attraction to the nucleus and are easier to remove. This means elements like boron and aluminum will have similar ionization energy patterns which show them to be in the same group, but boron will have a higher first ionization energy than aluminum.
So without actually providing the ionization energies for all the group 13 elements, they could say that the element has the second highest first ionization energy in its group, which would be aluminum. But even that wouldn’t work well since gallium (the element beneath aluminum) has about the same first ionization energy as aluminum. The general rule is that ionization energy decreases down a group, but the pattern isn’t as identifiable as the pattern across a period.
Most questions wouldn’t bother with all this nuance and would simply just say which period the element is in.
Hope that helps.(3 votes)
- Why is there a difference between the first, second, and third ionization energies? For the first three electrons removed, they were all valence electrons so shouldn't they all have been equally as easy to remove?(1 vote)
- Subsequent ionization energies of all elements have a general increasing trend because it becomes progressively harder to remove additional electrons. When we remove the first electron from a neutral atom with the first ionization energy, it becomes a cation, or a positively charged ion. Having an overall positive charge means the remaining electrons feel a greater force of attraction to the atom. Essentially by removing one electron, the remaining electrons have one less repulsive force (because electrons repel each other with their negative charge) but the same attractive force (from the positive protons) resulting in a net increase of attractive force. Removing even more electrons will result in even higher third, fourth, fifth, etc. ionization energies because the magnitude of the positive charge continues to increase.
Hope that helps.(3 votes)
- Can we say N is atom that have a same ionazitaion energy, because it show same ionazitaon ratio, or we vant say because N can not make 5.th ionazition?(1 vote)
- Can you rephrase that? It doesn't make sense.(2 votes)
- So I understand that it has three valence electrons, but wouldn't the other elements in that column also have 3? so why is Al?(1 vote)
- Sal says a third period (third row) element at1:42.(1 vote)
- [Instructor] We are told that the first five ionization energies for a third period element are shown below. What is the identity of the element? So pause this video and see if you can figure it out on your own and it'll probably be handy to have a periodic table of elements. So before I even look at a periodic table of elements, let's make sure we understand what this table is telling us. This is telling us that if we start with a neutral atom of this mystery element, it would take 578 kilojoules per mole to remove that first electron to turn that atom into an ion with a plus one positive charge. And then, it would take another 1,817 kilojoules per mole to remove a second electron. So to make that ion even more positive. And then after that it would take another 2,745 kilajoules per mole to remove the third electron. And then to remove the fourth electron, it takes a way larger amount of energy. It takes 11,578 kilojoules per mole. And then the fifth electron takes even more, 14,842 kilojoules per mole. And so, for the first, second, and third you do have an increase in ionization energy, but when you go to the fourth the energy required to remove those is way higher. So to me, these look like you're removing valence electrons and these look like you're removing core electrons. So one way to think about it is let's look on our periodic table of elements and look for a third period element that has three valence electrons. So we have our periodic table of elements. We want a third period element, so it's gonna be in this third row and which of these has three valence electrons? Well, sodium has one valence electron, magnesium has two valence electrons, aluminum has three valence electrons. So one way to think about it is that first electron, it's a reasonable ionization energy. Then the second one, a little higher. Then the third, a little bit higher than after that, but then the fourth, you're starting to go into the core. You're going to have to take an electron out of that full second energy shell, which takes a lot of energy. And so this is pretty clearly aluminum that is being described.