Worked example: Identifying an element from successive ionization energies | Khan Academy
Summary
TLDRIn this educational video, the instructor challenges viewers to identify a third period element based on its first five ionization energies. The script explains the concept of ionization energy and guides viewers through the process of elimination using a periodic table. The element in question, with three valence electrons and a significant increase in ionization energy for the fourth electron, is revealed to be aluminum.
Takeaways
- 🔍 The script discusses the ionization energies of a third period element's first five electrons.
- 📊 The first ionization energy is 578 kilojoules per mole, indicating the energy needed to remove the first electron from a neutral atom.
- 📈 The second ionization energy increases to 1,817 kilojoules per mole, and the third to 2,745 kilojoules per mole, showing a gradual increase.
- 🚀 The fourth ionization energy jumps significantly to 11,578 kilojoules per mole, suggesting a change in the type of electrons being removed.
- 🌟 The fifth ionization energy is even higher at 14,842 kilojoules per mole, further emphasizing the difficulty of removing inner electrons.
- 🔑 The increase in ionization energy for the first three electrons is expected as they are valence electrons.
- 🔬 The sharp increase after the third ionization suggests the removal of core electrons, which require more energy.
- 🧩 The element in question has three valence electrons, which is a key clue to its identity.
- 🌐 The periodic table is a valuable tool for identifying the element based on its ionization energies and electron configuration.
- 🎯 The element is identified as aluminum, which is in the third period and has three valence electrons.
- 📚 Understanding the periodic table and the concept of valence and core electrons is crucial for solving this problem.
Q & A
What is ionization energy?
-Ionization energy is the energy required to remove an electron from a neutral atom or ion to form a cation, and it is typically measured in kilojoules per mole (kJ/mol).
Why is it important to have a periodic table when trying to identify an element based on ionization energies?
-A periodic table is essential because it provides a visual representation of the elements' electron configurations, which can help in correlating the ionization energies with the number of valence electrons.
What does the first ionization energy of 578 kJ/mol suggest about the element?
-The first ionization energy of 578 kJ/mol suggests that the element has a relatively low energy barrier for losing its first electron, indicating it is likely in the earlier part of the third period of the periodic table.
How does the ionization energy trend for the first three electrons compare to the fourth and fifth?
-The ionization energy for the first three electrons increases gradually, but there is a significant jump for the fourth and fifth electrons, suggesting a change in the electron shell from which the electrons are being removed.
Why is there a large increase in ionization energy for the fourth electron compared to the third?
-The large increase in ionization energy for the fourth electron indicates that the electron is being removed from a more stable, fully filled second energy shell, which requires significantly more energy.
What is the significance of the term 'valence electrons' in the context of ionization energy?
-Valence electrons are the outermost electrons of an atom and are involved in chemical bonding. Their removal requires less energy compared to core electrons, which are more tightly bound to the nucleus.
Which third period element has three valence electrons?
-Aluminum (Al) is the third period element with three valence electrons, which aligns with the ionization energy pattern described in the script.
Why is aluminum the most likely element described in the script?
-Aluminum is the most likely element because its ionization energy pattern matches the script's description, with a gradual increase for the first three electrons and a significant increase for the fourth and fifth electrons.
What can we infer about the electron configuration of the mystery element from the ionization energy data?
-We can infer that the mystery element has three valence electrons and that the fourth and fifth ionization energies indicate the removal of electrons from a more stable, inner energy shell.
How does the concept of electron shells relate to ionization energy?
-Electron shells represent different energy levels within an atom. Electrons in higher energy shells (further from the nucleus) are easier to remove and thus require less ionization energy compared to those in lower energy shells.
What is the significance of the jump in ionization energy for the fourth and fifth electrons in identifying the element?
-The significant jump in ionization energy for the fourth and fifth electrons indicates that these electrons are being removed from a more stable, inner shell, which is a key clue in identifying the element as aluminum.
Outlines
🔬 Ionization Energy Trends of a Third Period Element
The instructor introduces a chemistry problem involving the ionization energies of a third period element. The first ionization energy is 578 kJ/mol, the second is 1,817 kJ/mol, and the third is 2,745 kJ/mol, showing a gradual increase. However, the fourth ionization energy jumps to 11,578 kJ/mol, and the fifth is even higher at 14,842 kJ/mol, indicating a significant change in energy required. This suggests that the first three electrons are being removed from the valence shell, while the fourth and fifth are core electrons. The instructor guides the viewer to use a periodic table to identify the element with three valence electrons in the third period, concluding that the element is aluminum, based on the ionization energy pattern described.
Mindmap
Keywords
💡Ionization Energy
💡Neutral Atom
💡Valence Electrons
💡Core Electrons
💡Periodic Table of Elements
💡Third Period Element
💡Sodium
💡Magnesium
💡Aluminum
💡Kilojoules per Mole
💡Electron Shell
Highlights
Introduction of a method to determine the identity of a third period element based on its ionization energies.
The first ionization energy of the mystery element is 578 kilojoules per mole.
The second ionization energy increases to 1,817 kilojoules per mole.
The third ionization energy further increases to 2,745 kilojoules per mole.
A significant jump in ionization energy to 11,578 kilojoules per mole for the fourth electron.
An even higher ionization energy of 14,842 kilojoules per mole for the fifth electron.
The pattern of increasing ionization energy suggests the removal of valence electrons for the first three electrons.
The sharp increase for the fourth and fifth ionizations indicates the removal of core electrons.
The instruction to use a periodic table to identify the element with three valence electrons in the third period.
Identification of sodium, magnesium, and aluminum as potential candidates based on their valence electrons.
Sodium has one valence electron, magnesium has two, and aluminum has three.
The reasonable ionization energy for the first electron suggests the start of valence electron removal.
The slight increase in ionization energy for the second and third electrons supports the valence electron removal theory.
The significant energy required for the fourth electron implies the beginning of core electron removal.
The conclusion that the described element is aluminum based on its ionization energy pattern.
The importance of understanding the periodic table and ionization energy trends for element identification.
The practical application of ionization energy trends in distinguishing between valence and core electron removal.
Transcripts
- [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.
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