When it comes to defining, explaining, expressing, discussing and telling what ionization energy is, many give variety of defination, explaination, expression, discussion and teaching from different angles but still arriving at the same point thereby confusing some learners like me at the end of the whole show. My question to this great honourable forum is this:
What is the exact and precise defination to the term ionization energy?
Another question:
It is known and widely accepted fact that ionization energy generally decreases steadily down a given group and increases from left to right, what does that mean? How can I visualize it (the statement) for a better understanding?
So folks, what do you have to say about the above questions?

When it comes to defining, explaining, expressing, discussing and telling what ionization energy is, many give variety of defination, explaination, expression, discussion and teaching from different angles but still arriving at the same point thereby confusing some learners like me at the end of the whole show. My question to this great honourable forum is this:
What is the exact and precise defination to the term ionization energy?
Another question:

definition (http://chemistry.about.com/od/chemistryglossary/a/ionizationenerg.htm)


It is known and widely accepted fact that ionization energy generally decreases steadily down a given group and increases from left to right, what does that mean? How can I visualize it (the statement) for a better understanding?

Periodic table and ionization energy (http://chemistry.about.com/od/periodicitytrends/a/ionization-energy.htm)

When it comes to defining, explaining, expressing, discussing and telling what ionization energy is, many give variety of defination, explaination, expression, discussion and teaching from different angles but still arriving at the same point thereby confusing some learners like me at the end of the whole show. My question to this great honourable forum is this:
What is the exact and precise defination to the term ionization energy?
Another question:
It is known and widely accepted fact that ionization energy generally decreases steadily down a given group and increases from left to right, what does that mean? How can I visualize it (the statement) for a better understanding?
So folks, what do you have to say about the above questions?



I believe is contrary to what you mentioned , As you move from group l to group ll the ionization strength increase Say Li ionization 5. ev Be 9.2 ev F 17 ev.
As you go down in group th ionization decreases fron Li 5. ev Rb 4.1 ev
in this case the distance from the nucleus shows its effect.
Hope it helps

I believe is contrary to what you mentioned , As you move from group l to group ll the ionization strength increase Say Li ionization 5. ev Be 9.2 ev F 17 ev.
As you go down in group th ionization decreases fron Li 5. ev Rb 4.1 ev
in this case the distance from the nucleus shows its effect.
Hope it helps

which is contrary to what I mentioned?

which is contrary to what I mentioned?

"ionization energy generally decreases steadily down a given group and increases from left to right


First electron to ionize ( Li 5.0 electovolt) F 17 .0 electrovolt this is going from group l to group Vll

Now if you say on the same group l from lithium to Cs. then there is a decrease in ionization strength. Uf this is what you meant I agree with you . In this case my explanation would be : there is a steady increase in electrons in the element from on period to the next . So two things affect the distance from the nucleus and the other is the electron cloud density which increases with period and that will reduce the bond strength the outer electron to the nucleus.

It's mostly because electrons don't do a very good job shielding other electrons with the same principal quantum number, so as you add more electrons to orbitals with the same principal quantum number (meaning going left to right across a period), the valence electrons "see" more positive charge from the nucleus, and so are more strongly attracted, meaning the ionization energy goes up.

When it comes to defining, explaining, expressing, discussing and telling what ionization energy is, many give variety of defination, explaination, expression, discussion and teaching from different angles but still arriving at the same point thereby confusing some learners like me at the end of the whole show. My question to this great honourable forum is this:
What is the exact and precise defination to the term ionization energy?
it's the voltage needed to strip electrons from their parent atoms.
it's probably directly related to the dielectric constant of the material in question.

it's the voltage needed to strip electrons from their parent atoms.
it's probably directly related to the dielectric constant of the material in question.

I wouldn't have thought of that, because if I was working in pure theory, in the periodic table, or just looking at the orbitals, and the electron potential at each level, I would have overlooked that the material is what I really have to work with. But the material itself has organic properties - like the intrinsic conductivity, and this, the dielectric constant, too. And there's a field theory behind getting it to go into emission, in which even the shape affects where the field breaks down, so this is yet another level for analysis. Hmmm. So that's why they have a lab section to go with the lecture.

Those are the really tough questions, though, where you try to account for the empirical result not matching your theoretical prediction.

When it comes to defining, explaining, expressing, discussing and telling what ionization energy is, many give variety of defination, explaination, expression, discussion and teaching from different angles but still arriving at the same point thereby confusing some learners like me at the end of the whole show. My question to this great honourable forum is this:
What is the exact and precise defination to the term ionization energy?
Another question:
It is known and widely accepted fact that ionization energy generally decreases steadily down a given group and increases from left to right, what does that mean? How can I visualize it (the statement) for a better understanding?
So folks, what do you have to say about the above questions?

So do you understand the basic concept of ionization energy? You sound like you are looking for a very basic understanding. Do you think you have a handle on it now?

So do you understand the basic concept of ionization energy? You sound like you are looking for a very basic understanding. Do you think you have a handle on it now?

Well, I think am catching it gradually but in the course of my research, I will forward in more questions and contributions concerning the thread. Thank you!

Since noble gases have large atomic radius, are they not suppose to have low ionization energy? The reason being that thier valence electron will be very far from the nucleuos attraction. That means only small amount of energy will be required in removing it (the valence electron).

So do you understand the basic concept of ionization energy? You sound like you are looking for a very basic understanding. Do you think you have a handle on it now?

I was going through my past question and answer, I then came across this:
"The noble gases have the highest ionization energies in each period".
I then wondered why and how the above statement is a reality. How can the above statement be true when the noble gases have large atomic radi?

I know that Ionization energy, is the energy required to remove the outermost electron from a gasous atom.
If you say that the noble gases has the highest Ionization energies, how true is that when you know that the noble gases have a relatively large atomic radius?
I belive that the noble gases should have the lowest ionization energy since they have large atomic radius. This (lowest ionization energy) is possible because the larger the atomic radius the farer the distance of the outermost electron from the attraction of the nucleuos, thereby making it easier for outermost electron to escape. In such circustance, the outermost electron will leave with a very small energy.
From my statement above, I have it that the noble gases have the lowest and not the highest ionization energy.
If you the person viewing this my thread has a contrary view to mine on this concept on discusion feel free to share your own opinion. Let's see how we can collectively solve this problem.

If you look at a graph of the first ionization energy as a function of atomic number it is not a straight line, as one might think at first, and there are sharp peaks at each of the noble gasses.

In fact, the reason they are even noble gasses is directly related to these peaks. It takes so much energy to ionize them, that they tend to remain inert (pure atoms, not forming compounds). And this is because there shells are completely closed (full of electrons). This means they have achieved electrical neutrality, so they don't already tend one way or the other (to accept of donate electrons).

If you look at their names you will also recognize some of them as the names of flourescent lighting. This is because the energy level to ionize them is sharp enough that they will tend to fall back into stability and emit a photon in the process. Basically, the flourescent lamp puts them into emission, but they don't stay energized, they fall back, emitting an electron, then are quickly re-energized by the heat in the bulb and this they oscillate back and forth between high and low energy states, constantly emitting photons.

As you will see, nearly all the properties of materials arise from the character of the outer shell. For the noble gasses, it's a tendency to stay electrically neutral, with the shells full, to resist ionization, and to avoid bonding, which is precisely why they exist as gasses in nature.

Yes it turns out that atomic radius in itself is not nearly as important as whether the atoms have achieved electrical neutrality.

If you look at a graph of the first ionization energy as a function of atomic number it is not a straight line, as one might think at first, and there are sharp peaks at each of the noble gasses.

In fact, the reason they are even noble gasses is directly related to these peaks. It takes so much energy to ionize them, that they tend to remain inert (pure atoms, not forming compounds). And this is because their shells are completely closed (full of electrons). This means they have achieved electrical neutrality, so they don't already tend one way or the other (to accept of donate electrons).

If you look at their names you will also recognize some of them as the names of flourescent lighting. This is because the energy level to ionize them is sharp enough that they will tend to fall back into stability and emit a photon in the process. Basically, the flourescent lamp puts them into emission, but they don't stay energized, they fall back, emitting an electron, then are quickly re-energized by the heat in the bulb and this they oscillate back and forth between high and low energy states, constantly emitting photons.

As you will see, nearly all the properties of materials arise from the character of the outer shell. For the noble gasses, it's a tendency to stay electrically neutral, with the shells full, to resist ionization, and to avoid bonding, which is precisely why they exist as gasses in nature.

If you look at a graph of the first ionization energy as a function of atomic number it is not a straight line, as one might think at first, and there are sharp peaks at each of the noble gasses.

In fact, the reason they are even noble gasses is directly related to these peaks. It takes so much energy to ionize them, that they tend to remain inert (pure atoms, not forming compounds). And this is because there shells are completely closed (full of electrons). This means they have achieved electrical neutrality, so they don't already tend one way or the other (to accept of donate electrons).

If you look at their names you will also recognize some of them as the names of flourescent lighting. This is because the energy level to ionize them is sharp enough that they will tend to fall back into stability and emit a photon in the process. Basically, the flourescent lamp puts them into emission, but they don't stay energized, they fall back, emitting an electron, then are quickly re-energized by the heat in the bulb and this they oscillate back and forth between high and low energy states, constantly emitting photons.

As you will see, nearly all the properties of materials arise from the character of the outer shell. For the noble gasses, it's a tendency to stay electrically neutral, with the shells full, to resist ionization, and to avoid bonding, which is precisely why they exist as gasses in nature.

Is because all noble gases has attain octet structure and cannot loose electron from their outer most shell that they all have high ionization energies?

If you look at a graph of the first ionization energy as a function of atomic number it is not a straight line, as one might think at first, and there are sharp peaks at each of the noble gasses.

In fact, the reason they are even noble gasses is directly related to these peaks. It takes so much energy to ionize them, that they tend to remain inert (pure atoms, not forming compounds). And this is because there shells are completely closed (full of electrons). This means they have achieved electrical neutrality, so they don't already tend one way or the other (to accept of donate electrons).

If you look at their names you will also recognize some of them as the names of flourescent lighting. This is because the energy level to ionize them is sharp enough that they will tend to fall back into stability and emit a photon in the process. Basically, the flourescent lamp puts them into emission, but they don't stay energized, they fall back, emitting an electron, then are quickly re-energized by the heat in the bulb and this they oscillate back and forth between high and low energy states, constantly emitting photons.

As you will see, nearly all the properties of materials arise from the character of the outer shell. For the noble gasses, it's a tendency to stay electrically neutral, with the shells full, to resist ionization, and to avoid bonding, which is precisely why they exist as gasses in nature.

Nice post.:)

@Origin

Coming from you this truly a compliment. Thanks. :)

@chikis

I goofed in the middle of my last post when I said they emit an electron. I meant to say photon. First comes electron emission then an electron is captured to fill the vacancy and this causes a photon to be emitted.



Is because all noble gases has attain octet structure and cannot loose electron from their outer most shell that they all have high ionization energies?

They all have an octet except of course Helium. I think it will help more to think of this in terms of electrical charge. These are electrically neutral atoms. When you strip one away from eight, it leaves 7 charges behind which is highly negative. You must overcome the difference between neutrality (zero net charge) and this highly negative condition, so that takes more energy than any neighboring atom might require to strip one away.

@Origin

Coming from you this truly a compliment. Thanks. :)

@chikis

I goofed in the middle of my last post when I said they emit an electron. I meant to say photon. First comes electron emission then an electron is captured to fill the vacancy and this causes a photon to be emitted.




They all have an octet except of course Helium. I think it will help more to think of this in terms of electrical charge. These are electrically neutral atoms. When you strip one away from eight, it leaves 7 charges behind which is highly negative. You must overcome the difference between neutrality (zero net charge) and this highly negative condition, so that takes more energy than any neighboring atom might require to strip one away.

Is helium not a noble gas? Why are you puting it as exception?

Yes, it is a noble gas which is why I mentioned it. I was just pointing out that it stands alone (in the upper right of the periodic table) because it's too small to have 8 electrons. Its atomic number is 2, so it only has 2 electrons.

Good lord...

I was going through my past question and answer, I then came across this:
"The noble gases have the highest ionization energies in each period".
I then wondered why and how the above statement is a reality. How can the above statement be true when the noble gases have large atomic radi?
No. Noble gases have SMALLER atomic radii. Which is part of why they have such high ionization energies. http://www.grandinetti.org/Teaching/Chem121/Lectures/AtomicRadii

Yes it turns out that atomic radius in itself is not nearly as important as whether the atoms have achieved electrical neutrality.Ionization energy is proportional to charge, but it's proportional to the square of the radius...so a change in atomic radius generally matters more than a change in charge. It takes less energy to go from Mg+1 to Mg+2 than it does to go from neutral Ar to Ar+, because Ar has a smaller radius.

It takes so much energy to ionize them, that they tend to remain inert (pure atoms, not forming compounds).Really? It takes more energy to ionize Cl than to ionize Xe, yet Cl is very reactive and Xe is not...clearly ionization energy is not a great predictor of reactivity.
And this is because their shells are completely closed (full of electrons). This means they have achieved electrical neutrality, so they don't already tend one way or the other (to accept of donate electrons).You appear to be saying that having a full shell means that an atom is electrically neutral, which is wrong. But perhaps you simply meant that they are both electrically neutral and have a full shell.

If you look at their names you will also recognize some of them as the names of flourescent lighting. This is because the energy level to ionize them is sharp enough that they will tend to fall back into stability and emit a photon in the process. Basically, the flourescent lamp puts them into emission, but they don't stay energized, they fall back, emitting an electron, then are quickly re-energized by the heat in the bulb and this they oscillate back and forth between high and low energy states, constantly emitting photons. No. The photons in a fluorescent light are emitted by mercury atoms, not the noble gases. The purpose of the noble gases in a fluorescent bulb is to completely lose electrons, providing electrons that will be accelerated by the voltage in the lamp and slam into the mercury atoms so that the mercury atoms emit light. If the electrons on the excited noble gases were prone to falling back to their parent noble gas nucleus by emitting a photon (rather than being stripped away to fly off and hit a Hg atom), the lamp wouldn't work because there wouldn't be any free electrons to fly around and bang into the mercury atoms.

Nice post.:)
No it wasn't.
I goofed in the middle of my last post when I said they emit an electron. I meant to say photon. First comes electron emission then an electron is captured to fill the vacancy and this causes a photon to be emitted.Still wrong...

Edit: fixed typo

Good lord...

No. Noble gases have SMALLER atomic radii. Which is part of why they have such high ionization energies. http://www.grandinetti.org/Teaching/Chem121/Lectures/AtomicRadii
Ionization energy is proportional to charge, but it's proportional to the square of the radius...so a change in atomic radius generally matters more than a change in charge. It takes less energy to go from Mg+1 to Mg+2 than it does to go from neutral Ar to Ar+, because Ar has a smaller radius.
Really? It takes more energy to ionize Cl than to ionize Xe, yet Cl is very reactive and Xe is not...clearly ionization energy is not a great predictor of reactivity. You appear to be saying that having a full shell means that an atom is electrically neutral, which is wrong. But perhaps you simply meant that they are both electrically neutral and have a full shell.
No. The photons in a fluorescent light are emitted by mercury atoms, not the noble gases. The purpose of the noble gases in a fluorescent bulb is to completely lose electrons, providing electrons that will be accelerated by the voltage in the lamp and slam into the mercury atoms so that the mercury atoms emit light. If the electrons on the excited noble gases were prone to falling back to their parent noble gas nucleus by emitting a photon (rather than being stripped away to fly off and hit a Hg atom), the lamp wouldn't work because there wouldn't be any free electrons to fly around and bang into the mercury atoms.

No it wasn't.Still wrong...

Edit: fixed typo

Go back to the periodic table and study it carefully. You will see that ionization increases from left to right across a period. It decreases down a given group.
As for atomic radi, you will see that it decreases across a period from left to right but as you reach towards the noble gases, you will see that their atomic radi are larger. In such circustance, the ionization energies of noble gases should be lower but it won't because they have already attained their octet structure and will hardly loose electron. Therefore, their ionization energy are high.

Yes, it is a noble gas which is why I mentioned it. I was just pointing out that it stands alone (in the upper right of the periodic table) because it's too small to have 8 electrons. Its atomic number is 2, so it only has 2 electrons.

So are you saying because helium have only two electron filling it shell, is no longer octet?

Go back to the periodic table and study it carefully. You will see that ionization increases from left to right across a period. It decreases down a given group.Sure. Because radius decreases left to right and bottom to top.

As for atomic radi, you will see that it decreases across a period from left to right but as you reach towards the noble gases, you will see that their atomic radi are larger.You're just wrong about this. http://environmentalchemistry.com/yogi/periodic/atomicradius.html

Note how Ne<F, Ar<Cl, etc.

@Nasor,
I see what you have in the link:

http://environmentalchemistry.com/yogi/periodic/atomicradius.html (http://environmentalchemistry.com/yogi/periodic/atomicradius.html")

Wait, don't be any hurry, I will paste in a link to support my point.

Sure. Because radius decreases left to right and bottom to top.
You're just wrong about this.

Note how Ne<F, Ar<Cl, etc.

Except that you're wrong. I had wanted to post an image of the periodic table showing relative atomic radii for a clearer illustration, but apparently my low post count prohibits that. A strange way to counter spam bots, but no matter. Look it up and you'll see what I mean.


So are you saying because helium have only two electron filling it shell, is no longer octet?

It doesn't have an octet because it only has a single s orbital available to it. However, that orbital is still completely filled in He in exactly the same way as the 2s and 2p orbitals are filled in the later noble gases. As was said in another post, the fact that they have particularly high ionization energy is because these atoms simply don't want to have any of their electrons removed. They are already in a very stable, inert state as it is.

If you have a look at the electron configurations of the atoms, the periodic trend becomes quite understandable and as much as I hate personifying chemistry, for the purposes of your question I'll make a few concessions. Atoms are happiest when their highest energy orbitals are filled (see the octet rule). Sodium, for instance, has a single valence electron. To have a complete highest energy orbital it can either pick up 7 electrons or lose 1 - I think it's pretty obvious which one it would rather do. On the other end of the periodic table we have the halogens, which have 7 valence electrons. To have a full set of orbitals, it can either lose 7 or gain 1; again, the answer is pretty obvious as to which it would rather.

Going back to your original question, I think that the easiest way to think about it is by considering the definition of what ionization energy actually is. Simply, it describes the energy that is needed in order to remove a single electron from an atom. Given what I've just said and with the examples of the halogens and the alkali metals in mind, you should now be able to understand why sodium, which needs to lose 1 electron to gain a full outer orbital, has a much lower ionization energy (i.e. the energy required to remove a single electron) than say, chlorine, which would need to lose 7.

http://www.joshualovett.com/MDME/MEMmods/MEM30007A/classification/classification_files/periodic_table_diameters.jpg

The reason for this departure from the planet model, is connected to magnetic field additions. When electron move at velocity=V, they set up a magnetic field. Using the right hand rule (curl the fingers and point the thumb in the direction of motion), there are three vectors used to describe the motion. The first is the velocity, the second is the magnetic field direction and the third is a force vector. The orbital shapes reflect the best shape needed to get all the forces created by the electrons, to attract. The octet stability is due to the best magnetic force addition.

Do you have any evidence for this or is this just your idea. The orbital shape is generally discussed in PChem where the overridding mecanism is the wave like aspects of the electron. The Schrodinger Equation as it is applied to electons orbitals is the heart of the discussion. No where, that I am aware of, are there references to magnetic fields shaping the orbitals. I am at a loss to understand how the 'left hand rule' could be applied to the non-particle aspects of an electron orbital.

Good lord...

No. Noble gases have SMALLER atomic radii. Which is part of why they have such high ionization energies.

I want you to click on the link below and watch that pictorial representation of atomic radi of elements and tell me what you noticed.

http://www.crystalmaker.com/support/tutorials/crystalmaker/atomicradii/resources/VFI_Atomic_Radii.jpg

It is true that atomic radi decreases from left to right across a period. One would except that trend to extend to the noble gases. If that happenes, a more justification will be given for the high ionization energies exihibited by the noble gases. The otherwise is the case from the pictorial representation of atomic radi of elements which is in the link I pasted in. You will see that from that picture, that the atomic radi began decreasing from each period, (left to right) of the periodic table.

If you watch that table well, you will notice the following:
In the first period, hydrogen has an atomic radi which is smaller than that of helium and helium is a noble gas

In the second period, Litium has large atomic radi but you will notice that after litium, the next elements after litium as you move from left to right, say from berylium, their atomic radi start getting smaller till Neon the noble gas which have atomic radi which is almost the same size to that of litium.

When you get to period three, as usuall, the atomic radi of sodium is large, as you move from sodium through magnesium, their atomic radi start getting smaller up till chlorine, then to the argon whose atomic radius is large almost to the size of sodium.

As you move through the proceeding periods, you will see how from the begining of each period how the akali metal has large atomic radi. But if you move from left to the right, the atomic radi start getting smaller till the extreme end. The noble at each end is always large and having it atomic radius size equal to the akali metal in that period.

From the above ilustration, I can tell you that the noble gases exhibit high ionization energies for the simple fact that they have atained their octet structure and will hardly ionize.

Is not wise to judge it high ionization energy using the properties of it atomic radius since the may be other sources having information about the atomic radi of element which may be opposite to what is in the link I pasted in thereby causing unecessary confusion.

I want you to click on the link below and watch that pictorial representation of atomic radi of elements and tell me what you noticed.That's a picture of "VFI" atomic radii, which is based on data related to the length of the covalent bonds that the elements form when they're part of a molecule. The VFI method of determining atomic radii works fairly well for elements that actually take part in covalent bonding, but is not a useful way to determine the atomic radius for noble gases, since they only form very very weak covalent bonds. The stronger a covalent bond, the shorter it will be. Strong, short covalent bonds normally go hand-in-hand with short atomic radii, which is why the VFI method of measuring radius works well for elements other than noble gasses. Any attempt to determine the atomic radius of a noble gas with the VFI method will be skewed by the fact that noble gasses only form very very weak (and therefore very long) covalent bonds.

Note that this was partly explained in the text at the bottom of your picture: "These data are based on interatomic distances in the structures of the elements."

Pete's chart, and probably whatever hypervalent_iodine wanted to link to, suffer from the same problem. Although Pete's chart doesn't actually say how the radius values were calculated, I would bet money that it is also VFI radii, or some similar method that depends on bond length.

Here is another link from the same source as your graphic:http://www.crystalmaker.com/support/tutorials/crystalmaker/atomicradii/index.html

Go down to the bottom and look at the actual numbers for atomic radii. You'll see that the radii decrees all the as way across the row from left to right, with no discrepancy for the noble gases.

Edit: If anyone here believes that the noble gasses really have a smaller atomic radius (by which I mean the average distance from the nucleus to the valence electrons), I defy you to provide a coherent explanation of why that would be, especially considering Slater's rules...

Do you have any evidence for this or is this just your idea. The orbital shape is generally discussed in PChem where the overridding mecanism is the wave like aspects of the electron. The Schrodinger Equation as it is applied to electons orbitals is the heart of the discussion. No where, that I am aware of, are there references to magnetic fields shaping the orbitals. I am at a loss to understand how the 'left hand rule' could be applied to the non-particle aspects of an electron orbital.
Don't listen to wellwisher; he has a demonstrated history of saying things in the chemistry forum that are not just wrong, but borderline insane. He appears to have little to no actual education in chemistry, and bases his understanding of chemistry on a strange mixture of the Bohr model of the atom and a highschool-level understanding of electromagnetism. Notice how it apparently doesn't occur to him that the last two electrons are hard to remove because they are the two 1s electrons...

Don't listen to wellwisher; he has a demonstrated history of saying things in the chemistry forum that are not just wrong, but borderline insane. He appears to have little to no actual education in chemistry, and bases his understanding of chemistry on a strange mixture of the Bohr model of the atom and a highschool-level understanding of electromagnetism. Notice how it apparently doesn't occur to him that the last two electrons are hard to remove because they are the two 1s electrons...When you say "last two" are you implying they are the two with the highest energy requirement or are you saying they are always the last to be removed. Is it like knocking the easy ones off first, and only then the "last two", or if you were to use the right energy frequencies you can knock the "last two" off the atom to begin with?

When you say "last two" are you implying they are the two with the highest energy requirement or are you saying they are always the last to be removed. Is it like knocking the easy ones off first, and only then the "last two", or if you were to use the right energy frequencies you can knock the "last two" off the atom to begin with?
They are definitely the two that require the most energy to remove. I don't believe that they would necessarily be the last two to be removed if you were to ionize the atom; I think you might knock a "core" electron off before a valence electron, depending on your ionization scenario, although I'm not sure how the odds of removing a 1s compares to a 2s if you're doing something like, say, bombarding the atom with hard x-rays that have enough energy to ionize any of the inner electrons.

They are definitely the two that require the most energy to remove. I don't believe that they would necessarily be the last two to be removed if you were to ionize the atom; I think you might knock a "core" electron off before a valence electron, depending on your ionization scenario, although I'm not sure how the odds of removing a 1s compares to a 2s if you're doing something like, say, bombarding the atom with hard x-rays that have enough energy to ionize any of the inner electrons.
In the process of refreshing on ionization came across this Shielding Effect.

Shielding effect From Wikipedia, The shielding effect describes the decrease in attraction between an electron and the nucleus in any atom with more than one electron shell. It is also referred to as the screening effect or atomic shielding.
So it is always easier to knock the valence shell electrons off first.

So it is always easier to knock the valence shell electrons off first.
It is "easier" in terms of taking less energy, but I don't know about the relative odds of knocking off a valence electron vs. a core electron if the atom is hit by a photon that has more than enough energy to knock off either.

It is "easier" in terms of taking less energy, but I don't know about the relative odds of knocking off a valence electron vs. a core electron if the atom is hit by a photon that has more than enough energy to knock off either.It would be like body guards, more likely to hit them than the inner circle.

Except that you're wrong. I had wanted to post an image of the periodic table showing relative atomic radii for a clearer illustration, but apparently my low post count prohibits that. A strange way to counter spam bots, but no matter. Look it up and you'll see what I mean.



It doesn't have an octet because it only has a single s orbital available to it. However, that orbital is still completely filled in He in exactly the same way as the 2s and 2p orbitals are filled in the later noble gases. As was said in another post, the fact that they have particularly high ionization energy is because these atoms simply don't want to have any of their electrons removed. They are already in a very stable, inert state as it is.

If you have a look at the electron configurations of the atoms, the periodic trend becomes quite understandable and as much as I hate personifying chemistry, for the purposes of your question I'll make a few concessions. Atoms are happiest when their highest energy orbitals are filled (see the octet rule). Sodium, for instance, has a single valence electron. To have a complete highest energy orbital it can either pick up 7 electrons or lose 1 - I think it's pretty obvious which one it would rather do. On the other end of the periodic table we have the halogens, which have 7 valence electrons. To have a full set of orbitals, it can either lose 7 or gain 1; again, the answer is pretty obvious as to which it would rather.

Going back to your original question, I think that the easiest way to think about it is by considering the definition of what ionization energy actually is. Simply, it describes the energy that is needed in order to remove a single electron from an atom. Given what I've just said and with the examples of the halogens and the alkali metals in mind, you should now be able to understand why sodium, which needs to lose 1 electron to gain a full outer orbital, has a much lower ionization energy (i.e. the energy required to remove a single electron) than say, chlorine, which would need to lose 7.


It doesn't have an octet because it only has a single s orbital available to it.
In other to make things more clearer, define that word "octet" in precise term as used in this concept.

In other to make things more clearer, define that word "octet" in precise term as used in this concept.

Duet = Two.
Quartet = Four.
Given that an 8 sided shape is an octagon, how many do you think 'Octet' is?

The answer is eight.

In highschool you get taught the Octet rule. That is. Among the first twenty elements, with the exception of Hydrogen and Helium, the most electrons an element can have in its valence shell is eight.

From the octet rule, and the aufbau principle we derive lewis dot diagrams which provide a reasonably good model for the stoichometry, but not the shape of the first twenty elements.

When I went through Highschool we also had to learn the first twenty elements.
Harry He Likes Beer By Cups, Not Over Frothy, Never Nasty Mugs, All Sisters Please Stop Climbing Around Kinky Caves.

Somewhere between the end of highschool and the beginning of university, should you choose to pursue chemistry, you learn the reason for this. There are these things called orbitals. Chemists label these S, P, D, & F (there's also a hypothetical G).

The reason that the octet rule works is because there is one S orbital and three P orbitals, with each orbital holding two electrons of opposing spin. This gives us the total of eight electrons observed in the first twenty elements.

It's mostly because electrons don't do a very good job shielding other electrons with the same principal quantum number, so as you add more electrons to orbitals with the same principal quantum number (meaning going left to right across a period), the valence electrons "see" more positive charge from the nucleus, and so are more strongly attracted, meaning the ionization energy goes up.


It's mostly because electrons don't do a very good job shielding other electrons with the same principal quantum numbe
Define that word "shielding" in a more simple term as it relates to the topic on discusion.

Define that word "shielding" in a more simple term as it relates to the topic on discusion.
Canceling out some of the attractive force between another electron and the nucleus.

Canceling out some of the attractive force between another electron and the nucleus.

Thanks man, I think I now have a better understanding on it.
I can also picture it this way:
the word shielding originated from the word shield which can be defined as a broad piece of metal or another suitable material held by straps or handle attached to one side which is used as protection against missiles or blows.
Now that the word has being applied to this subject, it can be viewed as a protection against the attraction of the outer electron to towards the nucleus. This is because, the shell or shells before the outermost electron serve as a shield, shieding (protecting) the outermost electron from the attraction of the nucleuos.
Therefore the more shells before an outer electron, the more the shielding effect. How about that?

That's a picture of "VFI" atomic radii, which is based on data related to the length of the covalent bonds that the elements form when they're part of a molecule. The VFI method of determining atomic radii works fairly well for elements that actually take part in covalent bonding, but is not a useful way to determine the atomic radius for noble gases, since they only form very very weak covalent bonds. The stronger a covalent bond, the shorter it will be. Strong, short covalent bonds normally go hand-in-hand with short atomic radii, which is why the VFI method of measuring radius works well for elements other than noble gasses. Any attempt to determine the atomic radius of a noble gas with the VFI method will be skewed by the fact that noble gasses only form very very weak (and therefore very long) covalent bonds.

Note that this was partly explained in the text at the bottom of your picture: "These data are based on interatomic distances in the structures of the elements."

Pete's chart, and probably whatever hypervalent_iodine wanted to link to, suffer from the same problem. Although Pete's chart doesn't actually say how the radius values were calculated, I would bet money that it is also VFI radii, or some similar method that depends on bond length.

Here is another link from the same source as your graphic:http://www.crystalmaker.com/support/tutorials/crystalmaker/atomicradii/index.html

Go down to the bottom and look at the actual numbers for atomic radii. You'll see that the radii decrees all the as way across the row from left to right, with no discrepancy for the noble gases.

Edit: If anyone here believes that the noble gasses really have a smaller atomic radius (by which I mean the average distance from the nucleus to the valence electrons), I defy you to provide a coherent explanation of why that would be, especially considering Slater's rules...

I now realise my mistake, you are very correct. Am begining to come to my senses. Initialy whenever the word radi is mentioned, what comes to my mind is nothing but atomic radi. I never knew there are diffrences. I used to think that ionic, vandawal, atomic and convalent and metallic radi are all the same. But in this thread am discussing atomic radi. All the same, I say thank you for the awareness.
I now accept that atomic radi decreases left to right across a period thus the nobles gases being at the extrem right of each row should have the smallest atomic radi. Thanks am now cleared.

the thing about noble gases having a larger atomic radius is in our book too. infact, they claimed noble gases have almost twice the radius as the halogen before them. hat i think they mean is that it is that way when you take inter molecular (atomic distance) distance between them... but it still doesnt make sense because they clearly mentioned that the reason for their big size is inter-electron repulsion because there are just "too many" electrons in there and they dont like each others' company much.
can anyone plz tell me more on this?
thanks :)

the thing about noble gases having a larger atomic radius is in our book too. infact, they claimed noble gases have almost twice the radius as the halogen before them. hat i think they mean is that it is that way when you take inter molecular (atomic distance) distance between them... but it still doesnt make sense because they clearly mentioned that the reason for their big size is inter-electron repulsion because there are just "too many" electrons in there and they dont like each others' company much.
can anyone plz tell me more on this?
thanks :)

I think if you dig into this a little deeper you will begin to understand that the idea of a "radius" comes from the old idea that atoms are little spheres which isn't correct. The shell is a region of probable occupation by the electron, not an actual surface of definitive shape.

Check out the various ways we refer to atomic radius under "Definitions" at

http://en.wikipedia.org/wiki/Atomic_radius