Does the size of the shell in 2s change when it has 2 electrons?

[STAii]

Greetings.
I am having problems understanding some things about shells.
Actually, i am not even sure of if i am talking about shells (i am not sure this is the term in English, but i think it is).
I have been taught that a shell is a place of space round the nuclei with a highest probability of having an electron.
I also have been taught that the size of a shell gets bigger as its quantum number n gets bigger.
That is, a shell in 2s is bigger than a shell in 1s.
My problem became really clear when i started to study the atomic radius.
Let's take two imaginary atoms, ₃X:1s²2s¹ and ₄Y:1s²2s².
We know that the atom radius of Y is smaller than the atomic radius of X (it has been explained to me that the protons in Y are more effective in attracting the electrons of 2s than the protons in X).
If this is true, then does this mean that the size of the shell in 2s is smaller when it is has 2 electrons than when it has only one ?

I am also unable to understand how we say that a certain atom ending with a shell in p has an atomic radius, if the shells in p are not spherical, how can they have a constant radius ?

Any help would be appreciated, since i feel really lost !
Thanks in advance.

 

[FZ+]

If this is true, then does this mean that the size of the shell in 2s is smaller when it is has 2 electrons than when it has only one ?

Yes, in terms of distance to the nucleus. Realise though that the electron clouds represented by the shells corresponds to the area of probability that the electron can be in. It says nothing about how many electrons are in there.

I am also unable to understand how we say that a certain atom ending with a shell in p has an atomic radius, if the shells in p are not spherical, how can they have a constant radius ?

Ah true! The atomic radius idea comes from an assumption atoms are little balls, which is of course not true. What we measure as atomic radius is a measure of averages.

Though I could be wrong...

 

[STAii]

Originally posted by FZ+
Yes, in terms of distance to the nucleus. Realise though that the electron clouds represented by the shells corresponds to the area of probability that the electron can be in. It says nothing about how many electrons are in there.

I don't really understand what you mean.
If the shell says nothing about how many electrons are in there then how come your answer is Yes (in terms of disance to nucleus)

Can you explain it more please ?
Thansk FZ+.

 

[FZ+]

I mean that when we talk about the "size" of an electron shell, we mean something like the average distance from the nucleus an electron in the shell has.(and hence each electron has a higher energy) We don't actually mean there is a larger electrical charge in the shell, or more electrons.

 

[radagast]

It does make sense that the average size of the 2s² orbital is smaller that the 2s¹, when you think in terms that there is a higher charge (thus higher electrostatic force) attracting the electron toward the nucleus. Just because there is also another electron doesn't mean the other electron is able to screen all the added attraction from the nucleus. That's why you tend to find the atomic radii of group I metals tends to be distinctly higher than the corrosponding halogen of the same series, even though they have many more electrons.

 

[Zargawee]

It says nothing about how many electrons are in there.

but the electron cloud has more density when there's 2 electrons than the case when there's only one.
so ,doesn't that create more repulsion between the two electrons , therefore , the size of the shell must be bigger ?

 

[radagast]

But it isn't. They may repulse each other, but they can avoid each other easier than they can get further from the growing (as the number of protons increases) positive charge of the nucleus.

 

[Another God]

Originally posted by radagast
But it isn't. They may repulse each other, but they can avoid each other easier than they can get further from the growing (as the number of protons increases) positive charge of the nucleus.

But if the number of protons was kept constant, and you had 2s¹ vs 2s², by growing the shell ever so slightly, the electrons would have A LOT more room to move, and would thus be able to avoid each other more easily.

it makes sense to me also to believe that the two negative forces, in repelling each other, would on average tend to move out further (take up more space) in order to avoid the repulsive force. The single electron could in fact, under this understanding, be able to move wherever it wants without needing to avoid any repulsive forces, so should theoretically take up much less space. (But of course, there is still the 1s shell which is repulsive to it)

 

[radagast]

Originally posted by Another God
But if the number of protons was kept constant, and you had 2s¹ vs 2s², by growing the shell ever so slightly, the electrons would have A LOT more room to move, and would thus be able to avoid each other more easily.

it makes sense to me also to believe that the two negative forces, in repelling each other, would on average tend to move out further (take up more space) in order to avoid the repulsive force. The single electron could in fact, under this understanding, be able to move wherever it wants without needing to avoid any repulsive forces, so should theoretically take up much less space. (But of course, there is still the 1s shell which is repulsive to it)

Assuming you are adding electrons without adding protons, you are correct. Anions have larger shells, in general, than the neutral atom of the element in question. But as the atomic number goes up (say from Oxygen to Chlorine) the nuclear charge increases by one and the electron shell decreases in size.

If you mean increase the size of the shell without increasing the positive or negative charge, then that takes energy. Just like moving an object in orbit to a higher orbit - energy is required.

[Assuming ground state] The electrons are in the orbitals they are in because it takes the least energy for them to be there.