Why does the atomic radii get smaller from the left to right?

[WK95]

Why does the atomic radii get smaller from the left to right of a period but get bigger from top to bottom of a group?

Wikipedia says in an article about electron shielding
"Next we take Beryllium, Be as an example. It has 2 electrons in the 2s shell and thus, these electrons will repel each other as well. Since the repulsion is a vector, the vector may point in the direction of the nucleus at some times and thus the shielding provided by the 1s electrons cannot counteract this vector that well. Thus we know that the Be atomic radius is smaller than the Li atomic radius. From left to right across a period of elements, the atomic radius decreases because of this counteracting of the electron shielding effect."

I think I get that electron shielding is the decrease in an electron's attraction to the nucleus due to electrons of lower subshells and shells or rather, electrons that are closer. However, I'm not sureabout Wikipedia's explanation.

 

[B4ssHunter]

as you go from left to right in the periodic table , the amount of protons increase and also the amount of electrons increase within an a sublevel or an energy level , thus leading to more attraction force and thus less radius , notice that this does not apply to all elements , take flourine for example , it has a bigger radius that Cl , that happens because electrons are forced into a small place which leads to repulsion and this increasing the radius

 

[DrClaude]

Wikipedia says in an article about electron shielding
"Next we take Beryllium, Be as an example. It has 2 electrons in the 2s shell and thus, these electrons will repel each other as well. Since the repulsion is a vector, the vector may point in the direction of the nucleus at some times and thus the shielding provided by the 1s electrons cannot counteract this vector that well. Thus we know that the Be atomic radius is smaller than the Li atomic radius. From left to right across a period of elements, the atomic radius decreases because of this counteracting of the electron shielding effect."

What article is that? Because this sounds to me like utter nonesense.

The main reason that the radii decreases is that the nuclear charge increases, therefore electrons are closer to the nucleus. Electron shielding mitigates that effect somewhat (the effective charge felt by an electron is lower because of the other electrons). The article has this backwards.

 

[DrClaude]

take flourine for example , it has a bigger radius that Cl

Both the ionic and the covalent radii of fluorine are smaller than those of chlorine. See http://chemistry.about.com/library/blperiodictable.htm

 

[WK95]

What article is that? Because this sounds to me like utter nonesense.

The main reason that the radii decreases is that the nuclear charge increases, therefore electrons are closer to the nucleus. Electron shielding mitigates that effect somewhat (the effective charge felt by an electron is lower because of the other electrons). The article has this backwards.

It's from teh article entitled "shielding effect"

 

[WK95]

What article is that? Because this sounds to me like utter nonesense.

The main reason that the radii decreases is that the nuclear charge increases, therefore electrons are closer to the nucleus. Electron shielding mitigates that effect somewhat (the effective charge felt by an electron is lower because of the other electrons). The article has this backwards.

I think I'm starting to get it but if the nuclear charge increases doesn't that mean that the number of electrons also increase so don't the positive and nuclear charges cancel each other out resulting in a neutrally charged atom? If in a lone atom there is always as many electrons as there are protons, I don't quite get how an increasing nuclear charge attracts electrons more strongly.

This being the case, why, when one jumps down an period, does the radii jump up at the start before decreasing again from there?

 

[DrClaude]

I think I'm starting to get it but if the nuclear charge increases doesn't that mean that the number of electrons also increase so don't the positive and nuclear charges cancel each other out resulting in a neutrally charged atom? If in a lone atom there is always as many electrons as there are protons, I don't quite get how an increasing nuclear charge attracts electrons more strongly.

Yes, the atoms remain neutral (although sometimes it is more interesting to compare atoms with the same number of electrons).

Consider the helium atom. With one electron, you have a hydrogenic atom with a nucleus of charge ##Z=2##, so the 1s orbital will be much closer to the nucleus than in hydrogen. When you had the second electron, also in a 1s orbital, you do get a repulsion between the electrons, but much less than the extra charge in the nucleus. The electrons might not "feel" the full ##Z=2##, but a bit less, which is still a much greater charge than the ##Z=1## of hydrogen, so both electrons are closer to the nucleus.

This being the case, why, when one jumps down an period, does the radii jump up at the start before decreasing again from there?

Because you change orbital. As you go from, for example, neon to sodium, you go from 2p electrons to 3s electrons. That leads to a shift of the electronic wave function for the valence electron(s) away from the nucleus.

 

[B4ssHunter]

I think I'm starting to get it but if the nuclear charge increases doesn't that mean that the number of electrons also increase so don't the positive and nuclear charges cancel each other out resulting in a neutrally charged atom? If in a lone atom there is always as many electrons as there are protons, I don't quite get how an increasing nuclear charge attracts electrons more strongly.

This being the case, why, when one jumps down an period, does the radii jump up at the start before decreasing again from there?

you are missing a point here, the charge of the atom itself is neutral yes , but the attraction increases
positive attracts negative , and negative attracts positive
such as * this is just an analogy to explain this particular point in a classical view *
two massive bodies attract each other , increases the mass of each of them and they attract more
1 positive attracts 1 negative
now add 2 more positive and 2 more negatives for example
you now have not only 1 positive attracting another negative , there are 3 positives attracting 3 negatives

again , positives and negatives do not cancel out
they attract each other , while they do cancel the charge of the atom , but its not like they cancel out so there is no more attraction between them !
they attract each other , the more the attraction , the less the radius !

 

[B4ssHunter]

Both the ionic and the covalent radii of fluorine are smaller than those of chlorine. See http://chemistry.about.com/library/blperiodictable.htm

damn our education system here -_-
my chemistry book says that flourine has a larger radius because electrons have to fit in a smaller area . i think it was talking about the normal radius not that in ionic or covalent , however i am not sure how it is possible to measure such a thing !

 

[DrClaude]

flourine has a larger radius because electrons have to fit in a smaller area

This I don't understand. If the electrons must fit in a smaller radius, wouldn't that mean the the radius is smaller also?

 

[K^2]

This being the case, why, when one jumps down an period, does the radii jump up at the start before decreasing again from there?

Quantum physics happens. 4s shells are larger than 3s shells, for example, because the radial component of the wave function ends up having more nodes. In general, the further down the periodic table you go, the more complicated the radial function gets. This corresponds to higher energy of the particle. The classical analog, for the L=0 states, is a pendulum whose average kinetic energy increases resulting in wider swings. Similarly, the average radius of the more energetic electrons is also larger. But if you want proper understanding of this, you really have to learn a few things about the wave functions and Schrodinger Equation.

 

[B4ssHunter]

This I don't understand. If the electrons must fit in a smaller radius, wouldn't that mean the the radius is smaller also?

oh i apologize , i re read the chemistry book and it was the electron affinity ,
flourine's radius is smaller than that of chlorine , so the repulsion decreases the electron affinity