1. Oct 19, 2012

### krackers

First, it says in my textbook that in an anion, the atomic size will be larger than that of the neutral atom. It says this is because the extra electrons causes more repulsion and then increases the atomic radius. In the next chapter, it says as you move across a period [from left to right] in the periodic table, the atomic radius gets smaller. However, wouldn't it get bigger because of the extra protons + neutrons, and the extra electrons creating repulsion?

2. Oct 20, 2012

### Staff: Mentor

To some extent it does - when you move down the column, radius definitely increases. But when you move across the period electrons fill the same orbitals and the effect on the radius is much smaller and easily countered by increased attraction from the higher charge of the nucleus.

3. Oct 20, 2012

### chill_factor

protons and neutrons take up almost zero volume. they're at 10^-15 meters. that's to a nano-human, what a micrometer is to us AKA totally invisible and negligible. Nano is what chemistry cares about. Everything smaller than nano is useless in chemistry and engineering. so the increased numbers of protons/neutrons has zero effect on the size of the atom. but there will be an increased positive charge. this positive charge will cancel out the repulsion from the negative charge.

you may think, why would this be? its because of a result in classical electrodynamics that outside a shell covering a sphere of zero net charge, the force it exerts on a charged particle outside is zero. so you can roughly imagine the inner shell electrons as being the sphere that covers the nucleus and has a tiny net positive charge. correspondingly the few electrons that are outside the inner shell will be attracted by that tiny net positive charge, and will have less repulsion from the inner shell than you might think.

4. Oct 20, 2012

### krackers

Makes sense, but in the case of an ion, the extra electron gained also fills the same shell, so why does the radius increase then?

5. Oct 20, 2012

### Staff: Mentor

General rules are about general trends - doesn't mean they are always perfectly obeyed.

That's the problem with chemistry - exact rules often deal with esoteric properties, which are not easy to translate into easily measurable properties (they can be sometimes calculated from these esoteric ones, but calculations are pretty difficult). Easy to apply rules are just approximations, so they are not always right.

That being said - according to wikipedia atomic radius of iron is 126 ppm, atomic radius of manganese (element with one electron less) is 127 ppm, so definitely the radius decreases, not increases.

6. Oct 21, 2012

### AGNuke

The atomic radii can also be explained by "Shielding effect of electrons" and "Effective Atomic Number". Every electron shields the effect of electrostatic attraction on other electrons by exerting its own repulsive force. When we add another electron to the same energy level to form an ion, it too provides the shielding effect, for which the ions compensates by increasing the volume of its electron cloud. If we move to right in periodic table, the addition of another proton in the nucleus outweighs the effect of small shielding of electron, therefore decreasing the atomic radii.

So, the atomic trend for ions, Cl->Ar>K+. As for atoms, the trend is generally true. (Leaving aside Noble gases, they aren't so noble to tell us their actual atomic radii)

In the next chapter, you may read that 6th period elements have equal or smaller atomic radii than 5th period elements. That too is explained by Effective Atomic Number and Shielding effect.

Last edited: Oct 21, 2012
7. Oct 27, 2012

### krackers

So let me see if I have this right. This can be explained through electrostatic attraction right? If you take the coulumbs law,

$\frac { { Q }_{ 1 }{ Q }_{ 2 } }{ 4\pi { \varepsilon }_{ 0 }{ r }^{ 2 } }$

then as you go towards the right in a period, both Q1 (the charge of the nucleus) and -Q2 (charge of the electron cloud, negative because electrons are negatively charged) increases because you are adding protons and electrons respectively. Because $4\pi e_{0}$ is pretty much a constant, and when you move towards the right of a period, the added electrons just fill up the existing shell, not create a new one, the r^2 remains a constant. Thus, because Q1 and Q2 increase, the net force between them increases, and the electrons are pulled close. This can be related to having two small magnets, and then two large ones. The two large ones are pulled together much more than two small ones.

When you go down a group, you are adding a new shell and that increases the atomic size.

Now is the part I am not clear about.
When you form an anion, you add an electron to a given atom. That should make the atomic size increase. However, using coulumbs law, wouldn't the force between them be greater?

8. Oct 28, 2012

### AGNuke

You forgot accounting for repulsions. That too is acting between electrons, easily competing with attractive force of nucleus. That's what I mentioned as "Shielding Effect".