2
$\begingroup$

I've asked a similar question before, but now I want to find out the real rule. Pleas tell me if my interpretation is correct: Cd, Ag , Au (increasing ray order). This happens because although we know that period and group effect the radius length, most important one is the group. So firstly we should list the elements according to their group and then list them in the same group... For example Zn, Cd (same group), then Pd, Pt, (same group), then Ph, ir (same group). From our lesson we know that from Cd to Hg the ray doesn't change but I think when we have to list them we should consider that it grows a little. Are these statements right? If not, please help me clear the concept. I'm not a native speaker of english, so I may not express correctly in some occasions

$\endgroup$
1
$\begingroup$

This is the radius general trend on the periodic table:

enter image description here

The main reason that explains how it changes is the nuclear effective charge experimented by the electrons when atracted by the nucleos. Briefly speaking, the more electrons you add as you go down a group (in different energy levels), the lower the nuclear effective charge that the electrons will experiment, so the electron denisity will be less "compresed", resulting in a larger radius. Going from the right to the left, you add one by one, a new electron that experiments almost the same nuclear effective charge, so it will be atracted with a similar force, and also, you are adding another proton to the nucleus, so the overall atraction will be higher, the electron density will decrease and the atomic radius will decrease.

| improve this answer | |
$\endgroup$
0
$\begingroup$

The OP's understanding is not correct. The Periodic Table does a very poor job of predicting radii, nor was such a use ever intended for it. There is no obvious reason that the radius, r, of the elements should show periodic or systematic patterns.

As it turns out, however, the Table does capture certain systematic trends, as the answer by Altered State indicates. Refer to the covalent (measured) radii from https://en.wikipedia.org/wiki/Atomic_radius and you will see that there are numerous exceptions to the general trends. For instance:

  1. P, S and Cl all have roughly the same radii
  2. Ru < Rh < Pd < Ag is opposite to what you'd predict from the general trend, even though they're in the same row and adjacent, and finally
  3. Pt, Au, & Hg are all smaller than the elements directly above them (i.e. same groups) in the table.

We can conclude that covalent radii cannot be confidently predicted using the Periodic Table. It is beyond the scope of this answer to explain the details behind various measurements of radii, but three common ways radii are tabulated is using covalent, ionic, or van der Waals measurements, as well as calculated values (see Wikipedia link, above). The results depend on the way the values are measured. Because of the atomic nature of the elements, and because atoms are held together principally by electrostatic forces (ignoring how the nucleus is held together), a change in the electric field due to valence and chemical bonding (either ionic or covalent) has a dramatic effect on an atom's radius.

Since the valence of an element varies, it is also incorrect to believe that an element has a fixed (or nearly fixed) radius. Iron will have a substantially different radius in FeCl3 compared to that in Fe2O3. It may be of some interest to compare the radii of the elements in their 0 valence state as solids, but even then the crystal structures (hence the electronic interactions between atoms) vary. For instance in comparing the radius of Fe in bcc iron with N, do we cool the N2 to form the hexagonal crystal (down to -210 °C) and search for values for Iron at that temperature? Worse, do we compare the N-N molecular triple bond distance or the van der Waal distances? And then there's He, we need to cool He down to 1 K and also subject it to 2.5 MPa pressure to get a solid. I'm sure it would be quite difficult to obtain radii of the other elements under these conditions. ("Difficult" in the sense of finding reliable measurements in the literature.) In estimating the comparative size of two atoms, it is generally only useful to do so when they are in the same environment. This usually means with the same bonding, physical state, etc. In these situations, periodic trends are not very useful; detailed analysis of the quantum mechanics, especially the orbital (shell) occupancies, is required.

We all learned in physics that like electrical charges repel and opposites attract. How does this help us understand the trend in atom size with increasing atomic number? An increase in atomic number of 1 increases the number of protons and electrons by 1 (and adds variable number of neutrons). Radius is frequently assumed to be the distance from the (point-like) nucleus to the distance at which the valence electron(s) are 75%, 85%, or 90% contained. (This is arbitrary.) This distance depends on the repulsion from all of the other electrons that the valence electron(s) experience as will as the attraction between in and the positive nucleus. Since all of the electrons spend the majority of their time in positions other than directly between a valence electron and the nucleus, the shielding by those other electrons felt by the valence electrons is less than 100%, meaning it feels more attraction than repulsion (if this were not true, the electron would wander (or be pushed) away.)

What this means for the radius is that while yes the radius increases with increasing number of electrons (and protons), but the increase is not as great as you might expect since the outer-most electrons are attracted a little bit more than you'd expect. There are effects from half-full shells, as well as the differences in shielding of the various orbitals (s, p, d, f). This makes it quite difficult to confidently predict sizes once you get below row 2 in the Periodic Table.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.