Attached is a graph depicting atomic radius size across the first 20 elements. Why does the radius decrease in size when going from left to right across a period, and why is the decrease exponential, and not the same increment each time?
[Physics] Reason for exponentially(?) decreasing atomic radius
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You are not correct in your latter part of the analysis; the chemical properties (which is mostly what matters in ordinary matter) almost only depend on the electron shell, and in particular the outermost electrons (called the valence electrons).
So more protons mean more electrons and a different electron shell, meaning different chemical properties.
Why there is such a diversity of properties just by changing around the electron shell, is one of the wonders of chemistry! Due to quantum mechanics, the electrons don't simply spin around the nucleus like planets around the sun, but arrange themselves in particular, complicated patterns. By having different patterns, you can achieve a lot of different atom<->atom binding geometries, at a lot of different energies. This is what gives the diversity of chemical properties of matter (see the periodic table).
You can add or remove electrons to an atom to make the electron shells look more like the shells of another atom (with a different number of protons), but then the atom as a whole is then no longer electrically neutral, and due to the strength of the electromagnetic force, the resulting ion does not imitate the other atom type very well (I'm not a chemist - I'm sure there are properties that indeed could become similar).
Many physical properties are also mostly due to the electron shells, like photon interactions including color. Mass obviously is almost only due to the nucleus though, and I should add that in many chemical processes the mass of the atoms are important for the dynamics of processes, even if it isn't directly related to the chemical bindings.
This was just a small introduction to chemistry and nuclear physics ;)
- This is could be considered chemistry question and might receive a more interesting response on Chem.SE. But it's probably fine here.
- According to ChemWiki at UC Davis, "Atomic size gradually decreases from left to right across a period of elements. This is because, within a period or family of elements, all electrons are added to the same shell. However, at the same time, protons are being added to the nucleus, making it more positively charged. The effect of increasing proton number is greater than that of the increasing electron number; therefore, there is a greater nuclear attraction. This means that the nucleus attracts the electrons more strongly, pulling the atom's shell closer to the nucleus. The valence electrons are held closer towards the nucleus of the atom. As a result, the atomic radius decreases.
- Nuclear charge increases due to protons NOT electrons. Greater number of electrons doesn't increase the strength of the nucleus. There are no electrons in the nucleus! As you move across the periodic table, the number of protons increases, increasing the charge of the nucleus by $+1$ for each proton added. For atoms with neutral charge, this implies same number of protons and electrons. And since the periodic table lists atoms in their neutral state, this has the effect that the number of electrons happens to correlate with the number of protons and thus the charge of the nucleus. But if I remove two electrons from a calcium atom, I still have a calcium atom because the nucleus is the same. Also, just to be thorough, two atoms can have the same number of protons and electrons and still be different in the number of neutrons. They would still be the same element, but we call them different "isotopes."
- There's a difference between nuclear charge and effective nuclear charge. In a hydrogen atom, the electron experiences the full charge of the positive nucleus, which is just a proton. A hydrogen atom is thus like two point charges, so the effective nuclear charge can be calculated from Coulomb's law. However, in an atom with many electrons the outer electrons are simultaneously attracted to the positive nucleus and repelled by the negatively charged electrons. Each electron (in the n-shell) experiences both the electromagnetic attraction from the positive nucleus and repulsion forces from other electrons in shells from 1 to n. (Remember, electricity has both attraction and repulsion!) This causes the net force on electrons in outer shells to be significantly smaller in magnitude. Thus, these electrons aren't as strongly bonded to the nucleus as electrons closer to the nucleus. This is known as the 'shielding effect.'
- There are many interesting periodic trends. You can learn more here.
EDIT:
I was able to narrow down an answer I think. We define effective nuclear charge as $Z_{\mathrm{eff}} = Z - S$, where $Z$ is number of protons and $S$ is the average number of electrons between the nucleus and the electron in question. Only the 1s orbital electrons have $Z_{\mathrm{eff}} = Z + 0 = Z$, ie $S = 0$ only for neutral hydrogen and helium. For all other standard elements, we have additional orbitals like 2s and 2p and 3s, etc and these all experience $S \neq 0$ so $Z_{\mathrm{eff}} < Z$.
Apparently, there is something called the Slater's rules. The shielding constant for each group is formed as the sum of the following contributions:
So for iron, here is the effective nuclear charge for different electrons.
Now let me try to address the radius issue according to Slater's rules. I make no guarantee Slater's rules are foolproof. You should investigate that yourself. I'm just going to take these rules and apply them. Let's consider fluorine. I like fluorine as an example because I always think of it as the hungriest of the elements. It wants an electron to fill it's outer shell badly. Why? So for the 2nd row, the effective nuclear charge increases as you go across the periodic table as follows:
- $Z_{\mathrm{eff,Li}} = 3 - (0.85 \times 2 + 0.35\times 1) = 0.95$
- $Z_{\mathrm{eff,Be}} = 4 - (0.85 \times 2 + 0.35\times 2) = 1.60$
- $Z_{\mathrm{eff,B}} = 5 - (0.85 \times 2 + 0.35\times 3) = 2.25$
- $Z_{\mathrm{eff,C}} = 6 - (0.85 \times 2 + 0.35\times 4) = 2.90$
- $Z_{\mathrm{eff,N}} = 7 - (0.85 \times 2 + 0.35\times 5) = 3.55$
- $Z_{\mathrm{eff,O}} = 8 - (0.85 \times 2 + 0.35\times 6) = 4.20$
- $Z_{\mathrm{eff,F}} = 9 - (0.85 \times 2 + 0.35\times 7) = 4.85$
So the effective nuclear charge does increase across the table! And thus, the electrons in the outer orbital experience a greater nuclear charge for elements on the left than on the right.
The basic idea is,
- Electrons in the same orbital don't shield as well as those of lower energies. So lower energy orbitals contribute more to shielding.
- Different orbitals contribute different shielding amounts, which presumably breaks down to the intricacies of the chemistry and physics.
and the result is the radii decrease as you go from right to left as a result because effective nuclear charge is increasing.
EDIT 2:
I learned on Chem SE that Slater's rules are just an approximation.
Best Answer
The cycles of rise/fall you see are a reason the periodic table is called ... periodic.
At the core of atoms is a nucleus, which attracts (because it has a positive charge) negative charges, electrons. Electrons can only be stable for extended times when they are in one of the allowed orbitals, and... for H and He, with one or two electrons, the lowest energy cost is the 1S orbital. Any electrons in higher orbitals will eventually emit energy and decay to 1S.
He (having two positive charges) pulls its electrons to a nearer distance than H (having only one).
Li through Ne, however, have to put electrons 3 and up into unfilled spaces, of which the 1S orbital has none. 2S or 2P orbitals (two allowed in 2S, six allowed in 2P) are the lowest available. It isn't exactly exponential, but the higher atomic number within the shell means a higher positive nuclear charge, and thus a smaller orbit, modulated by the repulsion of the electrons... so with a crowded 2S and 2P shell, the size creeps up.
These rules can be related to quantum mechanical principles, and to a lot of chemistry and other phenomena. Radius is only the beginning of what a periodic table can illuminate.