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The explanation that I have heard states that when we move horizontally across the periodic table, the number of electrons increases leading to a greater force of attraction from the nucleus.

For instance, let's take the example of Lithium (atomic radius = 152pm) and Carbon (atomic radius = 77). The size of Lithium is greater than that of Carbon is because in Carbon, the number of electrons increases leading to a greater force of attraction than in Lithium. However, when the number of electrons in Carbon increases, the number of proton increases as well, right?

If in Lithium we had 3 electrons and 3 protons, then there is no net charge. Similarly in Carbon, there are 6 electrons and 6 protons, again, no net charge. Then why is it that the effective charge in Carbon is somehow greater leading to a smaller atomic radius?

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    $\begingroup$ just because there is no net charge in two different situations does not mean the force of attraction will be the same: two charges +10Q and -10Q will feel a force of attraction 100 times that of +Q and -Q $\endgroup$ – danimal Mar 12 '15 at 10:49
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  1. This is could be considered chemistry question and might receive a more interesting response on Chem.SE. But it's probably fine here.
  2. 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.
  3. 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."
  4. 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.'
  5. 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:

enter image description here

So for iron, here is the effective nuclear charge for different electrons.

enter image description here

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.

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  • $\begingroup$ I got all of your point (the thing about isotopes, shielding effect etc.) I still don't get it. You mentioned that for each of the proton added, the force of attraction experienced by the electrons would increase, right? However, when we add a proton (when we take another element), we have an additional electron as well. Does that not cancel out the effect of that additional proton? $\endgroup$ – Always Learning Forever Mar 12 '15 at 18:38

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