Why Atomic Radii Decrease From Left To Right

Atomic radius patterns are observed throughout the periodic table. 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.

1 Answer 1 Sorted by:

  • This is could be considered chemistry question and might receive a more interesting response on Chem.SE. But its 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 atoms 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 doesnt 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.”
  • Theres 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 Coulombs 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 arent 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 $S eq 0$ so $Z_{mathrm{eff}} < Z$.

    Apparently, there is something called the Slaters 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 Slaters rules. I make no guarantee Slaters rules are foolproof. You should investigate that yourself. Im just going to take these rules and apply them. Lets 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 its outer shell badly. Why? So for the 2nd row, the effective nuclear charge increases as you go across the periodic table as follows:

  • $Z_{mathrm{eff,Li}} = 3 – (0.85 times 2 + 0.35times 1) = 0.95$
  • $Z_{mathrm{eff,Be}} = 4 – (0.85 times 2 + 0.35times 2) = 1.60$
  • $Z_{mathrm{eff,B}} = 5 – (0.85 times 2 + 0.35times 3) = 2.25$
  • $Z_{mathrm{eff,C}} = 6 – (0.85 times 2 + 0.35times 4) = 2.90$
  • $Z_{mathrm{eff,N}} = 7 – (0.85 times 2 + 0.35times 5) = 3.55$
  • $Z_{mathrm{eff,O}} = 8 – (0.85 times 2 + 0.35times 6) = 4.20$
  • $Z_{mathrm{eff,F}} = 9 – (0.85 times 2 + 0.35times 7) = 4.85$
  • 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.

  • Electrons in the same orbital dont 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 Slaters rules are just an approximation.

    Thanks for contributing an answer to Physics Stack Exchange!

  • Please be sure to answer the question. Provide details and share your research!
  • Asking for help, clarification, or responding to other answers.
  • Making statements based on opinion; back them up with references or personal experience.
  • Use MathJax to format equations. MathJax reference.

    To learn more, see our tips on writing great answers. Draft saved Draft discarded

    This is a very important periodic phenomenon: the contraction of atomic radii across the period. While as we add to Z (the number of protons in the nucleus), we also add another electron (and charge is therefore kept neutral), the increased nuclear charge acts disproportionately on the valence electrons, and contracts this shell.

    Nuclear charge dominates whilst adding electrons to the same shell. The net result is atomic contraction across a period.

    EXCEPTIONS: Because the electrons added in the transition elements are added in the inner electron shell and at the same time, the outer shell remains constant, the nucleus attracts the electrons inward. The electron configuration of the transition metals explains this phenomenon. This is why Ga is the same size as its preceding atom and why Sb is slightly bigger than Sn.

    When a covalent bond is present between two atoms, the covalent radius can be determined. When two atoms of the same element are covalently bonded, the radius of each atom will be half the distance between the two nuclei because they equally attract the electrons. The distance between two nuclei will give the diameter of an atom, but you want the radius which is half the diameter.

    An anion, on the other hand, will be bigger in size than that of the atom it was made from because of a gain of an electron. This can be seen in the Figure 4 below. The gain of an electron adds more electrons to the outermost shell which increases the radius because there are now more electrons further away from the nucleus and there are more electrons to pull towards the nucleus so the pull becomes slightly weaker than of the neutral atom and causes an increase in atomic radius.

    If we were able to determine the atomic radius of an atom from experimentation, say Se, which had an atomic radius of 178 pm, then we could determine the atomic radius of any other atom bonded to Se by subtracting the size of the atomic radius of Se from the total distance between the two nuclei. So, if we had the compound CaSe, which had a total distance of 278 pm between the nucleus of the Ca atom and Se atom, then the atomic radius of the Ca atom will be 278 pm (total distance) – 178 pm (distance of Se), or 100 pm. This process can be applied to other examples of ionic radius.

    Determining the atomic radii is rather difficult because there is an uncertainty in the position of the outermost electron – we do not know exactly where the electron is. This phenomenon can be explained by the Heisenberg Uncertainty Principle. To get a precise measurement of the radius, but still not an entirely correct measurement, we determine the radius based on the distance between the nuclei of two bonded atoms. The radii of atoms are therefore determined by the bonds they form. An atom will have different radii depending on the bond it forms; so there is no fixed radius of an atom.

    FAQ

    Why does atomic radius decrease as you go left to right?

    On the periodic table, atomic radius generally decreases as you move from left to right across a period (due to increasing nuclear charge) and increases as you move down a group (due to the increasing number of electron shells).

    Why do atomic radii get smaller in going from left to right in a row of the periodic table why does it go up as one goes down in a column?

    This is because in going down a column you are jumping up to the next higher main energy level (n) and each energy level is further out from the nucleus – that is, a bigger atomic radius. Atoms get smaller as you go across a row from left to right.

    Why does the atomic radii decrease across a period?

    Across a period, effective nuclear charge increases as electron shielding remains constant. A higher effective nuclear charge causes greater attractions to the electrons, pulling the electron cloud closer to the nucleus which results in a smaller atomic radius.

    Why atomic radius increases from top to bottom and decreases from left to right?

    In conclusion we say that the atomic radius decreases when we go from the bottom left to the upper right of the periodic table due to two factors the effect of effective nuclear charge (left to right) and reduction of shell (from bottom to top).

    Related Posts