Trends in the Periodic Table
Atomic Radii
In order to talk about the radius of an atom, we have to make an arbitrary decision about where the edge of the atom is. It is arbitrary because the electron orbitals do not end sharply.
Nevertheless, we can do like we did with the 3D contour plots of the orbitals and just arbitarily choose the radius that the electron spends 90% of its time inside.
The electrons spend 90% of the time inside the black line.
Using this definition consistently, we can look at the trends of the atomic radii as a function of position in the periodic table.
That trend is...
In general the size of the atom depends on how far the outermost valence electron is from the nucleus. With this in mind we understand two general trends...
- Size increases down a group:
- The increasing principle quantum number of the valence orbitals means larger orbitals and an increase in atomic size.
- Size generally decreases across a period from left to right:
- To understand this trend it is first important to realize that the more strongly attracted the outermost valence electron is to the nucleus then the smaller the atom will be. While the number of positively charged protons in the nucleus increases as we move from left to right the number of negatively charged electrons between the nucleus and the outer most electron also increases by the same amount. Thus you might expect there to be no change in the radius of the outermost electron orbital since the increasing charge of the nucleus would be canceled by the electrons between the nucleus and the outermost electron. In reality, however, this is not quite the case. The ability of an particular inner electron to cancel the charge of the nucleus for the outermost electron depends on the orbital of that inner electron. Remember that electrons in the s-orbital have a greater probability of being near the nucleus than a p-orbital, so the s-orbital does a better job of canceling the nuclear charge for the outermost electron than an electron in a p-orbital. Likewise, an electron in a p-orbital is does a better job than a d-orbital. Thus, as we move across a given period the ability of the inner electrons to cancel the increasing charge of the nucleus diminishes and the outermost electron is more strongly attracted to the nucleus. Hence the radius decreases from left to right.
Ionization Energy
The ionization energy of an atom is the amount of energy required to remove an electron from the gaseous form of that atom or ion.
1st ionization energy - The energy required to remove the highest energy electron from a neutral gaseous atom.
For Example:
| Na(g) → Na+(g) + e- | I1 = 496 kJ/mole |
Notice that the ionization energy is positive. This is because it requires energy to remove an electron.
2nd ionization energy - The energy required to remove a second electron from a singly charged gaseous cation.
For Example:
| Na+(g) → Na2+(g) + e- | I2 = 4560 kJ/mole |
The second ionization energy is almost ten times that of the first because the number of electrons causing repulsions is reduced.
3rd ionization energy - The energy required to remove a third electron from a doubly charged gaseous cation.
For Example:
| Na2+(g) → Na3+(g) + e- | I3 = 6913 kJ/mole |
The third ionization energy is even higher than the second!
Succesive ionization energies increase in magnitude because the number of electons, which cause repulsion, steadily decrease. This is not a smooth curve There is a big jump in ionization energy after the atom has lost its valence electrons. An atom that has the same electronic configuration as a noble gas is really going to hold on to its electrons. So, the amount of energy needed to remove electrons beyond the valence electrons is significantly greater than the energy of chemical reactions and bonding. Thus, only the valence electrons (i.e., electrons outside of the noble gas core) are involved in chemical reations.
The ionization energies of a particular atom depend on the average electron distance from the nucleus and the effective nuclear charge
These factors can be illustrated by the following trends:
1st ionization energy decreases down a group.
This is because the highest energy electrons are, on average, farther from the nucleus. As the principal quantum number increases, the size of the orbital increases and the electron is easier to remove.
Examples:
I1(Na) > I1(Cs)
I1(Cl) > I1(I)
1st ionization energy increases across a period.
This is because electrons in the same principal quantum shell do not completely shield the increasing nuclear charge of the protons. Thus, electrons are held more tightly and require more energy to be ionized.
Examples:
I1(Cl) > I1(Na)
I1(S) > I1(Mg)
The graph of ionization energy versus atomic number is not a perfect line because there are exceptions to the rules that are easily explained.
Filled and half-filled subshells show a small increase in stability in the same way that filled shells show increased stability. So, when trying to remove an electron from one of these filled or half-filled subshells, a slightly higher ionization energy is found.
Example 1:
I1(Be) > I1(B)
It's harder to ionize an electron from beryllium than boron because beryllium has a filled "s" subshell.
Example 2:
I1(N) > I1(O)
Nitrogen has a half-filled "2p" subshell so it is harder to ionize an electron from nitrogen than oxygen.
Which element has a higher ionization energy, Zinc or Gallium?
