The strength of any model is in it ability to explain experimental observations. The Quantum Mechanical model of the atom can 'tested' by looking at the experimental data of atomic radius and ionization energy. The first property to explore is atomic radius. The immediate question is what is an atomic radius. Our quantum mechanical description of an atom suggests a very broad region for finding the electron. It is difficult to define a sharp boundary for distance between the electrons in any particular atom and the nucleus. So some approximations are made in determining this parameter. For example, the distance between the two chlorine atoms in Cl₂ is known to be 1.988 Å. To get the atomic radius we assume the distance between the two nuclei is the sum of two chlorine atomic radii. Therefore the atomic radius of chlorine is 0.994 Å. Additional distances can be obtained from other distances between atoms. Care must be taken in these types of determinations however.

After collecting large amounts of data generally accepted atomic radii are known for most elements in the periodic table. Atomic radii determine this way are also called covalent radii. (we'll discuss the term covalent in the next chapter) Atomic radii for metals can be estimated from the distance between metal atoms in the pure solids.

Looking at the table above notice from lithium to cesium, going down the group, the atomic radius increases. This is as expected as the valence electron is located further and further from the nucleus. Each of the elements in a group have the same number of valence electrons. In the case of Group I there is one valence electron. In lithium this single valence electron is found in the second level, in sodium the single valence electron is located in the third level. The third level is higher in energy AND further away from the nucleus. Therefore a sodium atom is larger than a lithium atom. The argument is the same for other higher members of the group.

Also notice that going across a period the atomic radius decreases. We can understand this trend by considering that as we go across a period each electron added is going into the same level. For example look at the electron configuration for the eight elements in the third period.

Sodium has one valence electron and as we move across to argon with eight valence electrons. Remember, and this is important, the valence electrons are in the outer most level of the atom (the furthest from the nucleus). Remember also that as we move from sodium to argon, the nuclear charge, the number of protons in the nucleus, is also increasing. Sodium has 11 protons and argon has 18 protons. Each time we add a proton to a the nucleus the electron (which are negatively charged) feel a greater attraction to the nucleus. Since valence electrons are all in the same level, they feel a greater attraction to the nucleus as we move across the period. If the valence electrons feel a greater attraction they are pulled in closer to the nucleus and the atom gets smaller. So the one valence electron in sodium feels a certain attraction to the nucleus. Magnesium has two valence electrons, both in the same level, and both of these electrons feel a greater attraction to the nucleus because the nuclear charge has increased from going from sodium to magnesium.

Another way to explain the trend in atomic radius across a period is in terms of the effective nuclear charge experienced by the valence electrons. Since the valence electrons are furthest from the nucleus (outer most level) the inner core electrons (shown in red in the table above) shield the outer electrons from some of the positive charge on the nucleus. The valence electrons do not experience the total positive charge on the nucleus but an effective nuclear charge. The effective nuclear charge experienced by the valence electrons is calculated by subtracting the number of inner core electrons from the nuclear charge on the element. In the table above we see the effective nuclear charge increases as we proceed from sodium to argon. Since the valence electrons experience a great effective nucclear charge as be proceed from sodium to argon, the valence electrons feel a greater attraction to the nucleus and the atoms get smaller.

You might ask why doesn't the radius get smaller as we go from sodium to potassium? Potassium has a greater nuclear charge compared to sodium. You're right potassium does have a greater nuclear charge compared to sodium. The difference is the valence electron in potassium is in a different and HIGHER level compared to the valence electron in sodium. So the valence electron in potassium is further away and can not feel the higher nuclear charge. When going across a period each added electron goes into the SAME level. Since the nuclear charge is also increasing the valence electron feel a greater attraction to the nucleus and the atom's size gets smaller.

Another important trend which is related to the periodic table is the ionization energy for an element. It is related to the energy required to remove an electron from an atom. Clearly a multi-electron atom would have many ionization energies. So by definition the first ionization energy is defined as the energy required to remove the outer most electron from a neutral atom in the gas phase. An equation can be written to show this definition,

M(g) -> M⁺(g) + 1e^-

The second ionization energy is the energy required to remove the next outer electron from the singly charged ion.

M⁺(g) -> M²⁺(g) + 1e^-

Subsequent removal of electrons can also be written and determined.

As you would expect each successive removal of an electron requires more energy. As electrons are removed the remaining electrons experience a greater attraction to the nucleus. As shown in the table the ionization energies increase with each successive removal. Notice that there occur major changes as successive electrons are removed. For magnesium there is a large change between the second and third electrons, for aluminum between the third and fourth, for silicon between the fourth and fifth. Magnesium has the electron configuration of 1s²2s²2p⁶3s². The first two electrons are removed from the third level. The third electron is removed from the second level. Electrons in lower levels feel a greater attraction to the nucleus and are more difficult to remove.

Looking at the table suggests the valence electrons, the outer most electrons, require considerably less energy to remove. This suggests that the outer electrons are much easier to remove from the atom compared to the inner level electrons. The inner level electrons are too tightly bound to be involved in chemical bonding.

TABLE Successive values of ionization energies (I) for the elements sodium through argon (kJ/mol)a

When we look at first and second ionization energy for sodium there is a significant difference. The second ionization is almost ten time learger than the first. What is going on? To explain this all we need to do is consider which electrons are being ionized and what level are those electrons coming from. So lets look at the electron configuration for sodium,

1s²2s²2p⁶3s¹

The first electron ionized it the electron in the highest energy level...the 3s. That electron requires some amount of energy to remove. When the second electron is removed it comes from the 2p sublevel, as that is the sublevel with the next highest energy. Why do it take so much more energy to remove the second electron in sodium? It has to do with the fact that the second electron feels a greater attraction to the nucleus compared to the first electron removed. The effective nuclear charge is a direct measure of the attraction an electron feels to the nucleus.

The effective nuclear charge experienced by the valence (3s) electron in the neutral sodium atom is +1. We get this number by subtracting the inner core electrons (10) from the total nuclear charge (11). After removing the 3s electron we have Na⁺, which has the electron configuration of 1s²2s²2p⁶. The next electron removed comes from the 2p sublevel. To calculate the effective nuclear charge on a 2p electron we subtract the inner core electrons (2) from the nuclear charge (11) and we get +9. Clearly an electron in the 2p sublevel experiences a great effective nuclear charge compared to the 3s electron. That is why it takes more energy to remove an electron from the 2p sublevel in sodium compared to the 3s sublevel.

Now consider the variation in the first ionization potentials and position of the element in the periodic table. If we look at a plot of 1st ionization versus atomic number certain trends are evident. The figure below shows the 1st ionization energies.

Notice the general trend is for the ionization energy to increase going across a period. This makes sense if we recall the atomic radius decreases going across the period. If the atom gets smaller it means the electrons feel a greater attraction to the nucleus and it is more difficult to remove the electron.

When we look more closely at the trend in ionization energies we see two deviations. Boron's ionization energy is lower than beryllium's and oxygen's is lower than nitrogen's. What gives? In the first case, lets look at the energy level diagram for boron and beryllium.

Notice the energy for the 2s and the 2p sublevels. The 2s sublevel is lower in energy compared to the 2p sublevel. So when we try to remove the first electron in boron it is easier because it is higher in energy. Therefore the the first ionization energy for boron drops when compared to beryllium. As we continue across to carbon and then to nitrogen the electrons are going into the same sublevel and we see the ionization energy increasing.

The second occurs from nitrogen to oxygen. What is going on in this case? Again we need to look at the energy level diagrams for the two atoms. In nitrogen all of the 2p orbitals are half-filled. When we go next to oxygen the electron must be placed in one of the half-filled orbitals. But we know that it requires some extra energy to place two electrons into the same orbital. Since it takes slightly more energy to pair the electrons it takes less energy to remove the paired electron in oxygen. So we see a small decrease in the first ionization energy when going from nitrogen to oxygen.