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 . 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 . 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. Also notice that going across a period the atomic radius decreases. How do we understand these two trends?

Explanations of trends in atomic radii that do not work perfectly:

We might argue that going from hydrogen through the elements that the number of electrons increase. Since electrons repel each other due to the like charges, and the size of the atom will increase. The more electrons the larger the atom. We might be able to use that argument to explain the trend within a group, but when we look at the atomic radius trend going across a period the radius decreases.

So we could argue that going from hydrogen through the elements the number of protons increases. Since electrons are attracted to protons we might expect that the more protons the smaller the radius. That might explain the trend in atomic radii going across a period but it does not explain the trend going down a group. Cesium has a large number of protons, yet it is one of the largest atoms, not the smallest.

Here is our lecture discussion on this issue.

To develop an explanation of the trends in atomic radii we need to discuss several new ideas; valence electrons, inner core electrons, effective nuclear charge and shielding.

To get started we must consider the electron configuration for the eight elements in the third period. Here is the portion of lecture which develops the table below.

Element

Nuclear
Charge

Electron
Configuration

Inner Core
Electrons

Valence
Electrons

Effective Nuclear
Charge

Na

11

1s22s22p63s1

10

1

+1

Mg

12

1s22s22p63s2

10

2

+2

Al

13

1s22s22p63s23p1

10

3

+3

Si

14

1s22s22p63s23p2

10

4

+4

P

15

1s22s22p63s23p3

10

5

+5

S

16

1s22s22p63s23p4

10

6

+6

Cl

17

1s22s22p63s23p5

10

7

+7

Ar

18

1s22s22p63s23p6

10

8

+8

 

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.

How would we explain why the atomic radius of a phosphorus atom is smaller than the atomic radius of a magnesium atom?

How would we explain why the atomic radius of a sodium atom is smaller than the atomic radius of a cesium atom?

Here is the explanation from lecture.

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

Element

I1

I2

I3

I4

I5

I6

I7

Na

490

4560

         

Mg

735

1445

7730

       

Al

580

1815

2740

11,600

     

Si

780

1575

3220

4350

16,100

   

P

1060

1890

2905

4950

6270

21,200

 

When we look at first and second ionization energy for sodium there is a significant difference. The second ionization is almost ten times larger 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.

Element

Electron
configuration

Electron
of interest

Inner Core/
valence electrons

Effective
Nuclear Charge
on valence electrons

Na

1s22s22p63s1

3s

1s22s22p63s1

+1

Na+

1s22s22p6

2p

1s22s22p6

+9

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.

Ionic Radii

Before continuing we should consider the radii of the ions (common cations/common anions) formed when electrons are lost or gained by neutral atoms. The cation formed when one or more electrons are lost has a smaller ionic radius compared to the neutral atom. The loss of valence electrons yields an cation with the outer most electrons consisting of inner core electrons. Consider sodium as shown in the table below. In the neutral atom it has one valence electron that experiences an effective nuclear charge of +1. When that electron is ionized it leave a sodium +1 cation. Now the outer most electrons are those electrons in the 2nd level. Those electrons experience an effective nuclear charge of +9, so they are very attractted to the nucleus and the radius of is Na+ very small.

Adding of one or more electrons to a neutral atom, to form an anion, causes the ionic radius to increase compared to the neutral atomic radius. This happens becasue of the higher electron-electron repulsions. For isoelectronic ions the greater the positive charge the smaller the ionic radius. While for the isoelectronic ions the greater the negative charge the larger the atomic radius.

Nuclear Charge

Element

Electronic Configuration

Effective Nuclear Charge

Valence Electrons

Core Electrons

11+

Na

1s22s22p63s1

1+

1

10

11+

Na+

1s22s22p6

9+

8

2

8+

O

1s22s22p4

+6

6

2

8+

O2-

1s22s22p6

+6

8

2

9+

F

1s22s22p5

7+

7

2

9+

F-

1s22s22p6

7+

8

2

Notice in the case of O2- and F- that both have the same number of valence electrons. But the valence electrons in O2- experience a lower effective nuclear charge compared to the valence electrons in F-. Therefore O2- is larger than F-.

Some elements which form ionic compounds are too far removed from a noble gas to readily achieve the rare gas configuration. Members of the transition metals series would have to lose a large number of electrons to become isoelectronic with the nearest noble gas. Most transition metals form cations of 2+ and 3+. This suggests that stable ions can exist which do not duplicate the electronic arrangements of the noble gases. Some of the ions can be characterized by filled subshells, for example, Zn2+ [Ar]3d10, Ag+ [Kr]3d10 and Pb2+ [Xe]6s25d10. It is therefore more difficult to predict most stable ions in the transition metals. Instead their chemistry is a collection of a large variety of ionic compounds.

Electron Affinity

Electron affinity is defined according to the equation below

M(g) + 1e- -> M-(g)

Here a neutral atom is gaining an electron. Looking at the graph below a trend in electron affinity is not immediately obvious. However, if we look carefully we can find some interesting observations. Fluorine (atomic number 9) has a very negative electron affinity, energy is released when is gains an electron. This is not too much of a surprise considering florine is one electron away from a noble gas configuration (filled shell). Oxygen also like to gain an electron. However, lithium and beryllium do not want to gain electrons. Beryllium is particularly uninterested. Adding an electron to beryllium we must place the electron into a higher energy level (2p), so takes energy to add that electron to beryllium. Ionization energies are always positive. electron affinity energies can be positive or negative.