In this section we will explore what happens when multiple electrons are present around different nuclei. The manner in which two electrons interact with each other is a major part of the story. Another piece of the picture that will come out of this exploration will be learning about the different orbitals that the electrons adopt as they are added. So far, we've seen that H and hydrogen-like ions utilize the spherical 1s orbital, but there are other orbitals with different shapes.

HYDROGEN. We will start with the simplest case. We've already looked at the neutral H atom, consisting of one proton and one electron. We can represent H as shown below:

6.1

The blue-tinted circle represents the 1s orbital on H, and the dot indicates that it is occupied with one electron. The dot does not represent the position of the electron! This model reminds us that the H 1s orbital is spherical. Neutral H holds only one electron in its orbital. The dot represents only the number of electrons and says nothing about its position (remember that we cannot know the position and speed of electrons in atoms).

We could also use a colored sphere to represent H, like this:

6.2

where we will use yellow to denote a singly occupied orbital where the electron is spin up and blue to denote a singly occupied orbital where the electron is spin down.

What happens when a second electron is added to the H atom? The net charge will be -1, so this would be the H anion, H^–. There are two ways to add an electron to H: so that both electrons have the same spin (↑↑ or ↓↓) or opposite spins (↑↓ or ↓↑). The result depends very much on which combination we choose. Let's look at this using both representations for H:

6.3

If the electrons have opposite spins, H^– is formed. In the second case, when the electron spins are the same, nothing happens. This unusual behavior is a consequence of the Pauli principle, which can be stated as follows: an orbital can only hold two electrons, and it can only hold two electrons if they have opposite spins. Two electrons always repel each other, but if they have different spins they can tolerate each other's presence in the same space in a way that two electrons of the same spin cannot. This fact will be absolutely critical when we discuss bonding between two atoms in a later section.

You may have noted that a gray sphere is used in the 3D version of the equation above. We will use gray throughout the module to indicate a pair of electrons when using the 3D representations of atoms.

Below we show the 1s orbitals of H and H- in their lowest energy states:

6.4

Remember that H consists of one proton and one electron, while H^– has one proton and two electrons. They are both using the 1s orbital, but it is singly occupied in H and doubly occupied in H^–, or 1s¹ and 1s², respectively. (Chemists frequently omit the superscript if an orbital is singly occupied, but we will not do so in this module.) You can see that the two orbitals look somewhat different. The outer contour in H^– is a bit larger, and the inner contours are less intensely colored. The net effect is that H^– is more spread out than H, which is a direct consequence of the repulsion between the two electrons. Two electrons need more space than one electron. But it's actually remarkable that the change is as small as it is. This is something we will see again in larger atoms. The difference between singly and doubly occupied orbitals can be even less obvious than it is here.

Before we move on to larger nuclei than H, we will consider one other possibility: what happens if we try to add a third electron to H^–? We find that H^–2 does not form. Three electrons are too much for a single proton to hold onto. One of the reasons H^–2 cannot form is because the third electron cannot be placed in the 1s orbital due to the Pauli principle. It would need to go into one of the higher orbitals, such as the 2s orbital, and that is not very favorable.

Here's a review question for you to think about:

The answer is H⁺, H, and H^–. Although H⁺ does not have any electrons, it is still a form of hydrogen because the nucleus has one proton.

HELIUM. The helium nucleus has two protons. We looked at He⁺ in the previous section when we considered the hydrogen-like ions. Remember that the 1s orbital of He⁺ is somewhat smaller than the 1s orbital of H (see Figure 5.1), and the single electron has more energy (four times as much). What do you think happens if we add a second electron to He⁺? The equations below indicate the two possibilities. Only one of them results in the formation of the He atom in its ground state.

6.5

You're correct if you chose answer (a). Although exceptions certainly occur—we'll run across a case or two later in the module—the rules we deduce in chemistry are often repetitive. Adding a second electron to He⁺ is very similar to adding a second electron to the H atom. It only yields the neutral He atom in its most stable form if the electrons have opposite spins. How big do you think the doubly occupied 1s orbital of He is compared to the singly occupied 1s orbitals of H and He⁺? The figure below depicts the orbitals of H, He⁺, and He and their occupations on the same scale:

6.6

We see that both He⁺ and He are smaller than H. The doubly occupied orbital for He is larger than the singly occupied orbital of He⁺, but the difference is not very great. Two electrons occupy more space than one electron, but they often do not need much more space if they are sharing an orbital.

As with the H nucleus, we can try to add a third electron to He, but the He nucleus cannot hold onto a third electron. In other words, He^– is not stable.

One of the things we'll do in the module is to build a table of the elements. We can start with H and He:

6.7

These two elements define the first row of the periodic table. They are the only two elements where the most stable states use only the 1s orbital. Heavier nuclei need higher orbitals to accomodate more than three electrons, starting with the 2s and 2p orbitals.

NEON.We're going to skip ahead momentarily to the Z = 10 nucleus, neon. As we add electrons to the nuclei with three through ten protons—the elements lithium through neon—the first two electrons always go into the 1s orbital, which gets smaller and smaller as Z increases. The next figure shows that the 1s² pair of Ne is smaller smaller than that of He due to the much greater attraction of its nucleus.

6.8

The next orbital that accepts electrons after the 1s is filled is also spherical and is called the 2s orbital. It is bigger because the filled 1s orbital forces the electrons in the 2s orbital to be farther away from the nucleus. Like all orbitals, the 2s orbital can hold two electrons. The next type of orbital that accepts electrons is no longer spherical. They have a lobe on either side of the nucleus. There are three of them, each 90° away from each other so they are aligned with one of the cartesian axes. They are called 2p orbitals. The three 2p orbitals are thus designated 2p_x, 2p_y, and 2p_z. The 2s and three 2p orbitals hold up to eight electrons and are sufficient to account for Li through Ne.

The next figure shows the doubly occupied 2s and two of the doubly occupied 2p orbitals of Ne, as well as all three 2p orbitals superimposed together as they are in the actual atom. The figure also shows the 2D and 3D representations we will use for these orbitals.

6.9

While s orbitals look a bit like onions in cross-section, p orbitals are more complex. Each of their lobes do have onion-like structure, but they are offset from the nucleus. Note that when all there 2p orbitals are superimposed, they become roughly spherical.

The 2D and 3D symbolic representations of p orbitals shown in the second and third rows of the figure above reflect the lobes, their displacement from the nucleus, and their alignment with the cartesian axes. In the 2D representation, the out-of-plane orbital is represented by the small circle. In case this isn't clear, here is a figure that shows both representations side-by-side with labels.

6.10

Here we've left all of the orbitals unoccupied, while for Ne in the previous figures we used two dots (2D model in row 2 above) or gray color (3D model in row 3 above) to indicate that all of the orbitals are doubly occupied.

All of the elements from Li through Ne use some or all of the 2s and 2p orbitals that Ne uses. Except for Ne, the orbitals will not be fully occupied. We will now investigate the other second row elements to see what we can learn about how many electrons each nuclei can hold onto and the order in which the orbitals are filled.

LITHIUM and BERYILLIUM. The next two elements have nuclei with three and four protons, respectively. In both cases, the first two electrons go into the 1s orbital as always and yield 1s² pairs. The third electron must go somewhere else. Let's look at the orbitals for Li and Be⁺, with He included for comparison: First, we see that the doubly occupied 1s orbitals become increasingly smaller as the number of protons increases, just as we saw earlier when they were singly occupied in the hydrogen-like atoms Li⁺² and Be⁺³. The third electron added to the Li and Be nuclei goes into the first new orbital, 2s. This new orbital is clearly considerably larger than the 1s orbital, which occupies the region of space closest to the nucleus. The 2s orbital for Li is more spread out than the one for Be.

6.11

The Li and Be nuclei can each hold onto one more electron, to form Li^– and Be. Based upon what we saw when we added the second electrons to H and He⁺, what do you think is the relationship between the spins of the electrons in the 2s² pairs of Li^– and Be? How large are they with respect to the singly occupied orbitals of Li and Be⁺? The configurations for Li and Be⁺ are 1s² 2s¹, while those for Li^– and Be are 1s² 2s².

Like He, we find that Be cannot hold onto a fifth electron to form an anion.

Beginning with Li, we distinguish between valence and core electrons. The 1s² pairs of Li and beyond do not participate in chemistry. They provide the foundation of what we call the electronic structure of the atoms but are too tightly bound to their nuclei to do anything else. This is why they are called core electrons. The remaining electrons are called valence electrons because they determine how many bonds each atom can form with other atoms. We'll look at bonding extensively in Chapter 2 of the module.

When we represent atoms with symbols, we only show the valence electrons. Neutral Li and Be have one and two valence electrons, respectively, and we will represent them (for now) with these diagrams:

6.12

The valence configurations of Li and Be are 2s¹ and 2s², which we have indicated with one and two dots in the appropriate circle of the 2D representation and with yellow and gray in the 3D representation. The empty 2p orbitals are shown.

BORON, CARBON, NITROGEN, OXYGEN, and FLUORINE. The B, C, N, O, and F nuclei consist of five to nine protons and can capture and hold onto more electrons than the nuclei with fewer protons. The first four electrons fill the 1s and 2s orbitals in all five elements. Once we add a fifth electron to the boron nucleus or to one of the heavier nuclei, it uses one of the available 2p orbitals. The singly occupied 2p orbitals of B and C⁺ are shown below, along with their 2s orbitals. The 2p orbital of C⁺ is smaller than the 2p¹ orbital of B (Z = 5) due to the greater nuclear charge of C (Z = 6).

6.13

Let's add a sixth electron to B or C nucleus to form B^– and neutral C. The new electron could be added to more than one place in the positions available in the 2p orbitals. Several combinations are possible and allowable for the orbital assigments and spins of the two electrons in the 2p orbitals.

Only one of these possibilities is not allowed, (b). It has two electrons with the same spin in the same orbital, which violates the Pauli principle. While (a) and (d) are possible, the most stable occupation occurs when the two electrons are in different 2p orbitals but have the same spin, case (c) in the list. The orbitals for this are shown above in Figure 6-13, second row. The configuration for C atom in its ground state is 1s² 2s² 2p_x¹ 2p_y¹. Carbon had four valence electrons.

The preference for two electrons to be in different orbitals (of the same type) rather than in the same one is clear enough that it merits a box:

The boron nucleus can hold on to up to six electrons (forming B^–) the same number as in the ground state of the C atom. The carbon and nitrogen atoms can each hold onto seven electrons, which means that the C^– anion and the N atom each have three 2p electrons. Based on the rules given in the box, see if you can choose the right answer to the following question:

The correct answer is (c) because rule (i) tells us it is better to put each of the three electrons in a different 2p orbital and because rule (ii) tells us it is better for them to all have the same spin. Response (a) violates rule (i), while response (d) violates rule (i). Response (b) violates rule (i), but more importantly, it violates the Pauli principle.

For O, F, and Ne, at least one 2p orbital must be doubly occupied. In O, there are two ways we could occupy the 2p orbitals, but the rules from above tell us that the most stable configuration is the one with one doubly occupied and two singly occupied 2p orbitals. In F and Ne, there is only way in each case to occupy the 2p orbitals. The F atom has two doubly occupied and one singly occupied 2p orbital. In Ne, all of the 2s and 2p orbitals are fully occupied. The figure below summarizes the most stable occupations for B through Ne.

6.14

It's straightforward to convert each of these configurations into 2D and 3D orbital diagrams:

6.15

We now have ten atoms in our abbreviated periodic table:

6.16

The elements with blue borders only have s electrons in their outer shells, while the elements with red borders have both s and p electrons in their outer shells.

SODIUM through ARGON. The next eight elements are very straightforward extensions of the second row of the table above. As before, we will start with the largest element in the row, argon. It has a total of 18 electrons. The first two electrons go in the 1s orbital, like He. The next eight electrons go in the 2s and 2p orbitals, like Ne. The final eight electrons go in the next two orbitals, the 3s and 3p, which look very much like their second row counterparts. The figure below shows the occupied orbitals for the He, Ne, and Ar atoms at the same scale. We see the same trends in the orbitals above that we've seen elsewhere. As Z increases, the size of the 1s orbital decreases. The same is true of the 2s and 2p orbitals. Finally the 3s and 3p orbitals are larger yet, although the 3s orbital of Ar is not really very much larger than the 1s orbital of He.

6.17

The atomic configuration for Ar atom in its ground state is 1s² 2s² 2p_x² 2p_y² 2p_z² 3s² 3p_x² 3p_y² 3p_z². The core electrons are the ten in the 1s, 2s, and 2p orbitals, and the valence electrons are the eight in the 3s and 3p orbitals.

Since we are filling s and p orbitals as we did for Li through Ne, the other atoms in the third row, Na through Cl, behave very similarly to their second row counterparts. The ground states have analogous configurations, with a Ne-like 1s² 2s² 2p⁶ core and valence 3s and 3p orbitals rather than valence 2s and 2p orbitals. Here is the periodic table for the first 18 elements:

6.18

Beyond the Third Row: ZINC, a sample transition metal. Building atoms of the fourth row elements, potassium (K) through krypton (Kr), is somewhat less straightforward. After filling the 1s² 2s² 2p⁶ 3s² 3p⁶ Ar-like core, there are 4s and 4p orbitals. There is also a new type of orbital we have not yet encountered, the 3d orbitals. In the case of the transition metals scandium (Sc) though zinc (Zn), the electrons are found to fill the 4s and 3d orbitals in an unpredictable ordering. In the Zn atom, however, the 4s and 3d orbitals are entirely filled in the ground state. Below we show cross-sections of the five 3d orbitals as well as an isodensity surface plot of all five orbitals superimposed together.

6.19

Except for the first orbital, all of the 3d orbitals have four lobes (where p orbitals have two lobes and s orbitals have no lobes). The combined plot on the right shows that one again the five orbitals together are roughly spherical.