The following notes were written for the previous IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

IB syllabus for first examinations 2016

Atomic theory - Electronic configuration


12.1.1: Explain how evidence from first ionization energies accounts for the existence of the main energy levels and sub-levels in atoms.

The ionisation energy is defined as 'the energy required to remove 1 mole of electrons from 1 mole of gaseous atoms forming 1 mole of singly charged ions'.

Example:

Mg(g) Mg+(g) + 1e

The numerical value of the ionisation energy can be used as a measure of how tightly the outer electron is held to the nucleus of the atom. Large value = strong force, small value = small force

This force of attraction is electrostatic in nature and depends on three factors:

  1. The size of the positive nucleus, i.e. how many protons it contains
  2. The distance between the electron and the nucleus
  3. The repulsion of the other electrons

The larger the positive charge the greater the energy. The energy is directly porportional to the nuclear charge. The greater the distance the lower the energy. The energy is inversely proportional to the distance. The interelectron repulsion decreases the energy needed but only by small amounts. It could be considered an attenuating factor (i.e. it is only responsible for 'fine-tuning' the energy value)

In summary:

Energy = constant x nuclear charge/distance

Hence, if we know the charge on the nucleus we can use the First Ionization Energies to come to conclusions regarding the distance of the electron from the nucleus and possible interelectronic repulsion effects.

Hydrogen can be used as our baseline, as it has only one proton in the nucleus and one electron. It's 1st ionisation energy is 1312 kJ mol-1. Helium has two protons in the nucleus and so according to our equation:

Energy = constant x nuclear charge/distance

If the outer electron is the same distance from the nucleus as in hydrogen then we would expect the 1st ionisation energy of helium to be twice as large as that of hydrogen. In fact it is 2372 kJ mol-1. It is about 250 kJ lower than expected. We can put this decrease down to the repulsion of the second electron.

We conclude that the electrons in hydrogen and helium are at similar distances from the nucleus. We say that they are located in an orbital.

Moving on to lithium, there are now 3 protons in the nucleus, so following the same logic as above the 1st ionisation energy should be much larger than that of hydrogen, possibly a factor of three times with perhaps a slight lowering due to inter-electron repulsions. However, the experimental value is only 520 kJ mol-1.

There can only be one explanation for this low value; that the outer electron is much further from the nucleus.

Beryllium is the next atom. It has four protons in the nucleus and we would expect the energy to be 4/3 times greater than that of lithium. In fact it is 899 kJ mol-1. This is a little larger than expected, but not by very much. This means that the outer electron is in a similar region to that of lithium, but that the extra attraction of the nucleus has shrunk this region relative to lithium a little.

Boron has 5 protons in the nucleus so we expect a value of 5/4 times that of beryllium (with perhaps a slightly higher value due to 'shrinkage'. However, the 1st ionisation energy of boron is only 800 kJ mol-1. The only possible explanation for this is that the electron is further from the nucleus than expected. It is in a second orbital, which is more diffuse (spread out further) than the electrons in Li and Be. We call this second type of orbital a 'p' orbital.

Carbon holds no surprises and the expected value (6/5 times that of boron) is 1086 kJ mol-1. The slightly large value can be ascribed to 'shrinkage' once again.

Nitrogen also follows the pattern as it first ionisation energy is 1420 kJ mol-1. This is a little larger than 7/6 times the value for carbon.

Oxygen, however, has an unexpectedly low first ionisation energy of 1314 kJ mol-1. This low value is caused by electrons pairing up in the relatively small 'p' orbitals for the first time. Seeing as this has happened for the first time, we can conclude that there are three 'p' orbitals in the energy level being filled.

Fluorine follows the pattern with an increase of 9/8 times that of oxygen, with a small increase due to 'shrinkage' - 1st ionisation energy 1681 kJ mol-1

Neon also follows the pattern - 1st ionisation energy 2080 kJ mol-1

Sodium, however breaks the pattern completely. It low value of 496 kJ mol-1 is lower than even lithium. The only explanation is that the outer electron is much, much further freom the nucleus. It seem that a new energy shell altogether has been started.

The remaining elements of the third period then follow the pattern of the second period.


12.1.2: Explain how successive ionization energy data is related to the electronic configuration

More information on successive ionisation energies  

12.1.3: State the relative energies of s, p, d and f orbitals.

Energies of sub-shells : s < p < d < f (in a given energy level)

More information about sub-shells  

12.1.4: State the maximum number of orbitals at each energy level.

Number of orbitals at each level : s=1, p=3, d=5, f=7

More information about the energies of orbitals  

12.1.5: Draw the shape of an s orbital and the shapes of the px, py and pz orbitals.

Shapes of orbitals:

  • s orbital is a sphere around the nucleus.
  • p orbitals are shaped like a figure 8 (and there are 3 of them at 90 degrees around the nucleus.
More information about the shapes of orbitals  

12.1.6: Apply the Aufbau principle, Hund's rule and the Pauli exclusion principle to write electronic configurations up to z=54.

The Aufbau Principle

Move diagonally down and left through each diagonal...ie 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d...

1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f

Pauli's exclusion principle says that there can only be 2 electrons in each orbital (with opposite spins).

Hund's rule says that each orbital should be half filled before any is completely filled (since there is less repulsion if all electrons have the same spin). Electrons will therefore fill the lowest energy levels (ie 1 then 2 and so on) with two going in each orbital, but only doubling up when all orbitals in the level are filled.

Systematically fill the orbitals as shown above up to Z = 56.

This can be abbreviated by writing [x] where x is a noble gas.

The periodic table

The small double column on the left (groups 1 and 2) is the s shell being filled.

The block on the right is the p shell being filled (groups 3 to 8)

The d block (in the middle) is the d shell being filled.

More information about the Aufbau principle  

Resources

atomic theory

periodic table for interactive configurations.

mass spectra of common elements




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