An ion is an atom that has gained or lost electrons so that it is now charged – losing electrons forms cations which are positively charged and gaining electrons forms anions which are negatively charged.
Before we go any further you should complete the true / false quiz from the Royal Society of Chemistry so that you can challenge any misconceptions you might have around atoms and ions – the first couple of pages are the answers so scroll past to page 3 – no cheating! If you have any misconceptions, spend some time really thinking through the answers because it is absolutely vital that you begin your A level on a solid foundation.
Removing an electron from an atom is an endothermic process (we need to put in energy to overcome the attraction of that electron to the positively charged protons in the nucleus). We can define the first ionisation enthalpy for an atom as
the energy needed to remove one electron from every atom in a mole of gaseous atoms to form a mole of gaseous ions with a single positive charge
Li(g) ⇾ Li+(g) + e–
Trends in ionisation enthalpies are evidence of the electronic structure of atoms.
- As we move down a group in the Periodic Table, 1st ionisation enthalpies (IE) decrease – the outermost electron being removed is further from the nucleus and increasingly shielded from the positive nuclear charge by inner shell electrons. This means less energy is needed for ionisation.
| Element | 1st IE / kJ mol-1 |
| Li | 520 |
| Na | 496 |
| K | 419 |
- As we move along a period, 1st IE generally increases because electrons are being added to the same shell as more protons are added to the nucleus. The effective nuclear charge felt by the valence (outer shell) electrons is stronger, the electrons are held more tightly and so it takes more energy to remove a valence electron.
| Element | 1st IE / kJ mol-1 |
| Li | 520 |
| Be | 900 |
| B | 801 |
| C | 1086 |
| N | 1402 |
| O | 1314 |
| F | 1691 |
But can we explain the blips? The answer lies in looking more closely at how electrons are arranged in the orbitals.
- the 2p valence electron lost in boron is in a slightly higher energy orbital so it takes less energy than expected to remove, despite the increase in effective nuclear charge, hence the slightly lower 1st IE.
- electron-electron repulsion is higher between spin paired than parallel electrons, so the reduction in this repulsion on losing a valence electron is greater for oxygen than for nitrogen (again, despite the increase in effective nuclear charge in oxygen compared with nitrogen). The result is a slightly lower than expected value for the 1st IE of oxygen.
- Ionisation enthalpies increase each time we remove another electron from an atom.
| Mg(g) ⇾ Mg+(g) + e– | 736 kJ mol-1 |
| Mg+(g) ⇾ Mg2+(g) + e– | 1450 kJ mol-1 |
| Mg2+(g) ⇾ Mg3+(g) + e– | 7740 kJ mol-1 |
| Mg3+(g) ⇾ Mg4+(g) + e– | 10500 kJ mol-1 |
- the first two electrons removed from a magnesium atom are from the n=3 energy level (3s2) and then the third electron is removed from the n=2 energy level (2p6), which is closer to the nucleus. This electron experiences a far greater effective nuclear charge because the proton : electron ratio is greater and there is less shielding of this attraction from inner shell electrons.
Practice questions
- Elements in the Periodic Table have been arranged so that predictions can be made about their behaviour. Many properties, such as first ionisation energy, show a gradual change across the period.
(a) Write an equation, with state symbols, to represent the first ionisation energy of magnesium
(b) Explain the general increase in first ionisation energies across Period 3 (sodium to argon).
2. Referring to the graph below, explain each of these changes in first ionisation energy:
(a) Between hydrogen and helium
(b) Between helium and lithium
(c) Between beryllium and boron
(d) Between nitrogen and oxygen
(e) Along the peaks for the noble gases
(f) Along the lowest points for the alkali metals
Answers
1 (a) Mg(g) ⇾ Mg+(g) + e– (2)
(b) as we move across Period 3 electrons are being removed from the same shell (1) so there is the same amount of shielding from electrons in the first and second shells (1) but the number of protons / nuclear charge is increasing which means the outermost electrons are more strongly attracted, hence takes more energy to remove one (1)
2 (a) He has one more proton than H, increased nuclear attraction to electrons (1), which are in the same shell, so more difficult to remove = He has a higher 1st IE (1)
(b) the outer electron in Li is in the 2s orbital so despite increased nuclear charge compared with He, this electron is actually easier to remove … shielding of nuclear attraction by 1s electrons (1) and the 2s electron is further from the the nucleus (1)
(c) the outer electron in B is in a 2p orbital so is more effectively shielded from increased nuclear charge by inner electrons (1) and the 2p orbital is at a slightly higher energy than the 2s (where the outer electron in Be is located) and so is slightly further from the nucleus (1)
(d) the three 2p electrons in N are unpaired, each occupying its own sub-orbital, which limits mutual repulsion between them (1) but in O two of the 2p electrons are spin paired = more repulsion = less energy needed to remove one (1)
(e) each noble gas has the highest 1st IE in its period and 1st IE decreases down group because as we move down the group the outer shell electrons are further from nucleus (1) and more effectively shielded from increasing nuclear charge (1), so takes less energy to remove an electron (1)
(f) each alkali metal has the lowest 1st IE in its group and 1st IE decreases down group because as we move down the group the outer shell electrons are further from nucleus (1) and more effectively shielded from increasing nuclear charge (1), so takes less energy to remove an electron (1)