Trend of 1st Ionisation Energy Across Period 2:
- Comparing the 1st Ionisation Energies of He (1s²) and Li (1s² 2s¹):
- Li has an extra proton in its nucleus (3) but two inner-shell electrons.
- These inner-shell electrons cancel out the charge of two of the protons, reducing the effective nuclear charge on the 2s electron to +1.
- This is lower than the effective nuclear charge on the He 1s electrons, +2 and so the electrons are less strongly held to and easier to remove.
- The first ionisation enery of Li is thus lower than that of He.
- Comparing the first ionisation energies of Li (1s² 2s¹) and Be (1s² 2s²):
- Be has one more proton in the nucleus than Li, and no extra inner-shell electrons, so the effective nuclear charge on Be is higher and the Be electrons are more strongly attarcted to the nucleus.
- The first ionisation energy of Be is thus higher than that of Li.
- In general, the first ionisation energy increases across a period because the nuclear charge increases but the shielding stays the same.
- Comparing the first ionisation energies of Be (1s² 2s²) and B (1s² 2s² 2p¹):
- B has one more proton in the nucleus than Be but there are also 2 extra inner sub-shell electrons.
- These cancel out the charge of two moer of the protons, leaving an effective nuclear charge of only +1.
- This is less than Be (+2) so the electrons are less strongly attracted to the nucleus and thus less difficult to remove.
- The first ionisation energy of B is thus lower than that of Be.
- Ionisation energies decrease from group 2 to group 3 because in group 3 the electrons are removed from a p-orbital, so it is shielded by the s-electrons in the outer shell. Thus, the effective nuclear charge decreases.
- Comparing the first ionisation energies from B (1s² 2s² 2p¹) to N (1s² 2s² 2p²):
- The proton number increases, but the number of electrons shielding the nuclear charge remains the same at 4.
- Thus, the effective nuclear charge increases from B to N and the electrons become progressively harder to remove.
- The first ionisation energy thus increases from B to N.
- So far the concepts of effective nuclear charge and shielding have been used to explain the trend in first ionisation energies for the first 7 elements.
- However, the fall between N (1s² 2s² 2p³) and O (1s² 2s² 2p^4) cannot be explained due to these concepts as the fall between N and O is caused by something else entirely.
- Comparing the first ionisation energies of N (1s² 2s² 2p³) and O (1s² 2s² 2p^4):
- When studying the electronic configuration of N and O carefully, you should notice that in N the electron is removed from an unpaired orbital, but in O it is removed from a paired orbital.
- In a paired orbital, the two electrons share a confined space and so repel each other.
- They are therefore less stable and easier to remove.
- This repulsion effect outweighs the higher effective nuclear charge in O.
- The first ionisation energy of O is thus lower than that of N.
- First ionisation energies decrease from group 5 to group 6 since the electron removed from the group 6 atom is paired, so there is more repulsion between the electrons and the electron is easier to remove.
- Comparing the first ionisation energies from O to Ne:
- The first ionisation energies increase as expected from O to Ne, due to the increase in effective nuclear charge.
Summary of the Trend of 1st Ionisation Energy Across Period 2:
- There is a general increase across the period as the nuclear charge increases and the shielding remains the same.
- There is a drop from Be to B becasue in B a 2p electron is being removed and the extra shielding from the 2p subshell actually causes a fall in the effective nuclear charge.
- There is also a drop from N to O because the electron in O is being removed from a paired orbital. The repulsion of the electrons in this orbital makes them less stable and easier to remove.
- The same trend can be found in Period 3 (Na - Ar). There is a general increase, but a drop between Mg and Al and also between P and S.