Nuclear Charge: As the number of protons in the nucleus increases, the positive charge increases. This results in a stronger electrostatic attraction for the outer electrons, which generally increases the ionisation energy required to remove them.
Atomic Radius: The distance between the nucleus and the outer electrons significantly impacts the force of attraction. According to Coulomb's Law, as the distance increases, the electrostatic attraction decreases rapidly, making it easier to remove an electron from an atom with a larger radius.
Shielding (Screening): Inner shell electrons repel the outer electrons and 'shield' them from the full attractive force of the nucleus. An increase in the number of inner shells increases the shielding effect, thereby reducing the effective nuclear charge felt by the valence electrons and lowering the ionisation energy.
Generally, the first ionisation energy increases across a period from left to right. This occurs because the nuclear charge increases (more protons) while the shielding remains relatively constant as electrons are added to the same principal energy level.
As the effective nuclear charge increases, the atomic radius decreases because the outer electrons are pulled closer to the nucleus. The combination of a higher nuclear charge and a smaller radius results in a much stronger attraction for the valence electrons, requiring more energy for their removal.
The first ionisation energy decreases as you move down a group in the periodic table. Although the nuclear charge increases significantly, its effect is outweighed by the addition of new principal energy levels (shells).
Each new shell increases the atomic radius, placing the outermost electrons further from the nucleus. Additionally, the increased number of inner shells provides more shielding, which reduces the net attractive force on the valence electrons, making them easier to remove.
| Feature | Group 2 to Group 13 Dip | Group 15 to Group 16 Dip |
|---|---|---|
| Reason | Subshell change ( to ) | Electron-pair repulsion |
| Mechanism | The -electron is in a higher energy subshell and is slightly shielded by the electrons. | Electrons in the same -orbital repel each other, making one easier to remove. |
| Example | Beryllium vs. Boron | Nitrogen vs. Oxygen |
The dip between Group 2 and Group 13 occurs because the electron being removed from Group 13 is in a -orbital, which is at a higher energy level and further from the nucleus than the -orbital electron in Group 2. This extra distance and the shielding provided by the electrons make the -electron easier to remove.
The dip between Group 15 and Group 16 occurs due to spin-pair repulsion. In Group 15, the -subshell is half-filled with one electron in each orbital; in Group 16, the fourth electron must pair up in an orbital already containing one electron. The repulsion between these two negatively charged electrons in the same orbital makes it easier to remove one of them.
Identify the Jump: When presented with a list of successive ionisation energies, look for the largest ratio increase between values. A sudden, massive jump (e.g., from 2nd to 3rd IE) indicates that the electron is being removed from a new, inner shell closer to the nucleus, which identifies the element's group.
State Symbols Matter: Always ensure equations for ionisation energy include for all species. Examiners frequently penalize the omission of state symbols as the definition is strictly based on the gaseous phase.
Three-Point Explanation: When explaining trends, always mention all three factors: Nuclear Charge, Atomic Radius, and Shielding. Even if one factor is constant (like shielding across a period), stating that it remains constant shows a complete understanding of the mechanism.
Check the Subshell: If a question asks why an element's IE is lower than expected based on the general trend, check if it involves a move from an to a subshell or the start of electron pairing in a -orbital.