Chemistry (9701) | Ionisation energy

Welcome to the World of Ionisation Energy!

Hi there! This chapter is really important because it helps us understand why atoms behave the way they do—why some lose electrons easily to form positive ions, and others hold onto them tightly. Essentially, mastering ionisation energy is key to unlocking the mysteries of the Periodic Table trends and bonding.
Don't worry if the name sounds intimidating; it's just about the energy needed to kick an electron out of an atom. Let's break it down!

1. Defining Ionisation Energy (IE)

What is First Ionisation Energy (IE) ?

The First Ionisation Energy (\(IE_1\)) is the minimum energy required to remove one mole of electrons from one mole of atoms in the gaseous state to form one mole of unipositive ions, also in the gaseous state.

It is crucial that the atoms are in the gaseous state because this ensures they are isolated and no intermolecular forces are involved.

The Equation:

For any element \(X\):

\(X\text{(g)} \longrightarrow X^+\text{(g)} + e^-\)

Since energy must be supplied to remove an electron (it requires breaking the electrostatic attraction to the nucleus), ionisation is always an endothermic process. Therefore, the value for \(\Delta H\) (the IE) is always positive.

Subsequent (Successive) Ionisation Energies

What happens if you remove a second, third, or fourth electron? These are the Successive Ionisation Energies (\(IE_2\), \(IE_3\), etc.).

Second Ionisation Energy (\(IE_2\)):

The energy required to remove the second electron from a unipositive gaseous ion:

\(X^+\text{(g)} \longrightarrow X^{2+}\text{(g)} + e^-\)

Third Ionisation Energy (\(IE_3\)):

The energy required to remove the third electron from a dipositive gaseous ion:

\(X^{2+}\text{(g)} \longrightarrow X^{3+}\text{(g)} + e^-\)

Key Takeaway: Successive ionisation energies always increase. Why?

When you remove the first electron, you form a positive ion (\(X^+\)). When you try to remove the second electron, you are pulling it away from an already positive species. The remaining electrons are held more tightly by the same number of protons. More energy is required to overcome this increased attraction.

2. Factors Influencing Ionisation Energy

The magnitude of the ionisation energy is determined by how strongly the nucleus attracts the electron being removed. This depends on three main factors (and one important subtlety):

Factor 1: Nuclear Charge (Number of Protons)

The Rule: A higher nuclear charge (more protons) means a stronger pull on the electrons.
The Effect: Higher Nuclear Charge \(\implies\) Higher IE.

Simple explanation: More protons in the nucleus create a greater positive charge, increasing the electrostatic attraction exerted on all electrons, making them harder to remove.

Factor 2: Atomic/Ionic Radius (Distance)

The Rule: The further the outer electron is from the nucleus, the weaker the attraction (Coulomb's Law).
The Effect: Larger Radius \(\implies\) Lower IE.

Simple explanation: Electrons in shells further away are easier targets for removal because the nucleus's pull drops off significantly with distance.

Factor 3: Shielding by Inner Electrons

The Rule: Inner shell electrons repel the outer electrons, reducing the net positive attraction felt by the valence electron. This is called the shielding effect or screening effect.
The Effect: More Inner Shells (More Shielding) \(\implies\) Lower IE.

Simple explanation: Think of the inner electrons acting like a protective chemical 'sunscreen' or 'shield' for the outermost electron, blocking the full attractive force of the nucleus.

Factor 4: Sub-shell Type and Spin-Pair Repulsion (The Nuances)

Electrons in different sub-shells (s, p, d, f) have slightly different energies and shapes:

• Penetration: S-orbital electrons penetrate closer to the nucleus than p-orbital electrons in the same shell. This means s-electrons experience less shielding and are generally harder to remove than p-electrons.
• Spin-Pair Repulsion: When an orbital (like a p-orbital) contains two electrons, the mutual repulsion between these two paired electrons makes them slightly easier to remove compared to an electron in a half-filled or singly-occupied orbital.

This subtle difference explains some of the exceptions to the general trends we are about to look at.

3. Periodic Trends in Ionisation Energy

Trend 1: Moving Down a Group (e.g., Group 1 or Group 17)

As you move down a group (e.g., Li to Na to K):

1. Nuclear Charge increases (more protons). This should increase IE.
2. Atomic Radius increases (new principal quantum shells, n).
3. Shielding Effect increases dramatically (more inner shells).

The increase in Shielding (Factor 3) and the increase in Atomic Radius (Factor 2) are far more dominant than the increase in nuclear charge.

Conclusion: First Ionisation Energy decreases down a group.

Encouragement: This is the easiest trend to explain! Just mention that the outermost electron is further away and heavily shielded by the increasing number of inner electron shells.

Trend 2: Moving Across a Period (e.g., Period 3: Na to Ar)

As you move across a period (from left to right):

1. Nuclear Charge increases steadily (one more proton each time).
2. Atomic Radius decreases (protons pull the electrons in closer).
3. Shielding Effect remains relatively constant (electrons are added to the same main outer shell).

Because the shielding is similar, the electron feels the full effect of the increasing nuclear charge (Factor 1), while simultaneously being held closer (Factor 2).

General Conclusion: First Ionisation Energy generally increases across a period.

The Exceptions to the Trend (The "Wobbles")

While the IE generally increases across a period, there are two important dips in the graph for elements in Period 2 and 3 that you must be able to explain using Factors 3 and 4 (sub-shell effects):

Exception 1: The Dip from Group 2 to Group 13 (e.g., Mg to Al)

Example: IE of Magnesium (Group 2, \(1s^2 2s^2 2p^6 3s^2\)) is higher than IE of Aluminium (Group 13, \(1s^2 2s^2 2p^6 3s^2 3p^1\)).

Explanation:

• Magnesium removes an electron from the 3s sub-shell.
• Aluminium removes an electron from the 3p sub-shell.
• The 3p orbital has a slightly higher energy and less penetrating power than the 3s orbital.
• The 3p electron is better shielded by the inner 3s electrons (as well as the full inner shells) compared to the 3s electron in Mg.

Therefore, less energy is required to remove the 3p electron from Al than the 3s electron from Mg, causing the IE to drop.

Exception 2: The Dip from Group 15 to Group 16 (e.g., P to S)

Example: IE of Phosphorus (Group 15, \([\text{Ne}] 3s^2 3p^3\)) is higher than IE of Sulfur (Group 16, \([\text{Ne}] 3s^2 3p^4\)).

Explanation:

• Phosphorus has a half-filled 3p sub-shell (3 electrons, one in each orbital: \(p_x, p_y, p_z\)). This arrangement is particularly stable.
• Sulfur has one paired orbital in the 3p sub-shell (4 electrons: \(p_x^2, p_y^1, p_z^1\)).
• The electron being removed from Sulfur comes from the paired orbital.
• This paired electron experiences spin-pair repulsion (Factor 4) from its partner electron, which overcomes the slightly increased nuclear charge of Sulfur.

Because of this repulsion, the paired electron in Sulfur requires less energy to remove than the singly-occupied electron in the stable half-filled sub-shell of Phosphorus.

4. Using Successive Ionisation Energies to Deduce Structure

Successive IE data provides direct evidence for the existence of electron shells (principal quantum numbers, n).

Identifying Electron Shells

When you plot the successive IE values for an element, you observe huge jumps in energy. These jumps occur when the electron being removed comes from an electron shell closer to the nucleus (i.e., a new, lower principal quantum number, n).

Think of it like peeling an onion: The outer layers come off easily, but once you hit a new layer, it's firmly attached, requiring a massive jump in energy to break through.

Deducing Group Number

The largest jump in energy occurs after all the valence (outermost shell) electrons have been removed. The number of electrons removed *before* this massive jump is equal to the element’s Group Number in the Periodic Table.

Step-by-Step Deduction Example

Consider an imaginary element \(Y\) with the following successive ionisation energies (in \(\text{kJ mol}^{-1}\)):

\(IE_1\): 580
\(IE_2\): 1815
\(IE_3\): 2740
\(IE_4\): 11580
\(IE_5\): 14820

Analysis:

1. The ratio \(\frac{IE_2}{IE_1}\) is about 3.1.
2. The ratio \(\frac{IE_3}{IE_2}\) is about 1.5.
3. The ratio \(\frac{IE_4}{IE_3}\) is about 4.2! This is the biggest jump.

The huge jump occurs between the removal of the 3rd and 4th electron. This tells us that the first three electrons came from the outermost shell, and the fourth electron was removed from the next inner shell.

Deduction: Since \(Y\) has 3 valence electrons, it belongs to Group 13 (or Group III in older notations).

Deducing Electronic Configuration

The data also allows us to determine the arrangement of electrons in shells and sub-shells. Following the example above, if the jump occurs after the 3rd electron:

• 3 electrons in the outer shell (Valence Shell).
• 8 electrons in the inner shell (Shell 2). (Since IE4 and IE5 are still relatively low compared to the 1s core).
• 2 electrons in the innermost shell (Shell 1).

The shell structure would be 2, 8, 3. The total number of electrons is 13, meaning the element is Aluminium (Al).

Key Takeaway from Ionisation Energy

Ionisation energy is the energy fingerprint of an element. It is governed by the delicate balance between the attractive force of the nucleus and the repulsive forces of the inner electrons (shielding). Understanding how these factors change across the Periodic Table allows us to predict and explain the reactivity and physical properties of elements.