Ionisation Energy Trends

Introduction to Ionisation Energy

Ionisation Energy: The energy required to remove an electron from a neutral gaseous atom. It is a fundamental concept in chemistry, essential for understanding the reactivity and stability of elements. Lower ionisation energies indicate higher reactivity, especially among metals.

Fundamental Concepts of Ionisation Energy

First Ionisation Energy

• First ionisation energy: The energy required to remove the first electron from a neutral atom.
• This metric reflects an element's reactivity and stability.
• Each successive electron removal necessitates more energy due to the increased positive charge on the cation.

Successive Ionisation Energies

• Ionisation energy increases with each electron removed, as a result of the heightened nuclear charge.
• Example: Sodium (Na):
  • First electron removal: Na to Na+
  • Additional electrons require increasing energy.
• Attaining a noble gas configuration enhances stability, complicating further ionisation.

Formula for Ionisation Energy

Ionisation energy can be determined using the following equation:

$IE = E_{\text{atom}} - (E_{\text{ion}} + E_{\text{electron}})$

• IE = Ionisation Energy
• It represents the difference between the initial energy and the energy post-ionisation.

Numeric Example

• Initial energy of atom ($E_{\text{atom}}$) = 500 kJ/mol
• Energy of resulting ion plus electron ($E_{\text{ion}} + E_{\text{electron}}$) = 200 kJ/mol

Calculation:

$IE = 500 - 200 = 300 \text{ kJ/mol}$

This shows the energy necessary to remove an electron.

Trends in Ionisation Energy

Across a Period

• Increases from left to right due to an increased nuclear charge, leading to a stronger attraction of electrons.
• Atomic Radius: Decreases across a period, resulting in stronger electron-nucleus interactions.
• Notable Exceptions:
  • Be vs. B: The filled s-subshell in Be requires more energy than in B.
  • N vs. O: Electron pairing in O decreases the energy compared to N.

Down a Group

• Decreases down a group because:
  • Increasing atomic size places electrons further from the nucleus.
  • Enhanced electron shielding: Additional inner electrons reduce the nuclear attraction to outer electrons.

Exceptional Cases

Beryllium vs. Boron

• Despite being to the right of beryllium, boron demonstrates lower ionisation energy.
• The extra electron enters a higher-energy p-subshell.

Nitrogen vs. Oxygen

• Oxygen exhibits lower ionisation energy than nitrogen due to electron pairing, which increases repulsion.

Factors Influencing Ionisation Energy

Key Influencing Factors

1. Atomic Size:

   • A larger size corresponds with lower ionisation energy.
   • Electrons that are more distant from the nucleus are more easily removed.

2. Nuclear Charge:

   • A higher nuclear charge increases the energy required to remove an electron.

3. Electron Shielding:

   • Inner electrons diminish the nuclear pull on outer electrons.
   • Greater shielding results in lower required energy.

Ionisation Energy in Practical Applications

• Industrial Relevance: Metals with low ionisation energy, such as sodium and potassium, are highly reactive.
• Comparison Table:

Metal	First Ionisation Energy (kJ/mol)	Reactivity Level
Lithium	520.2	Low
Sodium	495.8	Medium
Potassium	418.8	High

Calculation Methods

1. Spectroscopy:

   • Emission lines reflect energy levels.

2. Photoelectron Spectroscopy (PES):

   • It measures the kinetic energy of electrons after ejection, providing precise ionisation energy assessment.

3. Einstein's Equation:

   • Calculate ionisation energy using $KE_{\text{max}} = hf - IE$

Key Study Points

• Trends: Ionisation energy increases across a period and decreases down a group.
• Essential Concepts:
  • Atomic Radius affects electron removal ease.
  • Nuclear Charge & Shielding are influential factors.
• Exceptions: Observe energy variations due to electron shell configurations.

Practice Problems

Problem 1: Period Trend

• Arrange elements by increasing first ionisation energy within Period 2.
  • Solution: Li < Be < B < C < N < O < F < Ne (Elements are arranged from lowest to highest ionisation energy, following the general trend across Period 2)

Problem 2: Group Trend

• Compare ionisation energies in Group 1: Li > Na > K.
  • Solution: This ordering is correct. Ionisation energy decreases down Group 1, with Li having the highest and K the lowest due to increasing atomic radius and electron shielding.

Calculation Exercise

• Given frequency $f = 6 \times 10^{14}$ Hz, compute ionisation energy using Planck's constant.
  • Solution: Using $KE_{\text{max}} = hf - IE$, where $h = 6.626 \times 10^{-34}$ J·s
  • $IE = hf - KE_{\text{max}}$
  • If $KE_{\text{max}} = 0$ (threshold frequency case), then $IE = hf = 6.626 \times 10^{-34} \times 6 \times 10^{14} = 3.98 \times 10^{-19}$ J per atom
  • Converting to kJ/mol: $3.98 \times 10^{-19} \text{ J/atom} \times 6.022 \times 10^{23} \text{ atoms/mol} \times 10^{-3} \text{ kJ/J} \approx 239.7 \text{ kJ/mol}$