Successive Ionisation Enthalpies

Introduction

First ionisation enthalpy removes one electron from a neutral atom. A second removes one from the resulting positive ion, and so on. The sequence generally increases. A particularly large jump can reveal when removal has moved from outer electrons to a much more tightly bound inner shell.

Core explanation

Write the first process as X(g) → X⁺(g) + e⁻, the second as X⁺(g) → X²⁺(g) + e⁻, and the third as X²⁺(g) → X³⁺(g) + e⁻. Each value is an energy change per mole for its specified gaseous starting species. The second value is not obtained by removing two electrons at once from neutral X; it is the next step after the first ion has formed.

Successive ionisation enthalpies usually rise because the same nucleus attracts fewer remaining electrons, and a positive ion holds its electrons more tightly. A dramatic increase appears when all relatively accessible valence electrons have been removed and the next electron must come from an inner shell. Inner electrons are often closer to the nucleus and less shielded, so removing one costs much more energy.

Magnesium illustrates the pattern. Neutral Mg has [Ne]3s². The first and second ionisations remove its two 3s electrons. Mg²⁺ then has the neon-like core; the third ionisation must remove an electron from that inner n = 2 structure. The jump between IE₂ and IE₃ is therefore large. In an introductory main-group interpretation, the position of the major jump suggests two outer electrons and supports magnesium's group-2 assignment.

For sodium, [Ne]3s¹, the first removal reaches the neon-like core, so the large jump occurs between IE₁ and IE₂. For aluminium, [Ne]3s²3p¹, removal of three outer electrons precedes a large core-removal jump between IE₃ and IE₄. These examples show the pattern, but one should inspect actual data and configurations rather than mechanically assume every high oxidation state is stable in compounds.

The numerical ratios can be striking, but there is no universal threshold such as “ten times larger” that identifies a shell boundary for every element. Plot the values, locate a jump much larger than neighboring increments, and relate it to the electron configuration. If several subshells within the same principal level have different binding energies, smaller jumps can occur before the major core jump.

Successive ionisation data help infer valence structure, not the full thermodynamics of compounds. Forming Mg³⁺ as an isolated gas ion is energetically costly; a particular chemical environment might alter stabilization, but Mg²⁺ is the ordinary ion in simple salts. The data support this common behavior without proving every possible compound formula from the jump alone.

Step-by-step reasoning

1. Label IE₁, IE₂ and later values with their correct gaseous starting ions. 2. Arrange the values in order and identify the largest relative increase. 3. Count how many electrons were removed before that jump. 4. Check a ground-state configuration to see whether the next electron is core. 5. Infer a likely main-group valence count, stating limits for complex elements.

Visual explanation

Draw a bar chart of IE₁ through IE₄ for magnesium. Show two moderate-height bars, then a much taller IE₃ bar. Above the first two bars draw two 3s electron dots leaving; above the tall bar draw a dot leaving the inner neon-like core.

Real-world analogy

Removing objects from an outer coat pocket is easier than opening a locked inner compartment. Once the accessible pocket is empty, the next removal costs much more effort. The analogy represents a shell boundary, though electron energies are continuous quantum differences rather than literal locks.

Real-world example

Suppose an unknown main-group element has successive ionisation values with a pronounced jump after the third. A chemist may infer three accessible outer electrons and look for group-13 behavior, then test compound formulas and other properties to confirm the assignment.

Why?

Why does each removal usually become harder even before a shell boundary? The ion becomes more positively charged while the nucleus retains the same proton count. Remaining electrons generally experience stronger net attraction and less electron–electron repulsion.

Common misconception

“The third ionisation enthalpy means removing three electrons from one neutral atom in a single step.” It specifically describes removing one electron from a gaseous 2+ ion. The total energy to reach 3+ from neutral includes IE₁ + IE₂ + IE₃.

Worked example

An unknown element has IE values of 590, 1140, 4900 and 6500 kJ mol⁻¹, rounded for this exercise. The conspicuous jump is after the second value, from about 1140 to 4900. Infer that two electrons are relatively outer and the third begins core removal. A group-2 main-group atom is a plausible interpretation; the data alone do not identify the element uniquely without atomic number or more measurements.

Quick check

1. Which step removes an electron from Mg²⁺(g)? Answer: The third ionisation, Mg²⁺(g) → Mg³⁺(g) + e⁻.

Exam focus

Write successive equations accurately and locate the jump after the valence-electron count. A large jump after IE₂ suggests two relatively accessible outer electrons, not a common +3 ion.

Advanced insight

The total energy to form Xⁿ⁺(g) from X(g) is the sum of the first n ionisation enthalpies. In a real ionic solid this large cost can be offset partly by lattice formation, but the balance depends strongly on charge, ion size and other steps. The jump is structural evidence, not a complete prediction of thermodynamic stability.

Summary

Successive ionisation enthalpies remove electrons one by one from increasingly positive gaseous ions. They usually rise, and a major jump marks a move to a more tightly bound inner shell. For simple main-group atoms, the jump position helps infer outer-electron count.

Practice questions

1. A series has a large jump between IE₁ and IE₂. How many easily removed outer electrons are suggested? Answer: One; the second removal likely enters an inner core. 2. Write the second ionisation process for aluminium. Answer: Al⁺(g) → Al²⁺(g) + e⁻. 3. If IE₁ = 500 and IE₂ = 1000 kJ mol⁻¹, what energy per mole is needed for X(g) → X²⁺(g) + 2e⁻? Answer: 1500 kJ mol⁻¹, the sum of the first two successive ionisation enthalpies.