Ionization Energy Trend Across a Period — Why It Increases

Question

Explain why first ionization energy generally increases as we move from left to right across a period. Also explain the anomalies observed in Period 2 — specifically why IE₁(Be) > IE₁(B) and IE₁(N) > IE₁(O).

Solution — Step by Step

Step 1: Define Ionization Energy

First ionization energy (IE₁) is the minimum energy required to remove the outermost electron from a neutral gaseous atom:

M(g) → M⁺(g) + e⁻ ; ΔH = IE₁

Higher IE₁ = the electron is held more tightly = harder to remove.

Step 2: Identify the Factors That Control IE

Three factors determine how strongly an electron is held:

1. Nuclear charge (Z): More protons pull outer electrons more strongly → higher IE
2. Atomic radius: Smaller atom → outer electron is closer to nucleus → stronger pull → higher IE
3. Shielding effect: Inner electrons partially offset the nuclear pull. More shielding → lower effective nuclear charge → lower IE

Step 3: Apply Across a Period

Moving left to right across Period 2 (Li → Ne):

• Nuclear charge increases: Li (Z=3) → Be (4) → B (5) → C (6) → N (7) → O (8) → F (9) → Ne (10)
• All outer electrons are added to the same shell (n=2) → shielding from inner (1s²) electrons stays roughly constant at σ ≈ 2
• Atomic radius decreases due to increasing nuclear pull
• Net effect: Effective nuclear charge (Z_eff = Z − σ) increases steadily

Since Z_eff increases, the outer electrons are pulled closer and held more tightly → IE₁ generally increases across the period.

Step 4: Understand the Anomalies

Anomaly 1 — IE₁(Be) > IE₁(B):

Configuration of Be: [He] 2s² — fully filled 2s subshell. Configuration of B: [He] 2s² 2p¹ — one electron in the 2p subshell.

The 2p orbital is higher in energy than 2s (p electrons are farther from the nucleus on average). The 2p¹ electron in B is also partially shielded by Be's 2s² electrons. So despite B having one more proton, its 2p¹ electron is easier to remove than one of Be's 2s² electrons.

IE₁(Be) = 900 kJ/mol > IE₁(B) = 801 kJ/mol ✓

Anomaly 2 — IE₁(N) > IE₁(O):

Configuration of N: [He] 2s² 2p³ — half-filled 2p subshell. Configuration of O: [He] 2s² 2p⁴ — one 2p orbital has a paired electron.

Nitrogen's half-filled 2p subshell has extra stability: all three 2p electrons are in separate orbitals with parallel spins — maximum exchange energy, symmetric distribution. Oxygen must pair one electron into an already occupied 2p orbital, creating electron–electron repulsion that makes that electron easier to remove.

IE₁(N) = 1402 kJ/mol > IE₁(O) = 1314 kJ/mol ✓

Why This Works

The general trend (increasing IE across a period) = increasing Z_eff effect. The anomalies = subshell stability overriding the Z_eff effect at specific points.

Two stability boosts to remember:

• Fully filled subshell (ns²): Be, Mg, Zn — harder to remove than expected
• Half-filled subshell (np³): N, P, As — harder to remove than expected

These stability boosts occur because fully/half-filled configurations have:

• Higher exchange energy (electrons of the same spin in the same subshell stabilise each other)
• Symmetric electron distribution (reduces electron–electron repulsion)

Alternative Method — Reading the Graph

Plot IE₁ vs atomic number for Period 2:

The graph rises overall from Li to Ne, with two downward dips: at B (from Be) and at O (from N). Recognise these dips, know their cause, and you can answer any arrangement question.

Common Mistake