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IB DP Chemistry – S1.3.6 Ionization energy – Study Notes

IB DP Chemistry - S1.3.6 Ionization energy - Study Notes - New Syllabus - 2026, 2027 & 2028

IB DP Chemistry – S1.3.6 Ionization energy – Study Notes – New Syllabus

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Structure 1.3.6 — Ionization Energy, Spectral Convergence, and Related Calculations

Structure 1.3.6 — Ionization Energy, Spectral Convergence, and Related Calculations

Emission Spectrum and Convergence Limit

As electrons transition to lower energy levels, atoms emit radiation at specific frequencies this forms a line emission spectrum.

  • The limit of convergence is where the lines get closer at high frequency (or short wavelength).
  • This limit represents the energy required to remove an electron from the ground state completely the first ionization energy.

Ionization Energy from Spectral Data

We can calculate the energy of ionization using spectral data at the convergence limit.

Step 1: Use Planck’s equation

\( E = h \times f \)

  • \( E \) = energy (Joules)
  • \( h = 6.63 \times 10^{-34} \, \text{J·s} \)
  • \( f \) = frequency (Hz)

Step 2: If given wavelength instead of frequency

\( c = \lambda f \Rightarrow f = \frac{c}{\lambda} \)

Substitute into Planck’s equation:

\( E = \frac{h c}{\lambda} \)

  • \( c = 3.00 \times 10^8 \, \text{m/s} \)
  • \( \lambda \) = wavelength (in meters!)

Step 3: Convert to molar ionization energy

  • Multiply \( E \) (in J) by Avogadro’s number: \( N_A = 6.02 \times 10^{23} \)
  • Convert to kilojoules per mole by dividing by 1000
  • \( \text{IE (kJ/mol)} = \frac{E \times N_A}{1000} \)

Trends in First Ionization Energy

Across a Period (left → right):

  • First ionization energy increases
  • More protons → increased nuclear charge
  • Electrons are added to the same shell → stronger attraction → harder to remove
  • Atomic radius decreases

Down a Group:

  • First ionization energy decreases
  • Electrons are added to higher energy levels → further from the nucleus
  • More inner electrons → increased shielding → weaker nuclear attraction
  • Atomic radius increases

Discontinuities in Periodic Trends (Exceptions)

(a) Group 2 to Group 13 (e.g. Be to B or Mg to Al):

  • Ionization energy drops
  • The outermost electron in Group 13 is in a p orbital, which is higher in energy and less tightly held than an s electron

(b) Group 15 to Group 16 (e.g. N to O or P to S):

  • Ionization energy drops
  • Electron pairing in the same p orbital causes repulsion, making it easier to remove one of the paired electrons

Example

Calculate the first ionization energy of hydrogen in kJ/mol if the convergence limit of the Lyman series is at a wavelength of 91.2 nm.

▶️ Answer/Explanation

Convert wavelength to meters
\( \lambda = 91.2\, \text{nm} = 91.2 \times 10^{-9}\, \text{m} \)

Use Planck’s equation:
\( E = \frac{h \times c}{\lambda} \)
\( = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{91.2 \times 10^{-9}} \)
\( = 2.18 \times 10^{-18} \, \text{J} \) (per atom)

 Convert to kJ/mol
\( \text{IE} = \frac{2.18 \times 10^{-18} \times 6.02 \times 10^{23}}{1000} \)
\( = 1312 \, \text{kJ/mol} \)

Final Answer: 1312 kJ/mol

Example 

Explain why the first ionization energy of aluminum (Z = 13) is lower than that of magnesium (Z = 12), despite the general trend of increasing IE across a period.

▶️ Answer/Explanation

Magnesium: Electron removed from a 3s orbital
Aluminum: Electron removed from a 3p orbital

The 3p orbital in aluminum is higher in energy and further from the nucleus than the 3s orbital in magnesium. The 3p electron also experiences more shielding, making it easier to remove.

Final Point: The electron in aluminum is less tightly held, so its ionization energy is lower.

Example 

Explain why the first ionization energy decreases from beryllium (Be) to barium (Ba) in Group 2 of the periodic table.

▶️ Answer/Explanation

As you move down Group 2:

  • Atomic radius increases
  • More inner electron shells → more shielding
  • Weaker attraction between nucleus and outermost electron

Even though nuclear charge increases, the effect is outweighed by increased shielding and distance, making it easier to remove an electron.

Final Conclusion: First ionization energy decreases down the group.

Example 

The frequency corresponding to the convergence limit in the hydrogen emission spectrum is \( 3.29 \times 10^{15} \, \text{Hz} \). Calculate the first ionization energy of hydrogen in kJ mol⁻¹.

▶️ Answer/Explanation

Use the Planck equation
\( E = h \times f \)
\( = 6.63 \times 10^{-34} \, \text{J·s} \times 3.29 \times 10^{15} \, \text{Hz} \)
\( = 2.18 \times 10^{-18} \, \text{J (per atom)} \)

Convert to per mole using Avogadro’s constant
\( E_{\text{mol}} = 2.18 \times 10^{-18} \, \text{J} \times 6.02 \times 10^{23} \, \text{mol}^{-1} \)
\( = 1.312 \times 10^6 \, \text{J/mol} \)

Step 3: Convert to kJ/mol
\( = \frac{1.312 \times 10^6}{1000} = 1312 \, \text{kJ/mol} \)

Final Answer: \( 1312 \, \text{kJ mol}^{-1} \)

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