MolSynapsequestion bank
MolSynapse / Question bank / Periodic Table and Periodicity / Ionisation enthalpy
JEE Advanced levelSingle correct optionDifficulty 3 of 3pt146

The first ionisation energy of is 24.6 eV and that of hydrogen is 13.6 eV. Take 1 eV per atom = . The total energy to remove both electrons from one mole of gaseous atoms is closest to

  1. A
  2. B
  3. C
  4. D

Answer: A

The second step is the ionisation of : 13.6 x 4 = 54.4 eV. Total = 24.6 + 54.4 = 79.0 eV per atom; 79.0 x 96.485 = 7622 kJ/mol. 5249 is only the second step and 2373 only the first.
Practice Periodic Table and Periodicity More on Ionisation enthalpy

More questions on Ionisation enthalpy