Energy of an electron in the ground state of the hydrogen atom is - 2 . 18 × 10 - 18 J . Calculate the ionization enthalpy of atomic hydrogen in terms of J mol - 1 .

The energy of an electron in the ground state of the hydrogen atom is – 2 . 18 × 10 - 18 J . Therefore, the energy required to remove that electron from the ground state of hydrogen atom is 2 . 18 × 10 - 18 J . Ionization enthalpy of atomic hydrogen = 2 . 18 × 10 - 18 J . Hence, ionization enthalpy of atomic hydrogen in terms of J mol - 1 = 2 . 18 × 10 - 18 × N A . = 2 . 18 × 10 - 18 × 6 . 023 × 10 23 J mol - 1 = 1 . 31 × 10 6 J mol - 1 .

Ionization enthalpy 1 IE 1 of O is less than that of N -Explain.

The electronic configuration of nitrogen ( N , Z = 7 ) is 1 s 2 2 s 2 2 p 3 and that of oxygen ( O , Z = 8 ) is 1 s 2 2 s 2 2 p 4 . Nitrogen atom acquires extra stability due to half- filled 2 p - orbitals while oxygen can acquire half-filled orbital and extra stability by losing one electron. Thus, oxygen can easily lose one electron. Nitrogen requires more energy to remove an electron due to extra stability. Therefore, first ionisation enthalpy of oxygen is less than that of nitrogen.