c) Consider the following table listing electron affinity and first ionization e

Question

c) Consider the following table listing electron affinity and first ionization energy Element EA kJ/mole Er K/mole +5208020153 Li -59 Be 0 +899 Ne 0 328 +801 1086 1402 14 +1086+1402 i. Predict the value for IE of He: choose from +3009 kJ/mole or +1200 kJ/mole, explain your choice (2 Mark) ii. Give reasons why the electron affinity of B, N and Ne are 0 (zero) (3 Marks) iii. Give reasons why ionization energy of B is less than that of Be (1.5 Marks) iv. Give reasons why is the ionization of O less than that of N? (1.5 Marks)

Explanation / Answer

II. The stable half-filled electronic configuration in case of N forces it to not lose its stability by adding another electron and hence, has a zero EA. Ne has a stable full octet configuration and does not wish to add any more electrons to its octet and hence, has a zero EA. Be has an electronic configuration with a complete 2s shell and does not wish to gain electrons to avoid destabilising its stable configuration.

NOTE: There is a mis-print in the question as the third element with a zero EA is Be and not B.

III. Although, from left to right the Ionisation Energy increases, the IE of B is lesser than that of Be because B has only one electron in the p orbital and p orbital has higher energy than s orbital, so this electron in the p orbital tends to go to a lower energy level or lose in order to make the atom more stable.

IV. The IE of O is less than that of N because in nitrogen p - subshell is half filled . According to Hund , half filled & full filled orbitals are more stable . Hence it is difficult to remove an electron from 2p. So more amount of energy is required to remove an electron, making the IE of N more than that of O.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.