Question

SECTION B
2(a)(i) Define the first ionization energy of an element.
(ii) Consider the following table:
Element: Li, Be, B, C, N, O, F, Ne
Atomic No: 3,4,5,6,7,8,9,10
1st I.E./kJ/mol: 520,900,801,1086,1402,1314,1681,2081
Explain briefly why the first ionization energy of B is less than that of Be despite the fact that the atomic number of B is greater than that of Be.

Accepted Answer

(a)(i) The first ionization energy of an element is the minimum energy required to remove one electron from the valence shell of a gaseous atom of the element.

(ii) Be has the electronic configuration 1s²2s². The electron to be removed from Be is in the 2s orbital which is closer to the nucleus and thus more strongly attracted to the nuclear (greater nuclear attraction). Hence its ionization energy is greater. B has the electronic configuration 1s²2s²2p¹, and the electron to be removed is in the 2p orbital which is farther away from the nucleus and less strongly attracted (lower nuclear attraction), hence its first ionization energy is smaller than that of Be.

WAEC Chemistry 2022 Theory — Question 2

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