A) \[[Ne]\,\,3{{s}^{1}}\]
B) \[[Ne]\,\,3{{s}^{2}}3{{p}^{3}}\]
C) \[[Ne]\,\,3{{s}^{10}},4{{s}^{2}},4{{p}^{3}}\]
D) \[[Ne]\,\,3{{s}^{2}}3{{p}^{4}}\]
Correct Answer: B
Solution :
Key Idea In a period, from left to right ionisation energy of element generally increases and in a group from top to bottom ionization energy of element decreases. The ionisation energy of V A group is greater than VI A group due to extra stabilility of half-filled p-subshell. [Ne] \[3{{s}^{1}}\] is present in IA group and third period. [Ne] \[3{{s}^{2}}3{{p}^{3}}\] is present in V A group and third period. [Ne] \[3{{d}^{10}},4{{s}^{2}}4{{p}^{3}}\] is present in V A group and fourth period. [Ne] \[3{{s}^{2}},3{{p}^{4}}\] is present in VI A group and third period. We know that IE of V A group is greater than VI A group. So, the IE of [Ne] \[3{{s}^{2}}3{{p}^{3}}\] is greater than [Ne] \[3{{s}^{2}}3{{p}^{4}}\] Further, [Ne] \[3{{d}^{10}}4{{s}^{2}}4{{p}^{3}}\] is also placed in V A group but it is placed below the [Ne] \[3{{s}^{2}}3{{p}^{3}}\]. As the IE in a group from top to bottom decreases therefore, the ionisation energy of [Ne] \[3{{s}^{2}}3{{p}^{3}}\] is greatest in given elements due to extra stability of half-filled p-sub-shell.
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