A) He
B) B
C) Li
D) Na
Correct Answer: A
Solution :
Key Idea: Electron affinity is the amount of energy given out when an electron is added to an isolated gaseous atom. Its value depends on the electronic configuration of valence shell. Write and compare the electronic configuration of given elements to find which of them has electron affinity value closer to Be. \[Be(4)=1{{s}^{2}},2{{s}^{2}}\] \[He(2)=1{{s}^{2}}\] \[B(5)=1{{s}^{2}},2{{s}^{2}},2{{p}^{1}}\] \[Li(3)=1{{s}^{2}},2{{s}^{1}}\] \[Na(11)=1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{1}}\] \[\because \] electronic configuration of valence shell of Be and He is same \[\therefore \] electron affinity of Be is similar to that of He.
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