A) \[1{{s}^{2}}\,\,2{{s}^{2}}\]
B) \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}\]
C) \[3{{d}^{10}}4{{s}^{2}}4{{p}^{2}}\]
D) \[1{{s}^{2}}2{{s}^{2}}\,2{{p}^{2}}3{{s}^{1}}\]
Correct Answer: D
Solution :
When 2p orbital is completely filled then electron enter in the 3s. The capacity of 2p orbital containing \[{{e}^{-}}\]is 6. So \[1{{s}^{2}},\,2{{s}^{2}}2{{p}^{2}}3{{s}^{1}}\] is a wrong electronic configuration the write is \[1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}.\]
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