A) \[{{E}_{2s(H)}}>{{E}_{2s(Li)}}>{{E}_{2s(Na)}}>{{E}_{2s(K)}}\]
B) \[{{E}_{2s(H)}}>{{E}_{2s(Na)}}>{{E}_{2s(Li)}}>{{E}_{2s(K)}}\]
C) \[{{E}_{2s(H)}}>{{E}_{2s(Na)}}={{E}_{2s(K)}}>{{E}_{2s(Li)}}\]
D) \[{{E}_{2s(K)}}<{{E}_{2s(Na)}}<{{E}_{2s(Li)}}>{{E}_{2s(H)}}\]
Correct Answer: A
Solution :
The electronic configuration of the given atoms is as \[H=1{{s}^{1}}\] \[Li=1{{s}^{2}},2{{s}^{1}}\] \[Na=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{1}}\] \[K=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}},4{{s}^{1}}\] Thus, the increasing order of energy of 2s orbitals is \[{{E}_{2s(K)}}<{{E}_{2s(Na)}}<{{E}_{2s(Li)}}<{{E}_{2s(H)}}\]
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