A) \[1{{s}^{2}},2{{s}^{2}}2{{p}^{5}}\]
B) \[1{{s}^{2}},2{{s}^{2}}2{{p}^{4}},3{{s}^{1}}\]
C) \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{1}}3{{p}^{1}}\]
D) \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{5}}\]
Correct Answer: A
Solution :
[a] \[1{{s}^{2}},2{{s}^{2}},2{{p}^{5}},2,\,7\] (\[\because \] it has capacity to accept electron therefore, it is electronegative.) [b] \[1{{s}^{2}},2{{s}^{2}},2{{p}^{4}},3{{s}^{1}}=2,\,6,\,1\] (configuration not correct \[(2{{p}^{4}})\] [c] \[1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{1}},3{{p}^{5}}=2,\,8,\,6\] (configuration not correct \[3{{s}^{1}}\]) [d] \[1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}},3{{p}^{5}}=2,\,8,\,7\] (\[\because \] it has capacity to accept electron therefore, it is electronegative) Smaller the size, greater will be electronegativity. Since, element in choice [a] is smaller in size, it will be more electronegative than [d]. In choice [a] the atomic number of element is 9, which is of fluorine and it is the most electronegative element of the periodic table.
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