A) \[1{{s}^{2}},2{{s}^{2}}{{p}^{6}}.......n{{s}^{2}}{{p}^{3}}\]
B) \[1{{s}^{2}},2{{s}^{2}}{{p}^{6}}.......n{{s}^{2}}{{p}^{6}}{{d}^{3}},(n+1){{s}^{2}}\]
C) \[1{{s}^{2}},2{{s}^{2}}{{p}^{6}}.......n{{s}^{2}}{{p}^{6}}{{d}^{10}},(n+1){{s}^{2}}{{p}^{1}}\]
D) \[1{{s}^{2}},2{{s}^{2}}{{p}^{6}}.......n{{s}^{2}}{{p}^{6}}\]
Correct Answer: B
Solution :
\[1{{s}^{2}},\,2{{s}^{2}}{{p}^{6}}.........n{{s}^{2}}{{p}^{6}}{{d}^{3}},\,(n+1){{s}^{2}}\] as last electron enters d-subshell.
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