A) 3
B) 7
C) 5
D) 9
Correct Answer: B
Solution :
\[n=4\,\to 1{{s}^{2}},\,2{{s}^{2}},\,2{{p}^{6}},\,3{{s}^{2}},\,3{{p}^{6}},\,3{{d}^{10}},\,4{{s}^{2}},\,4{{p}^{6}},\,4{{d}^{10}},\,4{{f}^{14}}\] So \[l=(n-1)=4-1=3\] which is f orbit contain 7 orbital.
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