A) third
B) fourth
C) fifth
D) second
Correct Answer: B
Solution :
No. of period = No. of outermost shell \[(3sBr=1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{10}},4{{s}^{2}}4{{p}^{5}})\] Because the outer most shell of bromine is 4th hence, its period is fourth.
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