A) \[P{{b}^{207}}\]
B) \[B{{i}^{209}}\]
C) \[P{{b}^{208}}\]
D) \[P{{b}^{206}}\]
Correct Answer: C
Solution :
Series of an element is decided by dividing its atomic mass by 4. In case of 4n series, the remainder is zero. This is true only in case of \[P{{b}^{208}}\]. Thus, \[P{{b}^{208}}\] belongs to 4n series.
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