A) \[5,1,1,+\frac{1}{2}\]
B) \[6,0,0,+\frac{1}{2}\]
C) \[5,0,0,+\frac{1}{2}\]
D) \[5,1,0,+\frac{1}{2}\]
Correct Answer: C
Solution :
\[_{37}Rb{{=}_{36}}[Kr]\,5{{s}^{1}}\]Its valence electron is \[5{{s}^{1}}\]. So, \[n=5\] \[l=0\] (For s orbital) \[m=0\] (As \[m=-l\] to \[+l\]) \[s=+\frac{1}{2}\]
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