A) 5, 0, 0, \[+\frac{1}{2}\]
B) 5,1, 0, \[+\frac{1}{2}\]
C) 5, 1, 1, \[+\frac{1}{2}\]
D) 6, 0, 0, \[+\frac{1}{2}\]
Correct Answer: A
Solution :
Electronic configuration of Rb (Z = 37) is\[1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}},3{{p}^{6}},3{{d}^{10}},4{{s}^{2}},4{{p}^{6}},5{{s}^{1}}\] So, for the valence shell electron \[(5{{s}^{1}})\] \[n=5\], \[l=0\], \[m=0\], \[s=+\frac{1}{2}\]
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