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 :
The electronic configuration of \[{}_{37}Rb=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{10}},4{{s}^{2}}4{{p}^{6}},5{{s}^{1}}\square 1\] Hence, the correction quantum number of last electron are \[n=5,\] \[l=0,\] \[m=0,\] \[s=+\frac{1}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
