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 \[m=-\text{ }l\]to \[+l\]) |
| \[s=+\frac{1}{2}\] |
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