A) \[2,0,0,+\frac{1}{2}\]
B) \[2,1,-1,+\frac{1}{2}\]
C) \[3,1,-1,\pm \frac{1}{2}\]
D) \[3,0,0,\pm \frac{1}{2}\]
Correct Answer: C
Solution :
\[_{17}Cl=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{5}}\] \[\therefore \]One unpaired electron is present in 3p orbital. \[n=3\] \[l=1,\] \[m=-1,\] \[s=\pm \frac{1}{2},\]You need to login to perform this action.
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