A) \[n=2,\,l=0,\,m=0,\,s=+\frac{1}{2}\]
B) \[n=4,\,l=0,\,m=0,\,s=+\frac{1}{2}\]
C) \[n=3,\,l=1,\,m=1,\,s=+\frac{1}{2}\]
D) \[n=4,\,l=2,\,m=-1,\,s=+\frac{1}{2}\]
Correct Answer: B
Solution :
\[{{K}_{19}}=1{{s}^{2}},\,2{{s}^{2}}\,2{{p}^{6}},\,3{{s}^{2}}\,3{{p}^{6}},\,4{{s}^{1}}\] for \[4{{s}^{1}}\]electrons. \[n=4,\,l=0,\,m=0\] and \[s=+\frac{1}{2}.\]
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