A) \[n=4,\,\,l=0,\,\,m=0,\,\,{{m}_{s}}=+\frac{1}{2}\]
B) \[n=3,\,\,l=1,\,\,m=-1,\,\,{{m}_{s}}=+\frac{1}{2}\]
C) \[n=4,\,\,l=1,\,\,m=\pm 1,\,\,{{m}_{s}}=-\frac{1}{2}\]
D) \[n=3,\,\,l=2,\,\,m=\pm 2,\,\,{{m}_{s}}=-\frac{1}{2}\]
Correct Answer: A
Solution :
\[Sc(21)=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{1}}{{s}^{2}}\] 19th electron goes into\[4s\] Hence,\[n=4\], \[l=0\] \[m=0\] \[{{m}_{s}}=+\frac{1}{2}\]
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