question_answer

Write down all the four quantum numbers for (i) 19th electron of \[{}_{24}Cr\] (ii) 21st electron for\[{}_{21}Sc.\]

Answer:

(i) 19th electron of \[{}_{24}Cr\]. \[Cr(Z=24)\,1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}\,3{{s}^{2}}3{{p}^{6}}4{{s}^{1}}3{{d}^{5}}\] The 19th electron is present in the 4s orbital. For the 4s electron \[n=4,\,\,l=0,\,\,m=0,\,\,s=+\frac{1}{2}\] or\[-\frac{1}{2}\]. (ii) 21st electron for \[{}_{21}Sc.\] \[Sc(Z=21)\,1{{s}^{2}}\,2{{s}^{2}}\,2{{p}^{6}}\,3{{s}^{2}}\,3{{p}^{6}}\,4{{s}^{2}}\,3{{d}^{1}}\]. The 21st electron is present in the 3d orbital. For 3d electron, \[n=3,\,l=2,m=-2,-1,0,+1,+2,\]\[~s+\frac{1}{2}\,\,\,\text{or}\,\,-\frac{1}{2}\]