A) \[n=5,\,l=0,\,m=0,\,s=+\,1/2\]
B) \[n=5,\,l=1,\,m=1,\,s=+\,1/2\]
C) \[n=5,\,l=1,\,m=1,\,s=+\,1/2\]
D) \[n=6,\,l=0,\,m=0,\,s=+\,1/2\]
Correct Answer: A
Solution :
Key Idea: Write configuration Rb and then find quantum numbers of valence electron. \[\text{Rb}\]Atomic number is 37, so configuration is \[1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}},3{{p}^{6}},4{{s}^{2}},3{{d}^{10}},4{{p}^{6}},5{{s}^{1}}\] \[\therefore \] Last electron (valence electron) is \[5{{s}^{1}}\] \[\therefore \]\[n=5\] (\[\because \]electron enters 5 energy level) \[l=0\] (\[\because \]it is a sub-shell). \[m=0\] \[s=1/2\]
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