A) \[Na\]
B) \[Cl\]
C) \[Cr\]
D) \[Rb\]
Correct Answer: C
Solution :
\[n=4,l=0,m=0\]and \[s=+\frac{1}{2}\]set represents that the last electron enters in a \[4s\] orbital. Among the given options, only Cr is the element, last electron of which enters in a 4s orbital. \[Cr=[Ar]\,3{{d}^{5}},4{{s}^{1}}\]You need to login to perform this action.
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