Question

The set of quantum numbers $n=4,l=0$ and $s=+\frac{1}{2}$ correspond to the most loosely bound, ground state electron of which one of the following atoms?

• A. $Na$
• B. $Cl$
• C. $Cr$
• D. $Rb$

Answer 1

Answer: (C)

$Cr$

Answer 2

$n=4,l=0,m=0$ 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.
$Cr=[Ar]\,3d^{5},4s^{1}$