Question

a ground state atom of ar could not have any electrons with which of the following configurations?
answer:
a $n = 3, l = 1, m_{l}=0, m_{s}=+\frac{1}{2}$
b $n = 4, l = 2, m_{l}=0, m_{s}=+\frac{1}{2}$
c $n = 4, l = 0, m_{l}=0, m_{s}=+\frac{1}{2}$
d $n = 3, l = 2, m_{l}=0, m_{s}=+\frac{1}{2}$
e $n = 4, l = 1, m_{l}=0, m_{s}=+\frac{1}{2}$

Explanation:

Argon (\(Ar\)) has an atomic number of \(18\). Its electron configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\). The principal quantum number \(n = 4\) is higher than the highest occupied principal quantum number (\(n = 3\)) in the ground - state of \(Ar\). For \(n = 4\), electrons would be in an excited state. But for option \(B\) with \(n = 4,l = 2\) (which is the \(4d\) sub - shell), in the ground - state of \(Ar\), there are no electrons in the \(n=4\) shell.

Answer:

B. \(n = 4,l = 2,m_{l}=0,m_{s}=+\frac{1}{2}\)