Which of the following is correct set of 4 quantum numbers of 19th electron in Chromium (Atomic number = 24) in accordance with Aufbau principle?

According to the Aufbau (or $$n+l$$) rule, sub-shells are filled in the increasing order of their $$n+l$$ value; if two sub-shells have the same $$n+l$$ value, the one with lower $$n$$ is filled first.

Following this order we get the filling sequence
$$1s \; (2) \;\rightarrow\; 2s \; (2) \;\rightarrow\; 2p \; (6) \;\rightarrow\; 3s \; (2) \;\rightarrow\; 3p \; (6) \;\rightarrow\; 4s \; (2) \;\rightarrow\; 3d \; (10)\;\dots$$

Chromium has $$24$$ electrons. Up to the Argon core ($$18$$ electrons) the configuration is
$$1s^2\,2s^2\,2p^6\,3s^2\,3p^6 \qquad (\text{total } 18 \text{ e}^-)$$

The 19th electron is therefore the very next one to be accommodated. The next available sub-shell in the Aufbau sequence is $$4s$$.

For a $$4s$$ electron, the quantum numbers are:

• Principal quantum number $$n = 4$$ (fourth shell) 

• Azimuthal quantum number $$l = 0$$ (s-sub-shell)

• Magnetic quantum number $$m_l = 0$$ (only one orientation for $$l = 0$$)

• Spin quantum number $$m_s = +\frac{1}{2}$$ (the first electron in any orbital is taken with $$+\frac{1}{2}$$ spin)

Thus the correct set is $$n = 4, \; l = 0, \; m = 0, \; s = +\dfrac{1}{2}$$.

Option D $$=>$$ $$n=4, l=0, m=0, s=+\dfrac{1}{2}$$