If the Aufbau principle had not been followed,$Ca$ $(Z = 20)$ would have been placed in the

$A$ hydrogen atom has only one electron,so mutual repulsion between electrons is absent. However,in multielectron atoms,mutual repulsion between the electrons is significant. How does this affect the energy of an electron in the orbitals of the same principal quantum number in multielectron atoms?

Identify which of the following sets of quantum numbers are not possible and explain why.
$(a) n = 0, l = 0, m_l = 0, m_s = +1/2$
$(b) n = 1, l = 0, m_l = 0, m_s = -1/2$
$(c) n = 1, l = 1, m_l = 0, m_s = +1/2$
$(d) n = 2, l = 1, m_l = 0, m_s = -1/2$
$(e) n = 3, l = 3, m_l = 3, m_s = +1/2$
$(f) n = 3, l = 1, m_l = 0, m_s = +1/2$

Arrange the orbitals of $H$ atom in the increasing order of their energy:
$3p_x, 2s, 4d_{xy}, 3s, 4p_z, 3p_y, 4s$

Statement $A$: In a potassium atom,the $19^{th}$ electron enters the $4s$-orbital instead of the $3d$-orbital.
Reason $R$: The $(n + l)$ rule is followed to determine the minimum energy of an orbital.

How many electrons can be accommodated in a sub-shell for which $n = 3, l = 1$?