If an atom has $13$ electrons in the $M$ shell and $1$ electron in the $N$ shell,then the number of unpaired electrons in that atom is ...

For both $2p$ and $3s$ orbitals,the value of $(n + l)$ is equal to $3$. Which of these has lower energy and why?

The arrangement of orbitals on the basis of energy is based upon their $(n + l)$ value. Lower the value of $(n + l)$,lower is the energy. For orbitals having same values of $(n + l)$,the orbital with lower value of $n$ will have lower energy.
$(I)$ Based upon the above information,arrange the following orbitals in the increasing order of energy.
$(a)$ $1s, 2s, 3s, 2p$
$(b)$ $4s, 3s, 3p, 4d$
$(c)$ $5p, 4d, 5d, 4f, 6s$
$(d)$ $5f, 6d, 7s, 7p$
$(II)$ Based upon the above information,solve the questions given below:
$(a)$ Which of the following orbitals has the lowest energy? $4d, 4f, 5s, 5p$
$(b)$ Which of the following orbitals has the highest energy? $5p, 5d, 5f, 6s, 6p$

The electronic configuration of three elements $A$,$B$,and $C$ are given below. Stable form of $A$ may be represented by the formula:
$A : 1s^2, 2s^2, 2p^6$
$B : 1s^2, 2s^2, 2p^6, 3s^2, 3p^3$
$C : 1s^2, 2s^2, 2p^6, 3s^2, 3p^5$

The allowed set of quantum numbers for an electron in a hydrogen atom is

The atomic numbers of elements $X, Y$ and $Z$ are $19, 21$ and $25$ respectively. The number of electrons present in the $M$-shell of these elements follow the order: