For the $n = 2$ energy level,how many orbitals of all kinds are possible?

What determines the position of an element in the periodic table?

Correct statements for an element with atomic number $9$ are:
$A.$ There can be $5$ electrons for which $m_{s}=+\frac{1}{2}$ and $4$ electrons for which $m_{s}=-\frac{1}{2}$
$B.$ There is only one electron in $p_z$ orbital
$C.$ The last electron goes to an orbital with $n=2$ and $l=1$
$D.$ The sum of angular nodes of all the atomic orbitals is $1$
Choose the correct answer from the options given below:

The maximum number of electrons that can be accommodated in a shell with $n=4$ is:

Using $s, p, d, f$ notations,describe the orbital with the following quantum numbers:
$(a) n=2, l=1$
$(b) n=4, l=0$
$(c) n=5, l=3$
$(d) n=3, l=2$

Which of the following electronic configurations is not possible?