$4s^2$ is the configuration of the outermost orbit of an element. Its atomic number would be $:-$

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

What is the maximum number of electrons that can be associated with the following set of quantum numbers?
$n = 3, l = 1$ and $m = -1$

The shape of $p$-orbital is

For a hydrogen atom,the orbital$(s)$ with the lowest energy is/are:
$A$. $4s$
$B$. $3p_x$
$C$. $3d_{x^2-y^2}$
$D$. $3d_{z^2}$
$E$. $4p_z$
Choose the correct answer from the options given below:

Arrange the $1s$,$2s$,$4s$,$2p$,$3p$,$4p$,and $3d$ orbitals according to increasing order of energy.