The valence shell electronic configuration is $(n - 1)s^2 (n - 1)p^6 (n - 1)d^x ns^2$. If $n = 4$ and $x = 5$,what is the number of protons in the element?

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 atomic number of the element with the maximum number of unpaired $4p$ electrons?

If $n$ and $l$ represent the principal and azimuthal quantum numbers respectively, the formula used to determine the number of radial nodes possible for a given orbital is:

Arrange the following orbitals in decreasing order of energy:
$A$. $n = 3, l = 0, m = 0$
$B$. $n = 4, l = 0, m = 0$
$C$. $n = 3, l = 1, m = 0$
$D$. $n = 3, l = 2, m = 1$
The correct option for the order is:

The number of angular and radial nodes in $3p$ orbital respectively are