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?

The electronic configurations of four elements are given in brackets:
$L(1s^2, 2s^2\ 2p^1);$ $M(1s^2, 2s^2\ 2p^5)$
$Q(1s^2, 2s^2\ 2p^6, 3s^1);$ $R(1s^2, 2s^2\ 2p^2)$
The element that would most readily form a diatomic molecule is:

The number of unpaired electrons in the $p$-subshell of oxygen atom is

When the azimuthal quantum number has a value of $l = 0$,the shape of the orbital is

The atomic number of the element having the maximum number of unpaired electrons (in the ground state) is ........

The general configuration of the outermost and penultimate shell is $(n - 1)s^2 (n - 1)p^6 (n - 1)d^x ns^2$. If $n = 4$ and $x = 5$,then the number of protons in the nucleus will be: