The number of electrons in the outermost shell of an element with atomic number $15$ is . . . . . . .

The number of electrons present in all the completely filled subshells having $n=4$ and $s=+\frac{1}{2}$ is . . . . . . . (Where $n=$ principal quantum number and $s=$ spin quantum number)

Identify the element if its expected electronic configuration is $[Ar] 3d^{10} 4s^{2}$.

Two statements are given below:
Statement $I$: In $H$ atom, the energy of $2s$ and $2p$ orbitals is same.
Statement $II$: In $He$ atom, the energy of $2s$ and $2p$ orbitals is same.
The correct answer is:

Arrange the electrons represented by the following sets of quantum numbers in the decreasing order of energy.
$(i) \ n = 4, l = 0, m_l = 0, m_s = +\frac{1}{2}$
$(ii) \ n = 3, l = 1, m_l = 1, m_s = -\frac{1}{2}$
$(iii) \ n = 3, l = 2, m_l = 0, m_s = +\frac{1}{2}$
$(iv) \ n = 3, l = 0, m_l = 0, m_s = -\frac{1}{2}$

Maximum number of electrons in a subshell with $l = 3$ and $n = 4$ is