The correct statement about probability density (except at infinite distance from the nucleus) is:

The $71^{st}$ electron of an element $X$ with an atomic number of $71$ enters into the orbital:

The number of valence electrons in an atom with electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^3$ is

Select the correct statements for quantum numbers.
$(i)$ Magnetic quantum number $(m_l)$ gives information about the spatial orientation of orbitals with respect to the standard set of coordinate axes.
$(ii)$ Electron spin quantum number is represented by '$s$' and has values of $\pm \frac{1}{2}$.
$(iii)$ Principal quantum number $(n)$ determines the size of the orbitals and also,to a large extent,the energy of the orbitals.

Given below are the quantum numbers for $4$ electrons.
$A. \; n = 3, l = 2, m_l = 1, m_s = +1/2$
$B. \; n = 4, l = 1, m_l = 0, m_s = +1/2$
$C. \; n = 4, l = 2, m_l = -2, m_s = -1/2$
$D. \; n = 3, l = 1, m_l = -1, m_s = +1/2$
The correct order of increasing energy is

Arrange the following orbitals in increasing order of energy based on the given values of principal quantum number $(n)$ and azimuthal quantum number $(l)$:
$(1) n = 4, l = 1$
$(2) n = 4, l = 0$
$(3) n = 3, l = 1$
$(4) n = 3, l = 2$