Question

The correct set of four quantum number for the valence electrons of the rubidium atom (Z = 37) is:
(A) 5, 1, 1, $+\dfrac{1}{2}$
(B) 5, 0, 1, $+\dfrac{1}{2}$
(C) 5, 0, 0, $+\dfrac{1}{2}$
(D) 5, 1, 0, $+\dfrac{1}{2}$

Answer 1

Hint: There are four quantum numbers: principal quantum number, azimuthal quantum number, magnetic quantum number, and spin quantum number. The azimuthal quantum number for s-orbital is 0, for p-orbital is 1, for d-orbital is 2, and for f-orbital is 3.

Complete step by step solution:
There are four quantum numbers: principal quantum number, azimuthal quantum number, magnetic quantum number, and spin quantum number which tells the shape, size, and orientation of the orbital in the space.
(n) represents the principal quantum number and it tells the shell number of the orbital. n is always a natural number like 1, 2, 3, …….
(l) represents the azimuthal quantum number and it tells the number of subshells in the main shell. For a given value of n, l ranges from 0 to $n-1$. The azimuthal quantum number for s-orbital is 0, for p-orbital is 1, for d-orbital is 2, and for f-orbital is 3.
(m) represents the magnetic quantum and it tells the number of orbitals in the subshell. For a given value of l, $m=-l\text{ }to\text{ }+l$.
(s) represents the spin quantum number and it tells the magnetic property. It will be either $+\dfrac{1}{2}$ or $-\dfrac{1}{2}$ for an electron.
So, the atomic number is 37 and its configuration will be: $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{10}}4{{p}^{6}}5{{s}^{1}}$
So, the last electron is $5{{s}^{1}}$.
So, n will be 5, l will be 0 because the subshell is s-orbital, m will be 0, and s will be $+\dfrac{1}{2}$.

So, the correct answer is an option (C)- 5, 0, 0, $+\dfrac{1}{2}$.

Note: The magnetic quantum number for s-orbital will be 0, for p-orbital will be 3 (-1, 0, +1), for d-orbital will be 5 (-2, -1, 0, +1, +2), for f-orbital will be 7 (-3, -2, -1, 0, +1, +2, +3).