Question

The correct set of quantum numbers for valence electrons of $Rb$(atomic number-$37$):
a) $5,0,0,\dfrac{{ - 1}}{2}$
b) $5,1,0,\dfrac{1}{2}$
c) $6,0,1,\dfrac{1}{2}$
d) $5,1,1,\dfrac{1}{2}$

Answer 1

Proper knowledge of different quantum numbers is necessary. The relation between the different quantum numbers will be required to calculate the correct set. $Rubidium(Rb)$ Came in the $1st$ group of periodic table.

Complete solution:
As $Rubidium (Rb)$ came in the $1st$ group of periodic table , the electronic configuration is
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^1}$
Valence shell of an electron is the outermost shell having electrons known as valence electrons.
Quantum numbers are those numbers which tells us about the complete information of an electron in an atom.
Principal quantum number ($n$ ) specifies the shell number. It is a positive integer with value$n = 1,2,3.....$. The maximum value of the shell in an electronic configuration will be equal to the value of$n$.
According to the given question, the valence electron of $Rubidium(Rb)$ lies in the $5s$orbital.
$\therefore$ $n = 5$ For $Rubidium (Rb)$.
Azimuthal quantum number ($l$) specifies about the subshell. Its value lies between $0$ to$n - 1$, the following are the values for the respective orbitals, $l = 0;s:l = 1;p:l = 2;d:l = 3;f$ .
$\therefore$The valence orbital is in $Rubidium (Rb)$ for which the value of azimuthal quantum number will be $l = 0$.
Magnetic quantum number (${m_l}$) tells about the orientation in space of an orbital with $n$ and$l$. Its value lies between $- l,0, + l$ .
$\therefore$In $Rubidium(Rb)$ , ${m_l} = 0$ for $l = 0$
Spin quantum number gives us the orientation of electrons in space which can be either anticlockwise denoted by $s = \dfrac{{ - 1}}{2}or$ clockwise denoted by $s = \dfrac{1}{2}$ .
As only one electron is present in the valence shell it can acquire any orientation $s = \dfrac{{ - 1}}{2}or$ $s = \dfrac{1}{2}$ .
$\therefore n = 5;l = 0;{m_l} = 0;s = \pm \dfrac{1}{2}$

Hence the correct option is (a).

Note: $Rubidium(Rb)$ is an element from $1st$ group, thus it is called an alkali metal. Generally the alkali metals are mostly available in $+ 1$ oxidation state as by acquiring $+ 1$ the element has the most stable inert gas like configuration.