Question

Minimum number of electrons having ${{\text{m}}_{\text{s}}}\, = \, - 1/2$ in Cr is:

Answer 1

To answer this question we should know what ${{\text{m}}_{\text{s}}}$ is. ${{\text{m}}_{\text{s}}}$ is the spin quantum number. An electron can have two spins, upward and downward. So, an electron can have two spin quantum numbers, $- 1/2$ and $+ 1/2$. We have to write the electronic configuration to determine the number of electrons having ${{\text{m}}_{\text{s}}}\, = \, - 1/2$.

Complete step-by-step answer:
The chromium is a transition metal having atomic number $24$.
We use three rules in filling electrons.
Pauling rule: according to which two electrons cannot have the same value for all four quantum numbers.
Hund’s rule: according to this rule, electrons are filled singly first then get paired.
Aufbau rule: according to this electrons get filed in lower energy level first then in higher energy level.
By using all three rules we can write the electronic configuration of chromium as follows:
${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{d}}^4}$
As $4$s-orbital has lower energy then $3$d-orbital so, the above electronic configuration should be correct but the actual configuration is different because the transfer one electron from $4$s-orbital to $3$d-orbital gives the half-filled $3$d-orbital and the half-filled and fully filled electronic configuration are most stable so, the actual electronic configuration of chromium is,
${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^1}{\text{2}}{{\text{d}}^5}$
Now we will count the number of electrons having up spin and downspin. The electrons having up spin are $15$ and the electrons having down spin are $9$. If we consider, that the up spin of the electrons is denoted by $- 1/2$ then the number of ${{\text{m}}_{\text{s}}}\, = \, - 1/2$ will be $15$ and if we consider, that the down spin of the electrons is denoted by $- 1/2$ then the number of ${{\text{m}}_{\text{s}}}\, = \, - 1/2$ will be $9$.
We have to determine the minimum number of electrons having ${{\text{m}}_{\text{s}}}\, = \, - 1/2$ so, we can say the minimum number of electrons having ${{\text{m}}_{\text{s}}}\, = \, - 1/2$ are $9$ that are having downward spin.Therefore, the minimum number of electrons having ${{\text{m}}_{\text{s}}}\, = \, - 1/2$ in Cr is $9$.

Note: Generally the up spin of electron is denoted by ${{\text{m}}_{\text{s}}}\, = \, + 1/2$ and down spin is denoted by ${{\text{m}}_{\text{s}}}\, = \, - 1/2$. We can also determine the maximum number of electrons having ${{\text{m}}_{\text{s}}}\, = \, + 1/2$ that are $15$. Chromium disobeys the Aufbau rule. Similarly the copper also disobeys the Aufbau rule. The energy of the energy level is determined as ${\text{n}}\,{\text{ + }}\,{\text{l}}$. The l value for s-orbital is zero so, the energy of $4$s is $4$. The l value for p-orbital is $1$ so, the energy of $3$d is s $4$. As both have the same ${\text{n}}\,{\text{ + }}\,{\text{l}}$ value so, the energy is lowest for the one having low value of l that is s-orbital.