Question

How do you calculate the number of microstates a compound has?

Answer 1

Hint : Number of microstates is the number of possible arrangements of electrons in a compound. We can easily calculate those using formulas but let’s understand it as we have one electron in a p-orbital, so it can be arranged in 6 ways, with positive half spins in all the orbitals and also can be arranged in these with negative half spin.

Complete Step By Step Answer:
The number of arrangements are possible in many orbitals s, p, d etc. If we have electrons in same orbitals like if we have two electrons in p- orbital, then we can write the formula as- ${}^n{C_r}$ where $n$ is the total number of electrons which that orbital can possess and $r$ is the number of electrons it is having for that case. These numbers of arrangements are called microstates.
${}^n{C_r} = \,\dfrac{{n\,!}}{{r\,!\, \times \,(n - r)!}}$
Let’s take an example for your better understanding, we have two electrons in the p- orbitals so, in p- orbitals there can be a maximum of six electrons possible in it therefore we have value of $n = 6$ and we are talking about the possible arrangement for two electrons so, $r = 2$ .
Let’s calculate what comes when we solve it- ${}^6{C_2} = \,\dfrac{{6!}}{{2!\, \times \,4!}}$
$\dfrac{{6!}}{{2!\, \times \,4!}}\, = \dfrac{{6\, \times \,5\, \times \,4\, \times \,3\, \times \,2\, \times \,1}}{{(2\, \times \,1)\,(4\, \times 3\, \times 2\, \times 1)}}$ = $\dfrac{{6\, \times 5}}{2}$ = $15$
It means that there are a total of $15$ arrangements possible when we try to put these two electrons in three orbitals of p.

Note :
We know the formula for microstates when both electrons are in same orbital as ${}^n{C_r}$ where where $n$ is the total number of electrons which that orbital can possess and $r$ is the number of electrons it is having for that case.. But there is different formula when electrons are in different orbitals it will be like this $No.\,of\,microstates\, = \,{}^{{n_1}}{C_{{r_1}}}\, \times {}^{{n_2}}{C_{{r_2}}}$ .