Question

How do you draw the electron configuration diagram for Aluminum ?

Answer 1

The nuclear number of aluminum is $13$. We can keep the octet rule to compose the setup as per $K - shells$ can oblige $2{\text{ }}electrons$ $L - shell$ can oblige $8{\text{ }}electrons$ and the leftover $3{\text{ }}electrons$ will be in the $M - shell$.

Complete step by step answer:
To compose the Aluminum electron arrangement we first need to know the quantity of electrons for the $Al$ molecule $\left( {there{\text{ }}are{\text{ }}13{\text{ }}electrons} \right).$At the point when we compose the design we'll place each of the $13{\text{ }}electrons$ in orbitals around the core of the Aluminum molecule.
Recorded as a hard copy of the electron arrangement for Aluminum the initial two electrons will go during the $\;1s$ orbital. Since$\;1s$ can just hold two electrons, the next $2$ electrons for aluminum go during the $2s$ orbital. The following six electrons will go in the $\;2p$ orbital. The p orbital can grip up to six electrons. We'll place six in the $\;2p$ orbital and afterward put the following two electrons during the $3s$. Since the $3s$ is now full we'll move to the 3p where we'll put the leftover electron. Thusly the Aluminum electron design will be $1{s^2}2{s^2}2{p^6}3{s^2}3{p^1}$
The design documentation gives a simple method to researchers to compose and impart how electrons are organized around the core of an iota. This makes it more obvious and anticipates how molecules will collaborate to frame substance securities.

Note:
According to Pauli’s Exclusion Principle , each orbital can hold $2{\text{ }}electrons$ s-orbital set contains one orbital, along these lines can hold $2{\text{ }}electrons$. Essentially, the p-orbital set had three orbitals, along these lines can hold $6{\text{ }}electrons$ and d-orbital and f orbital have five and seven orbitals subsequently, can opening $10{\text{ }} and {\text{ }}14{\text{ }}electrons$ individually.