Consider the following electron configuration: (\sigma 3s)^{2} (\sigma 3s*)^{2} (\sigma 3p)^{2} (\pi 3p)^{4} (\pi 3p*)^{4} Give

Step 1
The given configuration starts with 3s-which shows they are valance configuration.
So, first of all the total number of valence electrons should be calculated from the given configuration as:
${\displaystyle 2+2+2+4+4=14}$.
Now as per the given configuration, 4 such species are required which have:
(a) 14 valence electrons
(b) under 3rd shell as the configuration is upto 3rd shell only.
Step 2
Depending upon the above two conditions, four species can be listed as:
(1) ${\displaystyle C{l}_2}$:
The chlorine molecule is formed by two Cl-atoms and each Cl-atom has 7 valance electrons.
So, total number of valance electrons are ${\displaystyle 2\times 7=14}$.
Also, the valance electronic configuration of Cl-atoms is ${\displaystyle 3s^{2}3p^5}$. They are under 3rd shell.
(2) Sulfide ion ${\displaystyle \left(S_2^{-2}\right)}$:
The sulfer has 6 valance electrons, so two sulfur atoms form ${\displaystyle 2\times 6=12}$ electrons while the two negative charge is added.
Hence, the total valance electrons over sulfide ion is ${\displaystyle =(2\times 6)+2=14}$.
Also, the valance electronic configuration for S-atoms is ${\displaystyle 3s^{2}3p^4}$. They are under 3rd shell.
(3) ${\displaystyle SC{l}^{-}}$ ion:
Here, the sulfer has 6 valance electrons,and Cl has 7 valence electrons. The ion also bears a negative charge which is added.
Hence, the total valance electrons over sulfide monochloride ion is ${\displaystyle =6+7+1=14}$.
Also, the valance electronic configuration for S-atoms is ${\displaystyle 3s^{2}3p^4}$ and of Cl-atoms is ${\displaystyle 3s^{2}3p^5}$. They are under 3rd shell.
(4) ${\displaystyle {\left(A{r}_2\right)}^{+2}}$ ion:
Here, each argon atom has 8 valance electrons. The ion also bears 2 positive charge which will be subtracted.
Hence, the total valance electrons over ${\displaystyle {\left(A{r}_2\right)}^{+2}ionis=8+8-2=14}$.
Also, the valance electronic configuration for Ar-atoms is ${\displaystyle 3s^{2}3p^6}$. They are under 3rd shell.

1) ${\displaystyle SC{l}^{-}\Rightarrow }$ sulfur has 6 valence electrons and chlorine has 7 valence electrons for a total of 13 valence electrons. but because of the negative charge, it gains one, so it has 14 valence electrons. THIS IS ONE OF OUR ANSWERS.
2) ${\displaystyle C{l}_2^{-}\Rightarrow }$ chlorine has 7 valence electrons. multiply by 2 to get 14 valence electrons. but because of the negative charge, there is 1 too many for this to be an answer
3) ${\displaystyle A{r}_2^2+\Rightarrow }$ since argon is a noble gas, it has 8 full valence electrons. multiply by two because there are 2 argons and you get 16 valence electrons. but because of the 2+ charge, it loses to valence electrons for a total of 14. THIS IS ANOTHER ANSWER
4) ${\displaystyle C{l}_2\Rightarrow 7}$ valence electrons, multiply by two ${\displaystyle =14}$ valence electrons.

$C{l}_2,S_2^{2-},S{O}^{2-},andA{r}^{2+}$ have the same number of
electrons. In theory. these four species would have the same electronic
configuration with $(\sigma 3s{)}^2(\sigma 3s\ast {)}^2(\sigma 3p{)}^2(\pi 3p{)}^4(\pi 3p\ast {)}^4$
Answer: $C{l}_2,S_2^{2-},S{O}^{2-},andA{r}^{2+}$