Consider the following electron configuration. (σ3s)2 (σ3s*)2 (σ3p)2 (π3p)4 (π3p*)4 select four species that, in theory, would have this electron configuration.

Explanation:

The first step is to get the total number of valence electrons. This is obtained by summing the exponents at the top in;

(σ3s)2 (σ3s*)2 (σ3p)2 (π3p)4 (π3p*)4

This gives us;

2+2+2+4+4=14.

We are pretty much looking for diatomic species (molecules or ions) with 14 electrons.

1) Cl₂ --> This molecule is made up of two chlorine aoms. Chlorine atoms belong to group 7, hence they have a valency of 7. The total valency in this diatomic molecule is given as; 2 * 7 = 14.

2) Ar₂²⁺ -->  This ion is made up of two argon atoms. Argon belongs to group 8, hence they have a valency of 8. The total valency in this diatomic molecule is given as; 2 * 8 = 16. The 2+ charge however, denotes that 2 electrons have been lost. The total number of valency electrons is now; 16 - 2 = 14.

3) S₂²⁻ --> This ion is made up of two sulphur atoms. Sulphur belongs to group 6, hence they have a valency of 6. The total valency in this diatomic molecule is given as; 2 * 6 = 12. The 2- charge however, denotes that 2 electrons have been gained. The total number of valency electrons is now; 12 + 2 = 14.

4)SCl⁻ --> This ion is made up of slphur and chlorine. Sulphur belongs to group 6, hence they have a valency of 6. Chlorine belongs to group 7, hence a valency of 7. The total valency in this diatomic molecule is given as; 6 + 7 = 13. The 1- charge however, denotes that 1 electron have been gained. The total number of valency electrons is now; 13 + 1 = 14.

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.