Consider the following electron configuration.

Question

Consider the following electron configuration.

(σ_3s)²(σ_3s*)²(σ_3p)²(π_3p)⁴(π_3p*)⁴

From this list, choose four species that, in theory, would have this electron configuration.

Cl₂⁻

S₂

S₂⁻

Ar₂

Cl₂

Ar₂²⁺

Cl⁻

SCl⁻

Ar₂⁺

S₂²⁻

Answer 1

This question is about MO Theory, combining Atomic Orbitals to form Molecular Orbitals. Assuming you know how to basically set up the MO diagram and know the difference between a sigma bonding and antibonding (*) orbitals, and the pi bonding and antibonding Molecular Orbitals. After the AO's overlap to form the MO's, we need to take the electrons that are being brought in by each contributing atom or ion and "dump them into the MO diagram" we have created.

The structure we are looking for has the 3s sigma bonding and antibonding MO's both filled, that's 4 e-. Then from the p orbitals overlapping, we have sigma bonding MO with 2 e-, 4 e- in the pi bonding MO's and 4 e- in the pi antibonding MO's. That's a total of 14 e- in the Molecular Orbitals for the molecule (or polyatomic ion). Therefore, we are looking for any of these species that have 14 total valence electrons being used by both atoms/ions in energy level 3.

This can be done several different ways - the most obvious is 7 valence electrons each, which would match up with Cl₂ molecule. Each Cl atom would bring in its electrons, and in the valence level n = 3, the configuration is 3s²3p⁵, which is 7 valence e- for each Cl atom. Two Cl atoms coming together to form a molecule would have 14 valence e- into the MO diagram. So that is one of your answers!

Look at the other atoms and how many valence electrons they would have. S has 6 val e-, so S₂ would only have 12 e- in the MO. That doesn't work. But if we put 2 more e- in there because of total negative charge on the species being -2, we can see that S₂^2- would work! Can you find the other two options that total 14 valence electrons? Hope that helps, good luck!

Answer 2

Watch this YouTube video to learn electron config: https://youtu.be/ououF9nHUhk

When finding the electron config of an element, each square in the periodic table represents an electron in the atom of that element. You stop the electron config process when you arrive at the element you are looking for. The orbitals s,p,d... are the orbital types where the electrons are located as they orbit the atom of that element.

ElECTRONS ARE NEGATIVE. When an element has plus + charge it lost an electron, so remove the last electron added in the config. If +2 remove 2... +3 remove 3. When an element has negative - charge an electron is added, so add electron to the config. If -2 add 2... -3 add 3.

Example: Al, Al⁺³

Al: 1s²2s²2p⁶3s²3p¹

Al⁺³: 1s²2s²2p⁶

Example: O, O^-2

O: 1s²2s²2p⁴

O^-2: 1s²2s²2p⁶

Example: F, F^-

F: 1s²2s²2p⁵

F^-: 1s²2s²2p⁶

Al⁺³, O^-2, F^- all have the same electron config. Hope this helps :)