Consider the following electron configuration.
(σ3s)2(σ3s*)2(σ3p)2(π3p)4(π3p*)4
Give four species that, in theory, would have this electron configuration.
SCl – Cl – Cl2– S2– Ar22+ S2 Ar2 Ar2+ Cl2 S22
1 Answer

YOU ARE IN LUCK. i just so happen to have this for homework as well:) assigned to me at 12 am today in fact.
this is how i did it:
first i counted the exponents at the top: 2+2+2+4+4=14.
the number 14 is the number of valence electrons that the species must have for the electron configuration to work.
1)SCl^ –> 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) Cl_2^ –> 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) Ar_2^2+ –>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)Cl_2 –> 7 valence electrons, multiply by two= 14 valence electrons. ANOTHER ANSWER
So, just keep doing that for the other species and you will get your final answer. this is all about knowing the valence electrons of each element and knowing what a positive charge and a minus charge do to the element. the charges just change the number of valence electrons.
the last one you need is S_2^2
Source(s): http://web.chem.ucsb.edu/~devries/chem1C/handouts/… i basically looked here for help and while looking at this, i realized a pattern and BAM! it hit me how i could do this.also what helped me was looking over my chemistry professors lecture notes and powerpoints. GO UCSC BANANA SLUGS!!!!!!!!!!!!!!!!! CLASS OF 2016!!