- #1
walker
- 28
- 0
Okay I think this question has been addressed slightly before however I seem to be having difficulty on it still.
The question asks: Arrange the following atoms or ions in order of increasing radius Cl, S2-, K, K+, O
Now I have them arranged as
O < Cl < K+ < S2- < K
The arrangement of K+ and S2- is where I am stumped. The question goes on to ask "Give an explanation for the position of S2- in relation to the atom or ion that comes just before and just after" Now I would only assume they're talking about the K+ and K atom/ion so I placed S2- in between. Now my best guess at why S2- is where it is, is due to the gain in electrons. But how do I prove this? Is it because the S atom has gained two electrons and the K atom has only lost one that the S2- ion becomes larger than the K+ ion?
Is there a certain method I should be using to determine the actual resultant size of these atoms when they gain or lose electrons?
I tried to keep the post as informative as possible I'm not trying to scam any answer out of anyone I'm trying to learn how to solve this problem.
The question asks: Arrange the following atoms or ions in order of increasing radius Cl, S2-, K, K+, O
Now I have them arranged as
O < Cl < K+ < S2- < K
The arrangement of K+ and S2- is where I am stumped. The question goes on to ask "Give an explanation for the position of S2- in relation to the atom or ion that comes just before and just after" Now I would only assume they're talking about the K+ and K atom/ion so I placed S2- in between. Now my best guess at why S2- is where it is, is due to the gain in electrons. But how do I prove this? Is it because the S atom has gained two electrons and the K atom has only lost one that the S2- ion becomes larger than the K+ ion?
Is there a certain method I should be using to determine the actual resultant size of these atoms when they gain or lose electrons?
I tried to keep the post as informative as possible I'm not trying to scam any answer out of anyone I'm trying to learn how to solve this problem.