What is the Lewis structure of #Cl_3#?

1 Answer
May 13, 2018

Answer:

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Explanation:

(Assuming that #Cl_3^-# was meant by #Cl_3#.)

  1. The total valence electrons for #Cl_3^-# is #7xx3+1=22#, since #Cl# has #7# valence electrons and there's a negative charge (which adds #1# electron).
  2. Let's draw a single bond between the #3# #Cl# atoms to begin:
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    Now, we have #22-4=18# electrons left to put on the diagram, since each single bond counts as #2# electrons.
  3. Then, we should complete the octets for all of the #Cl# atoms:
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  4. We still have #2# electrons left after placing all those valence electrons. #Cl# can have an expanded octet, though, so that's not too much of a problem.
  5. To determine where to put this extra electron pair, let's calculate the formal charges of all the situations:
  6. If the extra electron pair is placed on the left #Cl# atom, then the formal charge on that #Cl# atom would be #7-9=-2#.
  7. If it were placed on the right #Cl# atom, then that #Cl# atom would have a formal charge of #7-9=-2#.
  8. If it were placed on the central #Cl# atom, then that #Cl# atom would have a formal charge of #7-8=-1#.
  9. So, the best placement for that extra electron pair would be the central #Cl#, since that arrangement minimises formal charge.
  10. Let's draw that:
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  11. And add a bracket around everything, with a negative sign, to indicate that this is a negatively charged ion:
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  12. Finally, let's do a last double-check for the number of electrons. We should end up with #22# electrons. :)