How Do I Complete This Lewis Structure?

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To complete the Lewis structure for ClF4+, a total of 34 electrons must be accounted for, derived from one chlorine atom and four fluorine atoms, adjusted for the positive charge. The structure currently includes a lone pair on the chlorine atom, but there is uncertainty about placing the additional electrons without violating the octet rule. It is noted that chlorine, being in the third period, can utilize d orbitals, allowing for the accommodation of extra electrons. While there are no strict rules for these scenarios, having available orbitals is essential, and symmetry can guide the placement of electrons to avoid complex resonance structures. Understanding these principles can aid in constructing accurate Lewis structures for similar compounds.
JeweliaHeart
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Homework Statement


Draw the best Lewis Dot Structures the following species:
ClF4+

Homework Equations



none

The Attempt at a Solution



I calculated that there should be a total of 34 e- in the Lewis structure":
7 (1 chlorine atom) + 28 ( 7 from each of the four fluorine atoms) - 1(b/c the atom has a plus one charge)= 34 e-

My Lewis structure (what I have in it so far) is in the attachment below. I don't know how to add in the two other electrons w/o violating the octet rule or even where they should go if I did add them.
 

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The lone pair sits on the chlorine atom. You have to invoke higher-energy orbitals, so the octet rule is not followed.
 
Well, thank you.

And this must be possible b/c the chlorine atom is the third period so it can utilize d orbitals, right?

If so, that makes perfect sense to me, but just for future reference is there some hard & fast rule to apply whenever one comes across problems like these: For example, the extra electrons always goes to the atom that can support d orbitals (i.e. period 3 or higher elements)?
 
JeweliaHeart said:
And this must be possible b/c the chlorine atom is the third period so it can utilize d orbitals, right?
Right. I should've specified that when I said "higher-energy orbitals", it will only be stable if the difference in energy is not to big, and therefore the orbitals must come from the same shell.

JeweliaHeart said:
If so, that makes perfect sense to me, but just for future reference is there some hard & fast rule to apply whenever one comes across problems like these: For example, the extra electrons always goes to the atom that can support d orbitals (i.e. period 3 or higher elements)?
I wouldn't say that there are "hard & fast rules", especially since Lewis diagrams are actually gross simplifications of reality, and you have to do some quantum chemistry to really understand molecular structures. But you do need orbitals available to put in the extra electrons. And I think that symmetry is often a good guide also: were the F atoms able to gain extra electrons, that would result in a series of "ugly" resonance structures to account for all the possibilities.
 

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