Zinc Chloride: The Lewis Acid with Electrophilic Properties

[hariharan venkatasu]

In which case I feel 4 s is the lowest energy orbital

 

[chemisttree]

Yes. So what what was the difficulty understanding that the electrons would occupy the lowest energy orbitals?

 

[hariharan venkatasu]

Thank you for your reply.I was wondering how the electrons once left 4 s orbital leaving the zinc ion can rejoin the ion again?

 

[chemisttree]

Thank you for your reply.I was wondering how the electrons once left 4 s orbital leaving the zinc ion can rejoin the ion again?

Your original question/observation was that ZnCl₂ was a lewis acid because it accepted electrons into its 4d orbitals rather than the 4p orbitals. Are you now saying that you don’t understand how Zn⁺² accepts electrons at all?

 

[chemisttree]

Also, kinda iffy applying Aufbau to non-ground state atoms.

ZnCl₂ is only partially an ionic compound. The Zn-Cl bonds have significant covalent character.

Perhaps your question should have been, “Why is ZnCl₂ able to share additional electrons?” The s orbital is being used in the bond with chloride so the hand waving answer is that it can accept additional electron pairs (definition of Lewis Acid) to complete its octet. It has 4 electrons in its outer shell and would like an additional two pair. For electron pair donors, this would mean another two species capable of sharing two additional pairs are possible. Zinc is known to form tetrahedral complexes so that tracks with the analysis. In the case where zinc is tetrahedral it is due to the 4s and 4p orbitals hybridizing.

Zinc also forms trigonal bipyramid, square pyramid and octahedral complexes. In these complexes it is obvious that empty d orbitals are involved.

Your question was very general regarding zinc’s Lewis Acid character and the simplest answer to that question is that it is a cation that can accept an electron pair into its 4s.

 

[hariharan venkatasu]

Thanks for the excellent answer

 

[hariharan venkatasu]

In post number 35 it has been mentioned that"It has 4 electrons in its outer shell and it would like an additional two pair"How come this is so when it has only two electrons?Please clarify this.I am sorry for a late communication

 

[TeethWhitener]

In post number 35 it has been mentioned that"It has 4 electrons in its outer shell and it would like an additional two pair"How come this is so when it has only two electrons?Please clarify this.I am sorry for a late communication

https://en.m.wikipedia.org/wiki/18-electron_rule

 

[hariharan venkatasu]

Thank you for your information about the wikipedia article.However I don't find any clarification for my question.Could you please elaborate on this?

 

[TeethWhitener]

Thank you for your information about the wikipedia article.However I don't find any clarification for my question.Could you please elaborate on this?

The 18-electron rule is a general rule of thumb like the Octet rule for predicting stability in transition metal complexes. In the case of zinc chloride, you get 10 electrons from the $Zn^{2+}$ and 2 electrons each from the chlorides, giving 14 electrons. Thus this compound is electron deficient, and can accept four more electrons (or 2 pairs) to satisfy the 18-electron rule. This is the broad reason why zinc chloride is a good Lewis acid.

 

[hariharan venkatasu]

Thank you for the clarification.However it has kindled the following doubts.

--You have said that Zn2+ ion has 10 electrons and 2 electrons each from chlorides.These chlorides are the ones that left zinc chloride molecule
--Do they act as ligands and form coordinate bond?
--The configuration of zinc is 2,8,18,2.Does it mean zinc will have one more 18 electron?
--What will be the ultimate configuration of zinc chloride after accepting the extra electrons?
Could you please answer these also?

 

[hariharan venkatasu]

I shall be immensely thankful if you could reply to my post 41.I am eager to hear from you

 

[hariharan venkatasu]

Normally I used to see your reply quickly and promptly. There is an unusual delay in your replying.Could you kindly expedite your reply for post 41?Thanking you.

 

[hariharan venkatasu]

Could you please reply to my post 41?I am eagerly waiting for the answers for my queries.Thanking you.

 

[HAYAO]

Normally I used to see your reply quickly and promptly. There is an unusual delay in your replying.Could you kindly expedite your reply for post 41?Thanking you.

Could you please reply to my post 41?I am eagerly waiting for the answers for my queries.Thanking you.

Personally, I think they have done tremendous amount of job replying to you in a patient manner. I don't think reminding them like this would help you get any answer. Remember, they are doing a favor for you. They are not obliged in any way to answer your questions.

These chlorides are the ones that left zinc chloride molecule
--Do they act as ligands and form coordinate bond?

What do you mean "left" zinc chloride molecule? Chloride do act as a ligand.

--The configuration of zinc is 2,8,18,2.Does it mean zinc will have one more 18 electron?
--What will be the ultimate configuration of zinc chloride after accepting the extra electrons?
Could you please answer these also?

That kind of way of writing electron configuration are the ones you see in introductory high school textbooks (at least over here in Japan), but they are not very good in explaining the actual electronic configuration.
The thread has been about s, p, d, orbitals the entire time, so how about you write accordingly, eg 1s², 2s², 2p⁶, ...

If you can do that, then you should also be able to answer your other questions. Like they've been saying, use the Aufbau principle.

 

[hariharan venkatasu]

I am extremely sorry if I have hurt you any way by sending reminders.It was my curiosity that prompted me to do so.I owe apologies.
Regarding zinc chloride when it ionizes the two chlorine ions leaving the molecule which is what I meant by "left". The electron configuration is 1 2 s^ 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^10
Once again my excuses and thanks for all the efforts you have put into answer my questions.