I'm trying to get a sense of how $\ce{Cu^2+}$ and $\ce{NH3}$ are bonded together because it doesn't make any! I understand that $\ce{NH3}$ has a lone pair of electrons that can be donated (right?), but since there are four molecules of $\ce{NH3}$, then there are 4 pairs, and $\ce{Cu^2+}$ only needs two! the reaction in my textbook was written like this: $$\ce{CuSO4.4NH3 <=> [Cu(NH3)4]^2+ + SO4^2-}$$ I would appreciate any clarification.
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$\begingroup$ You are aware that element symbols start with capital letters, aren't you ? It is rather $$\ce{[Cu(H2O)4]^2+ + 4 NH3 <=> [Cu(NH3)4]^2+ + 4 H2O}$$You can find useful How can I format math/chemistry expressions on Chemistry Stack Exchange $\endgroup$– PoutnikCommented Feb 17, 2021 at 7:07
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1$\begingroup$ I assume you know what a coordination complex is, right? $\endgroup$– andselisk ♦Commented Feb 17, 2021 at 13:10
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2 Answers
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There are many ways to describe chemical bonds. Copper II ion binds to some anions with ionic bonds in solids, sometimes in covalent bonds (Ref 1).
In aqueous solution, copper II ions are surrounded (hydrated) with six water molecules, with the oxygen ends pointing in, toward the positive copper ion. Water is acting as a Lewis base. Ammonia is a stronger Lewis base - after all, it is basic in water. So an ammonia molecule will replace a water molecule in a hexa-aquated copper ion. Eventually, you get $Cu(NH_3)_4)^{2+}$.$(H_2O)_2$. Oh, I forgot, sometimes we leave out the water molecules, since we just assume that they are there - everybody knows that, since the teacher has explained that (unless the teacher hasn't explained that).
In the equation of the question, only four of the actual six water molecules are shown, because they are important for describing the interaction with $NH_3$, and the other two are just a complication, because I know you would ask, why don't they get replaced too, to give $Cu(NH_3)_6^{2+}$? After all, nickel, cobalt and chromium III do! Well, zinc doesn't; it forms a tetra-ammonia complex. And iron, manganese and aluminum don't even dissolve to form soluble ammonia complexes in alkaline solution. Reference 3 suggests that it is an equilibrium situation, but there are deeper explanations which you can explore later.
Ref 1. https://aip.scitation.org/doi/10.1063/1.1677079
Ref 2. https://www.chemguide.co.uk/inorganic/transition/copper.html
Ref 3. https://www.chemguide.co.uk/inorganic/complexions/aquanh3.html#top
answered Feb 17, 2021 at 14:28
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$\begingroup$ Answer too long, too complicated, a lot of unnecessary information. If you want to be precise, you have to describe it with orbitals and coordination spheres. Citing papers seems "intelligent" but doesn't say anything at all. $\endgroup$– deutiCommented Feb 18, 2021 at 15:58
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$\begingroup$ I think this attempts to answer the question but is not very well written. $\endgroup$– ringoCommented Feb 23, 2021 at 16:58
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Yes, you are right. This is not a covalent bond it is a coordination bond. The copper has a electron pair deficit (Lewis Base) and the Nitrogen (Lewis Acid) donates the elctrons pair which makes the molecule more stable in the end.
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1$\begingroup$ The answer is a a bit short, and unfortunately defines Lewis acids and bases incorrectly. Lewis bases have electron pairs; Lewis acids accept them. I give you credit for trying. $\endgroup$ Commented Feb 17, 2021 at 13:59
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$\begingroup$ I made it correct, no reason to downvote. $\endgroup$– deutiCommented Feb 18, 2021 at 15:59
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$\begingroup$ @deuti I think you misunderstood the correction from James' comment. Copper accepts electrons, so it would be a Lewis acid; Nitrogen donates electrons, so it would be a Lewis base. $\endgroup$– Tyberius ♦Commented Feb 18, 2021 at 22:43
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$\begingroup$ @deuti: I didn't downvote, but didn't feel an upvote was appropriate, so I gave you verbal credit for trying. So keep on - everybody makes misteaks. $\endgroup$ Commented Feb 19, 2021 at 15:23