Question

Ammonium formation - Clarifying the confusion?

I have watched your explanation for the ammonium formation in your youtube channel, which makes a lot of sense. In which you are saying in the formation of an ammonium ion, the ion will lose a valence electron. However, there is another explanation I heard which says that the lone pair in the NH3 will attract a hydrogen ion thus leaving NH4 molecule with a positive charge. (The no. of Ps will be more than e)

Can you please clarify this to me?

Answer 1

Well, we are flying blind here.... I take it you want to represent (or understand) the action of ammonia as a base?

Explanation:

Ammonia is a molecular species, the which we could represent as `NH_3`. For such a small molecule, and it is lighter than WATER...we got a relatively high normal boiling point of `-33.3` `""^@C`, and the rationale advanced for this is `"intermolecular hydrogen bonding"`, given that the ammonia molecule is quite polar, and this polarity, i.e. `stackrel(-delta)N-stackrel(delta+)H`, is a potent contributor to intermolecular force.

And both in the pure solvent, we could represent the basicity of the ammonia molecule...(this is the so-called `"ammonolysis reaction"`, the ammonia counterpart to `"autoprotolysis"`...)

`2NH_3(l) rightleftharpoons NH_2^(-) + NH_4^+`

...but this acid-base reaction operates in pure ammonia ONLY. More likely, we address the acid-base reaction that operates for solutions of ammonia in WATER:

`NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^+ + HO^-`

Here....`K_b=1.74xx10^-5`...this applies to the water solvent. Anyway, over to you. If you are not satisfied, you will have to qualify your question.