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The reaction of Zn2+ ion with NH3 draws some complication since NH3 has lone pairs and is considered as electron donor. On the other hand, Zn has vacant orbitals which can accommodate NH3 molecules as ligands.
The first reaction would has limited amount of NH3 molecules which can form Zn(OH)2. the reaction can be written as:
`Zn^(2+) + 2 NH_3 + 2 H_2O -> Zn(OH)_2 _(s) + 2 NH_4^(+)`
For the second reaction, there is an excess of NH3 which apparently can produce coordination and the product would be Zn[(NH3)4]^(2+). The reaction equation can be written as:
`Zn^(2+) _(aq) + 4 NH_3 (excess) -> Zn[(NH_3)_4]^(2+) _(aq) `
In the first reaction, precipitates of zinc hydroxide would be present. On the second reaction, the product formed is aqueous which means that they are soluble in the solution.
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