question archive The chemical equation for the formation of diaquabis(ethylenediamine)copper(II) iodide is (CH?COO)?Cu·H?O(aq) + 2H?NCH?CH?NH?(aq) + 2KI(aq) + H?O(l) → Cu(H?NCH?CH?NH?)?(H?O)?I?(s) + 2CH?COOK(aq) The product is a purple ionic solid, Cu(en)?(H?O)?²?, 2I?
The chemical equation for the formation of diaquabis(ethylenediamine)copper(II) iodide is (CH?COO)?Cu·H?O(aq) + 2H?NCH?CH?NH?(aq) + 2KI(aq) + H?O(l) → Cu(H?NCH?CH?NH?)?(H?O)?I?(s) + 2CH?COOK(aq) The product is a purple ionic solid, Cu(en)?(H?O)?²?, 2I?
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The chemical equation for the formation of diaquabis(ethylenediamine)copper(II) iodide is
(CH?COO)?Cu·H?O(aq) + 2H?NCH?CH?NH?(aq) + 2KI(aq) + H?O(l) → Cu(H?NCH?CH?NH?)?(H?O)?I?(s) + 2CH?COOK(aq)
The product is a purple ionic solid, Cu(en)?(H?O)?²?, 2I?.
Octahedral diaquabis(ethylenediamine) complexes can exist in two isomeric forms.
In the trans isomer, the bonds to the two water molecules are at 180° to each other.
In the cis isomer, the bonds to the two water molecules are at 90° to each other.
I am not an inorganic chemist, so I am not sure, but I believe the compound prepared in this procedure is the trans isomer.
