When [Ni(NH₃)₄]²⁺ is treated with concentrated HCl(aq),two compounds having the same formula,[Ni(NH₃)₄Cl₂],designated I and II,are formed.Compound I can be converted to compound II by boiling it in dilute HCl(aq).A solution of I reacts with oxalic acid,H₂C₂O₄,to form [Ni(NH₃)₄(C₂O₄)].Compound II does not react with oxalic acid.Compound II is: A)the cis isomer B)tetrahedral in shape C)the trans isomer D)the same as compound I E)octahedral in shape

Question

Quizplus · Accepted Answer

Compound I can be converted to compound II by boiling it in dilute HCl(aq), which suggests that the two compounds are different isomers. Compound I can react with oxalic acid to form a complex, but compound II cannot react with oxalic acid. This suggests that compound I has two different ligands that are cis to each other, whereas compound II has these ligands in a trans arrangement. Therefore, the best choice is C, which represents the trans isomer. The other options can be eliminated because they do not explain the observed behavior of the two compounds.

When [Ni(NH3)4]2+ Is Treated with Concentrated HCl(aq),two Compounds Having the Same

When [Ni(NH₃)₄]²⁺ Is Treated with Concentrated HCl(aq),two Compounds Having the Same