When [Ni(NH₃)₄]²⁺ is treated with concentrated HCl, 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. 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

The fact that compound I can be converted to compound II by boiling it in dilute HCl suggests that I and II are isomers. The fact that II does not react with oxalic acid suggests that the oxalate ligand is in a trans configuration in II. Therefore, II must be the trans isomer. Furthermore, because a change in shape from tetrahedral to octahedral would result in a change in formula, II must have the same shape as I (tetrahedral). Therefore, the best choice is C, the trans isomer.

When [Ni(NH3)4]2+ Is Treated with Concentrated HCl, Two Compounds Having

When [Ni(NH₃)₄]²⁺ Is Treated with Concentrated HCl, Two Compounds Having