Textbook Question 6D.11

[Eliana Carney 3E]

Hey guys! I'm unsure about how to write the chemical formulas for the aqueous solutions in textbook question 6D.11. Specifically, I don't understand how to write the equations for part E and F. If someone could explain where the answer key got the additional water molecules attached to the metal cations, that would be appreciated!

[David Jen 1J]

Me too. Also, does Al3+ make it's surrounding solution acidic because it can give off protons or is it accepting electrons?

[Lea Chamoun 2J]

So I think that Al3+ makes the surrounding solution acidic because first, it is forming bonds with H2O. When this happens, the O that forms the coordinate covalent bond with Al3+ now has 3 bonds. O is not very stable when it has 3 bonds, and the positive charge on the Al3+ delocalizes the electrons in the molecule and weakens the OH bond. The OH bond is weakened enough that the H+ will be given off to H2O, which will form H3O+. So by gaining electrons in the coordinate covalent bond with O, Al3+ helps weaken the bond between the OH so that the entire molecule Al(H2O) gives off a proton to water.
However, I am also unsure how to write the equation- specifically why the Al3+ forms 6 bonds with 6 different H2O molecules. For part f, the Cu also forms 6 bonds with H2O, so could someone also explain why? Is it just because the transition metals and Al like to form octahedral complexes?