[OTI]

I'm just wondering how you find the charges of elements in the D block.

My teacher gave us a sheet with the guide to charges and polyatomic ions, but the sheet didn't tell me how to find the charges of most of the D block elements.

[mikasaur]

Do you mean the common charges of d-block ions? I don't know that there is a hard and fast rule for determining them like you can with the alkali metals and halogens. Those are relatively easy because you can think of them wanting to get to the noble gas state so they'll give up (alkali metals) or gain (halogens) electrons really easily to become charged.

The d-block ions are tougher. I imagine it has something to do with the quantum state of the orbitals and trying to get to as low an energy level as possible. For instance Cu(I) kinda makes sense to me because the valence configuration of Cu is 3d¹⁰4s¹ and giving up an electron gets rid of that unpaired electron in the 3s orbital. Why Cu(II) is common, I'm not sure.

It may have something to do with molecular orbital theory? And the way in which the cation shares electrons with the anion it's bonded to? But I'm not sure how much MO theory pertains to mostly-ionic bonds.

Sorry for the rambling answer but I saw that you didn't get a response and thought I'd chime in. The short answer to your question is that I think a lot of those just have to be memorized.

[OTI]

Okay, guess I have a lot of memorization to do...

[Enthalpy]

Just a detail: remember to think cautiously with unpaired electrons. Pairing itself removes very little energy by magnetic interaction, but the nearer charges may raise the energy; pairing is rather a means to put two electrons on the same orbital.

If pairing permits a lower sum of energies, say in a covalent bond where both electron can be in a new favourable molecular orbital, fine. But if several orbitals have about the same energy, then the electrostatic repulsion tends to spread them unpaired on different orbitals, and this happens especially for transition metals - or in O₂ for instance.