[IBM]

Cu+ or Cu++ Which one is valid and why? Could you explain it, please? I am confused

[wildfyr]

They are different but both real chemical species. Cu++ would normally be written Cu²⁺

[IBM]

They are different but both real chemical species. Cu++ would normally be written Cu²⁺

But Cu has 1 electron in 4s if we see the electrons configuration, So it should only get the charge Cu+. Is it possible that it would be cu2+ and when?

[Enthalpy]

https://en.wikipedia.org/wiki/Transition_metal

It's not even limited to transitions metals. Carbon for example can make CO and CO₂, both stable.
https://en.wikipedia.org/wiki/Oxidation_state
https://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements
that table lists them all, but not even all bold figures are common.

[wildfyr]

Transition metals, lanthanides, and actinides can have several common transition states.

This is because atoms in these parts of the periodic table do not really adhere to the octet rule. One reason is because even when s electrons are the valence electrons of the 0 oxidation state for an atom, d orbital electrons can still participate in bonding.

[magician4]

But Cu has 1 electron in 4s if we see the electrons configuration, So it should only get the charge Cu+. Is it possible that it would be cu2+ and when?

sure? even with the "naked" copper atom having the configuration [Ar] 3d¹⁰ 4s¹  (instead of [Ar]4d⁹3s² , as derived from Madelung-population scheme? )

  if memory serves, configuration of the Cu⁺ - ion should be reported as [Ar]3d¹⁰ 4s⁰ , due to the special stability of the completly filled d-level
  this is one of those well-known like ~ 20% exceptions from the electronical population scheme

... but I am eager to learn about new insights, if you happen to have any scientific source with respect to this


regards

Ingo

[IBM]

Transition metals, lanthanides, and actinides can have several common transition states.

This is because atoms in these parts of the periodic table do not really adhere to the octet rule. One reason is because even when s electrons are the valence electrons of the 0 oxidation state for an atom, d orbital electrons can still participate in bonding.

'when s electrons are the valence electrons of the 0 oxidation state for an atom, d orbital electrons can still participate in bonding.' Could you simplify the point, please?