Ground State

[Yasmin Olvera 1D]

The ground state for Copper ion Cu+ was being asked for. I thought it would be [Ar]3d^(9)4s^(1)

However, the correct answer is [Ar]3d^(10).

Can someone explain the reasoning behind this?

[jlinwashington1B]

I'm not really sure, but I know that it is losing an electron (considering the "+") which means that it is taken out of the s orbital.

[Chem_Mod]

Yes, because the ground state configuration of Copper is [Ar]3d^(10)4s^(1), removing one electron to form the Copper ion gives you [Ar]3d^(10). Electrons are removed from the highest energy level, the 4s, before the 3d. Remember that the 4s orbital is higher in energy than the d orbitals after electrons have been added to the d orbitals.

[jlinwashington1B]

So (hypothetically) in the event that there is an electron configuration that exceeds 4s1, we would take an electron from the next orbital because the energy would be higher?

[Brian Kwak 1D]

So the ground state for the neutral copper is [Ar] 3d^(10) 4s^(1) and remember that if it is an ion like Cu+ you would have to remove the highest charged electron because it is easier to remove than a lower charge and in the configuration the highest in energy is 4s so it will be the easiest to remove. Therefore Cu+:[Ar] 3d^(10).

[Leah farhadi 1F]

Since it has a plus charge you would subtract an electron from your ground state configuration. If it were neutral that is when you would give the ground state configuration without subtracting or adding electrons.

[somyapanchal1D]

Remember that copper is one of the exceptions for electron configuration, along with Chromium. Copper has an electron configuration of [Ar]3d^(10)4s^(1) in its normal state. For the copper Cu+ ion, the electron configuration is [Ar]3d^(10). This is because electrons are removed from the 4s orbital before they are taken out of the 3d orbitals. Thus, this is just an exception that you will have to remember.

[Karolina herrera1F]

This would be right only because cu+ is an exception so instead of it being what you first assumed the 1 from the s would travel over to the d which would make it d^10 because of cu+ being an exception.

[Zachary Menz 1D]

When writing electron configurations of ions, you should write out the uncharged electron configuration first, then add and subtract from there. Since Cu is one of the exceptions, its uncharged electron configuration is [Ar]3d^(10) 4s^(1). Now to make the +1 cation you take off the last electron, making it simply [Ar]3d^(10).

[Radha Patel 4I]

So Copper can't have a d-orbital 3d^9 because any atom written as 3d^9 or 3d^4 has to have its electrons evened out in the d orbital subshell, so it becomes [Ar] 3d^10. If you set up a d orbital sub shell for 3d^10 , you will be able to fill 2 electrons for each of the 5 orbitals to make 10 electrons. If you only added 9 electrons instead of 10, you would notice that there is one orbital half full of electrons. To get this extra electron, the d orbital retrieves an electron from the s orbital to make the d orbital 3d^10 which explains why we have 4s^1 instead of 4s^2. Another example is Chromium, it can't be 3d^4 , it has to be 3d^5. The reason for this is because the d orbital can only be half full or completely full.

[simmoneokamoto3K]

Actually the configuration for Cu is [Ar}3d^9 4s^2 but because 3d^9 is so close to have a complete shell it steals an electron from 4s^2 taking away one electron and reducing it to [Ar]3d^10 4s^1. Lastly because Cu^+ has the plus sign it means you're removing an electron versus in the hypothetical if you had Cu^-3 you'd add three electrons.

[Yasmin Olvera 1D]

This would be right only because cu+ is an exception so instead of it being what you first assumed the 1 from the s would travel over to the d which would make it d^10 because of cu+ being an exception.

Oh okay, because when I counted it and went through the process I got 9 not 10 for the D orbital

[Yasmin Olvera 1D]

Actually the configuration for Cu is [Ar}3d^9 4s^2 but because 3d^9 is so close to have a complete shell it steals an electron from 4s^2 taking away one electron and reducing it to [Ar]3d^10 4s^1. Lastly because Cu^+ has the plus sign it means you're removing an electron versus in the hypothetical if you had Cu^-3 you'd add three electrons.

DOes that mean we do this same process of transferring to an "almost full shell" for other configurations.

[Chem_Mod]

In general, half filled d shells and fully filled d shells are typically preferred (ie slightly more stable). Some atoms will transfer the electrons, some won't, especially the heavier elements. It is mostly a case by case basis