d and s orbitals

[205743684]

1E.21 How many unpaired electrons are predicted for the ground-state configuration of each of the following atoms: (d) Ni?.

I thought the electron configuration of the ground state of Ni is (Ar)3d^84s^2 (because d is lower energy than s), yet the answer to this question is 2 unpaired electrons assuming it was looking at the d orbital, why? I thought we always had to write d in front of s in the electron configuration...'

[August_Blatney_1A]

You do need to write out d before s but thats not what the question is asking. When we fill the s orbital, 2 electrons fill the orbital leaving no electrons unpaired, but when we fill the d orbital, there are unpaired electrons. d has 5 orbitals with 2 electrons each. Using Hunds rule we fill all 5 orbitals with one electron then we start over to fill the orbitals with the remaining 3 electrons in this case. This means that 3 orbitals are full and 2 orbitals are not. In these 2 orbitals there is 1 electron each which makes 2 unpaired electrons.

[Ayla3H]

1E.21 How many unpaired electrons are predicted for the ground-state configuration of each of the following atoms: (d) Ni?.

I thought the electron configuration of the ground state of Ni is (Ar)3d^84s^2 (because d is lower energy than s), yet the answer to this question is 2 unpaired electrons assuming it was looking at the d orbital, why? I thought we always had to write d in front of s in the electron configuration...'

Ni has Z = 28. The ground state electron configuration of Ni: [Ar]3d^8 4s^2
As we can see from the electron configuration, the 4s-orbital already has two paired electrons and 0 unpaired electrons. Now, we have a remaining 8 valence electrons that must be in the d-orbitals. A d-subshell has 5 orbitals total, all of which can hold one pair of electrons. First we have to fill the orbitals singly before we start pairing electrons. We will use 5 electrons to fill the d-orbitals singly before we can start pairing them. After we have filled all of the 3d-orbitals singly, we can use our remaining 3 electrons to create pairs in 3 out of the 5 orbitals. This leaves us with 2 unpaired electrons in the d-orbital.

Also, our TA told us that we don't have to worry about the order in which we write electron configurations.

[Alexander Moroz 1B]

Then S subshell is full, leaving 8 electrons left in the configuration to fill subshell d. According to Hund's rule, it is more stable to first fill each orbital with only one electron and once each orbital is half full, begin filling orbitals completely. Since d has 5 orbitals, all five will have one electron, which leaves 3 electrons. The three electrons fill 3 of the 5 orbitals, which leaves 2 orbitals only half full - also considered unpaired electrons. This is how we get to 2 unpaired electrons.

[Sonia Virk 2A]

It seems like you understand that there is 2 unpaired electrons for the Ni configuration. The 4s subshell comes before the 3d in terms of number of electrons but it is written with 3d before 4s because electron configuration is typically written in order of the principle quantum number (n).

[Evan L 2B]

The way you wrote the electron configuration is correct, but when the question asks how many unpaired electrons there are we still need to look at the unfilled subshell. In this case, d is the subshell and it has 8 electrons. Due to Hund's rule and the Pauley exclusion principle, we know that all five of the d orbitals will have at least one electron and that 3 more of those orbitals will have a second electron that is spin paired. This means that there are 2 orbitals/electrons that are unpaired.

[Alexandra Gill]

The answer is two unpaired electrons because the 4s2 electrons are paired, and 3 of the 5 orbitals in the d subshell are paired since according to Hund's rule, electrons fill up each orbital before pairing. Because it is 3d^8, there will be 3 paired electrons and 2 unpaired electrons within the 5 orbitals.