In the ground-state electron configuration of Fe^{3+}, how many unpaired electrons are present?

An explanation would be great, thanks.

Electron configuration Fe is

26Fe = 1s2 2s2 2p6 3s2 3p6 3d6 4s2.
To make the 3+ ion we remove the two 4s electrons and one of the 3d6 to make it 3d5. Remember Hund's that electrons don't pair until the must AND that there are 5 3d orbitals, then all 5 of the 3d electrons must be unpaired.

Thanks so much, that really helped me solve other problems.

To determine the number of unpaired electrons in the ground-state electron configuration of Fe^3+, we need to first write out the electron configuration for Fe^3+.

1. Start with the electron configuration of the neutral atom, iron (Fe): [Ar] 4s^2 3d^6.

2. Fe^3+ means that the atom has lost three electrons. Remove the three electrons from the highest energy level, which is 4s^2, since it has a higher energy than the 3d orbitals.

3. The electron configuration for Fe^3+ is: [Ar] 3d^5.

Now, we can determine the number of unpaired electrons. In the 3d orbital, there are a total of 5 electrons. According to Hund's rule, electrons will only pair up when all the orbitals in the sublevel have at least one electron. Since there are five electrons in the 3d orbitals, we can conclude that there is one unpaired electron.

Therefore, in the ground-state electron configuration of Fe^3+, there is one unpaired electron.