Give the electron configuration for the following:
A) Ni^2+
B) Br^-
Ni is 1s2 2s2 2p6 3s2 3p6 3d8 4s2
Remove the 2 4s electrons to make the 2+ ion.
Now you do Br. Write the neutral atom then add an electron to the outside shell.
To determine the electron configuration for an ion, you'll first need to know the electron configuration of the neutral atom from which the ion is formed. The electron configuration is a way of representing the arrangement of electrons in an atom.
A) Ni^2+: Nickel (Ni) is the neutral atom with an atomic number of 28. The electron configuration for the neutral Ni atom is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8. When Ni loses two electrons to become Ni^2+, the two electrons are removed from the highest energy level, which in this case is the 4s orbital. Therefore, the electron configuration for Ni^2+ is 1s^2 2s^2 2p^6 3s^2 3p^6 3d^8.
B) Br^-: Bromine (Br) is the neutral atom with an atomic number of 35. The electron configuration for the neutral Br atom is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^5. When Br gains one electron to become Br^-, that extra electron is added to the highest available energy level, which in this case is the 4p orbital. Therefore, the electron configuration for Br^- is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6.