Question #2c7b6
1 Answer
Explanation:
Right from the start, you can tell that the answer will be
Simply put,
#1s^2 2s^2 2p^6 3s^2 3p^6 3d^3#
Write the electron configuration of a neutral chromium atom. Chromium,.
This means that the electron configuration of a neutral chromium atom must account for a total of
#"Cr: " 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1#
In order for the chromium(II) cation to form, the neutral chromium atom must lose three electrons. These electrons will come from the orbitals that are highest in energy.
In this case, the first electron will come from the 4s-orbital, since this orbital is highest in energy. The electron configuration for the
#"Cr"^(+): 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5#
Now, the second and third electrons will come from the 3d-orbitals. Since these orbitals hold a total of
#"Cr"^(3+): 1s^2 2s^2 2p^6 3s^2 3p^6 3d^3#
This matches the electron configuration given to you. At this point, it's clear that removing three more electrons in order to form the chromium(VI) cation will result in a completely different electron configuration
#"Cr"^(6+): 1s^2 2s^2 2p^6 3s^2 3p^6#
Therefore, you can say that