Question #2c7b6

Dec 13, 2015

${\text{Cr}}^{6 +}$

Explanation:

Right from the start, you can tell that the answer will be ${\text{Cr}}^{3 +}$ or ${\text{Cr}}^{6 +}$, since you can't have two ions of the same element with the same electron configuration.

Simply put, ${\text{Cr}}^{3 +}$ and ${\text{Cr}}^{6 +}$ cannot have the same electron configuration, so one of these ions will not match the electron configuration given to you

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{3}$

Write the electron configuration of a neutral chromium atom. Chromium,. $\text{Cr}$, is located in period 4, group 5 of the periodic table, and has an atomic number equal to $24$.

This means that the electron configuration of a neutral chromium atom must account for a total of $24$ electrons

$\text{Cr: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{5} 4 {s}^{1}$

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 ${\text{Cr}}^{+}$ cation looks like this

${\text{Cr}}^{+} : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{5}$

Now, the second and third electrons will come from the 3d-orbitals. Since these orbitals hold a total of $5$ electrons, losing two electrons will produce the following electron configuration

${\text{Cr}}^{3 +} : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{3}$

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

${\text{Cr}}^{6 +} : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6}$

Therefore, you can say that ${\text{Cr}}^{6 +}$ will not have the given electron configuration.