# What is the electron configuration of Cr 3+?

Nov 30, 2015

$C {r}^{3 +} : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{\textcolor{red}{0}} 3 {d}^{\textcolor{red}{3}}$

#### Explanation:

The atomic number of Chromium is $Z = 24$, therefore a $C r$ atom possesses $24$ electrons.

$C r : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{1} 3 {d}^{5}$

Note that it is $4 {s}^{1} 3 {d}^{5}$ and not $4 {s}^{2} 3 {d}^{4}$ because a half filled $d$ orbital is more stable than a partially filled $d$ orbital.

However, the chromium ion $C {r}^{3 +}$ possesses $24 {e}^{-} - 3 {e}^{-} = 21 {e}^{-}$ due to the loss of $3$ of its electrons.

Thus, the electron configuration of $C {r}^{3 +}$is:
$C {r}^{3 +} : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{\textcolor{red}{0}} 3 {d}^{\textcolor{red}{3}}$