# Write the electron configuration of "Cd"^(2+)?

Nov 4, 2015

$1 {s}^{2}$ $2 {s}^{2}$ $2 {p}^{6}$ $3 {s}^{2}$ $3 {p}^{6}$ $4 {s}^{2}$ $3 {d}^{\text{10}}$ $4 {p}^{6}$ $4 {d}^{\text{10}}$

or

[Kr] $4 {d}^{\text{10}}$

#### Explanation:

The atomic number of cadmium, $C d$, is 48. Thus, the ground state electron configuration of this element is

$1 {s}^{2}$ $2 {s}^{2}$ $2 {p}^{6}$ $3 {s}^{2}$ $3 {p}^{6}$ $4 {s}^{2}$ $3 {d}^{\text{10}}$ $4 {p}^{6}$ $5 {s}^{2}$ $4 {d}^{\text{10}}$

or

[$K r$] $4 {d}^{\text{10}}$ $5 {s}^{2}$

But since the problem is draw the electron configuration of the $C {d}^{\text{2+}}$, this means that the element lost $2 {e}^{-}$.

Thus, the electron configuration of $C {d}^{\text{2+}}$ is

$1 {s}^{2}$ $2 {s}^{2}$ $2 {p}^{6}$ $3 {s}^{2}$ $3 {p}^{6}$ $4 {s}^{2}$ $3 {d}^{\text{10}}$ $4 {p}^{6}$ $4 {d}^{\text{10}}$