# Question 19968

Apr 26, 2017

Mn(II)

#### Explanation:

From electron configurations
$F {e}^{o}$:[Ar]4s^2"3d_-2^2d_-1^1d_0^1d_1^1d_2^1# => 4 unpaired
$F e \left(I I\right) : \left[A r\right] 3 {d}_{-} {2}^{2} {d}_{-} {1}^{1} {d}_{0}^{1} {d}_{1}^{1} {d}_{2}^{1}$ => 4 unpaired
$F e \left(I I I\right) : \left[A r\right] 3 {d}_{-} {2}^{1} {d}_{-} {1}^{1} {d}_{0}^{1} {d}_{1}^{1} {d}_{2}^{1}$ => 5 unpaired
$M n \left(I I\right) : \left[A r\right] 3 {d}_{-} {2}^{1} {d}_{-} {1}^{1} {d}_{0}^{1} {d}_{1}^{1} {d}_{2}^{1}$ => 5 unpaired

Fe(III) & Mn(II) both have 5 unpaired electrons spread across the 5 orientations of the d-orbital system. Note: in ionization of transition metals, the 1st and 2nd ionizations occur at the ns valence level, n = the principle energy level. If a 3rd ionization is needed, then an electron is oxidized from the (n-1)d-orbital.

Recommend studying:
- How to write electron configuration summaries of electronic orbital systems for cations, anions and neutral elements from the elements position on the periodic table. (reference => Aufbau Principle, Aufbau Diagram & Hund's Rule).