Question #19968

1 Answer
Apr 26, 2017

Answer:

Mn(II)

Explanation:

From electron configurations
#Fe^o#:[Ar]#4s^2"3d_-2^2d_-1^1d_0^1d_1^1d_2^1# => 4 unpaired
#Fe(II):[Ar]3d_-2^2d_-1^1d_0^1d_1^1d_2^1# => 4 unpaired
#Fe(III):[Ar]3d_-2^1d_-1^1d_0^1d_1^1d_2^1# => 5 unpaired
#Mn(II):[Ar]3d_-2^1d_-1^1d_0^1d_1^1d_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).