## Ground-State Electron Configurations of Ions 3.11

Carrie Huang 2F
Posts: 13
Joined: Fri Sep 25, 2015 3:00 am

### Ground-State Electron Configurations of Ions 3.11

Question 3.11 states: Which M^3+ ions (where M is a metal) are predicted to have the following ground-state electron configurations:
a) [Ar]3d^6
b) [Ar]3d^5

The answers in the solutions manual are Co^3+ and Fe^3+, respectively. Why are the answers not Fe^3+ and Mn^3+?

Alina Avetisyan 1B
Posts: 25
Joined: Fri Sep 25, 2015 3:00 am

### Re: Ground-State Electron Configurations of Ions 3.11

a) The configuration for Co is [Ar] 3d^7 4s^2
To become C0^3+, it loses the two electrons in 4s, and then 1 electron from the d orbital and the final configuration will be [Ar]3d^6

b)The configuration for Fe is [Ar] 3d^6 4s^2
To become Fe^3+, Fe will lose the two electrons in 4s, and then lose one electron in the d orbital and the final configuration will be [Ar]3d^5
(You can also count the number of total electrons. An element with the configuration [Ar]3d^5 has 23 electrons. Fe has 26 electrons but when it loses 3 electrons, it will have 23 electrons.)

Kim Vu 2G
Posts: 29
Joined: Fri Sep 25, 2015 3:00 am

### Re: Ground-State Electron Configurations of Ions 3.11

a) the electrons in the element Argon equals 18 and adding 6 more electrons from 3d^6 would make it 24 electrons, which would be chromium. However, the question is looking for a metal with 3+ ion which is means that 3 electrons were reduced from it already, so to find the element, you would add 3 more electrons, making it 27 electrons and the element with the corresponding atomic number is Cobalt, therefore Co3+ is the correct answer. It cannot be Fe+3 because from Cr to Fe is only 2 electrons away instead of 3.

b)AR=18 electrons, so adding 5 more electrons would be 23 electrons which is Vanadium. Following the steps in a, you would then add 3 more electrons to 23, making it 26 electrons and the atomic number of 26 is Fe, therefore the correct answer is Fe^3+.