Write the electronic configuration of the following species Also and find the number of unpaired electron is each
`Fe,Fe^(2+),Fe^(3+)(Z of Fe = 26)`
b. `Br,Br^(Theta)(Z of Br = 35)`
`V,V^(3+)(Z of V = 23)`

Explanation

Follow the orde0r of increasing energy (authan rule)
`1s^(2)2s2p3s 4s 3p 4d 5s 4d….`
Note that `3d`1 per hand's rule

Cleraly , the number of unpaired electron is `4`
`Fe^(2+) (Z = 26)` [Number of electrons `= 24]`
White writing the electronic configuration of cation remove the desired number of electrons from the cutermost orbital
In `Fe^(2+)` rem,ove` 2e^(-)` from `4s^(2)` since `4s` orbit (through lower in energy than `3d)` is the outermost .Hence , the configuration of `Fe^(2+)` is ` 1s^(2)2s^(2)2p 2p^(6)3d^(6)4s^(0)`
Note that the number of unpaired electron remains same as that is `Fe i.e. 4`
`Fe^(3+)(Z = 26)` [Number of electron `= 23]`
Note remove `2s` from `4s^(2)` and `3d^(6)` to get the electronic configuration `1s^(2)s^(2)2p 2p^(6)3d^(6)4s^(0)`
Note the now all d orbital have an odd electron (i.e. are half filted)

Hence the number of unpaired electron in `Fe^(2+)` is `5`
b. `Br(Z = 35)`
Following nuthan rule , the unpaired configuration is `1s^(2)s^(2)2p^(6)3p^(6)3d^(10)4s^(2)4p^(5)`Clearly one of `4p^(2)` orbotal contain unpaired electron
`4p^(3)-= (##V_PHY_CHM_P1_C04_S01_328_S03.png" width="80%">
Orbital filled as pe r hund's rule
Hence Br has only one unpaired electron
`Br^(2)(Z = 36)`
Since anion (s) is (are) formedby adding electron (s) so simple write the electron configuration as per the total number of electron is
`1s^(2)2s^(2)2p^(6)3p^(6)3s^(23d^(10)4s^(2)4p^(6)`
Clearly there are no unpaired electron
Following the patterns in (a) we can write the electronic configuration for V as follows :
`V(Z = 23) :1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(8)`
`3d^(8) = (##V_PHY_CHM_P1_C04_S01_328_S04.png" width="80%">
Orbital filled as per hund's rule
There unpaired electron
`V^(3+)(Z = 20)`
Remve `3e` from the outemost orbital successsivelly i.e. `2e^(-)` from `4s^(2)` and i.e. from `3d^(5)` Hence the electronic configuration of `V^(3+)` is
`1s^(2)2s^(2)2p^(6)3s^(6)3p^(6)4s^(0)3d^(2)`

Orbital filled as per Hend's rule
Two unpaired electron