The maximum number of impaired electrons are in

Options

A. `Fe^(2+)`
B. `Fe^(3+)`
C. `Fe^(4+)`
D. Fe

Answer

B

Explanation

`._(26)Fe = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(6) 4s^(2)`
Unpaired electrons = 4
`Fe^(2+) = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(6)`
Unpaired electrons = 4
`Fe^(3+) = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(5)`
Unpaired electron = 5
`Fe^(4+) = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(4)`
Unpaired electrons = 4