Atomic numbers of `Cr` and `Fe` are respectively `24` and `26`. Which of the following is paramagnetic with the spin of the electron?

Options

A. (a) `[Cr(NH_3)_6]^(3+)`
B. (b) `[Fe(CO)_5]`
C. (c) `[Fe(CN)_6]^(4-)`
D. (d) `[Cr(CO)_6]`

Answer

A

Explanation

`[Fe(CN)_6]^(4-)` is diamagnetic (`Fe^(2+)` has `3d^6` configuration and the 6 electron pairs up in three d-orbitals followed by `d^2sp^3`-hybridization).
`[Cr(NH_3)_6]^(3+)` is paramagnetic (`Cr^(3+)` has `3d^3` configuration. Hybridization is `d^2sp^3`. Due to 3 unpaired electrons it is paramagnetic)
`[Cr(CO)_6:Cr(Z=25): [Ar]^(18)4s^1, 3d^5`.
The one `4s`-electron pairs up with five `3d`-electrons in three `d`-orbitals. This is followed by `d^2sp^3`-hybridization to give octahedral complex. No unpaired electron and hence complex is diamagnetic.
`Fe(CO)_5: Fe(Z=26): [Ar]^(18) 4s^2, 3d^6`.
The six electrons in d-subshell pairs up in three d-orbitals. This is followed by `d^2sp^3`-hybridization to give octahedral geometry with one vacant hybridized orbital. The resulting shape of the complex is square-based pyramid. As there is no unpaired electron, the complex is diamagnetic.