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. `[Cr(CO)_(6)]`
B. `[Fe(CO)_(5)]`
C. `[Fe(CN)_(6)]^(4-)`
D. `[Cr(NH_(3))_(6)]^(3+)`

Answer

D

Explanation

`[Fe(CN)_(6)]^(4-)` is diamagnetic `(Fe^(2+)` has `3d^(6)` configuration and the 6 electron pairs up in three d-orditals followed by `d^(2)sp^(3)`-hybridisation).
`[Cr(NH_(3))_(6)]^(3+)` is paramagnetic (`Cr^(3+)` has `3d^(3)` configuration . Hybridisation is `d^(2)sp^(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^(2)sp^(3)`-hybridisation to give octahedral complex . No unpaired electorn and hence complex is diamagnetic.
`Fe(CO)_(5)` : Fe(Z = 26) : `[Ar]^(18)4s^(2)3s^(6)` . The six electrons in d-subshell pairs up in three d-orbitals . This is followed by `d^(2)sp^(3)`-hybridisation to give octahedral geometry with one vacant hybridised orbital . The resulting shape of the complex is spuare based pyramid . As there is no unpaired electron, the complex is diamagnetic.