Which of the following is paramagnetic?

Options

A. `[Fe(CN)_(6)]^(4-)`
B. `[Ni(CO)_(4)]`
C. `[Ni(CN)_(4)]^(2-)`
D. `[CoF_(6)]^(3-)`

Answer

D

Explanation

In `[Fe(Cn)_(6)]^(4-)`, `Fe^(2+)` has configuration of `[Ar]^(18)3d^(6)`. After pairing up of six 3d-electrons in three d-orbitals, `d^(2)sp^(3)`-hybridisation given 6 empty hybridised orbitals for accommodating six pairs of electrons from six `CN^(-)` ions. Thus no unpaired electron is present and hence diamagnetic.
In `Ni(CO)_(4)`, Ni has configuration `[Ar^(18)]4s^(2)3d^(8)` Pairing up to two 4s-electrons with two unpaired 3d-electron followed up by `sp^(3)` hybridisation gives four empty hybrid orbitals to accommodate four pairs of electrons from CO ligands . Hence no unpaired electrons from complex is diamagnetic. In `[Ni(CN)_(4)]^(2-), Ni^(2+)` has configuration `[Ar^(18)]3d^(8)CN^(-)` is a strong ligand , 8 electrons pair up in four d-orbitals followed by `dsp^(2)`-hybridisation . No unpaired electron and hence diamagnetic.
In `[CoF_(6)]^(3-)`, `Co^(3+)` has `[Ar^(18)]3d^(6)` configuration `F^(-)` is a weak ligand no pairing up takes place. Hybridisation is `sp^(3)d^(2)` Due to the presence of four unpaired electrons it is paramagnetic.