Which of the following complex ions is dimagnetic ?

Options

A. `[Cr(H_(2)O)_(6)]^(2+)`
B. `[FeF_(6)]^(3-)`
C. `[Cu(NH_(3))_(6)]^(2+)`
D. `[Zn(NH_(3))_(6)]^(2+)`

Answer

D

Explanation

`Cr(3d^(5)4s^(1)) rarr Cr^(2+) (3d^(4)) rarr sp^(3)d^(2)` (four unpaired 3d electrons)
`Fe(3d^(6)4s^(2)) rarr F^(3+)(3d^(5)) rarr sp^(3) d^(2)` (five unpaired 3d electrons)
`Cu(3d^(10 4s^(1)) rarr Cu^(2+) (3d^(9)) rarr sp^(3)d^(2)` (one unpaired 3d electron)
`Zn(3d^(10) 4s^(2)) rarr Zn^(2+) (3d^(10)) rarr sp^(3) d^(2)` (zero unpaired 3d electrons)
Octahedral complexes of `Zn(+2)` are not very stable as there is no crystal field stabilization energy due to `d^(10)` configuration. Aqueous `Cu^(2+)` solutions form many complexes with ammonia and amines such as `[Cu(NH_(3))(H_(2)O)_(5)]^(2+)` , `[Cu(NH_(3))_(2)(H_(2)O)_(4)]^(2+)` , `[Cu(NH_(3))_(3) (H_(2)O)_(3)]^(2+)` and `[Cu(NH_(3))_(4) (H_(2)O_(2)]^(2+)`
It is difficult to add a fifth or sixth `NH_(3)` , thought it is possible to make `[Cu(NH_(3))_(6)]^(2+)` using liquid `NH_(3)` as solvent. The reason for this is the jahn-Tellar effect which distorts the octahedral complex and makes the fifth and sixth bonds long and weak. Note that weak field ligands such as `H_(2)O` and `F^(-)` can't reshuffle the 3d electron configuration to make two unoccupied 3d orbitals. Similar is the situtation with intermediate field ligands such as `NH_(3)` .