Question

The coordination number of `Ni^(2+)` is 4.
`NiCl_(2) + KCN ("excess") rarr A ("cyano comples")`
`NiCl_(2) + conc. HCl ("excess") rarr B ("chloro complex")`
Predict the magnetic nature of A and B.
A. Both are diamagnetic
B. A is diamagnetci and B is paramagnetic with one unpaired electron
C. A is diamagnetic and B is paramagnetic with two unparied electrons
D. Both are paramagnetic

Answer 1

Correct Answer - C
In cyano complex `K_(2)[Ni(CN)_(4)]`, complex ion is `[Ni(CN)_(4)]^(2-)` and Ni is present as `Ni^(2+)` or Ni(II) , so
`Ni^(2+)1s^(2),2s^(2),2p^(6),3s^(2)3p^(6)3d^(8),4s^(0)`

In it unpaired orbital is not present , so it is diamagnetic in character ( square planar shape ) .
In chloro complex `K_(2)[Ni(Cl)_(4)]`, complex ion is `[Ni(Cl_(4))]^(2-)` and Ni is present as `Ni^(2+)` or Ni(II), so `Ni^(2+)= 1s^(2),2s^(2)2p^(6),3s^(2)3p^(6)3d^(8),4s^(0)`
In `[Ni(Cl_(4))]^(2-)` ion, `Ni^(2+)` is present as follows due to weaker ligand character of `Cl^(-)` ion ( `Cl^(-)` is weak field ligand ) . So it is unable to pair up the electron and `Ni^(2+)` need four empty orbitals to accommodate four `Cl^(-)` ligand. Thus, `(NiCl_(4))^(2-)` shows `sp^(3)` - hybridisation ( Tetrahedral shape ).

Hence, due to presence of unpaired orbitals, it is paramagnetic in character.