Geometry, hybridisation and magnetic moment of the ions `[Ni(CN)_(4)]^(2-),[MnBr_(4)]^(2-)` and `[FeF_(6)]^(3-)` are .

Options

A. Tetrahedral , square planar, octahedral `sp^(3), dsp^(3), sp^(2)d^(2), 5.9,0,4.9`
B. Tetrahedral, square planar, octahedral `dsp^(2), sp^(3), sp^(3)d^(2), 0, 5.9, 4.9`
C. Square planar , tetrahedral , octahedral `dsp^(2), sp^(3), d^(2)sp^(3), 5.9 , 4.9, 0`
D. Square planar, tetrahedral, octahedral `dap^(2), sp^(3), sp^(3)d^(2), 0, 5.9, 4.9`

Answer

D

Explanation

`|Ni(CN_(4))|^(2-) : Ni^(2+)`

`CN^(-)` is a strong ligand 3d, electrons pair up . No unpaired electron is present . Magnetic moment `(mu)` = 0 . Hybridization will be `dsp^(2)`, Geometry will be square planar
` |MnBr_(4)|^(2-) : [Mn^(2+)]`

`Br^(-)` is a weak ligand. Electrons do not pair up. No. of unpaired electrons = 5
`mu = sqrt(n(n+3))=sqrt(5xx7)= 5.9 BM`
Hybridization will be `sp^(3)` Geometry will be tetrahedral.
`|FeF_(6)|^(3-) , Fe^(3+)[d^(4)]`

F is a weak ligand . No electron pairing takes place n = 4
` :.mu = sqrt(4xx6)= 4.9 BM`
Hybridization will be `sp^(3)d^(2)` . Geometry will be octahedral.