Match the complex ions given in column I with the hybridisation and number of unpaired electrons given in column II and assign the correct code

Options

A. `{:(A,B,C,D),(3,1,4,2):}`
B. `{:(A,B,C,D),(4,3,2,1):}`
C. `{:(A,B,C,D),(3,2,4,1):}`
D. `{:(A,B,C,D),(4,1,2,3):}`

Answer

A, B, C, D

Explanation

Formation of inner orbital complex and outer orbital complex determines hybridisation of molecule which inturn depends upon field strength of ligand and number of vacant d orbitals.
(i) Strong field ligand forms inner orbital complex with hybridisation `d^(2)sp^(3)`.
(ii) Weak field ligand forms outer orbital complex with hybridisation `sp^(3)d^(2)`
According to VBT, hybridisation and number of unpaired electrons of coordination compounds can be calculated as
A. `[Cr(H_(2)O)_(6)]^(3+)`
MOEC (Molecular orbital electronic configuration) of `Cr^(3+)` in `[Cr(H_(2)O)_(6)]^(3+)` is

Hybridisation `=d^(2) sp^(3)`
n (number of unpaired electrons) = 3
B. `[Co(CN)_(4)]^(2-)`
MOEC of `Co^(2+)` in `[Co(CN)_(4)]^(2-)` is

Hybridisation `= dsp^(2)`
`n=1`
C. `[Ni(NH_(3))_(6)]^(2+)`
MOEC of `Ni^(2+)` in `[Ni(NH_(3))_(6)]^(2+)` is

Hybridisation `=sp^(3) d^(2)`
`n=2`
D. `[MnF_(6)]^(4-)`
MOEC of `Mn^(2+)` in `[MnF_(6)]^(4-)` is

Hybridisation `=sp^(3) d^(2)`
`n=5`
Hence, correct choice can be represented by (a).