One the basis of `VBT` answer the following complex ions
(i) `[Ti(bpy)_(3)]^(Θ)`
(b) `[V(H_(2)O)_(6)]^(3+)`
(ii) `[V(H_(2)O)_(6)]^(3+)`
(III) `[Mn(CN)_(6)]^(4-)`
`[Mn(CN)_(6))]^(3-)`
(V) `[Ir(NH_(3))_(6)]^(3+)` .
Type of hybridisation involed
(b) Type of inner or outer orbital octahedral complex
(c ) Magnetic behaviour and `mu_(spin)` value .

Explanation

bpy or dipy is a strong field neutral bidentage ligand

(a) `d^(2)sp^(3)` hybridisation with `OH` shape
(b) Inner orbital complex
( c) Paramagnetic `n =1`
`mu_("spin") = sqrt3 BM = 1.732BM`
(II) `[V(H_(2)_(o)]^(3+)`
`V(Z =23)implies 3d^(3)4s^(2),V^(3+) = 3d^(2) 4s^(0)`
`CN^(Θ)` is a strong field ligand so pairing occurs

`d^(2)sp^(3)` hybridisation with `OH` shape
(b) Inner orbital complex
(c ) Paramagnetic , `n =2`
(d) `mu("spin") = sqrt8 BM= 2.828BM`
(III) `[Mn(CN)_(6)]^(4-)`
`Mn(Z=25) implies 3d^(5) 4s^(2),Mn^(2+) = 3d^(5 4s^(0)`
`CN^(Θ)` is a strong field ligand so pairing occurs

(a) `d^(2)sp^(3)` hybridisation with `OH` shape
(b) Inner orbital complex
(c) Paramagnetic `n =1`
(d) `mu_(spin) = sqrt3 BM = 1.732BM`
(IV) `[Mn(CN)_(6)]^(3-)`
`Mn(Z =25)implies 3d^(5) 4s^(2), Mn^(3+) = 3d^(4) 4s^(0)`
`CN^(Θ)` is a strong field ligand so only one pairing occurs

(a) `d^(2)sp^(3)` hybridisation with `OH` shape
(b) Inner orbital complex
(c ) Paramagnetic `n =2`
(d) `mu_("spin")=sqrt8 BM =2.828BM`
(V) `[Ir(NH_(3))_(6)]^(3+)`
ir(Z=7)implies5d^(7) 6s^(2), Ir^(3+) = 5d^(6) 6s^(0)`

(a) `d^(2)sp^(3)` hybridisation
(b) Inner orbital complex
(c ) Diamagnetic `mu_("spin")=0` .