1.

Compounds with spinonly magnetic moment equivalent to five unpaired electrons are

Answer»

`K_(4)[Mn(CN)_(6)]`
`[Fe(H_(2)O)_(6)]Cl_(3)`
`K_(3)[FeF(6)]`
`K_(4)[MnF_(6)]`

Solution :a. `K_(4)[Mn(CN)_(6)],Mn^(2+)=[Ar]3d^(5)4s^(0)`
`CN^(-)` is a strong field ligand, so pairing will TAKE place i.e. `t_(2g)^(2,2,1) e_(g)^(0,0)`
B. `[Fe(H_(2)O)_(6)]Cl_(3),Fe^(3+)=[Ar]3d^(5)4s^(0)`
`H_(2)O` is a weak field ligand so pairing will not occur.
i.e. `t_(2g)^(1,1,1), e_(g)^(1,1)`
c. `K_(3)[FeF_(6)],Fe^(3+)=[Ar]3d^(5) 4s^(0)`
`F^(-)` is a weak field ligand, so, pairing will not occur.
i.e `t_(2g)^(1,1,1) e_(g)^(1,1)`
d. `K_(4)[MnF_(6)], Mn^(2+)=[Ar]3d^(5)4s^(0)`
`F^(-)` is a weak field ligand so, pairing will not occur.
i.e. `t_(2g)^(1,1,1)e_(g)^(1,1)`
HENCE b, c and d complexes will SHOW spin only magnetic moment equivalent to five unpaired electrons.


Discussion

No Comment Found