Using valence bond theory, explain the following in relation to the complexes given below
`[Mn(CN)_(6)]^(3-), [Co(NH_(3))_(6)]^(3+), [Cr(H_(2)O)_(6)]^(3+), [FeCl_(6)]^(4-)`
(a) type of hybridisation
(b) Inner or outer orbital complex
(c) Magnetic behaviour
(d) Spin only magnetic moment value.

Explanation

(a) `[Mn(CN)_(6)]^(3-)`
`[Mn^(3+)=3d^(4)`

(i) `d^(2)sp^(3)` hybridisation
(ii) Inner orbital complex because `(n-1)d`-orbitals are used.
(iii) Paramagnetic moment `(mu)=sqrt(2(2+2))=sqrt(8)=2.82 BM`
(iv) Spin only magnetic moment `(mu)= sqrt(2(2+2))=sqrt(8)=2.82 BM`
(b) `[Co(NH_(3))_(6)]^(3+)`
`CO^(3+)=3d^(6) 4s^(0)`

(`NH_(3)` pair up the unpaired 3d electons.)
(i) `d^(2)sp^(3)` hybridisation
(ii) Inner orbital complex because of the involvement of `(n-1)` d-orbital in bonding.
(iii) Diamagnetic, as no unpaired electron is present.
(iv) `mu=sqrt(n(n+2))=sqrt(0(0+2))=0 ("Zero")`
(c) `[Cr(H_(2)O)_(6)]^(3+)`

(i) `d^(2) sp^(3)` hybridisation
(ii) Inner orbital complex (as `(n-1)` d-orbital take part.)
(iii) Paramagnetic (as three unpaired electrons are present.)
(iv) `mu=sqrt(n(n+2))=sqrt(3(3+2))=sqrt(15) =3.87 BM`
(d) `[Fe(Cl)_(6)]^(4-)`
`Fe^(2+)=3d^(6)`

(i) `sp^(3) d^(2)` hybridisation
(ii) Outer orbital complex because nd-orbitals are involved in hybridisation.
(iii) Paramagnetic (because of the presence of four unpaired electrons).
(iv) `mu=sqrt(n(n+2))=sqrt(4(4+2))=sqrt(24)=4.9 BM`