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Question

Using valence bond theory, explain the following in relation to the complexes given below
[Mn(CN)6]3,[Co(NH3)6]3+,[Cr(H2O)6]3+,[FeCl6]4
(a) Type of hybridisation
(b) Inner or outer orbital
(c) Magnetic behaviour
(d) Spin only magnetic moment value.

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Solution

(a) [Mn(CN)6]3



(i) d2sp3 hybridisation
(ii) Inner orbital complex because (n-1)d - orbitals are used.
(iii) Paramagnetic, as two unpaired electrons are present.
(iv) Spin only magnetic moment (μ)=2(2+2)=8=2.82BM

(b) [Co(NH3)6]3+
Co3+=3d64s0


(i) d2sp3 hybridisation
(ii) Inner orbital complex as(n-1)d-orbitals take part.)
(iii)Diamagnetic (as three paired electrons are present.)
(iv) μ= n(n+2) = 0(0+2) = 0 = 0BM

(c) [Cr(H2O)6]3+



(i) d2sp3 hybridisation
(ii) Inner orbital complex as(n-1)d-orbitals take part.)
(iii) Paramagnetic (as three unpaired electrons are present.)
(iv) μ=n(n+2)=3(3+2)=15=3.87BM

(d) [Fe(Cl)6]4
Fe2+=3d6



(i) sp3d2 hybridisation
(ii) Outer orbital complex because nd-orbitals are involved in hybridisation.
(iii) Paramagnetic (because of the presence of four unpaired electrons).
(iv) μ=n(n+2)=4(4+2)=24=4.9BM


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