Question

Using valence bond theory, explain the following in relation to the complexes given below
$\left[Mn\right(CN{)}_6{]}^{3-},\left[Co\right(N{H}_3{)}_6{]}^{3+},\left[Cr\right(H_{2}O{)}_6{]}^{3+},[FeC{l}_6{]}^{4-}$
(a) Type of hybridisation
(b) Inner or outer orbital
(c) Magnetic behaviour
(d) Spin only magnetic moment value.

Answer 1

(a) $\left[Mn\right(CN{)}_6{]}^{3-}$



(i) $d^{2}s{p}^3$ 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 $\left(\mu \right)=\sqrt{2(2+2)}=\sqrt{8}=2.82BM$

(b) $\left[Co\right(N{H}_3{)}_6{]}^{3+}$
$C{o}^{3+}=3d^{6}4s^0$


(i) $d^{2}s{p}^3$ hybridisation
(ii) Inner orbital complex as(n-1)d-orbitals take part.)
(iii)Diamagnetic (as three paired electrons are present.)
(iv) $\mu =\sqrt{n(n+2)}=\sqrt{0(0+2)}=\sqrt{0}=0BM$

(c) $\left[Cr\right(H_{2}O{)}_6{]}^{3+}$



(i) $d^{2}s{p}^3$ hybridisation
(ii) Inner orbital complex as(n-1)d-orbitals take part.)
(iii) Paramagnetic (as three unpaired electrons are present.)
(iv) $\mu =\sqrt{n(n+2)}=\sqrt{3(3+2)}=\sqrt{15}=3.87BM$

(d) $\left[Fe\right(Cl{)}_6{]}^{4-}$
$F{e}^{2+}=3d^6$



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