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Using valence bond theory, explain the following in relation to the complexes gives below:
$\left [ Mn(CN)_{6} \right ]^{3-},\left [ Co(NH_{3})_{6} \right ]^{3+},\left [ Cr(H_{2}O) _{6}\right ]^{3+},\left [ FeCl_{6} \right ]^{4-}$
(i) Type of hybridisation.
(ii) Inner or outer orbital complex.
(iii) Magnetic behaviour.
(iv) Spin only magnetic moment value.

Answers (1)

$\left [ Mn(CN)_{6} \right ]^{3-} :$

$Mn^{3+}=3d^{4}$

(i) Hybridisation - $d^{2}sp^{3}$

(ii) Inner orbital complex

(iii) Paramagnetic

(iv) Magnetic moment $\sqrt{2(2+2)}=2.87 \; B.M$

$\left [ Co(NH_{3})_{6} \right ]^{3+}:$

$Co^{3+}=3d^{6}$

(i) Hybridisation - $d^{2}sp^{3}$

(ii) Inner orbital complex

(iii) Diamagnetic

(iv) Magnetic moment=0

$\left [ Cr(H_{2}O) _{6}\right ]^{3+}$

$Cr^{3+}=3d^{3}$

(i) Hybridisation - $d^{2}sp^{3}$

(ii) Inner orbital complex

(iii) Paramagnetic

(iv) Magnetic moment= $\sqrt{3(3+2)}=3.87 \; B.M$

$\left [ FeCl_{6} \right ]^{4-}$

$Fe^{2+}=3d^{6}$

(i) Hybridisation - $sp^{3}d^{2}$

(ii) Outer orbital complex

(iii) Paramagnetic

(iv) Magnetic moment= $\sqrt{4(4+2)}=4.9 \; B.M$

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