Question

Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
$\left[Fe\right(C{N}_6){]}^{4-}$

$[Fe{F}_6{]}^{3-}$

$\left[Co\right(C_2{O}_4{)}_3{]}^{3-}$

$[Co{F}_6{]}^{3-}$

Answer 1

$\left[Fe\right(CN{)}_6{]}^{4-}$ In this complex Fe is present as $F{e}^2+$
$Fe=\left[Ar\right]3d^{6}4s^2$
Outer configuration of $F{e}^{2+}=3d^{6}4s^{\circ }$

$C{N}^{-}$ being strong field ligand, pair up the unpaired d electrons Thus, two 3d-orbital are now available for $C{N}^{-}$ ions.


Since, all the electrons are paired, the complex is diamagnetic. Moreover $n-1$ d-orbitals are involved in bonding, so, it is an inner orbital or low spin complex.

$[Fe{F}_6{]}^{3-}$ In this complex, the oxidation state of Fe is + 3.
$F{e}^{3+}=3d^{5}4s^0$

$F^{-}$ is not a strong field ligand. It is a weak field ligand, so pairing occurs. Thus, 3d-orbitals are not available to take part bonding.

Because of the presence of five unpaired electrons, the complex is paramagnetic.
Moreover, nd-orbitals are involved in bonding, so it is an outer orbital or high spin complex.

$\left[Co\right(C_2{O}_4{)}_3{]}^{3-}$ In this complex, the oxidation state of Co is + 3.
Outer configuration of Co = $3d^{7}4s^2$
$C{o}^{3+}=3d^{6}4s^0$

Oxalate ion being a strong field ligand pair up the 3d electrons, thus two out of the ifve 3d-orbitals are available for oxalate ions.

Since, all the electrons are paired, this complex is diamagnetic.
It is an inner orbital complex because of the involvement of (n-1) d-orbital for bonding.

$[Co{F}_6{]}^{3-}$ In this complex, Co is present as $C{o}^{3+}$.
Outer configuration of $C{o}^{3+}=3d^{6}4s^0$

$F^{-}$ is a weak field ligand, so no pairing occurs and thus, $F^{-}$ occupy nd orbitals as

Because of the presence of four unpaired electrons, the complex is paramagnetic.
Since, nd orbitals take part in bonding, it is an outer orbital complex or high spin complex.