Question

The ${\left[ {{\text{Fe}}{{\left( {{{\text{H}}_2}{\text{O}}} \right)}_6}} \right]^{ + 2}}$ and ${\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_6}} \right]^{ - 4}}$ differ in-
A.Geometry, magnetic moment
B.Geometry, hybridization
C.Magnetic moment, color
D.Hybridization, number of d electrons

Answer 1

Ligands with oxygen donor atoms are generally weak field ligands and on the other hand, ligands with donor atoms of carbon and nitrogen are strong field atoms. In presence of strong field ligands, before hybridization pairing of d electrons of central metal atoms occurs.

Complete answer:
According to valence bond theory, intermixing of orbital or overlapping of orbital occurs. Intermixing orbital are vacant orbital of central metal atom and orbital of ligand. Strength of ligand determines whether pairing of electron of central metal atom before hybridisation will occur or not.
In case of ${\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_6}} \right]^{ - 4}}$ , oxidation number of central metal atom, iron is $+ 2$ and coordination number is 6. As electronic configuration of Fe is $3{{\text{d}}^6}4{{\text{s}}^2}$ , so configuration of ${\text{F}}{{\text{e}}^{ + 2}} = 3{{\text{d}}^6}$ . Since cyanide is a strong field ligand, it will pair up as possible as an unpaired electron present in the d subshell of a central metal atom and make two 3d orbital vacant available. Since, coordination number is six, six orbits will hybridize and be used for making coordinate bonds. Thus, hybridization will be ${{\text{d}}^2}{\text{s}}{{\text{p}}^3}$ . As no free electron is present, it is a low spin complex and diamagnetic. Geometry will be octahedral. Magnetic moment is zero as there is no unpaired electron. It is a colorless complex.
Similarly, for ${\left[ {{\text{Fe}}{{\left( {{{\text{H}}_2}{\text{O}}} \right)}_6}} \right]^{ + 2}}$ , oxidation number of central metal atom, iron is $+ 2$ and coordination number is 6. As electronic configuration of Fe is $3{{\text{d}}^6}4{{\text{s}}^2}$ , so configuration of ${\text{F}}{{\text{e}}^{ + 2}} = 3{{\text{d}}^6}$ . Since water is a weak field ligand, there will be no pairing of unpaired electrons in the subshell of the central metal atom before hybridization. Since, coordination number is six, six orbits will hybridize and be used for making coordinate bonds. Thus, hybridization will be ${\text{s}}{{\text{p}}^3}{{\text{d}}^2}$ . As the number of unpaired electrons is 4, it is a high spin complex and paramagnetic. Magnetic moment is $\sqrt {12} {\text{ BM}}$ . Geometry will be octahedral. The color of this compound is pale green.

Thus, the correct option is C.

Note:
Oxidation number of central metal atom in coordination compounds can be calculated as $\left( {{\text{oxidation number}} = \left( {{\text{charge on coordination sphere}}} \right) - \left( {{\text{charge on ligand}}} \right)} \right)$ . Coordination number of the complex is calculated as $\left( {{\text{coordination number}} = {\text{denticity of ligand}} \times {\text{number of ligand}}} \right)$ .