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Q5.5(iv):
Specify the oxidation numbers of the metals in the following coordination entities: \(K_{3}[Fe(CN)_{6}]\)

Solution :

Step 1: Identification of Components

To determine the oxidation number of the central metal atom, iron ($Fe$), in the coordination entity $K_{3}[Fe(CN)_{6}]$, we must identify the charge of all constituent ions present in the coordination sphere and the outer coordination sphere.

  • Potassium ($K$): An alkali metal, consistently exhibits an oxidation state of $+1$ [Per the Group 1 periodic trends].
  • Cyanide ($CN^-$): A monoanionic ligand with a total charge of $-1$.
  • Iron ($Fe$): The central metal atom whose oxidation state is represented by $x$.

Step 2: Mathematical Formulation

The sum of the oxidation numbers of all atoms in a neutral chemical compound is equal to zero [Per the Principle of Electroneutrality]. The expression is structured as follows:

$3 \times (\text{Oxidation state of } K) + 1 \times (\text{Oxidation state of } Fe) + 6 \times (\text{Oxidation state of } CN^-) = 0$

Step 3: Algebraic Derivation

Substituting the known values into the equation:

$3(+1) + x + 6(-1) = 0$

Perform the arithmetic operations:

$3 + x - 6 = 0$

$x - 3 = 0$

$x = +3$

Step 4: Molecular Geometry Visualization

The complex ion $[Fe(CN)_6]^{3-}$ adopts an octahedral geometry where the $Fe^{3+}$ ion is coordinated to six cyanide ligands.

Fe³⁺ CN⁻ CN⁻ CN⁻ CN⁻

Final Analysis

The calculation confirms that for the neutral complex $K_{3}[Fe(CN)_{6}]$, the iron center must balance three potassium ions ($+3$ charge) and six cyanide ions ($-6$ charge). Thus, the oxidation state of iron is $+3$.

Final Solution: The oxidation number of Iron (Fe) in $K_{3}[Fe(CN)_{6}]$ is +3.


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