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Q5.24(i):
Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex: \(K[Cr(H_{2}O)_{2}(C_{2}O_{4})_{2}].3H_{2}O\)

Solution :

Step 1: IUPAC Nomenclature

To determine the IUPAC name, we analyze the components of the coordination entity $K[Cr(H_{2}O)_{2}(C_{2}O_{4})_{2}] \cdot 3H_{2}O$.

  • Counter ion: Potassium ($K^{+}$)
  • Ligands: Two aqua ligands ($H_{2}O$) and two oxalato ligands ($C_{2}O_{4}^{2-}$).
  • Central Metal: Chromium ($Cr$), written as 'chromate' because it is in an anionic coordination sphere.

Applying IUPAC rules: Potassium diaquadioxalatochromate(III) trihydrate. [Per IUPAC nomenclature rules for coordination complexes].

Step 2: Oxidation State Determination

Let the oxidation state of $Cr$ be $x$.

$1(+1) + x + 2(0) + 2(-2) = 0$ [Sum of oxidation states in a neutral complex equals zero].

$1 + x - 4 = 0 \implies x - 3 = 0 \implies x = +3$.

Oxidation state of $Cr$ is $+3$.

Step 3: Electronic Configuration

Atomic number of $Cr$ ($Z=24$): $[Ar] 3d^{5} 4s^{1}$.

For $Cr^{3+}$: Remove three electrons ($1$ from $4s$, $2$ from $3d$).

Configuration: $[Ar] 3d^{3}$.

Step 4: Coordination Number & Geometry

The coordination number ($CN$) is determined by the number of donor atoms attached to the central metal.

  • Aqua ($H_{2}O$) is monodentate ($2 \times 1 = 2$).
  • Oxalato ($C_{2}O_{4}^{2-}$) is bidentate ($2 \times 2 = 4$).
  • Total $CN = 2 + 4 = 6$.

A coordination number of 6 implies an octahedral geometry.

H₂O H₂O C₂O₄ C₂O₄ Cr³⁺

Step 5: Stereochemistry

The complex is of the type $[M(A)_{2}(B-B)_{2}]$. It exhibits geometrical isomerism (cis and trans) and optical isomerism (the cis-form is optically active as it lacks a plane of symmetry).

Step 6: Magnetic Moment Calculation

For $Cr^{3+}$ ($3d^{3}$), the electrons occupy the $t_{2g}$ orbitals singly: $t_{2g}^{3} e_{g}^{0}$.

Number of unpaired electrons ($n$) = 3.

Using the spin-only formula $\mu = \sqrt{n(n+2)} \text{ B.M.}$:

$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ B.M.}$

ParameterValue
IUPAC NamePotassium diaquadioxalatochromate(III) trihydrate
Oxidation State$+3$
Electronic Configuration$[Ar] 3d^{3}$
Coordination Number$6$
StereochemistryGeometrical and Optical Isomerism
Magnetic Moment$3.87 \text{ B.M.}$

Final Solution: The complex is Potassium diaquadioxalatochromate(III) trihydrate with a coordination number of 6, an oxidation state of +3, a configuration of $[Ar] 3d^{3}$, and a magnetic moment of 3.87 B.M.


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