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Q5.23(ii):
Give the oxidation state, d orbital occupation and coordination number of the central metal ion in the following complexes: cis-\([CrCl_{2}(en)_{2}]Cl\)

Solution :

Step 1: Decomposition of the Coordination Complex

The complex is given as $\text{cis-[CrCl}_{2}\text{(en)}_{2}\text{]Cl}$. To determine the oxidation state of the central metal ion, we must first dissociate the complex into its constituent ions:

$\text{[CrCl}_{2}\text{(en)}_{2}\text{]Cl} \rightarrow \text{[CrCl}_{2}\text{(en)}_{2}\text{]}^{+} + \text{Cl}^{-}$

[Per the Arrhenius theory of dissociation, the counter-ion dissociates in the ionization sphere].

Step 2: Calculating the Oxidation State of Chromium (Cr)

Let the oxidation state of $\text{Cr}$ be $x$.

  • $\text{Cl}^{-}$ (Chloride ion) has an oxidation state of $-1$.
  • $\text{en}$ (Ethylenediamine, $\text{H}_{2}\text{NCH}_{2}\text{CH}_{2}\text{NH}_{2}$) is a neutral bidentate ligand, so its oxidation state is $0$.
  • The net charge on the complex cation $\text{[CrCl}_{2}\text{(en)}_{2}\text{]}^{+}$ is $+1$.

Setting up the algebraic equation:

$x + 2(-1) + 2(0) = +1$

$x - 2 = +1$

$x = +3$

[By the Rule of Algebraic Sum of Oxidation States in a Complex Ion].

Step 3: Determination of Coordination Number

The coordination number is defined as the total number of $\sigma$-bonds formed by the ligands with the central metal ion.

  • Two $\text{Cl}^{-}$ ligands: Each forms $1$ bond $\rightarrow 2 \times 1 = 2$.
  • Two $\text{en}$ ligands: Each is bidentate (forms $2$ bonds) $\rightarrow 2 \times 2 = 4$.

Total Coordination Number $= 2 + 4 = 6$.

Step 4: Electron Configuration and d-orbital Occupation

The atomic number of Chromium ($\text{Cr}$) is $24$. The ground state electronic configuration is $[\text{Ar}] 3d^{5} 4s^{1}$.

For the $\text{Cr}^{3+}$ ion, we remove one electron from $4s$ and two from $3d$:

$\text{Cr}^{3+} = [\text{Ar}] 3d^{3}$

The $d$-orbital occupation is $t_{2g}^{3} e_{g}^{0}$.

dxy dyz dxz dx²-y² dz² Configuration: 3d³
Parameter Value
Oxidation State $+3$
Coordination Number $6$
d-orbital Occupation $t_{2g}^{3}$

Final Solution: The central metal ion $\text{Cr}^{3+}$ has an oxidation state of $+3$, a coordination number of $6$, and a d-orbital occupation of $t_{2g}^{3} e_{g}^{0}$.


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