The calculated magnetic moment for low spin $[\text{Ru}(\text{EDTA})]^-$ is
Show Hint
For second and third-row transition metals ($4d$ and $5d$ series like $\text{Ru}$, $\text{Rh}$, $\text{Pt}$), the crystal field splitting ($\Delta_o$) is exceptionally large. As a result, these complexes are always low-spin, regardless of whether the ligand is traditionally classified as weak-field or strong-field.
Step 1: Understanding the Question:
We need to calculate the spin-only magnetic moment ($\mu$) for the low-spin complex $[\text{Ru}(\text{EDTA})]^-$.
Step 2: Key Formula or Approach:
1. Find the oxidation state and $d$-electron configuration of the central metal ion (Ruthenium).
2. Determine the number of unpaired electrons ($n$) under low-spin octahedral conditions.
3. Use the spin-only magnetic moment formula:
\[ \mu = \sqrt{n(n + 2)}\text{ BM} \]
Step 3: Detailed Explanation:
Let us find the oxidation state of Ruthenium ($\text{Ru}$) in $[\text{Ru}(\text{EDTA})]^-$:
- EDTA is a hexadentate ligand with a charge of $-4$ ($\text{EDTA}^{4-}$).
- Let $x$ be the oxidation state of $\text{Ru}$:
\[ x + (-4) = -1 \implies x = +3 \]
Thus, Ruthenium is in the $+3$ oxidation state ($\text{Ru}^{3+}$).
Next, let us write the electronic configuration of $\text{Ru}^{3+}$:
- Ruthenium ($\text{Ru}$, atomic number 44) is located directly below Iron ($\text{Fe}$) in the second transition series ($4d$ series).
- The configuration of neutral $\text{Ru}$ is $[\text{Kr}] 4\text{d}^7 5\text{s}^1$ (total of 8 valence electrons).
- For $\text{Ru}^{3+}$, we remove 3 electrons, leaving a $4\text{d}^5$ configuration.
Since $\text{Ru}$ is a $4d$ transition metal, the crystal field splitting energy ($\Delta_o$) is very large. Consequently, all $4d$ complexes are invariably low-spin.
In a low-spin octahedral field, the five $d$-electrons populate the lower energy $t_{2g}$ orbitals:
\[ t_{2g}^5 e_g^0 \]
The pairing scheme in the $t_{2g}$ orbitals is:
\[ \uparrow\downarrow \ \ \uparrow\downarrow \ \ \uparrow \]
This configuration contains exactly $n = 1$ unpaired electron.
Now, let us calculate the spin-only magnetic moment ($\mu$):
\[ \mu = \sqrt{1(1 + 2)} = \sqrt{3}\text{ BM} \approx 1.73\text{ BM} \]
