Step 1: Understanding the Concept:
The magnetic moment ($\mu$) of a transition metal complex depends on the number of unpaired electrons ($n$) in its d-orbitals.
The spin-only formula is used to calculate this: $\mu = \sqrt{n(n+2)}$ Bohr Magnetons (BM).
For 4d and 5d transition metals (like Ruthenium), the crystal field splitting energy ($\Delta_o$) is significantly larger than for 3d metals. As a result, these complexes are almost always low-spin, regardless of the field strength of the ligand.
Step 2: Key Formula or Approach:
1. Determine the oxidation state of the central metal ion (Ru).
2. Identify the $d^n$ configuration of the ion.
3. Arrange electrons in low-spin octahedral splitting ($t_{2g}$ and $e_g$ orbitals).
4. Count unpaired electrons ($n$) and apply the formula $\mu = \sqrt{n(n+2)}$.
Step 3: Detailed Explanation:
1. Oxidation State: EDTA is a hexadentate ligand with a charge of $-4$.
Let $x$ be the oxidation state of Ruthenium in $[Ru(EDTA)]^-$.
$x + (-4) = -1 \rightarrow x = +3$.
So, we have the $Ru^{3+}$ ion.
2. d-electron configuration:
Ruthenium (atomic number 44) is below Iron in the periodic table.
Neutral Ru: $[Kr] 4d^7 5s^1$ (Total 8 valence electrons).
$Ru^{3+}$ (lose 3 electrons): $[Kr] 4d^5$.
3. Electron arrangement (Low Spin):
In a low-spin octahedral field, the five d-electrons fill the lower energy $t_{2g}$ orbitals as much as possible before any go to $e_g$.
$d^5$ low spin: $(\uparrow \downarrow) (\uparrow \downarrow) (\uparrow)$ in the $t_{2g}$ level, and $0$ in $e_g$.
This leaves exactly $1$ unpaired electron ($n = 1$).
4. Magnetic moment calculation:
$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732$ BM.
Step 4: Final Answer:
The calculated magnetic moment is 1.73 BM. The correct option is (B).