Question
The calculated magnetic moment for low spin $[Ru(EDTA)]^-$ is
This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
1.73 BM
The question asks for the calculated magnetic moment of the low spin complex \([Ru(EDTA)]^−\). To determine this, we follow these steps:
- Identify the oxidation state of Ru in the complex:
- EDTA is a hexadentate ligand with a charge of -4 (EDTA4-).
- Consider the whole complex: \([Ru(EDTA)]^−\).
- Let the oxidation state of Ru be 'x'. Thus, \(x - 4 = -1 \rightarrow x = +3\).
- Determine the electron configuration for Ru3+:
- The electronic configuration of neutral Ruthenium (Ru) is \([Kr] \, 4d^7 \, 5s^1\).
- Ru3+ loses 3 electrons, so: \([Kr] \, 4d^5\).
- Consider the ligand's effect:
- EDTA is a strong field ligand, causing pairing of electrons. This results in a low spin complex.
- In a low spin configuration, \(4d\) orbital will be: \(t_{2g}^5 e_g^0\).
- Calculate the magnetic moment:
- The formula for the magnetic moment (\(\mu\)) is: \(\mu = \sqrt{n(n+2)}\) Bohr Magnetons (BM), where 'n' is the number of unpaired electrons.
- With \(t_{2g}^5\), there is 1 unpaired electron.
- Therefore, the magnetic moment is: \(\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \, \text{BM}\).
Hence, the calculated magnetic moment for low spin \([Ru(EDTA)]^−\) is 1.73 BM.
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