Question

Which of the following complexes exhibit(s) magnetic moment close to 2 Bohr Magneton?
[\(\text{Fe}(\text{H}_2\text{O})_6\)](\(\text{NO}_3\))\(_2\), \(\text{K}_2\)[\(\text{MnCl}_4\)], \(\text{K}_4\)[\(\text{Mn}(\text{CN})_6\)], and [\(\text{Ni}(\text{CO})_4\)]

• A. Only \(\text{K}_4\)[\(\text{Mn}(\text{CN})_6\)]
• B. \(\text{K}_2\)[\(\text{MnCl}_4\)] and \(\text{K}_4\)[\(\text{Mn}(\text{CN})_6\)]
• C. [\(\text{Fe}(\text{H}_2\text{O})_6\)](\(\text{NO}_3\))\(_2\) and \(\text{K}_2\)[\(\text{MnCl}_4\)]
• D. \(\text{K}_4\)[\(\text{Mn}(\text{CN})_6\)] and [\(\text{Ni}(\text{CO})_4\)]

Hint:

If the magnetic moment is close to a whole number, that whole number minus one or the digit before the decimal point usually represents the number of unpaired electrons.
For example, a magnetic moment of 1.73 BM (\(\approx 2\) BM) indicates 1 unpaired electron.
A magnetic moment of 5.92 BM indicates 5 unpaired electrons.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks us to identify which of the given coordination complexes have a spin-only magnetic moment close to 2 Bohr Magneton (BM).

Step 2: Key Formula or Approach:
The spin-only magnetic moment (\(\mu\)) of a coordination complex depends on the number of unpaired electrons (\(n\)) in the central metal ion:
\[ \mu = \sqrt{n(n + 2)}\text{ BM} \]
Let's find the value of \(n\) that corresponds to a magnetic moment close to 2 BM:
- For \(n = 1\): \(\mu = \sqrt{1(3)} = \sqrt{3} \approx 1.73\text{ BM}\) (which is close to 2 BM).
- For \(n = 2\): \(\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83\text{ BM}\).
Thus, we need to find the complex(es) with exactly \(n = 1\) unpaired electron.

Step 3: Detailed Explanation:
Let's analyze the oxidation state, coordination number, d-electron configuration, and ligand field strength for each complex:
1. [\(\text{Fe}(\text{H}_2\text{O})_6\)](\(\text{NO}_3\))\(_2\):
- Central metal: \(\text{Fe}^{2+}\) (\(d^6\) configuration).
- Ligand: \(\text{H}_2\text{O}\) is a Weak Field Ligand (WFL), causing no pairing.
- Octahedral splitting: High-spin configuration is \(t_{2g}^4 e_g^2\).
- Number of unpaired electrons (\(n\)) = 4.
- \(\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90\text{ BM}\).
2. \(\text{K}_2\)[\(\text{MnCl}_4\)]:
- Central metal: \(\text{Mn}^{2+}\) (\(d^5\) configuration).
- Ligand: \(\text{Cl}^-\) is a Weak Field Ligand in a tetrahedral geometry.
- Tetrahedral splitting: High-spin configuration is \(e^2 t_2^3\).
- Number of unpaired electrons (\(n\)) = 5.
- \(\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92\text{ BM}\).
3. \(\text{K}_4\)[\(\text{Mn}(\text{CN})_6\)]:
- Central metal: \(\text{Mn}^{2+}\) (\(d^5\) configuration).
- Ligand: \(\text{CN}^-\) is a Strong Field Ligand (SFL), causing pairing.
- Octahedral splitting: Low-spin configuration is \(t_{2g}^5 e_g^0\).
- Number of unpaired electrons (\(n\)) = 1.
- \(\mu = \sqrt{1(3)} = \sqrt{3} \approx 1.73\text{ BM}\) (which is close to 2 BM).
4. [\(\text{Ni}(\text{CO})_4\)]:
- Central metal: \(\text{Ni}^0\) (\(d^{10}\) configuration after reorganization due to strong field \(\text{CO}\) ligand).
- Number of unpaired electrons (\(n\)) = 0.
- \(\mu = 0\text{ BM}\).

Step 4: Final Answer:
Only \(\text{K}_4\)[\(\text{Mn}(\text{CN})_6\)] has 1 unpaired electron, giving a magnetic moment of 1.73 BM (close to 2 BM). Thus, the correct option is (A).