Question

The calculated spin-only magnetic moments of \( K_3[Fe(OH)_6] \) and \( K_4[Fe(OH)_6] \) respectively are:

• A. 4.90 and 4.90 B.M.
• B. 5.92 and 4.90 B.M.
• C. 3.87 and 4.90 B.M.
• D. 4.90 and 5.92 B.M.

Hint:

For calculating the spin-only magnetic moment, determine the number of unpaired electrons using the electronic configuration of the metal in the complex.

Answer 1

The Correct Option is B

The spin-only magnetic moment is given by the formula: \[ \mu = \sqrt{n(n+2)} \, \text{B.M.} \] where \( n \) is the number of unpaired electrons. Based on the electronic configurations of \( Fe^{3+} \) (in \( K_3[Fe(OH)_6] \)) and \( Fe^{2+} \) (in \( K_4[Fe(OH)_6] \)), the number of unpaired electrons in these complexes is calculated. For \( Fe^{3+} \), there are 5 unpaired electrons and for \( Fe^{2+} \), there are 4 unpaired electrons, leading to magnetic moments of 5.92 and 4.90 B.M., respectively.

Answer 2

Step 1: Determine the oxidation states of Fe in both complexes.
For \( K_3[Fe(OH)_6] \):
Each \( OH^- \) group contributes a charge of –1, and there are 6 such groups. The overall charge on the complex ion is –3 (to balance 3 K⁺ ions).
Let the oxidation number of Fe be \( x \). Then:
\[ x + 6(-1) = -3 \Rightarrow x = +3 \] Hence, the oxidation state of Fe in \( K_3[Fe(OH)_6] \) is +3.

For \( K_4[Fe(OH)_6] \):
The overall charge on the complex ion is –4 (to balance 4 K⁺ ions).
Let the oxidation number of Fe be \( x \). Then:
\[ x + 6(-1) = -4 \Rightarrow x = +2 \] Hence, the oxidation state of Fe in \( K_4[Fe(OH)_6] \) is +2.

Step 2: Determine the electronic configurations.
For Fe (atomic number 26): ground-state configuration = [Ar] 3d⁶ 4s².

- For Fe³⁺: remove 3 electrons → [Ar] 3d⁵.
- For Fe²⁺: remove 2 electrons → [Ar] 3d⁶.

Step 3: Identify the number of unpaired electrons.
- \( Fe^{3+} \): 3d⁵ → five unpaired electrons.
- \( Fe^{2+} \): 3d⁶ → four unpaired electrons.

Step 4: Calculate spin-only magnetic moments.
The formula for spin-only magnetic moment is:
\[ \mu = \sqrt{n(n+2)} \, \text{B.M.} \] where \( n \) is the number of unpaired electrons.

For \( K_3[Fe(OH)_6] \): \( n = 5 \)
\[ \mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92 \, \text{B.M.} \]
For \( K_4[Fe(OH)_6] \): \( n = 4 \)
\[ \mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90 \, \text{B.M.} \]

Final Answer:
\[ \boxed{K_3[Fe(OH)_6] = 5.92 \, \text{B.M.}, \quad K_4[Fe(OH)_6] = 4.90 \, \text{B.M.}} \]