Question

The calculated spin-only magnetic moments of K$_3$ [Fe(OH)$_6$] and K$_4$ [Fe(OH)$_6$] respectively are :

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

Answer 1

Answer: (D)

5.92 and 4.90 B.M.

Answer 2

The spin-only magnetic moment ($\mu_s$) of a complex is given by the formula:

$\mu_s = \sqrt{n(n+2)}$ B.M.

where 'n' is the number of unpaired electrons.

For the complex K$_3$[Fe(OH)$_6$]:

1. Determine the oxidation state of Fe.
   The complex anion is [Fe(OH)$_6$]$^{3-}$. Let the oxidation state of Fe be x. The charge of the hydroxide ligand (OH$_-$) is -1.

   x + 6(-1) = -3
   x - 6 = -3
   x = +3
   So, iron is in the +3 oxidation state, Fe$^{3+}$.

2. Determine the electronic configuration of Fe$^{3+}$.
   The atomic number of Fe is 26. The electronic configuration of Fe is [Ar] 3d$^6$ 4s$^2$.
   Fe$^{3+}$ is formed by removing 3 electrons (2 from 4s and 1 from 3d).
   The electronic configuration of Fe$^{3+}$ is [Ar] 3d$^5$.

3. Determine the number of unpaired electrons (n).
   The complex is octahedral with OH$_-$ ligands. Hydroxide (OH$_-$) is considered a weak field ligand. In a weak field octahedral complex, the d-orbitals split into lower energy t$_{2g}$ set (3 orbitals) and higher energy e$_g$ set (2 orbitals). Electrons are filled according to Hund's rule.
   For Fe$^{3+}$ (3d$^5$) in a weak field:
   The 5 electrons are distributed as follows: t$_{2g}^3$ e$_g^2$.

   $\begin{array}{|c|c|c|} \hline \text{e}_g & \uparrow & \uparrow \\ \hline \text{t}_{2g} & \uparrow & \uparrow & \uparrow \\ \hline \end{array}$

   Number of unpaired electrons, n = 3 (in t$_{2g}$) + 2 (in e$_g$) = 5.

4. Calculate the spin-only magnetic moment.

   $\mu_s = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{5 \times 7} = \sqrt{35}$.

   $\sqrt{35} \approx 5.916$ B.M. (rounded to 5.92 B.M.).

For the complex K$_4$[Fe(OH)$_6$]:

1. Determine the oxidation state of Fe.
   The complex anion is [Fe(OH)$_6$]$^{4-}$. Let the oxidation state of Fe be y.

   y + 6(-1) = -4
   y - 6 = -4
   y = +2
   So, iron is in the +2 oxidation state, Fe$^{2+}$.

2. Determine the electronic configuration of Fe$^{2+}$.
   The electronic configuration of Fe is [Ar] 3d$^6$ 4s$^2$.
   Fe$^{2+}$ is formed by removing 2 electrons from 4s.
   The electronic configuration of Fe$^{2+}$ is [Ar] 3d$^6$.

3. Determine the number of unpaired electrons (n).
   The complex is octahedral with OH$_-$ ligands, which are weak field.
   For Fe$^{2+}$ (3d$^6$) in a weak field:
   The 6 electrons are distributed as follows: t$_{2g}^4$ e$_g^2$.

   $\begin{array}{|c|c|c|} \hline \text{e}_g & \uparrow & \uparrow \\ \hline \text{t}_{2g} & \uparrow\downarrow & \uparrow & \uparrow \\ \hline \end{array}$

   Number of unpaired electrons, n = 2 (paired electrons in t$_{2g}$ don't contribute to unpaired count) + 2 (unpaired electrons in t$_{2g}$) + 2 (unpaired electrons in e$_g$) = 4.

4. Calculate the spin-only magnetic moment.

   $\mu_s = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{4 \times 6} = \sqrt{24}$.

   $\sqrt{24} \approx 4.899$ B.M. (rounded to 4.90 B.M.).

The calculated spin-only magnetic moments for K$_3$[Fe(OH)$_6$] and K$_4$[Fe(OH)$_6$] are approximately 5.92 B.M. and 4.90 B.M. respectively.