Question

Arrange the following complexes in increasing order of the number of unpaired electrons present in the central metal ion:
I. \( [Fe(CN)_6]^{3-} \)
II. \( [Co(C_2O_4)_3]^{3-} \)
III. \( [Mn(CN)_6]^{3-} \) 

• A. II<III<I
• B. III<I<II
• C. II<I<III
• D. I<II<III

Hint:

The number of unpaired electrons in a complex depends on the ligand field strength. Strong field ligands (like CN\(^{-}\)) induce a low-spin configuration, minimizing unpaired electrons, while weak field ligands result in high-spin configurations with more unpaired electrons.

Answer 1

The Correct Option is C

To determine the number of unpaired electrons, we analyze each complex:

1. [{Fe(CN)}\(_6\)]\(^{3-}\) (Hexacyanoferrate(III)):
- Fe\(^{3+}\) (d\(^5\)): CN\(^{-}\) is a strong field ligand, leading to low-spin configuration.
- The low-spin d\(^5\) configuration results in 1 unpaired electron.

2. [{Co(C}\(_2\)O\(_4\))\(_3\)]\(^{3-}\) (Trioxalato cobalt(III)):
- Co\(^{3+}\) (d\(^6\)): \(C_2O_4^{2-}\) (oxalate) is a moderate field ligand, leading to a low-spin d\(^6\) configuration.
- The low-spin d\(^6\) configuration has 0 unpaired electrons.

3. [{Mn(CN)}\(_6\)]\(^{3-}\) (Hexacyanomanganate(III)):
- Mn\(^{3+}\) (d\(^4\)): CN\(^{-}\) is a strong field ligand, leading to a low-spin d\(^4\) configuration.
- The low-spin d\(^4\) configuration results in 2 unpaired electrons.

Thus, the increasing order of unpaired electrons is:
\[ [{Co(C}_2{O}_4)_3]^{3-} (0)<[{Fe(CN)}_6]^{3-} (1)<[{Mn(CN)}_6]^{3-} (2) \]