Question

Arrange the following in the increasing order of number of unpaired electrons present in the central metal ion: 
I. \([MnCl_6]^{4-}\) 
II. \([FeF_6]^{3-}\) 
III. \([Mn(CN)_6]^{3-}\) 
IV. \([Fe(CN)_6]^{3-}\) 
 

• A. \( IV<I<III<II \)
• B. \( I<III<II<IV \)
• C. \( IV<III<I<II \)
• D. \( I<II<III<IV \)

Hint:

The number of unpaired electrons depends on whether the ligand is strong field (low spin) or weak field (high spin). Strong field ligands like \( CN^- \) cause pairing of electrons, reducing the number of unpaired electrons.

Answer 1

The Correct Option is C

Step 1: Determining the number of unpaired electrons
- The number of unpaired electrons in a complex depends on the oxidation state of the metal and the ligand strength.
- Cyanide (\( CN^- \)) is a strong field ligand, leading to low spin configurations.
- Fluoride (\( F^- \)) and chloride (\( Cl^- \)) are weak field ligands, leading to high spin configurations. 
Step 2: Electron Configurations
- \([MnCl_6]^{4-}\) → High spin (\( Mn^{2+} \), \( d^5 \)) → 5 unpaired electrons
- \([FeF_6]^{3-}\) → High spin (\( Fe^{3+} \), \( d^5 \)) → 5 unpaired electrons
- \([Mn(CN)_6]^{3-}\) → Low spin (\( Mn^{4+} \), \( d^3 \)) → 3 unpaired electrons
- \([Fe(CN)_6]^{3-}\) → Low spin (\( Fe^{3+} \), \( d^5 \)) → 1 unpaired electron
Step 3: Arranging in Increasing Order - \( IV (1)<III (3)<I (5)<II (5) \) Thus, the correct order is: \[ IV<III<I<II \]