Question

Observe the following complex ions Identify the option in which the unpaired electrons in the complex ions are in correct increasing order

• A. C, A, B, D
• B. B, A, C, D
• C. D, A, B, C
• D. D, B, A, C

Hint:

The key to determining the number of unpaired electrons in an octahedral complex is the Spectrochemical Series, which dictates whether the complex is High Spin (Weak Field Ligand, small $\Delta_o$, electrons go to $e_g$) or Low Spin (Strong Field Ligand, large $\Delta_o$, electrons pair up in $t_{2g}$).

Answer 1

The Correct Option is D

To identify the correct order of unpaired electrons in the given complex ions, we need to analyze each complex's electronic configuration and determine the number of unpaired electrons. 

1. Complex [Mn(CN)₆]^3-:
   • Manganese in this complex is in the +3 oxidation state (Mn³⁺), giving it a configuration of 3d⁴.
   • CN^- is a strong field ligand, causing pairing of electrons. Therefore, all 3d electrons are paired, resulting in 0 unpaired electrons.
2. Complex [Fe(CN)₆]^3-:
   • Iron in the +3 oxidation state (Fe³⁺) has a configuration of 3d⁵.
   • CN^- is a strong field ligand, so it causes maximum pairing, resulting in 1 unpaired electron (low spin complex).
3. Complex [CoF₆]^3-:
   • Cobalt in the +3 oxidation state (Co³⁺) has a configuration of 3d⁶.
   • F^- is a weak field ligand, which does not cause pairing, resulting in 4 unpaired electrons (high spin complex).
4. Complex [Co(C₂O₄)₃]^3-:
   • Cobalt in the +3 oxidation state (Co³⁺) has a configuration of 3d⁶.
   • Oxalate (C₂O₄)^2- is a bidentate ligand and strong field, causing pairing, resulting in 0 unpaired electrons.

The correct increasing order of unpaired electrons is: D (0), B (1), A (0), C (4).

Thus, the correct option is D, B, A, C.