Question

w.o.f have $\Delta_o > P.E$? (SFL)

• A. Co(H_2O)_6^{+3}
• B. Fe(EDTA)^{-2}
• C. Pt(H_2O)_6^{+4}
• D. Co(NO_3)_6^{-3}

Answer 1

Answer: (A), (B), (C)

a, b, c

Answer 2

The condition $\Delta_o > P.E$ implies a low spin complex. A low spin complex occurs when the crystal field splitting energy ($\Delta_o$) is greater than the pairing energy (P.E.), meaning it is energetically more favorable for electrons to occupy the lower energy $t_{2g}$ orbitals before pairing up.

1. a) $Co(H_2O)_6^{+3}$:

   • Central metal ion: Cobalt ($Co^{+3}$)
   • $d$-electron configuration: $3d^6$
   • Ligand: $H_2O$ (Water) is a weak field ligand.
   • Analysis: For $Co^{+3}$ ($d^6$) with a weak field ligand, while $H_2O$ is weak, $Co^{+3}$ has a strong tendency to form low spin complexes. The $\Delta_o$ for this complex is approximately $22000 \, \text{cm}^{-1}$, and the pairing energy (P.E.) is around $17000 \, \text{cm}^{-1}$. Thus, $\Delta_o > P.E$.

2. b) $Fe(EDTA)^{-2}$:

   • Central metal ion: Iron ($Fe^{+2}$)
   • $d$-electron configuration: $3d^6$
   • Ligand: EDTA (ethylenediaminetetraacetate) is a strong field ligand, especially due to its chelating nature.
   • Analysis: With a strong field ligand like EDTA, the crystal field splitting energy ($\Delta_o$) is significantly large, overcoming the pairing energy (P.E.). For $Fe^{+2}$, P.E. is around $15000 \, \text{cm}^{-1}$. A strong ligand ensures $\Delta_o > P.E$.

3. c) $Pt(H_2O)_6^{+4}$:

   • Central metal ion: Platinum ($Pt^{+4}$)
   • $d$-electron configuration: $5d^6$
   • Ligand: $H_2O$ (Water) is a weak field ligand.
   • Analysis: Platinum is a 3rd-row transition metal. For 2nd and 3rd-row transition metals, $\Delta_o$ is significantly larger than for 1st-row metals. Even with a weak field ligand like $H_2O$, the $\Delta_o$ for $Pt^{+4}$ complexes is very large due to the metal's position in the periodic table and its high oxidation state. This ensures that $\Delta_o > P.E$.

4. d) $Co(NO_3)_6^{-3}$:

   • Central metal ion: Cobalt ($Co^{+3}$)
   • $d$-electron configuration: $3d^6$
   • Ligand: $NO_3^{-}$ (Nitrate) is a weak field ligand.
   • Analysis: For a $d^6$ ion with a weak field ligand like $NO_3^{-}$, the crystal field splitting energy ($\Delta_o$) is typically smaller than the pairing energy (P.E.). This leads to a high spin configuration, where $\Delta_o < P.E$.

Therefore, complexes a), b), and c) satisfy the condition $\Delta_o > P.E$. The correct options are a, b, and c.