Question

Identify the paramagnetic outer orbital complex ion from the following.

• A. \([Co(C_2O_4)_3]^{3-}\)
• B. \([CoF_6]^{3-}\)
• C. \([Fe(CN)_6]^{3-}\)
• D. \([Mn(CN)_6]^{3-}\)

Hint:

Weak-field ligands: \[ F^-,\; Cl^-,\; Br^-,\; I^- \] generally form \[ sp^3d^2 \] outer orbital complexes. Strong-field ligands: \[ CN^-,\; CO,\; NO_2^- \] usually form inner orbital complexes.

Answer 1

The Correct Option is B

Step 1: Know what is asked.
We need the ion that is both paramagnetic (has unpaired electrons) and outer orbital (high spin, weak-field ligand using outer $d$ orbitals).

Step 2: Find metal oxidation states and d count.
$[Co(C_2O_4)_3]^{3-}$ is $Co^{3+}$, $d^6$; $[CoF_6]^{3-}$ is $Co^{3+}$, $d^6$; $[Fe(CN)_6]^{3-}$ is $Fe^{3+}$, $d^5$; $[Mn(CN)_6]^{3-}$ is $Mn^{3+}$, $d^4$.

Step 3: Sort by ligand strength.
$CN^-$ and oxalate are strong-field ligands that pair electrons and give inner orbital complexes. $F^-$ is a weak-field ligand that does not pair, giving an outer orbital complex.

Step 4: Drop the strong-field ones.
The two cyanide complexes and the oxalate complex are inner orbital, so they are ruled out for outer orbital.

Step 5: Examine the fluoride complex.
$[CoF_6]^{3-}$ with weak $F^-$ keeps $d^6$ as high spin, giving $4$ unpaired electrons. It is paramagnetic and uses outer $d$ orbitals, so it is an outer orbital complex.

Step 6: State the answer.
The paramagnetic outer orbital ion is $[CoF_6]^{3-}$.
\[ \boxed{[CoF_6]^{3-}} \]