Question

The number of unpaired electrons in the paramagnetic complex ion \([FeF_6]^{3-}\) is

• A. 2
• B. 3
• C. 5
• D. 4

Hint:

Weak-field ligands such as \(F^-\), \(Cl^-\), and \(Br^-\) usually produce high-spin complexes with maximum possible unpaired electrons.

Answer 1

The Correct Option is C

Concept: To determine the number of unpaired electrons in a coordination complex, we first determine the oxidation state of the metal ion, then its electronic configuration, and finally whether the ligand is strong-field or weak-field. Fluoride ion (\(F^-\)) is a weak-field ligand and generally forms high-spin complexes.

Step 1: Determine the oxidation state of iron. For the complex: \[ [FeF_6]^{3-} \] Let oxidation state of Fe be \(x\). \[ x+6(-1)=-3 \] \[ x-6=-3 \] \[ x=+3 \] Thus, \[ Fe^{3+} \]

Step 2: Write the electronic configuration of \(Fe^{3+}\). Atomic number of iron: \[ Z=26 \] Electronic configuration of Fe: \[ [Ar]\,3d^6\,4s^2 \] For \(Fe^{3+}\): \[ [Ar]\,3d^5 \]

Step 3: Consider the effect of fluoride ligand. Since \(F^-\) is a weak-field ligand, pairing does not occur. Therefore, all five \(d\)-electrons remain unpaired. \[ \boxed{5\text{ unpaired electrons}} \] Hence, the correct option is \[ \boxed{(3)} \]