Question

According to the structural constraints of Valence Bond Theory (VBT), which of the following hybridization geometries and magnetic configurations accurately describes the low-spin coordination complex ion \( [\text{Fe}(\text{CN})_6]^{4- \)? (Atomic Number of \( \text{Fe} = 26 \))}

• A. \( \text{sp}^3\text{d}^2 \) hybridization and Paramagnetic
• B. \( \text{d}^2\text{sp}^3 \) hybridization and Diamagnetic
• C. \( \text{dsp}^2 \) hybridization and Diamagnetic
• D. \( \text{sp}^3\text{d}^2 \) hybridization and Diamagnetic

Hint:

Strong-field ligands like \( \text{CN}^- \) and \( \text{CO} \) drive electron pairing, which typically results in inner-orbital \( \text{d}^2\text{sp}^3 \) hybridizations for octahedral complexes. Weak-field ligands like \( \text{Cl}^- \) or \( \text{F}^- \) result in outer-orbital \( \text{sp}^3\text{d}^2 \) configurations.

Answer 1

The Correct Option is B

Concept: Valence Bond Theory determines the hybridization and geometry of coordination complexes by evaluating whether the incoming ligands are weak-field or strong-field, which dictates whether the d-orbital electrons pair up.

Step 1: Determine the oxidation state and electronic configuration of Iron.
Let the unknown oxidation state of Iron be \( x \). Cyanide is an anionic ligand (\(\text{CN}^-\)): \[ x + 6(-1) = -4 \quad \Rightarrow \quad x - 6 = -4 \quad \Rightarrow \quad x = +2 \] The electronic configuration of a neutral Iron atom (\( Z = 26 \)) is \( [\text{Ar}] \, 3\text{d}^6 \, 4\text{s}^2 \). For the divalent ion (\( \text{Fe}^{2+} \)), remove the two \(4\text{s}\) electrons: \[ \text{Fe}^{2+} = [\text{Ar}] \, 3\text{d}^6 \, 4\text{s}^0 \]

Step 2: Analyze the effect of the ligand on d-orbital electron pairing.
The cyanido (\(\text{CN}^-\)) ligand is a strong-field ligand. It forces the 6 electrons in the \(3\text{d}\) orbitals to pair up completely within the lower-energy \(t_{2g}\) orbitals: \[ 3\text{d} \text{ configuration after pairing} = \underline{\uparrow\downarrow} \,\, \underline{\uparrow\downarrow} \,\, \underline{\uparrow\downarrow} \,\, \underline{\,\,\,\,} \,\, \underline{\,\,\,\,} \] Because all 6 electrons are completely paired up, the complex contains zero unpaired electrons, making it Diamagnetic.

Step 3: Identify the vacant orbitals available for hybridization.
Pairing up the electrons leaves two internal \(3\text{d}\) orbitals completely empty. These two empty \(3\text{d}\) orbitals combine with the vacant \(4\text{s}\) orbital and three \(4\text{p}\) orbitals to form six equivalent hybrid orbitals: \[ \text{Hybridization} = \text{d}^2\text{sp}^3 \quad \text{(Inner orbital octahedral complex)} \]