Question

The hybridization and magnetic nature of $\left[\text{CoF}_6\right]^{3-}$ respectively are:}

• A. $\text{sp}^3\text{d}^2$ and paramagnetic
• B. $\text{sp}^3\text{d}^2$ and diamagnetic
• C. $\text{d}^2\text{sp}^3$ and paramagnetic
• D. $\text{d}^2\text{sp}^3$ and diamagnetic

Hint:

Weak field ligands (F$^-$, Cl$^-$, Br$^-$) → high spin → paramagnetic. Strong field ligands (CN$^-$, CO) → low spin → diamagnetic.

Answer 1

The Correct Option is A

Concept: This question is based on Valence Bond Theory (VBT) and Crystal Field Theory (CFT).
• Oxidation state of metal determines d-electron configuration.
• $\text{F}^-$ is a weak field ligand → high spin complex.
• Unpaired electrons determine paramagnetism/diamagnetism.

Step 1: Oxidation state of Co Let oxidation state of Co = $x$: \[ x + 6(-1) = -3 \Rightarrow x = +3 \] So Co$^{3+}$ = $3d^6$

Step 2: Hybridization Since $F^-$ is weak field ligand:
• No pairing in 3d orbitals
• Outer orbital complex formed Hybridization uses: \[ 4s + 4p + 4d \Rightarrow \text{sp}^3\text{d}^2 \]

Step 3: Magnetic nature $3d^6$ high spin → 4 unpaired electrons Hence: \[ \text{Paramagnetic} \]