Question

Which of the following complexes shows \( sp^3d^2 \) hybridization?

• A. \([Cr(NO_2)_6]^{3-}\)
• B. \([Fe(CN)_6]^{4-}\)
• C. \([CoF_6]^{3-}\)
• D. \([Ni(CO)_4]\)

Hint:

Weak ligands like \( F^- \) do not cause electron pairing, leading to high-spin octahedral complexes with \( sp^3d^2 \) hybridization.

Answer 1

The Correct Option is C

Step 1: Finding the hybridization of \([CoF_6]^{3-}\).
- The oxidation state of Co in \([CoF_6]^{3-}\) is: \[ x + 6(-1) = -3 \Rightarrow x = +3 \] - The electronic configuration of \( Co^{3+} \) is \( 3d^6 \).
Step 2: Determining hybridization.
- Fluoride (\( F^- \)) is a weak ligand and does not cause pairing of \( d \)-electrons.
- Hence, Co uses \( sp^3d^2 \) hybridization, resulting in octahedral geometry.
Step 3: Identifying the correct answer.
- Since \([CoF_6]^{3-}\) exhibits \( sp^3d^2 \) hybridization, the correct answer is (C).