Question:medium

Match the following complexes in List-1 with their hybridisation in List-2: {ll} LIST - 1 & LIST - 2
1. \( [\text{Ni}(\text{CO})_4] \) & a. \( \text{sp}^3\text{d}^2 \)
2. \( [\text{Ni}(\text{CN})_4]^{2-} \) & b. \( \text{d}^2\text{sp}^3 \)
3. \( [\text{Co}(\text{NH}_3)_6]^{3+} \) & c. \( \text{dsp}^2 \)
4. \( [\text{CoF}_6]^{3-} \) & d. \( \text{sp}^3 \)

Show Hint

To solve matching questions on hybridisation instantly, look for outer-orbital vs inner-orbital diagnostic markers: Fluoride (\( \text{F}^- \)) complexes with \( \text{Co}^{3+} \) are iconic classic examples of outer-orbital weak field complexes (\( \text{sp}^3\text{d}^2 \)). Identifying just this single link (\( 4 \rightarrow \text{a} \)) immediately eliminates two options!
Updated On: Jun 3, 2026
  • 1-c, 2-d, 3-a, 4-b
  • 1-d, 2-c, 3-a, 4-b
  • 1-d, 2-c, 3-b, 4-a
  • 1-c, 2-d, 3-b, 4-a
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
Hybridization depends on the coordination number (CN) and the field strength of ligands.
CN = 4 results in \(sp^3\) (tetrahedral) or \(dsp^2\) (square planar).
CN = 6 results in \(d^2sp^3\) (inner orbital octahedral) or \(sp^3d^2\) (outer orbital octahedral).
Strong field ligands (SFL) like \(CO, CN^-, NH_3\) cause electron pairing, while weak field ligands (WFL) like \(F^-\) do not.
Step 2: Detailed Explanation:
1. \([Ni(CO)_4]\): Ni is in 0 state (\(3d^8 4s^2\)). \(CO\) is a strong ligand, it forces \(4s\) electrons into \(3d\), resulting in a \(3d^{10}\) configuration. This leaves \(4s\) and \(4p\) vacant \(\implies sp^3\) (d).
2. \([Ni(CN)_4]^{2-\):} \(Ni^{2+}\) is \(3d^8\). \(CN^-\) is a strong ligand, pairs electrons in \(3d\), leaving one \(3d\), one \(4s\), and two \(4p\) vacant \(\implies dsp^2\) (c).
3. \([Co(NH_3)_6]^{3+\):} \(Co^{3+}\) is \(3d^6\). \(NH_3\) is a strong ligand here, pairs all 6 electrons into 3 orbitals. Two inner \(3d\), one \(4s\), and three \(4p\) vacant \(\implies d^2sp^3\) (b).
4. \([CoF_6]^{3-}\): \(Co^{3+}\) is \(3d^6\). \(F^-\) is a weak ligand, no pairing. Uses outer \(4d\) orbitals \(\implies sp^3d^2\) (a).
Step 3: Final Answer:
The matching is 1-d, 2-c, 3-b, 4-a, which is option (C).
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