Question:
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. \( \text{sp}^3 \)
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. \( \text{sp}^3 \)
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. \( \text{sp}^3 \)
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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: May 20, 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
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The Correct Option is C
Solution and Explanation
Concept:
According to Valence Bond Theory (VBT) for coordination complexes:
• Strong Field Ligands (SFL): Ligands like \( \text{CN}^- \) and \( \text{CO} \) cause pairing of unpaired electrons in the inner \( \text{d} \)-orbitals if necessary. \( \text{NH}_3 \) acts as a strong field ligand with higher oxidation states like \( \text{Co}^{3+} \).
• Weak Field Ligands (WFL): Ligands like \( \text{F}^- \) do not cause electron pairing; hence, outer \( \text{d} \)-orbitals are used for hybridisation.
• Coordination Number 4: Leads to either tetrahedral (\( \text{sp}^3 \)) or square planar (\( \text{dsp}^2 \)) geometry.
• Coordination Number 6: Leads to octahedral geometry, which can be inner orbital (\( \text{d}^2\text{sp}^3 \)) or outer orbital (\( \text{sp}^3\text{d}^2 \)).
Step 1: Analyzing complex 1: \( [\text{Ni}(\text{CO})_4] \).
The oxidation state of \( \text{Ni} \) in \( [\text{Ni}(\text{CO})_4] \) is \( 0 \). The electronic configuration of \( \text{Ni}(0) \) is \( [\text{Ar}] 3\text{d}^8 4\text{s}^2 \). Since \( \text{CO} \) is a very strong field ligand, it forces the two electrons from the \( 4\text{s} \) orbital into the \( 3\text{d} \) orbital to pair up completely. The configuration becomes \( 3\text{d}^{10} \). Since the \( 3\text{d} \) subshell is completely filled, the vacant \( 4\text{s} \) and three \( 4\text{p} \) orbitals hybridise to form \( \text{sp}^3 \) hybridisation. Thus, 1 matches with d.
Step 2: Analyzing complex 2: \( [\text{Ni}(\text{CN})_4]^{2-} \).
Let the oxidation state of \( \text{Ni} \) be \( x \): \( x + 4(-1) = -2 \Rightarrow x = +2 \). The electronic configuration of \( \text{Ni}^{2+} \) is \( [\text{Ar}] 3\text{d}^8 4\text{s}^0 \). \( \text{CN}^- \) is a strong field ligand, forcing the 8 electrons in the \( 3\text{d} \) orbital to pair up, leaving one inner \( 3\text{d} \) orbital vacant. The vacant orbitals involved in hybridisation are one \( 3\text{d} \), one \( 4\text{s} \), and two \( 4\text{p} \), resulting in \( \text{dsp}^2 \) hybridisation (Square Planar). Thus, 2 matches with c.
Step 3: Analyzing complex 3: \( [\text{Co}(\text{NH}_3)_6]^{3+} \).
The oxidation state of \( \text{Co} \) is \( +3 \). The electronic configuration of \( \text{Co}^{3+} \) is \( [\text{Ar}] 3\text{d}^6 4\text{s}^0 \). With \( \text{Co}^{3+} \), \( \text{NH}_3 \) behaves as a strong field ligand and causes pairing of the 6 electrons in the \( 3\text{d} \) subshell into three paired pairs (\( \text{t}_{2\text{g}}^6 \)). This leaves two inner \( 3\text{d} \) orbitals completely empty. These two vacant \( 3\text{d} \) orbitals, along with one \( 4\text{s} \) and three \( 4\text{p} \) orbitals, form \( \text{d}^2\text{sp}^3 \) hybridisation (Inner orbital octahedral). Thus, 3 matches with b.
Step 4: Analyzing complex 4: \( [\text{CoF}_6]^{3-} \).
The oxidation state of \( \text{Co} \) is \( +3 \). The electronic configuration of \( \text{Co}^{3+} \) is \( [\text{Ar}] 3\text{d}^6 4\text{s}^0 \). \( \text{F}^- \) is a weak field ligand and cannot cause electron pairing. The \( 3\text{d} \) configuration remains unpaired with maximum spin. Therefore, outer \( 4\text{s} \), three \( 4\text{p} \), and two \( 4\text{d} \) orbitals are used for hybridisation, resulting in \( \text{sp}^3\text{d}^2 \) hybridisation (Outer orbital octahedral). Thus, 4 matches with a. Combining all matches: 1-d, 2-c, 3-b, 4-a, which matches option (C).
• Strong Field Ligands (SFL): Ligands like \( \text{CN}^- \) and \( \text{CO} \) cause pairing of unpaired electrons in the inner \( \text{d} \)-orbitals if necessary. \( \text{NH}_3 \) acts as a strong field ligand with higher oxidation states like \( \text{Co}^{3+} \).
• Weak Field Ligands (WFL): Ligands like \( \text{F}^- \) do not cause electron pairing; hence, outer \( \text{d} \)-orbitals are used for hybridisation.
• Coordination Number 4: Leads to either tetrahedral (\( \text{sp}^3 \)) or square planar (\( \text{dsp}^2 \)) geometry.
• Coordination Number 6: Leads to octahedral geometry, which can be inner orbital (\( \text{d}^2\text{sp}^3 \)) or outer orbital (\( \text{sp}^3\text{d}^2 \)).
Step 1: Analyzing complex 1: \( [\text{Ni}(\text{CO})_4] \).
The oxidation state of \( \text{Ni} \) in \( [\text{Ni}(\text{CO})_4] \) is \( 0 \). The electronic configuration of \( \text{Ni}(0) \) is \( [\text{Ar}] 3\text{d}^8 4\text{s}^2 \). Since \( \text{CO} \) is a very strong field ligand, it forces the two electrons from the \( 4\text{s} \) orbital into the \( 3\text{d} \) orbital to pair up completely. The configuration becomes \( 3\text{d}^{10} \). Since the \( 3\text{d} \) subshell is completely filled, the vacant \( 4\text{s} \) and three \( 4\text{p} \) orbitals hybridise to form \( \text{sp}^3 \) hybridisation. Thus, 1 matches with d.
Step 2: Analyzing complex 2: \( [\text{Ni}(\text{CN})_4]^{2-} \).
Let the oxidation state of \( \text{Ni} \) be \( x \): \( x + 4(-1) = -2 \Rightarrow x = +2 \). The electronic configuration of \( \text{Ni}^{2+} \) is \( [\text{Ar}] 3\text{d}^8 4\text{s}^0 \). \( \text{CN}^- \) is a strong field ligand, forcing the 8 electrons in the \( 3\text{d} \) orbital to pair up, leaving one inner \( 3\text{d} \) orbital vacant. The vacant orbitals involved in hybridisation are one \( 3\text{d} \), one \( 4\text{s} \), and two \( 4\text{p} \), resulting in \( \text{dsp}^2 \) hybridisation (Square Planar). Thus, 2 matches with c.
Step 3: Analyzing complex 3: \( [\text{Co}(\text{NH}_3)_6]^{3+} \).
The oxidation state of \( \text{Co} \) is \( +3 \). The electronic configuration of \( \text{Co}^{3+} \) is \( [\text{Ar}] 3\text{d}^6 4\text{s}^0 \). With \( \text{Co}^{3+} \), \( \text{NH}_3 \) behaves as a strong field ligand and causes pairing of the 6 electrons in the \( 3\text{d} \) subshell into three paired pairs (\( \text{t}_{2\text{g}}^6 \)). This leaves two inner \( 3\text{d} \) orbitals completely empty. These two vacant \( 3\text{d} \) orbitals, along with one \( 4\text{s} \) and three \( 4\text{p} \) orbitals, form \( \text{d}^2\text{sp}^3 \) hybridisation (Inner orbital octahedral). Thus, 3 matches with b.
Step 4: Analyzing complex 4: \( [\text{CoF}_6]^{3-} \).
The oxidation state of \( \text{Co} \) is \( +3 \). The electronic configuration of \( \text{Co}^{3+} \) is \( [\text{Ar}] 3\text{d}^6 4\text{s}^0 \). \( \text{F}^- \) is a weak field ligand and cannot cause electron pairing. The \( 3\text{d} \) configuration remains unpaired with maximum spin. Therefore, outer \( 4\text{s} \), three \( 4\text{p} \), and two \( 4\text{d} \) orbitals are used for hybridisation, resulting in \( \text{sp}^3\text{d}^2 \) hybridisation (Outer orbital octahedral). Thus, 4 matches with a. Combining all matches: 1-d, 2-c, 3-b, 4-a, which matches option (C).
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