Match List I with List IILIST-I LIST-II A. K2[Ni(CN)4] I. sp3 B. [Ni(CO)4] II. sp3d2 C. [Co(NH3)6]Cl3 III. dsp2 D. Na3[CoF6] IV. d2sp3
Choose the correct answer from the options given below:
| LIST-I | LIST-II |
|---|---|
| A. K2[Ni(CN)4] | I. sp3 |
| B. [Ni(CO)4] | II. sp3d2 |
| C. [Co(NH3)6]Cl3 | III. dsp2 |
| D. Na3[CoF6] | IV. d2sp3 |
Choose the correct answer from the options given below:
- A-III, B-I, C-II, D-IV
- A-III, B-II, C-IV, D-I
- A-I, B-III, C-II, D-IV
- A-III, B-I, C-IV, D-II
The Correct Option is D
Approach Solution - 1
To solve this problem, we need to match the complexes in List-I with their corresponding hybridization states in List-II. Let's examine each complex step-by-step:
- A. K2[Ni(CN)4]
In the complex K2[Ni(CN)4], the nickel (Ni) is in the +2 oxidation state. The cyanide ion (CN-) is a strong field ligand, which causes the pairing of electrons. The electronic configuration of Ni2+ is 3d8.
Upon pairing, it forms the configuration t2g 6eg 2. Thus, the hybridization is dsp2, which is square planar.
Hence, it matches with III (sp). - B. [Ni(CO)4]
In this complex, nickel (Ni) is in the zero oxidation state. The carbon monoxide (CO) is a strong field ligand, which does not cause the pairing of electrons. Therefore, Ni stays in an unpaired state in its 3d10 configuration.
The hybridization is sp3, leading to a tetrahedral geometry.
Hence, it matches with I (sp3). - C. [Co(NH3)6]Cl3
Here, cobalt (Co) is in the +3 oxidation state. The ligand NH3 is a neutral and strong field ligand. The electronic configuration of Co3+ is 3d6. Due to NH3, electron pairing occurs within the d orbitals.
The hybridization is d2sp3, resulting in an octahedral geometry.
Hence, it matches with IV (d2sp3). - D. Na3[CoF6]
In this complex, cobalt (Co) is in the +3 oxidation state. The fluoride ion (F-) is a weak field ligand, which does not cause pairing of electrons. The electronic configuration is thus 3d6, without pairing.
The hybridization becomes sp3d2, leading to octahedral geometry.
Hence, it matches with II (sp3d2).
Therefore, the correct matching is:
- A - III (dsp2)
- B - I (sp3)
- C - IV (d2sp3)
- D - II (sp3d2)
This matches with the correct answer: A-III, B-I, C-IV, D-II.
Approach Solution -2
To solve this problem, we need to match the complexes provided in List-I with the correct hybridization states from List-II.
- K2[Ni(CN)4]:
- The complex ion [Ni(CN)4]2- has nickel in a +2 oxidation state.
- CN- is a strong field ligand causing pairing of electrons in nickel's d-orbitals.
- Nickel forms a square planar complex, leading to dsp2 hybridization.
- [Ni(CO)4]:
- Nickel in this complex is in a zero oxidation state.
- CO is a strong field ligand and the molecular geometry is tetrahedral.
- Thus, the hybridization is sp3.
- [Co(NH3)6]Cl3:
- In this complex, Co is in a +3 oxidation state.
- NH3 is a neutral, weak field ligand in such complexes but leads to a stable octahedral configuration.
- Forms an octahedral complex with d2sp3 hybridization.
- Na3[CoF6]:
- In this complex, Co is in a +3 oxidation state.
- F- is a weak field ligand, leading to no electron pairing in lower d orbitals.
- The geometry is octahedral involving the outer d-orbitals (4d), leading to sp3d2 hybridization.
By comparing the stated hybridizations, the correct matching is option: A-III, B-I, C-IV, D-II.
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