Question

Match List I with List II

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. A-III, B-I, C-II, D-IV
• B. A-III, B-II, C-IV, D-I
• C. A-I, B-III, C-II, D-IV
• D. A-III, B-I, C-IV, D-II

Answer 1

The Correct Option is D

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:

1. 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 t_2g ⁶e_g ². Thus, the hybridization is dsp², which is square planar. 
   Hence, it matches with III (sp).
2. 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).
3. 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).
4. 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.

Answer 2

To solve this problem, we need to match the complexes provided in List-I with the correct hybridization states from List-II.

1. 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.
2. [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.
3. [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.
4. 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.