Question

Match List - I with List - II.

Choose the correct answer from the options given below:

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

Hint:

For coordination compounds, use the valence bond theory (VBT) and crystal field splitting to determine hybridization.

Answer 1

The Correct Option is C

In the given problem, you need to match items from List - I with corresponding items in List - II. The correct matches have been given based on specific chemistry principles which relate to the properties, structures, and characteristics of the chemical compounds or concepts represented in each list.

After analyzing the options, the correct matching is:

1. (A)-(III): Matches the property or concept related to the item in List-I.
2. (B)-(II): Matches the property or concept related to the item in List-I.
3. (C)-(I): Matches the property or concept related to the item in List-I.
4. (D)-(IV): Matches the property or concept related to the item in List-I.

Therefore, the correct answer aligns with the option (A)-(III), (B)-(II), (C)-(I), (D)-(IV), matching each item correctly based on their chemistry principles.

Answer 2

Step 1: Given data.
We are asked to match the complexes in List–I with their respective hybridisations in List–II.

List–I (Complexes):
(A) [CoF₆]³⁻
(B) [NiCl₄]²⁻
(C) [Co(NH₃)₆]³⁺
(D) [Ni(CN)₄]²⁻

List–II (Hybridisation):
(I) d²sp³
(II) sp³
(III) sp³d²
(IV) dsp²

Step 2: Determine hybridisation for each complex.
(A) [CoF₆]³⁻
– Oxidation state of Co = +3
– Fluoride (F⁻) is a weak ligand → no pairing of electrons.
– The complex is outer orbital with sp³d² hybridisation.
Hence, hybridisation = sp³d² (outer orbital).
⇒ (A) → (III)

(B) [NiCl₄]²⁻
– Oxidation state of Ni = +2.
– Cl⁻ is a weak ligand → no pairing of electrons.
– 3d⁸4s² → two unpaired electrons remain → tetrahedral geometry.
– Hybridisation = sp³.
⇒ (B) → (II)

(C) [Co(NH₃)₆]³⁺
– Oxidation state of Co = +3.
– NH₃ is a strong ligand → pairing occurs.
– 3d⁶ configuration, after pairing → inner orbital complex.
– Hybridisation = d²sp³.
⇒ (C) → (I)

(D) [Ni(CN)₄]²⁻
– Oxidation state of Ni = +2.
– CN⁻ is a strong ligand → pairing occurs → dsp² hybridisation.
– Square planar complex.
⇒ (D) → (IV)

Step 3: Final Matching.
(A) [CoF₆]³⁻ → (III) sp³d²
(B) [NiCl₄]²⁻ → (II) sp³
(C) [Co(NH₃)₆]³⁺ → (I) d²sp³
(D) [Ni(CN)₄]²⁻ → (IV) dsp²

Final Answer:
\[ \boxed{(A)-(III), \, (B)-(II), \, (C)-(I), \, (D)-(IV)} \]