Given below are two statements:

Statement I: [Ni(CN)₄]^2– is square planar and diamagnetic complex, with dsp² hybridization for Ni but [Ni(CO)₄] is tetrahedral, paramagnetic and with sp³-hybridization for Ni.
Statement II: [NiCl₄]^2– and [Ni(CO)₄] both have same d-electron configuration, have same geometry and are paramagnetic.

In light of the above statements, choose the correct answer from the options given below :

Question

Match List - I with List - II.

List - I (Complex)	List - II (Hybridisation)
(A) \([\text{CoF}_6]^{3-}\)	(I) \( d^2 sp^3 \)
(B) \([\text{NiCl}_4]^{2-}\)	(II) \( sp^3 \)
(C) \([\text{Co(NH}_3)_6]^{3+}\)	(III) \( sp^3 d^2 \)
(D) \([\text{Ni(CN}_4]^{2-}\)	(IV) \( dsp^2 \)

Choose the correct answer from the options given below:

Answer 1

To determine the accuracy of the given statements, let's analyze each statement based on the properties of the nickel complexes involved.

Statement I: Analysis

The complex [\text{Ni}(\text{CN})_4]^{2-} is known to be square planar and diamagnetic. This arrangement occurs because the cyanide ion (\text{CN}^-) is a strong field ligand, causing pairing of electrons in the nickel ion (Ni²⁺), and it uses \text{dsp}^2 hybridization.

For [\text{Ni}(\text{CO})_4], the nickel is in a zero oxidation state (Ni⁰). Carbon monoxide (\text{CO}) is a neutral and strong field ligand but does not lead to electron pairing here. This complex is tetrahedral and diamagnetic, with \text{sp}^3 hybridization. Thus, the part about [Ni(CO)₄] being paramagnetic is incorrect, making this statement false.

Statement II: Analysis

The complex [NiCl₄]^2– has Ni²⁺, with a configuration of 3d^8. Chloride (\text{Cl}^-) is a weak field ligand, so it does not cause electron pairing in the d-orbitals, and the complex is tetrahedral and paramagnetic, with unpaired electrons.

On the other hand, as discussed above, [Ni(CO)₄] with Ni⁰ is tetrahedral and diamagnetic. Since they do not have the same electron configuration or magnetic properties, the statement is false.

Conclusion:

Both Statement I and Statement II have been found to be false based on the discussed coordination chemistry of the nickel complexes and their respective properties.

Answer 2

To determine the accuracy of the given statements, let's analyze each statement based on the properties of the nickel complexes involved.

Statement I: Analysis

The complex [\text{Ni}(\text{CN})_4]^{2-} is known to be square planar and diamagnetic. This arrangement occurs because the cyanide ion (\text{CN}^-) is a strong field ligand, causing pairing of electrons in the nickel ion (Ni²⁺), and it uses \text{dsp}^2 hybridization.

For [\text{Ni}(\text{CO})_4], the nickel is in a zero oxidation state (Ni⁰). Carbon monoxide (\text{CO}) is a neutral and strong field ligand but does not lead to electron pairing here. This complex is tetrahedral and diamagnetic, with \text{sp}^3 hybridization. Thus, the part about [Ni(CO)₄] being paramagnetic is incorrect, making this statement false.

Statement II: Analysis

The complex [NiCl₄]^2– has Ni²⁺, with a configuration of 3d^8. Chloride (\text{Cl}^-) is a weak field ligand, so it does not cause electron pairing in the d-orbitals, and the complex is tetrahedral and paramagnetic, with unpaired electrons.

On the other hand, as discussed above, [Ni(CO)₄] with Ni⁰ is tetrahedral and diamagnetic. Since they do not have the same electron configuration or magnetic properties, the statement is false.

Conclusion:

Both Statement I and Statement II have been found to be false based on the discussed coordination chemistry of the nickel complexes and their respective properties.

The Correct Option is B

Solution and Explanation

Statement I: Analysis

Statement II: Analysis

Conclusion:

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