Practicing Success

Target Exam

CUET

Subject

Chemistry

Chapter

Inorganic: D and F Block Elements

Question:

Match List-I with List-II

List-I

List-II

(A) \([Ni(CN)_4]^{2-}\)

(I) \(EDTA^{4-}\)

(B) Bidentate

(II) Paramagnetic

(C) Chelate ligand

(III) Diamagnetic

(D) \([Ni(Cl)_4]^{2-}\)

(iv) oxalate

Choose the correct answer from the options given below:

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct option is 1. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

(A) \([Ni(CN)_4]^{2-}\) - This is a complex ion with four cyanide ligands. It's a strong field ligand, so it's likely to cause pairing of electrons in the d-orbitals of the nickel ion.
(B) Bidentate - A bidentate ligand can bond to a central metal atom through two donor atoms.
(C) Chelate ligand - A chelate ligand is a type of ligand that can bond to a central metal atom through multiple donor atoms, forming a ring-like structure.
(D) \([Ni(Cl)_4]^{2-}\) - This is a complex ion with four chloride ligands. Chloride is also a strong field ligand.

Now, let's match these with the properties:
(I) \(EDTA^{4-}\) - EDTA is a hexadentate ligand, which means it can form six bonds with a central metal atom, making it a chelate ligand.
(II) Paramagnetic - Paramagnetic substances have unpaired electrons, typically found in compounds where the metal ion is not fully paired due to the ligands' influence.
(III) Diamagnetic - Diamagnetic substances have all paired electrons, typically found in compounds where the metal ion is fully paired due to the ligands' influence.
(IV) Oxalate - Oxalate is a bidentate ligand because it can form two bonds with a central metal atom.

Now, let's match the lists:
(A) \([Ni(CN)_4]^{2-}\) - (II) Paramagnetic
(B) Bidentate - (IV) Oxalate
(C) Chelate ligand - (I) \(EDTA^{4-}\)
(D) \([Ni(Cl)_4]^{2-}\) - (III) Diamagnetic

So, the correct option is: (1) (A)-(II), (B)-(IV), (C)-(I), (D)-(III)