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:
Answer & explanation
Correct answer: option 1
The correct answer is option 1. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
|
List-I |
List-II |
|
(A) \([Ni(CN)_4]^{2-}\) |
(III) Diamagnetic |
|
(B) Bidentate |
(iv) oxalate |
|
(C) Chelate ligand |
(I) \(EDTA^{4-}\) |
|
(D) \([Ni(Cl)_4]^{2-}\) |
(II) Paramagnetic |
(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 us 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.
So, the correct option is: (1) (A)-(II), (B)-(IV), (C)-(I), (D)-(III)