NEETChemistryCoordination Compounds
Match List-I with List-II: List-I List-II (A) C ₂ O ₄²⁻ (I) Hexadentate (B) SCN ⁻ (II) Unidentate (C) EDTA ⁴⁻ (III) Ambidentate (D) H ₂ O (IV) Didentate Choose the correct answer from the options given below:
Options
- A(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
- B(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
- C(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
- D(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Correct answer
D. (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Step-by-step solution
C ₂ O ₄²⁻ (oxalate ion) coordinates through two oxygen atoms simultaneously, making it a didentate ligand. SCN ⁻ (thiocyanate ion) can coordinate through either the sulfur or nitrogen atom, making it an ambidentate ligand. EDTA ⁴⁻ (ethylenediaminetetraacetate ion) can bind through two nitrogen and four oxygen atoms, making it a hexadentate ligand. H ₂ O coordinates through a single oxygen atom, making it a unidentate ligand. Thus, the correct matching is (A)-(IV), (B)-(III), (C)-(I), (D)-(II). Answer: (A)-(IV), (B)