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) \([NiCl_4]^{2-}\) |
(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)
A complex is diamagnetic when all its electrons are paired and paramagnetic when it has unpaired electrons. A bidentate ligand binds through two donor atoms; a chelate ligand binds through two or more donor atoms and forms a ring with the metal. Ni (Z = 28) is [Ar]3d84s2; removing 4s2 gives \( \ce{Ni^{2+}} \): 3d8, [↑↓][↑↓][↑↓][↑][↑].
(A) \( \ce{[Ni(CN)4]^{2-}} \) — This matches (III) diamagnetic. \( \ce{CN^{-}} \) is a strong field ligand; it pairs the two unpaired electrons, [↑↓][↑↓][↑↓][↑↓][ ], and the complex is dsp2, square planar, with 0 unpaired electrons.
(B) bidentate — This matches (IV) oxalate. \( \ce{C2O4^{2-}} \) binds through two oxygen atoms.
(C) chelate ligand — This matches (I) EDTA4−. EDTA4− is hexadentate (two N and four O donors) and wraps round the metal ion forming several rings.
(D) \( \ce{[NiCl4]^{2-}} \) — This matches (II) paramagnetic. \( \ce{Cl^{-}} \) is a weak field ligand; the two unpaired electrons remain, the complex is sp3 (tetrahedral) with 2 unpaired electrons.
Shortcut: (A)-(III) leaves Options (1) and (3); (B)-(IV) then leaves only Option (1).