Match the complexes given in List-I with the geometry given in List-II
|
List-I Complex |
List-II Geometry |
|
(A) \( [NiCl_4]^{2-} \) |
(I) Tetrahedral, paramagnetic |
|
(B) \( [Ni(CN)_4]^{2-} \) |
(II) Square planar, diamagnetic |
|
(C) \( [CoF_6]^{3-} \) |
(III) Octahedral, diamagnetic |
|
(D) \( [Co(NH_3)_6]^{3+} \) |
(IV) Octahedral, paramagnetic |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(I), (B)-(II), (C)-(IV), (D)-(III)
By valence bond theory, the geometry follows from the hybridisation, and the magnetic behaviour from whether unpaired electrons remain. Weak field ligands (\( \ce{Cl-} \), \( \ce{F-} \)) leave the d electrons unpaired, so outer orbitals are used (sp3 for four ligands, sp3d2 for six). Strong field ligands (\( \ce{CN-} \), and \( \ce{NH3} \) with \( \ce{Co^3+} \)) pair the electrons and free inner d orbitals (dsp2, d2sp3).
(A) \([NiCl_4]^{2-}\) — This matches (I) Tetrahedral, paramagnetic. x + 4(−1) = −2 gives Ni(+2); Ni (Z = 28) is \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \), so \( \ce{Ni^2+} \) is \( \mathrm{3d^{8}} \). Weak field \( \ce{Cl-} \) causes no pairing: sp3, tetrahedral, with 2 unpaired electrons.
(B) \([Ni(CN)_4]^{2-}\) — This matches (II) Square planar, diamagnetic. Again \( \ce{Ni^2+} \), \( \mathrm{3d^{8}} \); strong field \( \ce{CN-} \) pairs the eight electrons into four orbitals, freeing one 3d orbital for dsp2 hybridisation: square planar, no unpaired electron.
(C) \([CoF_6]^{3-}\) — This matches (IV) Octahedral, paramagnetic. x + 6(−1) = −3 gives Co(+3); Co (Z = 27) is \( \mathrm{[Ar]\,3d^{7}\,4s^{2}} \), so \( \ce{Co^3+} \) is \( \mathrm{3d^{6}} \). Weak field \( \ce{F-} \) leaves 4 unpaired electrons: outer orbital sp3d2, octahedral.
(D) \([Co(NH_3)_6]^{3+}\) — This matches (III) Octahedral, diamagnetic. \( \ce{Co^3+} \), \( \mathrm{3d^{6}} \); \( \ce{NH3} \) pairs all six electrons into three orbitals, and the two free 3d orbitals give d2sp3: inner orbital, octahedral, no unpaired electron.
Shortcut: (A)-(I) appears in Options (3) and (4); (C)-(IV) then leaves only Option (4).