Match the complex given in List-I with the number of unpaired electrons given in List-II
|
List-I Complex |
List-II Number of unpaired electrons (n) |
|
(A) \( [NiCl_4]^{2-} \) |
(I) 5 |
|
(B) \( [Co(NH_3)_6]^{3+} \) |
(II) 1 |
|
(C) \( [FeF_6]^{3-} \) |
(III) 2 |
|
(D) \( [Fe(CN)_6]^{3-} \) |
(IV) 0 |
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)
The number of unpaired electrons of a complex follows from the d electron count of the metal ion and the strength of the ligand field. Weak field ligands such as \( \ce{Cl-} \) and \( \ce{F-} \) do not force pairing, so the electrons stay as unpaired as possible (high spin). Strong field ligands such as \( \ce{CN-} \), and \( \ce{NH3} \) with \( \ce{Co^3+} \), make the electrons pair up in the lower orbitals (low spin).
(A) \([NiCl_4]^{2-}\) — This matches (III) 2. x + 4(−1) = −2 gives Ni(+2). Ni (Z = 28) is \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \), so \( \ce{Ni^2+} \) is \( \mathrm{3d^{8}} \). With weak field \( \ce{Cl-} \) the complex is tetrahedral (sp3) and the 3d8 set keeps 2 unpaired electrons.
(B) \([Co(NH_3)_6]^{3+}\) — This matches (IV) 0. \( \ce{NH3} \) is neutral, so cobalt is +3. Co (Z = 27) is \( \mathrm{[Ar]\,3d^{7}\,4s^{2}} \), so \( \ce{Co^3+} \) is \( \mathrm{3d^{6}} \). \( \ce{NH3} \) pairs all six electrons into the three lower orbitals (d2sp3), leaving no unpaired electron.
(C) \([FeF_6]^{3-}\) — This matches (I) 5. x + 6(−1) = −3 gives Fe(+3). Fe (Z = 26) is \( \mathrm{[Ar]\,3d^{6}\,4s^{2}} \), so \( \ce{Fe^3+} \) is \( \mathrm{3d^{5}} \). Weak field \( \ce{F-} \) gives an outer orbital (sp3d2) complex with 5 unpaired electrons.
(D) \([Fe(CN)_6]^{3-}\) — This matches (II) 1. Again \( \ce{Fe^3+} \), \( \mathrm{3d^{5}} \); strong field \( \ce{CN-} \) pairs the five electrons into the three lower orbitals as two pairs and one unpaired electron (d2sp3).
Shortcut: \( \ce{[FeF6]^3-} \), the classic high spin d5 complex, has 5 unpaired electrons; (C)-(I) appears only in Option (1).