Arrange the following complexes in decreasing order of number of unpaired electrons on the metal ion
(A) \( [Co(NH_3)_6]^{3+} \)
(B) \( [CoF_6]^{3-} \)
(C) \( [FeF_6]^{3-} \)
(D) \( [Fe(CN)_6]^{3-} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (C), (B), (D), (A)
The number of unpaired electrons in an octahedral complex depends on the d electron count of the metal ion and on the ligand field strength. A weak field ligand (\( \ce{F^{-}} \)) gives a high spin complex, in which the electrons stay unpaired as far as possible; a strong field ligand (\( \ce{CN^{-}} \), and \( \ce{NH3} \) with \( \ce{Co^{3+}} \)) gives a low spin complex, in which the electrons pair up in the three lower (t2g) orbitals.
(A) \( \ce{[Co(NH3)6]^3+} \): charge balance x + 6(0) = +3, so Co is +3. Co (Z = 27) is \( \mathrm{[Ar]\,3d^{7}\,4s^{2}} \); losing the two 4s electrons and one 3d electron gives \( \ce{Co^3+} \): \( \mathrm{3d^{6}} \). \( \ce{NH3} \) pairs all six electrons (\( \mathrm{t_{2g}^{6}} \), d2sp3, diamagnetic): 0 unpaired.
(B) \( \ce{[CoF6]^3-} \): x + 6(−1) = −3, so Co is +3, \( \mathrm{3d^{6}} \). \( \ce{F^{-}} \) is weak field (\( \mathrm{t_{2g}^{4}\,e_g^{2}} \), sp3d2, high spin): 4 unpaired.
(C) \( \ce{[FeF6]^3-} \): x + 6(−1) = −3, so Fe is +3. Fe (Z = 26) is \( \mathrm{[Ar]\,3d^{6}\,4s^{2}} \); losing the two 4s electrons and one 3d electron gives \( \ce{Fe^3+} \): \( \mathrm{3d^{5}} \). \( \ce{F^{-}} \) is weak field (\( \mathrm{t_{2g}^{3}\,e_g^{2}} \)): 5 unpaired.
(D) \( \ce{[Fe(CN)6]^3-} \): x + 6(−1) = −3, so Fe is +3, \( \mathrm{3d^{5}} \). \( \ce{CN^{-}} \) is strong field (\( \mathrm{t_{2g}^{5}} \), d2sp3, low spin): 1 unpaired.
Shortcut: The high spin d5 ion \( \ce{[FeF6]^3-} \) has the most unpaired electrons and the diamagnetic \( \ce{[Co(NH3)6]^3+} \) none, so the order starts with (C) and ends with (A); only Option (3) does both.