Select the high spin complexes from the following:
(A) \( [Ni(CN)_4]^{2-} \)
(B) \( [Fe(CN)_6]^{3-} \)
(C) \( [MnBr_4]^{2-} \)
(D) \( [Fe(H_2O)_6]^{2+} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (C) and (D) only
A complex is high spin when the electrons occupy as many d orbitals as possible before pairing. This happens when the splitting is small: with weak field ligands (halides, \( \ce{H2O} \)), and in nearly all tetrahedral complexes, whose splitting is only about 4/9 of the octahedral value. Strong field \( \ce{CN^{-}} \) forces pairing and gives low spin complexes.
(A) \( \ce{[Ni(CN)4]^2-} \) — Incorrect. x + 4(−1) = −2, so Ni is +2. Ni (Z = 28) is \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \); losing the two 4s electrons gives 3d8. \( \ce{CN^{-}} \) pairs all eight electrons (square planar, diamagnetic): low spin.
(B) \( \ce{[Fe(CN)6]^3-} \) — Incorrect. 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 3d5. Strong field \( \ce{CN^{-}} \) gives \( t_{2g}^{5} \) with one unpaired electron: low spin.
(C) \( \ce{[MnBr4]^2-} \) — Correct. x + 4(−1) = −2, so Mn is +2. Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \); losing the two 4s electrons gives 3d5. The complex is tetrahedral with weak field \( \ce{Br^{-}} \), so all five electrons stay unpaired: high spin.
(D) \( \ce{[Fe(H2O)6]^2+} \) — Correct. Fe is +2. Fe (Z = 26) loses the two 4s electrons to give 3d6. Weak field \( \ce{H2O} \) gives \( t_{2g}^{4}e_{g}^{2} \) with four unpaired electrons: high spin.