Question

which of the following octahedral complex ions will have the fewest number of unpaired electrons? 1) [FeF_6]^3 2)[Cr(H_2O)_6]^3+ 3) [Ni(NH_3))_6]^2+ 4) [RhCl_6]^3- 5)[V(H_2O)_6]^3+

Answer 1

The correct answer is 1.) The complex with the fewest number of unpaired electrons is [Fe(CN)6]^4-. Complexes with strong-field ligands like cyanide (CN-) produce a large crystal field splitting (Δ), causing the electrons to pair in the lower energy t2g orbitals. As a result, [Fe(CN)6]^4- is a low-spin complex with only one unpaired electron.

Step-by-step explanation:

The complex with the fewest number of unpaired electrons is [Fe(CN)6]^4-

Complexes with strong-field ligands like cyanide (CN-) produce a large crystal field splitting (Δ), which is the energy difference between the t2g and eg orbitals. In [Fe(CN)6]^4-, the strong field of six cyanide ligands produces a large Δ, causing the electrons to require less energy to pair in the lower energy t2g orbitals rather than populating the higher energy eg orbitals. As a result, [Fe(CN)6]^4- is a low-spin complex with only one unpaired electron.