Explain the following observations about complex ions of the transition metals. — Transition Metals and Coordination Chemistry Chemistry Question
Problem Context
Explain the following observations about complex ions of the transition metals.
The Cr(H2O)6 3+ ion is paramagnetic while the Sc(H2O)6 3+ ion is diamagnetic.
Model Answer
Cr(H2O)6 3+ has an odd number of electrons (Cr(III) has a d3 configuration) and thus must be paramagnetic.
Sc(H2O)6 3+ has a d0 metal ion, so all electrons are paired, giving a diamagnetic ion.
The CoF6 3– ion is paramagnetic while the Co(CN)6 3– ion is diamagnetic.
Model Answer
F– is a weaker field ligand than CN–, so the splitting between the two sets of d orbitals in an octahedral complex (the three so-called t2g or dπ orbitals at lower energy and the two eg or d* orbitals at higher energy) is lower in CoF6 3– than in Co(CN)6 3–. So the six d electrons occupy all five d orbitals in CoF6 3–, giving four unpaired electrons, while they are confined to the three t2g orbitals in Co(CN)6 3– and are thus all paired.
The NiCl4 2– ion is paramagnetic while the PtCl4 2– ion is diamagnetic.
Model Answer
NiCl4 2– is tetrahedral, and so all the d orbitals are similar in energy (three are at slightly higher energy and two at slightly lower energy, but the difference in energy is small). In its ground state, the eight d electrons in Ni(II) spread out as much as possible among the five d orbitals. This means that there are two half-filled d orbitals, so the compound has 2 unpaired electrons. In square planar PtCl4 2–, one d orbital (dx2–y2, if the chlorides are taken to lie along the x and y axes) is much higher in energy than any of the others. That d orbital is thus empty, and filling the remaining four d orbitals with eight electrons results in no unpaired electrons.
The CoCl4 2– ion is strongly colored while the ZnCl4 2– ion is colorless.
Model Answer
CoCl4 2– has d7 Co(II) and so has moderately intense d-d transitions where electrons are promoted from filled d orbitals to half-filled d orbitals. In contrast, ZnCl4 2– has a d10 metal center, so all the d orbitals are completely filled and no d-d transitions are possible.
The MnO4 – ion is strongly colored while the ReO4 – ion is colorless.
Model Answer
In MnO4 –, Mn(VII) is d0, so the color cannot arise from a d-d transition. It must be due to a ligand-to-metal charge transfer transition (LMCT), where an electron is promoted from a nonbonding orbital on oxygen to a Mn d orbital. This is possible in ReO4 – as well, but the transfer requires much more energy (Mn(VII) is much more oxidizing than Re(VII)) and occurs in the ultraviolet, giving rise to a colorless species.