d and f BLOCK ELEMENTS

Why do the complexes [Fe(CN)6]^4- and [Fe(CN)6]^3- have different colors in solution?

• Because CN- is a ligand.
• Because Fe has different oxidation states.
• Because they are both iron complexes.
• Because the different charges on the complexes ([Fe(CN)6]^4- vs [Fe(CN)6]^3-) lead to slightly different ligand field environments and crystal field splitting energies, absorbing different wavelengths of light.

Why do the first ionization enthalpies of the first transition series elements show a small overall increase but with notable dips at Cr and Cu?

• Because Cr and Cu are metals.
• Because Cr and Cu have high atomic numbers.
• Because Cr and Cu are located at the end of the series.
• Because the dips at Cr and Cu correspond to the removal of an electron from a more stable half-filled (Cr: 3d^5 4s^1) or fully-filled (Cu: 3d^10 4s^1) configuration, making the first IE lower than expected.

Why do the atomic radii of the elements immediately following the Lanthanoids (Hf, Ta, W, etc.) match those of the elements above them in the periodic table (Zr, Nb, Mo, etc.)?

• Because they are transition metals.
• Because the nuclear charge increases.
• Because the 4f electrons are added.
• Because the Lanthanoid contraction causes the elements following the Lanthanoids to be much smaller than expected, making their sizes similar to the corresponding elements in the previous period.

Why do the complexes [Cr(NH3)6]^3+ and [Mn(H2O)6]^2+ have different numbers of unpaired electrons despite both being high-spin octahedral complexes?

• Because Cr and Mn are different metals.
• Because NH3 and H2O are different ligands.
• Because they have different charges.
• Because Cr^3+ (d^3) and Mn^2+ (d^5) have different numbers of d-electrons, leading to different numbers of unpaired electrons even in high-spin configurations.

Why do the second and third ionization enthalpies of Zn show very large jumps compared to the first?

• Because Zn is a metal.
• Because Zn has a high melting point.
• Because Zn is less reactive.
• Because the first electron is removed from the 4s orbital, while the second and third electrons are removed from the much more stable and tightly bound 3d orbitals.