d and f BLOCK ELEMENTS

Which lanthanoid elements are known to commonly exhibit a +4 oxidation state, and what feature of their electronic configuration contributes to this?

• Only Lutetium (Lu); achieving a filled 4f^14 subshell.
• Ce (Cerium), Pr (Praseodymium), Nd (Neodymium), Tb (Terbium), and Dy (Dysprosium); losing electrons to achieve a stable [Xe] noble gas core or half-filled f-subshell.
• All lanthanoids can exhibit +4.
• Only Promethium (Pm); due to its radioactive nature.

What is the electronic configuration of the Ce^3+ ion, and how many unpaired electrons does it have according to Hund's rule?

• [Xe] 4f^2; 2 unpaired electrons.
• [Xe] 4f^0; 0 unpaired electrons.
• [Xe] 4f^1; 1 unpaired electron.
• [Xe] 5d^1; 1 unpaired electron.

Why are the 5f electrons in the actinoid series less effective at shielding the outer electrons from the nuclear charge compared to 4f electrons in lanthanoids?

• The 5f electrons are closer to the nucleus.
• The 5f electrons are completely filled.
• The 5f electrons themselves provide poor shielding from element to element in the series.
• The 5f electrons have lower energy than 4f electrons.

Why does Cu^+ disproportionate in aqueous solution?

• Because Cu^+ is unstable in any environment.
• Because the E0 value for the disproportionation reaction is favorable.
• Because Cu^+ is more stable than Cu^2+.
• Because Cu^+ is a strong oxidizing agent.

Why is Cr^2+ a stronger reducing agent than Fe^2+?

• Cr^2+ has fewer electrons than Fe^2+.
• Cr^2+ gets oxidized to the more stable d^3 configuration (Cr^3+), while Fe^2+ gets oxidized to the less stable d^5 configuration (Fe^3+).
• Fe^2+ has a higher atomic number.
• Cr^2+ is more electronegative than Fe^2+.