PAST PAPER Qs

Question 1

Explain why the halides of metals in low oxidation states (such as UF4) are ionic, while the halides of metals in high oxidation states (such as UF6) are covalently bonded.
[4 marks]

Answer 1

• It is possible to remove four valence electrons of U to form a U4+ ion. This energy cost is adequately compensated on formation of a UF4 ionic lattice with strong ionic bonding.
• Removal of six electrons from U atoms to form U6+ would require a very large amount of energy (and results in the formation of ions of very large charge density).
• Such a large energy cost is not compensated by the formation of an ionic lattice.

Thus UF6 adopts a covalent structure, preferred energetically.

Question 2

Explain the following observation:
The energies of the orbitals in a hydrogen atom depend only on the principal quantum number, n. However, the energies of orbitals in the lithium atom depend on both the principal and secondary quantum numbers, n and l.
[4 marks]

Answer 2

The multi-electron nature of Li changes the possible (approximate) solutions to the Schrödinger equation.
These approximate solutions are characterised by different values of the orbital angular momentum quantum number, l, and arise due to the differing interactions of an electron of interest with other electrons.
For a given n, orbitals with different “l” may experience a different effective nuclear change, Zeff. his may be rationalised qualitatively by teh differing ability of such orbitals to penetrate lower-lying orbitals. In turn, such orbitals differ in their ability to shield the nucleus from electrons in other orbitals.
Thus, in Li, a multielectron atom, the orbital energy depends on the vaye of both n and l, due to the electron-electron interactions described above.
For hydrogenic species, such electron-electron interactions are absent. The orbital energy depends on the value of n.

MULTIPLE ANSWERS ACCEPTABLE: for full marks students should explain accurately concepts such as ELECTRON-ELECTRON INTERACTIONS, SHIELDING, AND SOLUTION TO THE Schrödinger EQUATION WITH DIFFERENT VALUES OF l.

Question 3

Explain the following observation:
The atomic radii of the elements decrease as you go across the first row transition elements, from scandium to zinc.
[4 marks]

Answer 3

The monotonic decrease in ionic radius across the d block is known as the d-block contraction.
d-orbitals are poor at shielding the nucleus from other electrons, arising largely from the strong angular dependence of these orbitals (shapes).
Upon adding on additional electrons to the 3d-orbitals, this inefficient screening of the nuclear charge means that all 3s electrons experience a smaller increase in effective nuclear change, thus a smaller atomic/ionic radius.