6. [13%] The chemistry of beryllium (Be) has a number of interesting features. — Descriptive Chemistry / Bonding and Structure Chemistry Question
Problem Context
- [13%] The chemistry of beryllium (Be) has a number of interesting features.
a. Explain the differences between Be and its heavier congener, barium (Ba), with regard to the following properties.
i. Be has a higher ionization energy (900 kJ mol-1) than Ba (563 kJ mol-1).
Model Answer
The electron ionized in Be is a 2s electron, while Ba loses a 6s electron. The higher value of n corresponds to a higher energy, requiring less energy to remove.
ii. Adding an electron to a gas-phase Be atom requires energy, while adding an electron to a gas-phase Ba atom releases a small amount of energy (14 kJ mol-1).
Model Answer
Since both Be and Ba have an ns2 configuration, an added electron must enter a new subshell, which is higher in energy. The 2p subshell that would be occupied for Be is much higher in energy, but the next available orbital for Ba is a 5d orbital, which is closer in energy to the 6s subshell.
iii. BeCl2(s) has a more positive ∆Hºf (–496.2 kJ mol-1) than BaCl2(s) (–858.6 kJ mol-1)
Model Answer
Ba is more electropositive than Be, so it costs much less energy to remove its valence electrons to form the 2+ ion. This difference outweighs the somewhat higher lattice energy of BeCl2 than BaCl2.
iv. Solid BeCl2 adopts the structure shown on the left, while solid BaCl2 adopts the structure shown on the right (metal = black spheres, chlorine = gray spheres).
Model Answer
Be can only form 4 bonds, so it can only adopt the left structure. The larger Ba2+ ion can easily achieve the coordination number of 8 required by the right-hand structure.
b. In the vapor phase, BeCl2 exists as a mixture of BeCl2 monomers and Be2Cl4 dimers. Draw or clearly describe the geometries of these two gas-phase species.
c. At 800 K, Kp for dimerization of BeCl2(g) is 2.9.
2 BeCl2(g) ⇌ Be2Cl4(g) Kp = 2.9
Calculate the mole fraction of dimeric Be2Cl4 in BeCl2 vapor at a total pressure of 0.100 bar at 800 K.
Model Answer
Let PM = partial pressure of monomeric BeCl2(g)
PD = partial pressure of dimeric Be2Cl4(g)
PD/PM^2 = 2.9
0.100 – PM = 2.9PM^2
2.9PM^2 + PM – 0.100 = 0
Solving gives PM = 0.081 bar, and therefore PD = 0.100 bar – 0.081 bar = 0.019 bar. The mole fraction of the dimer is PD/0.100 bar = 0.19.