(A) Show that the atomic packing factor for HCP is 0.74 (B) For the HCP crystal structure, show that the ideal c/a ratio is 1.633. Acidities: Explain why: a) Propyne (pK a = 25) is much more acidic than either propene (pK a = 42) or propane (pK a ~54). b) 1,3-Cyclopentadiene (pK a = 15) is much more acidic than Propyne
The Correct Answer and Explanation is :
(A) Atomic Packing Factor for HCP
The Atomic Packing Factor (APF) is the fraction of volume occupied by atoms in a unit cell. To determine the APF for HCP (Hexagonal Close Packed) structure, follow these steps:
- HCP Unit Cell Geometry:
- The HCP structure has two atoms per unit cell.
- The unit cell consists of 12 atoms located at the vertices and 2 atoms within the unit cell (one at the center of each of the two hexagonal faces).
- Volume of the Unit Cell:
The unit cell consists of a hexagonal prism with a height (c) and base radius (r), with the ideal c/a ratio of 1.633.
- The volume of the unit cell (V_{cell}) is:
[
V_{cell} = \frac{\sqrt{3}}{2} a^2 c
]
where (a) is the side length of the hexagonal base, and (c) is the height.
- Atomic Volume:
The volume of the atoms in the unit cell is the sum of the volumes of the 2 atoms per unit cell.
- The volume of each atom is given by (V_{atom} = \frac{4}{3} \pi r^3), where (r) is the atomic radius.
- APF Calculation:
The packing factor is:
[
APF = \frac{\text{Total Atomic Volume}}{\text{Unit Cell Volume}}
]
After calculation, the APF for HCP is approximately 0.74, indicating that 74% of the unit cell is filled with atoms.
(B) Ideal c/a ratio for HCP
The ideal c/a ratio for the HCP structure is derived from the geometric relationship that maximizes atomic packing. The ratio is calculated based on the position of atoms in the unit cell. The ideal ratio is:
[
c/a = \frac{2 \sqrt{2}}{\sqrt{3}} = 1.633
]
This ratio minimizes the empty space between the atoms, thereby maximizing the packing efficiency.
Acidity Explanation
(a) Propyne vs Propene vs Propane:
Propyne (C₃H₄) is more acidic than both propene (C₃H₆) and propane (C₃H₈) because of its structure and the ability to stabilize the conjugate base. The pKa of propyne is 25, which is significantly lower than the pKa of propene (42) and propane (around 54). In propyne, the sp hybridized carbon (triple bond) bears the negative charge upon deprotonation, which is highly stabilized due to the increased s-character of the sp orbital. This means the negative charge is more effectively held closer to the nucleus, making it more stable and the proton easier to lose. On the other hand, propene and propane have sp² and sp³ hybridized carbons, which do not stabilize the negative charge as effectively, making them less acidic.
(b) 1,3-Cyclopentadiene vs Propyne:
1,3-Cyclopentadiene (pKa = 15) is more acidic than propyne because the conjugate base of 1,3-cyclopentadiene can delocalize the negative charge over the ring, stabilizing it. This resonance stabilization makes the loss of a proton much easier. Propyne, while acidic, does not benefit from resonance stabilization in its conjugate base, so it is less acidic compared to 1,3-cyclopentadiene, where the charge is delocalized and stabilized over the entire structure. Therefore, the stronger resonance stabilization in 1,3-cyclopentadiene leads to its lower pKa (higher acidity).