The lattice energy of sodium fluoride, NaF, is -908 kJ/mol. With the pairs Na+/Ca2+ and F/02 having roughly the same ionic radii, what would you expect the lattice energy of CaO to be?
1 -908 kJ/mol
2 +908 kJ/mol
3 -1777 kJ/mol
4 -3477 kJ/mol
The correct answer and explanation is :
The correct answer is 3 -1777 kJ/mol.
Explanation:
Lattice energy is the energy released when ions come together to form a solid lattice. It depends on two key factors: the charge on the ions and the ionic radii (or size of the ions). The lattice energy can be estimated using the Born-Haber cycle or the equation derived from Coulomb’s Law:
[
U = \frac{k \cdot |Q_1 \cdot Q_2|}{r}
]
Where:
- ( U ) is the lattice energy,
- ( k ) is a constant,
- ( Q_1 ) and ( Q_2 ) are the charges of the ions,
- ( r ) is the distance between the ions (which depends on their radii).
For Sodium Fluoride (NaF):
- The ions involved are Na(^+) and F(^-), each with a charge of ±1.
- The ionic radii of Na(^+) and F(^-) are relatively small, so the lattice energy of NaF is -908 kJ/mol.
For Calcium Oxide (CaO):
- The ions involved are Ca(^{2+}) and O(^{2-}), each with a charge of ±2.
- The problem states that the ionic radii of Na(^+)/Ca(^{2+}) and F(^-)/O(^{2-}) are roughly the same. This means the distance between the ions will be roughly similar for both compounds.
Given that the lattice energy is inversely proportional to the ionic radii and directly proportional to the product of the charges, we can predict that the lattice energy of CaO will be significantly higher (more negative) than that of NaF. This is because the charges of Ca(^{2+}) and O(^{2-}) are both larger (±2) compared to Na(^+) and F(^-), and since lattice energy increases with the square of the charge, the lattice energy of CaO will be approximately four times that of NaF.
Thus, the lattice energy of CaO is approximately four times that of NaF:
[
4 \times (-908 \text{ kJ/mol}) = -1777 \text{ kJ/mol}
]
Hence, the lattice energy of CaO is predicted to be -1777 kJ/mol.