The equilibrium spacing between the ions is 0.24 nm and the bonding energy is 5.3 eV.
The amount of energy required to separate the atoms forming a molecular bond into free atoms is known as bond energy, and it serves as a gauge of the strength of a chemical connection.
For Na⁺ and Cl⁻ ion pair, the attractive and repulsive energies are E(A) and E(R). The energies depend on the distance between the ions r.
We have,
E(A) = ( -1.436 / r )
E(R) = ( 7.32 × 10⁻⁶ / r⁸ )
(a) The plot for E(N), E(R), and E(A) versus r up to 1 nm is attached below.
(b) From the plot,
r(0) = 0.24 nm
E(0) = 5.3 eV
Hence, the equilibrium spacing r(0) between Na⁺ and Cl⁻ ions is 0.24 nm and the bonding energy is 5.3 eV.
Learn more about bonding energy here:
https://brainly.com/question/26662679
#SPJ9