Calculate the bond energy in the ionic compound Nacl if the bond length is 2.36x10^-10m

This question is incomplete, the complete complete question is;

Coulomb's law for the energy of interaction between two charged ions;

E = Ke × q1q2/r

Coulomb's constant Ke = 2.31 × 10⁻²⁸ J.m.

Calculate the bond energy in the ionic compound NaCl if the bond length is 2.36 × 10⁻¹⁰m

the bond energy is -9.788 × 10⁻¹⁹ J

Explanation:

Given the data in the question;

Ke = 2.31 × 10⁻²⁸ J.m which is the same as; Ke = 2.31 × 10⁻²⁸ Nm L/c²

ionic compound NaCl;

q1 = charge on Na in NaCl = Na⁺ = + 1C

q2 = charge on Cl in NaCl = Cl⁻ = - 1C

given that bond length r is 2.36 × 10⁻¹⁰m

E = Ke × q1q2/r

we substitute

E = 2.31 × 10⁻²⁸ × (+1 × -1) / 2.36 × 10⁻¹⁰

E = -2.31 × 10⁻²⁸ / 2.36 × 10⁻¹⁰

E = -9.788 × 10⁻¹⁹ J    { The negative sign means Attraction in ions }

Therefore, the bond energy is -9.788 × 10⁻¹⁹ J

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.

a corroding iron, if that's one of your choices