Question

Calcium and sulfur ions experience a force of -1.17x10^- N. Calcium ions have a charge of +2e, while sulfide ions have a charge of -2e. What is the distance between the 2 nuclei when bonded?

A) 1.17x10^-9 meters
B) 2.34x10^-9 meters
C) 3.51x10^-9 meters
D) 4.68x10^-9 meters

Answer 1

The distance between a calcium ion and a sulfide ion when bonded can be determined using Coulomb's Law, accounting for their charges and the given force. The formula takes into account Coulomb's constant for chemistry, which involves picometers and electron charge.

Step-by-step explanation:

The distance between the nuclei of a calcium ion and a sulfide ion when bonded can be calculated using Coulomb's Law. The force (F) between two charges (q1 and q2) at a distance (r) in a vacuum is given by F = k * |q1 * q2| / r^2, where k is Coulomb's constant. Given that calcium ions have a charge of +2e and sulfide ions have a charge of -2e, we can calculate the distance by rearranging the formula to r = sqrt(k * |q1 * q2| / |F|). The product of the charges in SI units is ((2e) * (-2e) = -4e^2), and we are provided the force as -1.17x10^-8 N. Using the proportionality constant for chemistry which incorporates picometers and the electron charge, the distance can be accurately determined. This calculation will yield the precise distance between the nuclei of calcium and sulfur ions when bonded.