Question

What is the electrostatic potential energy (Eel) of the ionic bond between a Ca2+ ion and a S2– ion in joules? Before you do the calculation, predict whether the result you get will be less than (more negative) or greater than (less negative) than the Eel value for KCl. The ionic radii of Ca2+ and of S2– are 100 pm and 184 pm, respectively.

Answer 1

Answer: the electrostatic potential energy is -3.26 × 10⁻¹⁸ J

Explanation:

Given the data in the question;

I think the value of electrostatic potential energy of calcium sulfide CaS formed will be less(more negative) than KCl because the Ca2+ and S2- ions has more charge that is +2 and -2 respectively when compared to Kcl which has +1 and -1 charge.so it will be more negative.

radius = r$_{ca+2}$ + r$_{s-2}$ = 100 + 184 = 284 pm = 2.84 × 10⁻¹⁰ m

we know that; k = 2.31 × 10⁻²⁸ J

q1 = 2  ( charge on Ca⁺² )

q2 = -2( charge on S⁻²)

so

Ep = k × q1q2/r

so we substitute

Ep = 2.31 × 10⁻²⁸ × (+2×-2) / 2.84 × 10⁻¹⁰

Ep = (2.31 × 10⁻²⁸ × -4) / 2.84 × 10⁻¹⁰

Ep = -9.24 × 10⁻²⁸ / 2.84 × 10⁻¹⁰

Ep = -3.26 × 10⁻¹⁸ J

Therefore, the electrostatic potential energy is -3.26 × 10⁻¹⁸ J