Question

Calculate the coulombic force of repulsion between nearest-neighbor o2- ions in cao. (note problems 2.15, 2.16, and 2.17.)

Answer 1

Answer:

F = -1.604x10⁻⁸ N

Explanation:

The Coulomb force (F) of repulsion between nearest-neighbor O²⁻ ions in CaO can be calculated using the next equation:    

$F = K frac{(Z_{Ca^{2+}})(Z_{O^{2-}})(e^{-})^{2}}{r^{2}}$

where K: is the coulomb's constant, Z: is the charge of the Ca²⁺ and O²⁻ ions, e⁻: is the electron's charge, and r: is the distance between the nuclei of the two ions.  

Having that:

• $r = r_{Ca^{2+}} + r_{O^{2-}} = 114 pm + 126 pm = 240 pm = 2.4\cdot 10^{-10} m$
• Z Ca²⁺ = 2+
• Z O²⁻ = 2-
• e⁻ = 1.602x10⁻¹⁹ C
• K = 9x10⁹ N*m²*C⁻²  

The Coulomb force (F) of repulsion is:

$F = 9\cdot 10^{9} N*m^{2}*C^{-2} \frac{(2+)(2-)(1.602\cdot 10^{-19}C)^{2}}{(2.4\cdot 10^{-10} m)^{2}} = -1.604 \cdot 10^{-8} N$

Hence, the Coulomb force of repulsion between the two ions in CaO is -1.604x10⁻⁸ N.

I hope it helps you!