What is the magnitude of the energy of a bond formed between a potassium (K+) cation and an iodide (I−) anion? The ionic radii of K+ and I−, are 152 pm and 206 pm, respectively. Assume the Born exponent n is 10. Please report your answer in joules.
To calculate the magnitude of the energy of a bond formed between a potassium (K+) cation and an iodide (I-) anion, we can use Coulomb's law and the Born-Lande equation.
Coulomb's law states that the energy of interaction between two charged particles is given by:
E = k * (q1 * q2) / r
Where:
- E is the energy of interaction
- k is the Coulomb constant
- q1 and q2 are the charges on the particles
- r is the distance between the particles
The Born-Lande equation describes the energy of an ionic bond:
E = (1 - 1/n) * (k * (q1 * q2) / r)
Where:
- n is the Born exponent
Let's calculate the energy of the bond:
First, convert the ionic radii from picometers (pm) to meters (m):
- r1 (potassium): 152 pm = 152 * 10^(-12) m
- r2 (iodide): 206 pm = 206 * 10^(-12) m
Next, calculate the charges on the particles:
- q1 (potassium): +1 charge (K+)
- q2 (iodide): -1 charge (I-)
Now we can substitute these values into the equation:
E = (1 - 1/10) * (k * (q1 * q2) / r)
The Coulomb constant k = 8.99 * 10^9 N m^2/C^2
Plug in the values and calculate the magnitude of the energy:
E = (1 - 1/10) * (8.99 * 10^9 * (1 * -1) / (152 * 10^(-12) + 206 * 10^(-12)))
Simplifying further:
E = (9/10) * (8.99 * 10^9 * -1 / (358 * 10^(-12)))
E = (9/10) * (-8.99 * 10^(-3) / 358)
E = (9/10) * (-0.0251)
E ≈ -0.0226 Joules
Therefore, the magnitude of the energy of the bond formed between a potassium cation (K+) and an iodide anion (I-) is approximately 0.0226 Joules.