What is 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
To determine the energy of a bond formed between a potassium cation (K+) and an iodide anion (I−), we can use the Born-Lande equation, which calculates the lattice energy of an ionic compound.
The Born-Lande equation is given by:
E = k * (Z+ * Z−) / r
Where:
E is the lattice energy,
k is the proportionality constant,
Z+ and Z− are the charges on the cation and anion, respectively,
r is the distance between the centers of the ions.
Given that the ionic radii of K+ and I− are 152 pm and 206 pm, respectively, we need to convert these values to the same units before calculating the energy.
1. Converting the ionic radii to Angstroms:
1 pm = 10^(-12) meters
1 A = 10^(-10) meters
Ionic radius of K+: 152 pm * (10^(-10) m/pm) = 15.2 Å
Ionic radius of I−: 206 pm * (10^(-10) m/pm) = 20.6 Å
2. Calculating the energy:
Using the given values:
Z+ = 1 (charge of K+)
Z− = -1 (charge of I−)
r = sum of ionic radii = 15.2 Å + 20.6 Å = 35.8 Å
Now, we can plug these values into the Born-Lande equation to calculate the energy of the bond formed:
E = k * (Z+ * Z−) / r
Since the Born exponent n is given as 10, the proportionality constant (k) can be determined using the formula:
k = 2.31 * 10^(-19) J * Å^n
Substituting the given values:
k = 2.31 * 10^(-19) J * Å^10
Then, we can calculate the energy (E):
E = (2.31 * 10^(-19) J * Å^10) * (1 * -1) / 35.8 Å
After canceling out the units Å (Angstrom):
E = (2.31 * 10^(-19) * 1 * -1) / 35.8 J
Simplifying:
E = -2.31 * 10^(-19) J / 35.8
E ≈ -6.45 * 10^(-21) J
Therefore, the energy of the bond formed between a potassium cation and an iodide anion is approximately -6.45 * 10^(-21) Joules.