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. Please report your answer in joules.
To determine the energy of the bond formed between a potassium cation (K+) and an iodide anion (I-), we can use the Born-Lande equation:
E = k * (Q1 * Q2) / (r * [1 - (1 / n)])
where:
- E is the energy of the bond
- k is a constant (8.9875 × 10^9 N m^2 C^-2)
- Q1 and Q2 are the charges of the cation and anion respectively (in Coulombs)
- r is the distance between the cation and anion (in meters)
- n is the Born exponent, which depends on the crystal structure and ranges typically from 5 to 12
First, we need to convert the ionic radii of potassium (K+) and iodide (I-) to meters. Since 1 pm (picometer) is equal to 1×10^-12 meters, we have:
Ionic radius of K+ = 152 pm = 152 × 10^-12 m
Ionic radius of I- = 206 pm = 206 × 10^-12 m
Next, we need to determine the charges of the cation and anion. The charge of a potassium cation is +1, denoted as K+ or simply 1+. The charge of an iodide anion is -1, denoted as I- or simply 1-.
Now we can substitute the values into the Born-Lande equation to calculate the bond energy:
E = (8.9875 × 10^9 N m^2 C^-2) * ((1+ * 1-) / (r * [1 - (1 / 10)]))
substituting the charges and radii:
E = (8.9875 × 10^9 N m^2 C^-2) * ((1 * -1) / ((152 × 10^-12 m + 206 × 10^-12 m) * (1 - (1 / 10))))
Calculating this equation will give us the energy of the bond between K+ and I- in joules.