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 calculate the energy of the bond between a potassium cation (K+) and an iodide anion (I-), we can use the Born-Lande equation. The equation is as follows:
E = k * (Q+ * Q-)/(r * n)
Where:
- E is the energy of the bond
- k is Coulomb's constant (8.99 * 10^9 N m² C^-2)
- Q+ and Q- are the charges on the respective ions (+1 for K+ and -1 for I-)
- r is the distance between the ions
- n is the Born exponent (given as 10)
First, we need to convert the ionic radii from picometers (pm) to meters (m):
Ionic radius of K+ = 152 pm = 152 * 10^-12 m
Ionic radius of I- = 206 pm = 206 * 10^-12 m
Next, we substitute the values into the equation:
E = (8.99 * 10^9 N m² C^-2) * [(+1) * (-1)] / [(152 * 10^-12 m + 206 * 10^-12 m) * 10]
Simplifying the equation gives us:
E = (8.99 * 10^9 N m² C^-2) * (-1) / [(358 * 10^-12 m) * 10]
E = -8.99 * 10^7 N m² C^-2 / (3580 * 10^-12 m)
E ≈ -2.512 * 10^-5 N m² C^-2
To convert from N m² C^-2 to joules, we use the conversion factor 1 N m² C^-2 = 1 J:
E ≈ -2.512 * 10^-5 J
Therefore, the energy of the bond formed between the potassium cation and iodide anion is approximately -2.512 * 10^-5 joules. The negative sign indicates that the energy is released when the bond is formed.