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.
x=
5.8033057e-19
To calculate 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 * n)
Where:
- E is the energy of the bond
- K is the electrostatic constant, approximately 8.99 x 10^9 Nm^2/C^2
- Q1 and Q2 are the charges of the ions (in coulombs)
- r is the distance between the ions (in meters)
- n is the Born exponent
First, we need to determine the charges of the ions. The potassium cation (K+) has a charge of +1, while the iodide anion (I-) has a charge of -1.
Next, we convert the ionic radii given in picometers to meters. The ionic radii of K+ and I- are 152 pm and 206 pm, respectively. This can be done by dividing the given values by 10^12.
r(K+) = 152 pm / 10^12 = 1.52 x 10^-10 m
r(I-) = 206 pm / 10^12 = 2.06 x 10^-10 m
Now we can substitute the values into the equation:
E = (8.99 x 10^9 Nm^2/C^2) * (1 * (-1)) / ((1.52 x 10^-10 m + 2.06 x 10^-10 m) * 10)
Simplifying the equation:
E = (8.99 x 10^9 Nm^2/C^2) * (-1) / (3.58 x 10^-10 m) * 10
E = -2.51 x 10^-18 Nm^2/C^2
Finally, we can convert this answer to joules by multiplying by the conversion factor 1 J = 1 Nm:
E = (-2.51 x 10^-18 Nm^2/C^2) * (1 J / 1 Nm)
E = -2.51 x 10^-18 J
The energy of the bond formed between a potassium cation and an iodide anion is approximately -2.51 x 10^-18 joules.