Calculate the amount of energy released in the formation of one mole of MgSe bonds (not lattice energy). The radius of the magnesium ion is 0.65 Å, and the radius of the selenide ion is 1.98 Å. Note that 1Å=10^−10m.

I know I need to use the formula E=k(q1*q2)/d

Honestly don't know where to begin. So far i got 2+|-2|=4meters and that would be the q1+q2. However, i'm not sure how to solve for D or where the 1Å=10^-10 is supposed to go... any help or direction would be greatly appreciated.

The formula you need to use is E = k*(q1*q2)/r, where k is Coulomb's constant (8.99*10^9 N*m^2/C^2), q1 and q2 are the charges of the ions (2+ and -2 for MgSe), and r is the distance between the ions. The distance between the ions is the sum of the radii of the ions, so r = 0.65 + 1.98 = 2.63*10^-10 m. Plugging these values into the equation gives E = 8.99*10^9 N*m^2/C^2 * (2+ * -2) / (2.63*10^-10 m) = -1.7*10^-18 J/mol.

To calculate the amount of energy released in the formation of one mole of MgSe bonds, you need to use the formula you mentioned: E = k(q1*q2)/d. Let's break down the steps to solve this problem:

Step 1: Obtain the charges of the magnesium (Mg) and selenide (Se) ions.
- Magnesium ion (Mg2+): The number 2+ indicates that the magnesium ion has a charge of +2.
- Selenide ion (Se2-): The number 2- indicates that the selenide ion has a charge of -2.
- The charges of the ions are q1 = +2 and q2 = -2.

Step 2: Calculate the distance between the Mg and Se ions.
- Given the radii of the magnesium and selenide ions: Mg radius = 0.65 Å and Se radius = 1.98 Å.
- The distance (d) between the centers of the ions is the sum of their radii.
- Convert the radii from angstroms (Å) to meters (m) using the conversion factor 1Å = 10^-10 m.

Let's calculate the distance (d) between the ions:
d = (Magnesium radius + Selenide radius) in meters
d = (0.65 Å + 1.98 Å) x (10^-10 m/Å) ≈ 2.63 x 10^-10 m

Step 3: Calculate the value of the electrostatic constant (k).
- The electrostatic constant, k, is a fundamental constant of nature with a value of approximately 8.99 x 10^9 Nm^2/C^2.

Step 4: Calculate the energy (E) released in the formation of one mole of MgSe bonds.
- Plug the values you obtained into the formula: E = k(q1*q2)/d
- For one mole of MgSe bonds, the value of q1*q2 will be (2)(-2) = -4.
- Convert the energy value to joules (J).

Finally, let's compute the energy:
E = (8.99 x 10^9 Nm^2/C^2) x (-4) / (2.63 x 10^-10 m)
E ≈ -1.36 x 10^-18 J

The negative sign indicates that energy is released during the formation of MgSe bonds.
Therefore, approximately -1.36 x 10^-18 Joules of energy are released in the formation of one mole of MgSe bonds.

To calculate the energy released in the formation of one mole of MgSe bonds, you can use Coulomb's law formula, which is E = k(q1*q2)/d.

Here's how you can proceed:

Step 1: Convert the given ionic radii from Ångstroms to meters. Remember that 1 Å = 10^-10 m.

Radius of magnesium (Mg) ion = 0.65 Å = 0.65 * 10^-10 m
Radius of selenide (Se) ion = 1.98 Å = 1.98 * 10^-10 m

Step 2: Calculate the distance (d) between the ions. In this case, the distance will be the sum of the radii, since the ions touch each other when they form the bond.

d = radius of cation + radius of anion
d = 0.65 * 10^-10 m + 1.98 * 10^-10 m

Step 3: Calculate the charges (q1 and q2) of the ions. Magnesium (Mg) forms a 2+ cation, and selenide (Se) forms a 2- anion. Therefore, q1 = 2 and q2 = -2.

Step 4: Substitute the values into the formula:

E = k(q1*q2)/d

where k is Coulomb's constant, which is approximately 8.99 * 10^9 Nm^2/C^2.

E = (8.99 * 10^9 Nm^2/C^2) * (2 * -2) / (0.65 * 10^-10 m + 1.98 * 10^-10 m)

Step 5: Simplify the expression and solve for E.

E = (8.99 * 10^9 Nm^2/C^2) * (-4) / (2.63 * 10^-10 m)
E ≈ -1.36 * 10^-18 J

Therefore, the energy released in the formation of one mole of MgSe bonds is approximately -1.36 * 10^-18 J (negative sign indicates energy release).