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.

Well, well, well, looks like we have a chemistry conundrum here! Don't worry, I'll try to help you out.

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

First, let's convert the given atomic radii to meters using the conversion factor you provided. The radius of magnesium (Mg) is 0.65 Å, which is 0.65 * 10^(-10) meters. Similarly, the radius of selenide (Se) is 1.98 Å, which is 1.98 * 10^(-10) meters.

Now we can substitute these values into the formula. The charges (q1 and q2) for the magnesium and selenide ions can be determined from their positions on the periodic table. Magnesium ion (Mg²⁺) has a charge of +2, while the selenide ion (Se²⁻) has a charge of -2. So, q1 = +2 and q2 = -2.

As for the separation distance (d), we can consider it as the sum of the radii of the magnesium and selenide ions. That's d = (0.65 + 1.98) * 10^(-10) meters.

Finally, we need to determine the value of the proportionality constant (k), which depends on the units used for charge (Coulombs) and distance (meters). Since the charges are given in terms of elementary charges, we can use the value of 2.31 * 10^(-28) J⋅m/C² for k.

Now, it's just a matter of plugging in the values and crunching the numbers to calculate the energy (E) released in the formation of one mole of MgSe bonds. Just remember to adjust the charges (q1 and q2) according to how many moles you want to consider.

I hope this helps you get started on solving the problem. And remember, chemistry is all about bonding with the elements, but sometimes it can be a bit electrifying!

To calculate the energy released in the formation of one mole of MgSe bonds, you can use the Born-Haber cycle and the formula for the formation energy. The formula you mentioned (E = k(q1*q2)/d) is unrelated to the calculation of energy released during bond formation.

Here's a step-by-step approach to solve this problem:

Step 1: Find the charges of the magnesium ion (Mg2+) and selenide ion (Se2-).
- The charge of the magnesium ion (Mg2+) is +2.
- The charge of the selenide ion (Se2-) is -2.

Step 2: Find the ionic radii of the magnesium ion and selenide ion.
- The radius of the magnesium ion (Mg2+) is given as 0.65 Å (1 Å = 10^-10 m).
- The radius of the selenide ion (Se2-) is given as 1.98 Å.

Step 3: Calculate the distance (d) between the centers of the ions using their radii.
- d = sum of the radii of the ions
- d = (radius of Mg2+ + radius of Se2-)

Step 4: Convert the distance from Angstroms (Å) to meters (m) for the calculation.
- Multiply the distance (d) by the conversion factor: 1 Å = 10^-10 m.

Step 5: Calculate the formation energy (ΔHf) using the formula:
- ΔHf = - [E1 + E2 + E3 + E4 + E5]
- E1: Ionization energy of Mg (Mg(g) → Mg+(g) + e-)
- E2: Sublimation energy of Mg (Mg(s) → Mg(g))
- E3: Electron affinity of Se (Se(g) + e- → Se-(g))
- E4: Lattice energy of MgSe (Mg+(g) + Se-(g) → MgSe(s))
- E5: Formation energy of MgSe (Mg(s) + Se(s) → MgSe(s))

Step 6: Substitute the known values and calculate the formation energy.

Note: The given equation E = k(q1*q2)/d is not applicable to this problem.

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

Let's break down the steps:

1. Determining the charges:
The charge of the magnesium ion (Mg2+) is +2, and the charge of the selenide ion (Se2-) is -2. This gives us q1 = +2 and q2 = -2.

2. Converting the distance:
The given radii are in angstroms (Å), but we need to convert them to meters for consistent units. As provided, 1 Å = 10^(-10) m.

The distance (d) required in the formula is the sum of the ionic radii. So,
d = (0.65 Å + 1.98 Å) * (10^(-10) m/Å)
= 2.63 Å * (10^(-10) m/Å)
≈ 2.63 * 10^(-10) m

3. Plugging values into the formula:
E = k(q1 * q2)/d

To simplify calculations, we should express q1 * q2 and d in terms of their magnitudes only:
q1 * q2 = (2) * (-2) = -4
d = 2.63 * 10^(-10) m

E = k(-4)/2.63 * 10^(-10) m

4. Determining the value of k:
The proportionality constant k is called the electrostatic constant, also known as Coulomb's constant. Its value is approximately 8.99 * 10^9 N m^2 C^(-2).

So, plugging in the values:
E ≈ (8.99 * 10^9 N m^2 C^(-2)) * (-4)/2.63 * 10^(-10) m

The units and C terms cancel out, leaving us with the unit of energy, joules (J).

5. Calculating the energy:
Evaluating the expression above will give us the amount of energy released in joules when one mole of MgSe bonds is formed.

Let's calculate it:

E ≈ (8.99 * 10^9 N m^2 C^(-2)) * (-4)/2.63 * 10^(-10) m
E ≈ -13.68 J

Therefore, the amount of energy released in the formation of one mole of MgSe bonds is approximately -13.68 joules. The negative sign indicates that energy is released during bond formation.