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.
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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.
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