Problem

The lattice energy of LiF is 1023 kJ/mol, and the…

03:53

Problem 78 Hard Difficulty

The reaction of a metal, M, with a halogen, X2, proceeds by an exothermic reaction as indicated by this
equation: M(s) + X2 (g) ? MX2 (s). For each of the following, indicate which option will make the reaction more exothermic. Explain your answers.
(a) a large radius vs. a small radius for $\mathrm{M}^{+2}$
(b) a high ionization energy vs. a low ionization energy for M
(c) an increasing bond energy for the halogen
(d) a decreasing electron affinity for the halogen
(e) an increasing size of the anion formed by the halogen

Answer

a) Hence, the reaction is more exothermic in case of small radius of $\mathrm{M}^{2+}$
b) Hence, the reaction is more exothermic in case of low ionization energy of M.
c) Hence, the reaction is less exothermic in case of the increase in bond energy
d) Therefore, for the decreasing electron affinity for the halogen, the reaction will become less
exothermic.
e) Therefore, as the radius of anion formed by halogen increases, the reaction will become less
exothermic.

Allea C.

University of Maryland - University College

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Video Transcript

this question was a little confusing because it initially asked, for each of the following indicate which option will make the reaction mawr exo thermic. And there's only two options. There's only two of the following that have options A and B. Everything else is just a statement of increasing or decreasing without an option. So I'm going to assume for the last three that if there's an option of increasing, there's also an option of decreasing and vice versa. The key to answering this question is to know what will make this process easier. What will require less energy for this process to occur And remember, energy is going to be expended in order to break this bond, create ions and then energy will be released when the ions come together. So if less energy is required to break this bond and create the ions, this will be more exo thermic. If more energy is released due to a stronger attraction between the cat I and and the anti on than Mawr, energy will be released. So a large radius versus a small radius of the metal, the smaller the radi i of the ions, the closer they get to each other. When they form the product, the stronger the bonds and the more energy will be released. So the smaller the radius, the better for be, it asks. Hi hi ionization energy or low ionization energy of the metal. Well, the ionization energy is the energy used to create the ion of the metal atom into the cat ion. So if there's less energy that is required, ah, low, smaller ionization energy than the process requires less energy and more energy will be released now for the last three, it says, an increasing bond energy of the halogen. Well, if we increase the bond energy, then that means it's going to take more energy to break this bond, which is not favorable for the reaction. So we want a decrease in the bond energy of the halogen in order to favor the reaction and make it more exa. Thermic. The next one says a decreasing electron affinity, affinity for the halogen. Well, if we decrease the electron affinity than once this bond is broken, then they'll be less attraction to make it and I on meaning more energy will be required to make it an ion, which is not favorable for the reaction. So we want a high electron affinity. We want that halogen atom to have a strong desire to become an ion which will allow for more of the energy to be released instead of consumed. Making the ion the last one is an increase in the size of the ion. Well, Justus, we've discussed up here on the smaller the atom, the greater the attraction. So if a larger radius makes it less excell excell thermic, this should probably be appear. The one that's going to make it mawr exa thermic is going to be a small radius. So just like down here, the smaller halogen will result in mawr energy released Mawr eggs a thermic vote to review smaller radius here low ionization energy So not a lot of energy is required to lionize the metal decreasing bond energy. So not a lot of energy is required to break this bond. Hi electron affinity so that it's very easy toe lionize the halogen once it's formed. And then a small halogen ion smaller, the radi I greater the attraction, the more energy that's released

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