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Ch.8 - Basic Concepts of Chemical Bonding

Chapter 8, Problem 102c

One scale for electronegativity is based on the concept that the electronegativity of any atom is proportional to the ionization energy of the atom minus its electron affinity: electronegativity = k1I - EA2, where k is a proportionality constant. (c) By using data in Chapter 7, determine the value of k that would lead to an electronegativity of 4.0 for F under this definition.

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everyone in this example, we're told that one way to give a relative value for election negativity is based on the ionization energy and electron affinity of a given element. We're told that the election negativity can be determined from the following equation where we take the rate constant K. Multiplied by the difference between ionization energy and electron affinity of given adam. We need to calculate the value of the rate constant K. Given the election negativity of chlorine as 3.0 from the periodic table. So because we need to find the value for K. We're going to isolate K in our given equation for electro negativity. So we're going to divide both sides by the difference between the ionization energy an electron affinity. And so that means we can simplify this so that we have the rate constant equal to the election negativity divided by the difference between the ionization energy and electron affinity. So what we're going to do is calculate based on that given election negativity value of chlorine. So we'll have that K is equal to The given election negativity from the prompt for Chlorine as 3.0. And this is going to be divided by the ionization energy of chlorine. Which we're going to want to look up in our textbooks. So we will see that the first ionization energy of chlorine in our textbooks is going to equal a value Of 1251 kg jewels Permal. And we also want to look up our electron affinity of chlorine. Which in our textbooks we would see is equal to a value of negative 349 kg joules Permal. So we're going to plug in those units above. So we would have 1251 minus the electron affinity of chlorine being negative 349. And we're going to close off our parentheses here. So what we would get for our value of rate constant is that it's equal to a value of 1.875 times 10 to the negative 3rd power. And this would be our final answer for our rate constant and our equation for electro negativity of chlorine. So this completes this example. If you have any questions, please leave them down below. Otherwise I'll see everyone in the next practice video.
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Consider the molecule C4H5N, which has the connectivity shown below. (a) After the Lewis structure for the molecule is completed, how many s and how many p bonds are there in this molecule?

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The electron affinity of oxygen is -141 kJ/mol, corresponding to the reaction O(g) + e- → O-(g). The lattice energy of K2O(s) is 2238 kJ/mol. Use these data along with data in Appendix C and Figure 7.10 to calculate the 'second electron affinity' of oxygen, corresponding to the reaction O-(g) + e- → O2-(g)

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Textbook Question

One scale for electronegativity is based on the concept that the electronegativity of any atom is proportional to the ionization energy of the atom minus its electron affinity: electronegativity = k1I - EA2, where k is a proportionality constant. (b) Why are both ionization energy and electron affinity relevant to the notion of electronegativity?

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Textbook Question

One scale for electronegativity is based on the concept that the electronegativity of any atom is proportional to the ionization energy of the atom minus its electron affinity: electronegativity = k1I - EA2, where k is a proportionality constant. (d) Use your result from part (c) to determine the electronegativities of Cl and O using this scale. Use your result to determine the electronegativity of Cl using this scale.

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Acetylene (C2H2) and nitrogen (N2) both contain a triple bond, but they differ greatly in their chemical properties. (b) By referring to Appendix C, look up the enthalpies of formation of acetylene and nitrogen. Which compound is more stable?

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Under special conditions, sulfur reacts with anhydrous liquid ammonia to form a binary compound of sulfur and nitrogen. The compound is found to consist of 69.6% S and 30.4% N. Measurements of its molecular mass yield a value of 184.3 g/mol. The compound occasionally detonates on being struck or when heated rapidly. The sulfur and nitrogen atoms of the molecule are joined in a ring. All the bonds in the ring are of the same length. (a) Calculate the empirical and molecular formulas for the substance.

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