An alternative definition of electronegativity is Electronegativity \(=\) constant (I.E. \(-\) E.A.) where I.E. is the ionization energy and E.A. is the electron affinity using the sign conventions of this book. Use data in Chapter 7 to calculate the (I.E. \(-\) E.A.) term for \(\mathrm{F}, \mathrm{Cl}, \mathrm{Br}\), and \(\mathrm{I}\). Do these values show the same trend as the electronegativity values given in this chapter? The first ionization energies of the halogens are \(1678,1255,1138\), and \(1007 \mathrm{~kJ} / \mathrm{mol}\), respectively. (Hint: Choose a constant so that the electronegativity of fluorine equals 4.0. Using this constant, calculate relative electronegativities for the other halogens and compare to values given in the text.)

Short Answer

Expert verified
Using the alternative definition of electronegativity, we first determine the constant by setting the electronegativity of fluorine to 4.0 using its I.E. and E.A. values. After determining this constant, we calculate the (I.E. - E.A.) term for each halogen (F, Cl, Br, and I) and multiply by our constant to find their respective electronegativity values. Finally, we compare these calculated values with the electronegativity values given in the chapter to determine if the trends are the same.

Step by step solution

01

Define electronegativity and gather the necessary data

The alternative definition of electronegativity is given as: Electronegativity = constant (I.E. - E.A.). We need to find this term for F, Cl, Br, and I. Firstly, we need to gather the ionization energies (I.E.) and electron affinities (E.A.) for each element. According to the problem, the ionization energies for F, Cl, Br, and I are 1678 kJ/mol, 1255 kJ/mol, 1138 kJ/mol, and 1007 kJ/mol, respectively. We will use Chapter 7 to find the electron affinities for these elements.
02

Choose a constant such that the electronegativity of fluorine equals 4.0

The hint provided suggests that we choose a constant such that the electronegativity of fluorine equals 4.0. To achieve this, we need to rearrange the formula and solve for the constant: \(constant = \frac{Electronegativity}{(I.E. - E.A.)}\). Using Fluorine's electronegativity of 4.0, its ionization energy of 1678 kJ/mol, and its electron affinity from Chapter 7, we can find the appropriate constant value.
03

Calculate the (I.E. - E.A.) term for each element

Now that we have found the constant value, we can calculate the (I.E. - E.A.) term for each element. For each element, we will subtract the electron affinity (E.A.) from its ionization energy (I.E.) and multiply the result by the constant calculated in step 2. This will give us the electronegativity value for each element according to the alternative definition.
04

Compare the electronegativity values

Finally, we need to compare the calculated electronegativity values using the alternative definition to those provided in the chapter. We will list both sets of electronegativity values for F, Cl, Br, and I and determine if the trends are the same. By following these steps, you would have successfully calculated the electronegativity values using the alternative definition and compare them to the values given in the text.

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Most popular questions from this chapter

Without using Fig. \(8.3\), predict the order of increasing electronegativity in each of the following groups of elements. a. \(\mathrm{Na}, \mathrm{K}, \mathrm{Rb}\) c. \(\mathrm{F}, \mathrm{Cl}, \mathrm{Br}\) b. \(\mathrm{B}, \mathrm{O}, \mathrm{Ga}\) d. \(\mathrm{S}, \mathrm{O}, \mathrm{F}\)

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