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 2 to calculate the (I.E. \(-\) E.A.) term for \(\mathbf{F}, \mathbf{C l}, \mathbf{B r}\) and 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.)

Step by step solution

01

Choose the constant value

Let's denote the constant value as \(k\). We want to make the electronegativity of fluorine equal to 4.0 using the given formula: Electronegativity \(= k (I.E. - E.A.)\) For fluorine, we know the electronegativity is 4.0; we need to find its I.E. and E.A. values. From the given values, the first ionization energy (I.E.) of fluorine is 1678 kJ/mol. From the Chapter 2, we find electron affinity (E.A.) of fluorine is -328 kJ/mol. We can now set up the equation: \(4.0 = k (1678 - (-328))\) Solving for \(k\), we get: \(k = \frac{4.0}{1678 + 328} = \frac{4.0}{2006}\) So, the chosen constant value is \(k = \frac{4.0}{2006}\)

02

Calculate the (I.E. - E.A.) term for F, Cl, Br, and I

We are given the first ionization energies of F, Cl, Br, and I. We need to find their electron affinities (E.A.) from Chapter 2. Then, we can calculate the (I.E. - E.A.) term for each element. - F: I.E. = 1678 kJ/mol, E.A. = -328 kJ/mol. So, (I.E. - E.A.) = 2006 kJ/mol - Cl: I.E. = 1255 kJ/mol, E.A. = -349 kJ/mol. So, (I.E. - E.A.) = 1604 kJ/mol - Br: I.E. = 1138 kJ/mol, E.A. = -324 kJ/mol. So, (I.E. - E.A.) = 1462 kJ/mol - I: I.E. = 1007 kJ/mol, E.A. = -295 kJ/mol. So, (I.E. - E.A.) = 1302 kJ/mol

03

Compute their electronegativity using the chosen constant value

We can now use the constant value \(k = \frac{4.0}{2006}\) to compute the electronegativity for each element: - F: Electronegativity \(= \frac{4.0}{2006} \times 2006 = 4.0\) - Cl: Electronegativity \(= \frac{4.0}{2006} \times 1604 \approx 3.20\) - Br: Electronegativity \(= \frac{4.0}{2006} \times 1462 \approx 2.91\) - I: Electronegativity \(= \frac{4.0}{2006} \times 1302 \approx 2.59\)

04

Compare the calculated values with the given ones

Comparing the calculated electronegativity values with the ones given in the text, we can see a similar trend. The text values for the electronegativity of F, Cl, Br, and I are 3.98, 3.16, 2.96, and 2.66, respectively. Our calculated values are approximately equal to the given values, which shows that the trend is the same using the alternative definition of electronegativity.

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