Problem 148 - Chemistry An Atoms First Approach

Question

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

Answer

The electronegativity values calculated using the alternative definition, Electronegativity \(= k (I.E. - E.A.)\), are approximately 4.0 for F, 3.20 for Cl, 2.91 for Br, and 2.59 for I. These values show a similar trend as the given electronegativity values in the text, which are 3.98, 3.16, 2.96, and 2.66, respectively. The trend remains the same using the alternative definition of electronegativity.

Step 1: 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 \(k = \frac{4.0}{2006}\)

Step 2: 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

Step 3: 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 \(= \frac{4.0}{2006} \times 1302 \approx 2.59\)

Step 4: 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.

Key Concepts

Ionization Energy

Ionization Energy (I.E.) is the amount of energy required to remove an electron from a gaseous atom or ion. It reflects the strength with which an atom holds its electrons.

Elements with high ionization energy usually do not lose electrons easily. As a result, these elements often have high electronegativity, meaning they attract electrons strongly.

This energy tends to increase:

• Across a period from left to right
• Up a group in the periodic table

This is because the nuclear charge increases, making it harder to remove an electron. For example, fluorine has a high ionization energy of 1678 kJ/mol. It requires a lot of energy to remove an electron, which contributes to its high electronegativity value.

Electron Affinity

Electron Affinity (E.A.) is the energy change that occurs when an electron is added to a neutral atom in the gas phase to form a negative ion. It measures the attraction of the atom for the added electron and is often expressed in kilojoules per mole (kJ/mol).

A negative E.A. value signifies that energy is released when an electron is added, indicating a greater tendency to gain electrons.

For instance, chlorine has an electron affinity of -349 kJ/mol, reflecting a strong attraction of its nuclear charge for additional electrons, similar to its high electronegativity. When examining periodic trends, electron affinity generally increases:

• Across a period from left to right, due to increased nuclear charge
• Down a group, although with less regularity compared to ionization energy

Halogens

Halogens are a group of elements located in Group 17 of the periodic table. This group includes fluorine (F), chlorine (Cl), bromine (Br), and iodine (I). They are known for their high reactivity, particularly due to their high electronegativity and electron affinity.

These properties make halogens effective at gaining electrons to form negative ions. Important characteristics of halogens include:

• They have the highest electronegativity values in their respective periods.
• Their electronegativity decreases down the group as seen in the decreasing values from fluorine to iodine.

Halogens participate in forming compounds with metals, often through ionic bonds, resulting in stable configurations by gaining a full valence electron shell.

Periodic Trends

Periodic trends refer to patterns in the properties of elements across different periods and groups of the periodic table. These trends are crucial for predicting the behavior of elements in chemical reactions.

Some key periodic trends include:

• Electronegativity: Increases across a period and decreases down a group. This trend explains why fluorine is more electronegative compared to iodine.
• Ionization Energy: Similar to electronegativity, it increases across a period and decreases down a group.
• Atomic Radius: Decreases across a period due to increased nuclear charge pulling electrons closer to the nucleus, and increases down a group as additional shells are added.

Understanding these trends assists in explaining the similarities and differences in the chemistry of various elements, particularly those in the same group.