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Chapter 8: Problem 1

Explain the electronegativity trends across a row and down a column of the periodic table. Compare these trends with those of ionization energies and atomic radii. How are they related?

Short Answer

Expert verified
Electronegativity, the ability of an atom to attract electrons in a chemical bond, increases across a row (left-to-right) and decreases down a column (top-to-bottom) in the periodic table due to changes in the effective nuclear charge. This trend is similar to ionization energy and contrasts with atomic radius trends, which are driven by the same underlying factors. Higher effective nuclear charge results in increased electronegativity and ionization energy, but decreased atomic radius, while lower effective nuclear charge yields opposite effects.

Step by step solution

01

Electronegativity Definition

Electronegativity is the ability of an atom to attract electrons towards itself in a chemical bond. Electronegativity increases towards the top-right corner of the periodic table. Electronegativity increases across a row (left-to-right) and decreases down a column (top-to-bottom) in the periodic table.
02

Ionization Energy Definition

Ionization energy is the amount of energy required to remove an electron from a gaseous atom. It increases across a row (left-to-right) and decreases down a column (top-to-bottom) in the periodic table.
03

Atomic Radius Definition

Atomic radius is the measure of the size of an atom. It decreases across a row (left-to-right) and increases down a column (top-to-bottom) in the periodic table.
04

Electronegativity Trend Across a Row

Moving across a row in the periodic table, more protons are added to the nucleus, leading to an increased positive charge. This increase in the positive charge draws the electrons in the atom closer, making it more difficult for other atoms to attract these electrons. As a result, electronegativity increases from left to right across a row in the periodic table.
05

Electronegativity Trend Down a Column

Moving down a column in the periodic table, there is an increase in the number of electron shells. This increase in electron shells causes the atomic radius to increase, resulting in the valence electrons being farther from the nucleus. Consequently, the ability of the atom to attract electrons decreases, and electronegativity decreases as we move down a column in the periodic table.
06

Comparison of Trends with Ionization Energies and Atomic Radii

The trends in electronegativity show similar patterns as those observed for ionization energy and atomic radius. Ionization energy increases across a row and decreases down a column, while atomic radius decreases across a row and increases down a column. This similarity in trends is due to the same underlying factors: the attractive force between the electrons and the nucleus, as dictated by the number of protons and electron shells.
07

Relationship Between Electronegativity, Ionization Energy, and Atomic Radius

Electronegativity, ionization energy, and atomic radius are interrelated because they all depend on the effective nuclear charge, which is the attractive force between the positively charged nucleus and the negatively charged electrons. Higher effective nuclear charge leads to an increase in electronegativity and ionization energy, while it causes a decrease in atomic radius. Conversely, a lower effective nuclear charge decreases electronegativity and ionization energy while increasing atomic radius.

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Most popular questions from this chapter

Use the following data to estimate \(\Delta H_{f}^{\circ}\) for potassium chloride. $$\mathrm{K}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{KCl}(s)$$ \(\begin{array}{l}{\text { Lattice energy }} & {-690 . \mathrm{kJ} / \mathrm{mol}} \\ {\text { Ionization energy for } \mathrm{K}} & \quad{419 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Electron affinity of } \mathrm{Cl}} & {-349 \mathrm{kJ} / \mathrm{mol}}\\\\{\text { Bond energy of } \mathrm{Cl}_{2}} & \quad {239 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Enthalpy of sublimation for } \mathrm{K}} & \quad {90 . \mathrm{kJ} / \mathrm{mol}}\end{array}\)

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The second electron affinity values for both oxygen and sulfur are unfavorable (endothermic). Explain.

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