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Chapter 3: 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 increases across a row (left to right) and decreases down a column (top to bottom) in the periodic table due to effective nuclear charge and electron shielding effects, respectively. Ionization energy follows a similar trend. Conversely, atomic radii decrease across a row and increase down a column. These trends are interconnected, as they influence an atom's reactivity, chemical bonding, and stability in compounds.

Step by step solution

01

Electronegativity Trends

Electronegativity is the measure of an atom's ability to attract shared electrons in a chemical bond. Across a row (from left to right) in the periodic table, electronegativity increases due to an increase in the effective nuclear charge experienced by the valence electrons. This is because the number of positively charged protons is increasing, making the attraction of negatively charged electrons even stronger. Going down a column (from top to bottom) in the periodic table, electronegativity decreases because the number of electron shells increases, causing a shielding effect which lessens the attractive force on shared electrons.
02

Ionization Energy Trends

Ionization energy is the amount of energy required to remove an electron from an atom in a gaseous state. Similar to electronegativity, ionization energy generally increases across a row (left to right) in the periodic table due to the increased effective nuclear charge experienced by the valence electrons. As you go down a column (top to bottom), ionization energy decreases because the increased number of electron shells results in a shielding effect, meaning more energy levels separate the negatively charged electrons from the positively charged nucleus, making the electrons easier to remove.
03

Atomic Radii Trends

Atomic radius refers to the size of an atom, which is determined by the distance between the nucleus and the outermost electron shell. As you move across a row (left to right) in the periodic table, atomic radii decrease due to the increased effective nuclear charge, which causes the electron cloud to be pulled closer to the nucleus. Going down a column (top to bottom) in the periodic table, atomic radii increase because the electron shielding effect caused by additional electron shells results in a weaker pull by the nucleus and larger atomic sizes.
04

Relationship Between Electronegativity, Ionization Energy, and Atomic Radii

Electronegativity, ionization energy, and atomic radii are all related through their dependence on the periodic table trends. As the effective nuclear charge increases across a row, both electronegativity and ionization energy tend to increase, while atomic radii decrease. Conversely, going down a column, as shielding effect increases due to a larger number of electron shells, electronegativity and ionization energy decrease while atomic radii increase. These three properties give us insights into an atom's reactivity, chemical bonding, and stability within specific compounds.

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Most popular questions from this chapter

An important observation supporting the concept of resonance in the localized electron model was that there are only three different structures of dichlorobenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}\right) .\) How does this fact support the concept of resonance (see Exercise 89)?

The major industrial source of hydrogen gas is by the following reaction: $$ \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) $$ Use bond energies to predict r this reaction.

Write Lewis structures that obey the octet rule for the following species. Assign the formal charge for each central atom. a. \(\mathrm{POCl}_{3}\) b. \(\mathrm{SO}_{4}^{2-}\) c. \(\mathrm{ClO}_{4}\) d. \(\mathrm{PO}_{4}^{3-}\) e. \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) f. \(\quad X \in O_{4}\) g. \(\mathrm{ClO}_{3}\) h. \(\mathrm{NO}_{4}^{3-}\)

Hydrogen has an electronegativity value between boron and carbon and identical to phosphorus. With this in mind, rank the following bonds in order of decreasing polarity: \(\mathrm{P}-\mathrm{H}\) \(\mathbf{O}-\mathbf{H}, \mathbf{N}-\mathbf{H}, \mathbf{F}-\mathbf{H}, \mathbf{C}-\mathbf{H}\)

Name the compounds in parts a-d and write the formulas for the compounds in parts e-h. a. \(\mathrm{NaBr}\) b. \(\mathrm{Rb}_{2} \mathrm{O}\) c. CaS d. \(\mathrm{AlI}_{3}\) e. strontium fluoride f. aluminum selenide g. potassium nitride h. magnesium phosphide
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