(a) What is the trend in electronegativity going from left to right in a row of the periodic table? (b) How do electronegativity values generally vary going down a column in the periodic table? (c) How do periodic trends in electronegativity relate to those for ionization energy and electron affinity?

Short Answer

Expert verified
(a) Going from left to right in a row (period) of the periodic table, the electronegativity generally increases due to the increasing number of protons in the nucleus, resulting in a stronger attractive force on the electrons shared in a bond. (b) Going down a column (group) in the periodic table, the electronegativity generally decreases due to the increase in atomic size and larger distance between the nucleus and the valence electrons. (c) Electronegativity, ionization energy, and electron affinity are related and generally follow the same trend in the periodic table. As electronegativity increases, ionization energy and electron affinity also generally increase.

Step by step solution

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Electronegativity

Electronegativity is a measure of the ability of an atom to attract electrons in a chemical bond. The electronegativity values increase from lower left to upper right in the periodic table.
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(a) Trend in electronegativity from left to right

Going from left to right in a row (period) of the periodic table, the electronegativity generally increases. This is because the number of protons in the nucleus increases as you go from left to right, which results in a stronger attractive force on the electrons shared in a bond.
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(b) Trend in electronegativity going down a column

Going down a column (group) in the periodic table, the electronegativity generally decreases. This can be explained by the increase in atomic size as you go down a group. The additional electron shells lead to a larger distance between the nucleus and the valence electrons, which decreases the effective nuclear pull on the valence electrons.
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(c) Relation to ionization energy and electron affinity

Electronegativity, ionization energy, and electron affinity are related since they all deal with the attraction of electrons. Ionization energy is the energy required to remove an electron from an atom, while electron affinity is the energy released when an electron is added to an atom. As electronegativity increases, ionization energy and electron affinity generally increase as well. The increase in electronegativity means that an atom attracts electrons more strongly, making it more difficult to remove an electron (higher ionization energy) and more favorable to add an electron (higher electron affinity). The periodic trends for ionization energy and electron affinity generally follow the same trend as electronegativity: they increase from left to right in a period and decrease going down a group in the periodic table.

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Most popular questions from this chapter

For each of the following molecules or ions of sulfur and oxygen, write a single Lewis structure that obeys the octet rule, and calculate the oxidation numbers and formal charges on all the atoms: \((\mathrm{a}) \mathrm{SO}_{2},(\mathrm{~b}) \mathrm{SO}_{3},(\mathrm{c}) \mathrm{SO}_{3}{ }^{2-},\) (d) Arrange these mol- ecules/ions in order of increasing \(\mathrm{S}-\mathrm{O}\) bond distance.

Which of the following bonds are polar: (a) \(\mathrm{B}-\mathrm{F},\) (b) \(\mathrm{Cl}-\mathrm{Cl}\), $$ \text { (c) } \mathrm{Se}-\mathrm{O} $$ (d) \(\mathrm{H}-\mathrm{I} ?\) Which is the more electronegative atom in each polar bond?

Write Lewis structures for the following: (a) \(\mathrm{H}_{2} \mathrm{CO}\) (both \(\mathrm{H}\) atoms are bonded to \(\mathrm{C}\) ), (b) \(\mathrm{H}_{2} \mathrm{O}_{2},\) (c) \(\mathrm{C}_{2} \mathrm{~F}_{6}\) (contains a \(\mathrm{C}-\mathrm{C}\) bond), \((\mathrm{d}) \mathrm{AsO}_{3}^{3-},(\mathrm{e}) \mathrm{H}_{2} \mathrm{SO}_{3}(\mathrm{H}\) is bonded to \(\mathrm{O}),(\mathrm{f}) \mathrm{C}_{2} \mathrm{H}_{2}\)

Use Table 8.4 to estimate the enthalpy change for each of the following reactions: (a) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{O}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{O}-\mathrm{Cl}(g)\) (b) \(\mathrm{H}_{2} \mathrm{O}_{2}(g)+2 \mathrm{CO}(g) \longrightarrow \mathrm{H}_{2}(g)+2 \mathrm{CO}_{2}(g)\) (c) \(3 \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12}(g)\) (the six carbon atoms form a six-membered ring with two \(\mathrm{H}\) atoms on each \(\mathrm{C}\) atom \()\)

Consider the following statement: "For some molecules and ions, a Lewis structure that satisfies the octet rule does not lead to the lowest formal charges, and a Lewis structure that leads to the lowest formal charges does not satisfy the octet rule." Illustrate this statement using the hydrogen sulfite ion, \(\mathrm{HSO}_{3}^{-}\), as an example (the \(\mathrm{H}\) atom is bonded to one of the \(\mathrm{O}\) atoms).

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