The radius trend and the ionization energy trend are exact opposites. Does this make sense? Define electronaffinity. Electron affinity values are both exothermic (negative) and endothermic (positive). However, ionization energy values are always endothermic (positive). Explain.

Short Answer

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The radius trend and ionization energy trend being exact opposites make sense because as atomic radius decreases, ionization energy generally increases due to the increased nuclear charge and greater electrostatic attraction between the nucleus and outermost electrons. Electron affinity is the energy change when an electron is added to a neutral atom, forming a negatively charged ion. Electron affinity values can be exothermic (negative) or endothermic (positive) due to stability of the resulting ion and differing energy requirements. Ionization energy values are always endothermic (positive) as removing an electron always requires energy input to overcome the electrostatic attraction to the nucleus.

Step by step solution

01

Define radius trend and ionization energy trend

The radius trend refers to the pattern observed in atomic radius sizes across the periodic table. In general, atomic radius decreases from left to right in a period and increases down a group. The ionization energy trend, on the other hand, refers to the energy required to remove an electron from an atom or ion. This trend generally increases from left to right in a period and decreases down a group.
02

Discuss the relationship between radius trend and ionization energy trend

Now that we understand both trends, let's explore the relationship between them. As the atomic radius decreases, the nuclear charge (the positive charge of protons in the nucleus) increases. Consequently, the outermost electrons experience a greater electrostatic attraction to the nucleus and are more tightly held. This results in an increase in ionization energy, as more energy is required to remove an electron from the atom or ion. Thus, it makes sense that the radius trend and ionization energy trend are exact opposites. As the atomic radius decreases, the ionization energy generally increases.
03

Define electron affinity

Electron affinity is defined as the energy change that occurs when an electron is added to a neutral atom, forming a negatively charged ion. In other words, it is a measure of how likely an atom is to gain an electron and form a more stable configuration.
04

Explain why electron affinity values can be exothermic and endothermic

When an atom gains an electron, the energy change can either be exothermic (negative), meaning energy is released when an electron is added, or endothermic (positive), meaning energy is absorbed when an electron is added. This depends on the stability of the resulting ion and how much energy is required to overcome any repulsion forces between incoming electrons and existing electrons.
05

Explain why ionization energy values are always endothermic

In contrast to electron affinity, ionization energy is always endothermic (positive) because energy must be absorbed to overcome the electrostatic attraction between the electron being removed and the positively charged nucleus. No matter the element or ion, removing an electron always requires an input of energy to break the attractive forces between the electron and the nucleus. In conclusion, the relationship between the radius trend and ionization energy trend makes sense, as they are influenced by atomic radius and nuclear charge. Additionally, electron affinity values can be both exothermic and endothermic due to the stability of the resulting ion and differing energy requirements, while ionization energy values are always endothermic because removing an electron always requires energy input to overcome the electrostatic attraction to the nucleus.

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Most popular questions from this chapter

Which of the following statements is(are) true? a. \(\mathrm{F}\) has a larger first ionization energy than does \(\mathrm{Li}\). b. Cations are larger than their parent atoms. c. The removal of the first electron from a lithium atom (electron configuration is \(1 s^{2} 2 s^{1}\) ) is exothermic - that is, removing this electron gives off energy. d. The He atom is larger than the \(\mathrm{H}^{+}\) ion. e. The \(\mathrm{Al}\) atom is smaller than the \(\mathrm{Li}\) atom.

For each of the following pairs of elements \((\mathrm{C}\) and \(\mathrm{N})\) \((\mathrm{Ar}\) and \(\mathrm{Br})\) pick the atom with a. more favorable (exothermic) electron affinity. b. higher ionization energy. c. larger size.

The successive ionization energies for an unknown element are \(I_{1}=896 \mathrm{~kJ} / \mathrm{mol}\) \(I_{2}=1752 \mathrm{~kJ} / \mathrm{mol}\) \(I_{3}=14,807 \mathrm{~kJ} / \mathrm{mol}\) \(I_{4}=17,948 \mathrm{~kJ} / \mathrm{mol}\) To which family in the periodic table does the unknown element most likely belong?

Give the maximum number of electrons in an atom that can have these quantum numbers: a. \(n=4\) b. \(n=5, m_{\ell}=+1\) c. \(n=5, m_{s}=+\frac{1}{2}\) d. \(n=3, \ell=2\) e. \(n=2, \ell=1\)

An ion having a \(4+\) charge and a mass of \(49.9\) amu has 2 electrons with principal quantum number \(n=1,8\) electrons with \(n=2\), and 10 electrons with \(n=3\). Supply as many of the properties for the ion as possible from the information given. (Hint: In forming ions for this species, the \(4 s\) electrons are lost before the \(3 d\) electrons.) a. the atomic number b. total number of \(s\) electrons c. total number of \(p\) electrons d. total number of \(d\) electrons e. the number of neutrons in the nucleus f. the ground-state electron configuration of the neutral atom

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