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Chapter 7: Problem 12

In going across a row of the periodic table, electrons are added and ionization energy generally increases. In going down a column of the periodic table, electrons are also being added but ionization energy decreases. Explain.

Short Answer

Expert verified
In summary, as we move across a row in the periodic table, ionization energy increases due to a stronger effective nuclear charge and a smaller atomic size, resulting in greater electrostatic attraction between the nucleus and valence electrons. Conversely, moving down a column of the periodic table, ionization energy decreases due to increased electron shielding and larger atomic size, weakening the electrostatic attraction between the nucleus and the outer electrons, making it easier for an electron to be removed.

Step by step solution

01

Define Ionization Energy

Ionization energy is the energy required to remove an electron from an atom or ion. More specifically, it is the minimum energy needed to remove a valence electron from a gaseous atom in its ground state. Generally, ionization energy gives an indication of how tightly an electron is held in an atom.
02

Discuss Trends Across a Period

When moving from left to right across a period in the periodic table, the atomic number increases, which means that there are more protons in the nucleus and more electrons around the nucleus. The atomic size decreases across a period due to an increase in the effective nuclear charge experienced by the valence electrons. This leads to a stronger electrostatic attraction between the electrons and the nucleus. As a result, the ionization energy increases because more energy is required to overcome this stronger attraction and remove an electron from the atom.
03

Discuss Trends Down a Column

When moving down a column in the periodic table, there is an increase in the number of electron shells or energy levels. Although there are also additional protons, which would tend to increase the ionization energy, the increased electron shielding due to the presence of more inner electrons outweighs the effect of the additional protons. Electron shielding reduces the attractive force between the nucleus and the outer electrons, making it easier for the outermost electron to be removed. Hence, the atomic size increases, and the ionization energy decreases down a group.
04

Finalize the Explanation

In summary, ionization energy generally increases across a row in the periodic table due to an increase in effective nuclear charge and a decrease in atomic size, resulting in a stronger attractive force between the electrons and the nucleus. In contrast, ionization energy decreases down a column because of increased electron shielding and a larger atomic size, which weakens the attractive force between the nucleus and the outermost electrons, making it easier for an electron to be removed.

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Most popular questions from this chapter

Answer the following questions assuming that \(m_{s}\) could have three values rather than two and that the rules for \(n, \ell\), and \(m_{\ell}\) are the normal ones. a. How many electrons would an orbital be able to hold? b. How many elements would the first and second periods in the periodic table contain? c. How many elements would be contained in the first transition metal series? d. How many electrons would the set of \(4 f\) orbitals be able to hold?

Consider the following approximate visible light spectrum: Barium emits light in the visible region of the spectrum. If each photon of light emitted from barium has an energy of \(3.59 \times 10^{-19} \mathrm{~J}\), what color of visible light is emitted?

The following numbers are the ratios of second ionization energy to first ionization energy: Na: \(\quad 9.2\) P: \(1.8\) Mg: \(\quad 2.0\) \(\mathrm{S}: \quad 2.3\) Al: \(\quad 3.1\) \(\mathrm{Cl}: \quad 1.8\) Si: \(\quad 2.0\) Ar: \(1.8\) Explain these relative numbers.

Which of the following sets of quantum numbers are not allowed? For each incorrect set, state why it is incorrect. a. \(n=3, \ell=3, m_{\ell}=0, m_{s}=-\frac{1}{2}\) b. \(n=4, \ell=3, m_{\ell}=2, m_{s}=-\frac{1}{2}\) c. \(n=4, \ell=1, m_{\ell}=1, m_{s}=+\frac{1}{2}\) d. \(n=2, \ell=1, m_{\ell}=-1, m_{s}=-1\) e. \(n=5, \ell=-4, m_{\ell}=2, m_{s}=+\frac{1}{2}\) f. \(n=3, \ell=1, m_{\ell}=2, m_{s}=-\frac{1}{2}\)

Cesium was discovered in natural mineral waters in \(1860 \mathrm{by}\) R. W. Bunsen and G. R. Kirchhoff using the spectroscope they invented in \(1859 .\) The name came from the Latin caesius ("sky blue") because of the prominent blue line observed for this element at \(455.5 \mathrm{~nm} .\) Calculate the frequency and energy of a photon of this light.
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