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Chapter 7: Problem 95

Use electron configurations to explain the following observations: (a) The first ionization energy of phosphorus is greater than that of sulfur. (b) The electron affinity of nitrogen is lower (less negative) than those of both carbon and oxygen. (c) The second ionization energy of oxygen is greater than the first ionization energy of fluorine. (d) The third ionization energy of manganese is greater than those of both chromium and iron.

Short Answer

Expert verified
(a) The first ionization energy of phosphorus is greater than that of sulfur due to the half-filled p orbitals in phosphorus ([Ne] 3s² 3p³), which provide extra stability compared to sulfur's electron configuration ([Ne] 3s² 3p⁴). (b) Nitrogen's electron affinity is less negative than those of carbon and oxygen due to the extra stability of half-filled p orbitals ([He] 2s² 2p³) in nitrogen, which makes it less likely to accept an additional electron. (c) The second ionization energy of oxygen is greater than the first ionization energy of fluorine because removing an electron from a half-filled p orbital in ionized oxygen ([He] 2s² 2p³) is more difficult than from fluorine's almost filled p orbital ([He] 2s² 2p⁵). (d) The third ionization energy of manganese is greater than those of chromium and iron due to the extra stability provided by half-filled d orbitals in manganese after losing two electrons ([Ar] 3d⁵), making it more difficult to remove an electron compared to chromium ([Ar] 3d⁴) and iron ([Ar] 3d⁶).

Step by step solution

01

(a) Comparing the first ionization energies of phosphorus and sulfur

To explain the difference in ionization energies, we need to compare the electron configurations of phosphorus and sulfur. Phosphorus has an electron configuration of [Ne] 3s² 3p³, while sulfur has an electron configuration of [Ne] 3s² 3p⁴. The first ionization energies generally increase across a period in the periodic table due to increasing effective nuclear charge, which leads to greater attraction between electrons and protons in the nucleus. However, the electrons in the 3p orbitals of phosphorus are less repelled than the electrons in the 3p orbitals of sulfur. This is due to the fact that phosphorus has exactly half-filled p orbitals, while sulfur has more than half-filled p orbitals. The additional electron in sulfur leads to increased electron-electron repulsion, making it easier to remove an electron from sulfur than it is from phosphorus. Therefore, the first ionization energy of phosphorus is greater than that of sulfur.
02

(b) Comparing the electron affinities of carbon, nitrogen, and oxygen

We need to compare the electron configurations of carbon, nitrogen, and oxygen to understand their electron affinities. Carbon has an electron configuration of [He] 2s² 2p², nitrogen has [He] 2s² 2p³, and oxygen has [He] 2s² 2p⁴. As we move across a period, the electron affinity tends to become more negative due to the increasing effective nuclear charge. However, nitrogen has a lower (less negative) electron affinity than both carbon and oxygen. This can be attributed to its half-filled p orbitals, which offer extra stability. Adding an electron to nitrogen would disturb the half-filled stability, making nitrogen less likely to accept an additional electron. Hence, nitrogen has lower electron affinity than both carbon and oxygen.
03

(c) Comparing the second ionization energy of oxygen and the first ionization energy of fluorine

To explain this observation, we need to look at the electron configurations of oxygen and fluorine after losing an electron. Oxygen's configuration after losing one electron is [He] 2s² 2p³ (after the first ionization), and fluorine's original configuration is [He] 2s² 2p⁵. The second ionization energy of oxygen involves removing an electron from a half-filled p orbital, which is relatively stable. On the other hand, the first ionization energy of fluorine involves removing an electron from a nearly filled p orbital. Although the effective nuclear charge increases across the period, making fluorine's electrons in general more difficult to remove, the extra stability of the half-filled p orbital in ionized oxygen outweighs this factor. Therefore, the second ionization energy of oxygen is greater than the first ionization energy of fluorine.
04

(d) Comparing the third ionization energies of manganese, chromium, and iron

We need to investigate the electron configurations of manganese, chromium, and iron after losing two electrons. Manganese's electron configuration after losing two electrons is [Ar] 3d⁵, chromium's is [Ar] 3d⁴, and iron's is [Ar] 3d⁶. The third ionization energy of each element involves removing an electron from a d orbital. The electron configuration of manganese after losing two electrons features half-filled d orbitals, which provides extra stability due to the even distribution of electrons within the orbitals. This stability makes it more difficult to remove an electron from manganese than from chromium or iron. Thus, the third ionization energy of manganese is greater than those of both chromium and iron.

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Most popular questions from this chapter

In the chemical process called electron transfer, an electron is transferred from one atom or molecule to another. (We will talk about electron transfer extensively in Chapter 20.) A simple electron transfer reaction is $$ \mathrm{A}(g)+\mathrm{A}(g) \longrightarrow \mathrm{A}^{+}(g)+\mathrm{A}^{-}(g) $$ In terms of the ionization energy and electron affinity of atom A, what is the energy change for this reaction? For a representative nonmetal such as chlorine, is this process exothermic? For a representative metal such as sodium, is this process exothermic?

Which will experience the greater effect nuclear charge, the electrons in the \(n=2\) shell in \(\mathrm{F}\) or the \(n=2\) shell in \(\mathrm{B}\) ? Which will be closer to the nucleus?

Using only the periodic table, arrange each set of atoms in order from largest to smallest: \((\mathbf{a}) \mathrm{Ar},\) As, \(\mathrm{Kr} ;\) (b) $\mathrm{Cd}, \mathrm{Rb}, \mathrm{Te} ;(\mathbf{c})$ \(\mathrm{C}, \mathrm{Cl}, \mathrm{Cu}\).

(a) Why does xenon react with fluorine, whereas neon does not? (b) Using appropriate reference sources, look up the bond lengths of Xe-F bonds in several molecules. How do these numbers compare to the bond lengths calculated from the atomic radii of the elements?

Mercury in the environment can exist in oxidation states \(0,\) \(+1,\) and \(+2 .\) One major question in environmental chemistry research is how to best measure the oxidation state of mercury in natural systems; this is made more complicated by the fact that mercury can be reduced or oxidized on surfaces differently than it would be if it were free in solution. XPS, X-ray photoelectron spectroscopy, is a technique related to PES (see Exercise 7.111 ), but instead of using ultraviolet light to eject valence electrons, X rays are used to eject core electrons. The energies of the core electrons are different for different oxidation states of the element. In one set of experiments, researchers examined mercury contamination of minerals in water. They measured the XPS signals that corresponded to electrons ejected from mercury's 4 forbitals at \(105 \mathrm{eV}\), from an X-ray source that provided \(1253.6 \mathrm{eV}\) of energy $\left(1 \mathrm{ev}=1.602 \times 10^{-19} \mathrm{~J}\right)$ The oxygen on the mineral surface gave emitted electron energies at $531 \mathrm{eV}\(, corresponding to the \)1 s$ orbital of oxygen. Overall the researchers concluded that oxidation states were +2 for \(\mathrm{Hg}\) and -2 for O. (a) Calculate the wavelength of the \(\mathrm{X}\) rays used in this experiment. (b) Compare the energies of the \(4 f\) electrons in mercury and the \(1 s\) electrons in oxygen from these data to the first ionization energies of mercury and oxygen from the data in this chapter. (c) Write out the ground- state electron configurations for \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); which electrons are the valence electrons in each case?
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