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Chapter 7: Problem 46

Write the electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Ru}^{3+}\), (b) \(\mathrm{As}^{3-}\), (c) \(\mathrm{Y}^{3+}\), (d) \(\mathrm{Pd}^{2+}\), (e) \(\mathrm{Pb}^{2+}\), (f) \(\mathrm{Au}^{3+}\).

Short Answer

Expert verified
The electron configurations for the given ions are as follows: (a) \(\mathrm{Ru}^{3+}\): \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^7}\) (b) \(\mathrm{As}^{3-}\): \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6}\) (c) \(\mathrm{Y}^{3+}\): \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6}\) (d) \(\mathrm{Pd}^{2+}\): \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^8}\) (e) \(\mathrm{Pb}^{2+}\): \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^10 5p^6 6s^2 4f^{14} 5d^{10} 6p^4}\) (f) \(\mathrm{Au}^{3+}\): \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^10 5p^6 6s^1 4f^{14} 5d^9}\) Out of these ions, only \(\mathrm{As}^{3-}\) and \(\mathrm{Y}^{3+}\) have noble-gas configurations.

Step by step solution

01

Write the electron configurations for the ions

To write the electron configurations, we'll follow the order of orbital filling, starting with 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p. Also, note that when an element loses electrons, it forms positive ions, and when it gains electrons, it forms negative ions. (a) \(\mathrm{Ru}^{3+}\): The atomic number of ruthenium is 44. Since it has 3+ charge, it has lost three electrons, leaving the configuration for 41 electrons. So, the electron configuration for \(\mathrm{Ru}^{3+}\) is \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^7}\). (b) \(\mathrm{As}^{3-}\): The atomic number of arsenic is 33. Since it has a 3- charge, it has gained three electrons, making it a configuration for 36 electrons. The electron configuration for \(\mathrm{As}^{3-}\) is: \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6}\). (c) \(\mathrm{Y}^{3+}\): The atomic number of yttrium is 39. Since it has a 3+ charge, it has lost three electrons, making it a configuration for 36 electrons. The electron configuration for \(\mathrm{Y}^{3+}\) is: \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6}\). (d) \(\mathrm{Pd}^{2+}\): The atomic number of palladium is 46. Since it has a 2+ charge, it has lost two electrons, making it a configuration for 44 electrons. The electron configuration for \(\mathrm{Pd}^{2+}\) is: \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^8}\). (e) \(\mathrm{Pb}^{2+}\): The atomic number of lead is 82. Since it has a 2+ charge, it has lost two electrons, making it a configuration for 80 electrons. The electron configuration for \(\mathrm{Pb}^{2+}\) is: \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^10 5p^6 6s^2 4f^{14} 5d^{10} 6p^4}\). (f) \(\mathrm{Au}^{3+}\): The atomic number of gold is 79. Since it has a 3+ charge, it has lost three electrons, making it a configuration for 76 electrons. The electron configuration for \(\mathrm{Au}^{3+}\) is \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^10 5p^6 6s^1 4f^{14} 5d^9}\).
02

Determine which ions have noble-gas configurations

Noble-gas configurations are achieved when an element has a filled outer electron shell. To determine if the given ions have noble-gas configurations, we'll find the nearest noble gas in the periodic table whose electron count is lesser than or equal to the ion's electron count and check if their electron configurations match. (a) \(\mathrm{Ru}^{3+}\): Nearest noble gas is \(\mathrm{Kr}\) with configuration \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^8}\). This does not match the configuration of \(\mathrm{Ru}^{3+}\), so it does not have a noble-gas configuration. (b) \(\mathrm{As}^{3-}\): Nearest noble gas is \(\mathrm{Kr}\) with the same configuration as above. The electron configuration of \(\mathrm{As}^{3-}\) matches with that of \(\mathrm{Kr}\), so it has a noble-gas configuration. (c) \(\mathrm{Y}^{3+}\): Nearest noble gas is also \(\mathrm{Kr}\). The electron configuration of \(\mathrm{Y}^{3+}\) matches with that of \(\mathrm{Kr}\), so it has a noble-gas configuration. (d) \(\mathrm{Pd}^{2+}\): Nearest noble gas is \(\mathrm{Kr}\) with the same configuration as mentioned above. The electron configuration of \(\mathrm{Pd}^{2+}\) does not match that of \(\mathrm{Kr}\), so it does not have a noble-gas configuration. (e) \(\mathrm{Pb}^{2+}\): Nearest noble gas is \(\mathrm{Xe}\) with configuration \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^10 5p^6 6s^2 4f^{14} 5d^{10}\). This does not match the configuration of \(\mathrm{Pb}^{2+}\), so it does not have a noble-gas configuration. (f) \(\mathrm{Au}^{3+}\): Nearest noble gas is \(\mathrm{Xe}\) with the electron configuration mentioned above. The electron configuration of \(\mathrm{Au}^{3+}\) does not match that of \(\mathrm{Xe}\), so it does not have a noble-gas configuration. In conclusion, \(\mathrm{As}^{3-}\) and \(\mathrm{Y}^{3+}\) have noble-gas configurations, while the other ions do not.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Noble Gas Configuration
The noble gas configuration is a way of simplifying how we represent electron configurations. Instead of writing out all of the electron orbitals, we use the noble gases, which are group 18 elements, as a shorthand because they have full outer electron shells. Understanding noble gas configurations is important because they are especially stable due to their full valence shells. For instance, the noble gas neon (Ne) has a configuration of \(1s^2 2s^2 2p^6\), often represented by its symbol \[\text{[Ne]}\]. When writing the electron configuration for an element, we begin by writing the noble gas preceding the element in brackets, indicating that the element has the same electron configuration as the noble gas up to that point. For example, for chlorine, instead of \(1s^2 2s^2 2p^6 3s^2 3p^5\), we use \[\text{[Ne]} 3s^2 3p^5\].

Using noble gas configuration not only makes writing electron configurations more manageable but also allows students to focus on valence electrons, which are most important in chemical reactions.
Ion Electron Configuration
The concept of ion electron configuration becomes relevant when we talk about atoms gaining or losing electrons to become ions. When atoms become ions, the electron configuration changes because electrons are either removed or added. For positive ions (cations), electrons are removed. This usually occurs with metals and results in a positive charge. For example, when a neutral atom of sodium (\(\text{Na}\)) loses one electron, it becomes \(\text{Na}^+\), with an electron configuration of \(1s^2 2s^2 2p^6\), resembling that of the earlier noble gas neon. Meanwhile, for negative ions (anions), electrons are added to the valence shell. Non-metals commonly gain electrons to achieve a stable configuration. For instance, a chlorine atom gains one electron to become \( ext{Cl}^鈭� \), configured as \[1s^2 2s^2 2p^6 3s^2 3p^6\], achieving the noble gas configuration of argon.

Ion electron configurations thus reflect how ions attempt to attain stable electron arrangements, typically resembling the nearest noble gas.
Orbital Filling Order
Orbital filling order refers to the sequence in which electrons fill orbitals around an atom's nucleus. This is determined by two key principles: the Aufbau principle and Hund's rule. According to the Aufbau principle, electrons occupy the lowest energy orbitals available, building up from there. For instance, the 1s orbital fills before the 2s orbital because it has a lower energy. Known as "building-up," this order is essential for writing correct electron configurations, as seen in the sequence: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, and so on. Hund鈥檚 rule adds that electrons will fill degenerate orbitals (orbitals at the same energy level) singly before pairing up, to minimize electron repulsion. This principle affects the arrangement of electrons in orbitals like 2p and 3p, where spreading electrons across Orbitals before pairing promotes stability.

Remembering this order helps when writing configurations for elements and ions by ensuring electrons fill in an energetically favorable, predictable manner.

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Most popular questions from this chapter

Which of the following is the expected product of the reaction of \(\mathrm{K}(s)\) and \(\mathrm{H}_{2}(g)\) ? (i) \(\mathrm{KH}(s)\), (ii) \(\mathrm{K}_{2} \mathrm{H}(s)\), (iii) \(\mathrm{KH}_{2}(s)\), (iv) \(\mathrm{K}_{2} \mathrm{H}_{2}(s)\), or (v) \(\mathrm{K}(s)\) and \(\mathrm{H}_{2}(g)\) will not react with one another.

Some ions do not have a corresponding neutral atom that has the same electron configuration. For each of the following ions, identify the neutral atom that has the same number of electrons and determine if this atom has the same electron configuration. If such an atom does not exist, explain why. (a) \(\mathrm{Cl}^{-}\), (b) \(\mathrm{Sc}^{3+}\), (c) \(\mathrm{Fe}^{2+}\), (d) \(\mathrm{Zn}^{2+}\), (e) \(\mathrm{Sn}^{4+}\).

Arrange each of the following sets of atoms and ions, in order of increasing size: (a) \(\mathrm{Se}^{2-}, \mathrm{Te}^{2-}, \mathrm{Se} ;\) (b) \(\mathrm{Co}^{3+}, \mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}\); (c) \(\mathrm{Ca}^{2 \mathrm{Ti}^{4+}}, \mathrm{Sc}^{3+}\); (d) \(\mathrm{Be}^{2+}, \mathrm{Na}^{+}, \mathrm{Ne}\).

Write balanced equations for the following reactions: (a) potassium oxide with water, (b) diphosphorus trioxide with water, (c) chromium(III) oxide with dilute hydrochloric acid, (d) selenium dioxide with aqueous potassium hydroxide.

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, \(\mathrm{X}\) 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 \(4 f\) orbitals at \(105 \mathrm{eV}\), from an X-ray source that provided \(1253.6 \mathrm{eV}\) of energy. 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 \(\mathrm{O}\). (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? (d) Use Slater's rules to estimate \(Z_{\text {eff }}\) for the \(4 f\) and valence electrons of \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); assume for this purpose that all the inner electrons with \((n-3)\) or less screen a full \(+1\).
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