Question

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

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

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 \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^10 5p^6 6s^1 4f^{14} 5d^9}\).

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.

Key concepts

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 \( ext{Cl}^− \), 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’s 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.