Question

Identify at least two ions that have the following ground-state electron configurations: (a) \([\mathrm{Ar}] ;\) (b) \([\mathrm{Ar}] 3 d^{5}\); (c) \([\mathrm{Kr}] 5 s^{2} 4 d^{10}\)

Short answer

The short answer is: (a) K+ and Cl- both have the electron configuration [Ar]. (b) Mn and Fe+3 both have the electron configuration [Ar]3d^5. (c) Cd+2 and In+3 both have the electron configuration [Kr] 5s^2 4d^10.

Step-by-step solution

a) [Ar]

To find the ion with the electron configuration [Ar], we first need to identify the element that has the same configuration as Argon (Ar) in its neutral state. Argon has 18 electrons, so the element in question must have a neutral state with 18 electrons. We find that this is Ar itself. Now, we need to identify at least two ions that can have the same configuration as Argon. Since Argon is a noble gas, it usually doesn't form ions. However, neighboring elements from the periodic table are known to form ions by gaining or losing electrons to achieve this electron configuration. The two ions can be: 1. K+ (Potassium ion): A neutral potassium atom has 19 electrons (\([Ar]4s^1\)). When it loses one electron, it forms the K+ ion with an electron configuration of [Ar]. 2. Cl- (Chloride ion): A neutral chlorine atom has 17 electrons (\([Ar]3p^5\)). When it gains one electron, it forms the Cl- ion with an electron configuration of [Ar].

b) [Ar] 3d^5

To find ions with this electron configuration, we need to look for elements that can either gain or lose electrons to achieve this. Since [Ar] represents 18 electrons and 3d^5 adds five more, we are looking for ions with a total of 23 electrons. Two such ions are: 1. Mn (Manganese ion): A neutral manganese atom, with 25 electrons (\([Ar]4s^2 3d^5\)), can lose 2 electrons from its 4s orbital to end up with an electron configuration of [Ar]3d^5. 2. Fe+3 (Ferric ion): A neutral iron atom, with 26 electrons (\([Ar]4s^2 3d^6\)), can lose 3 electrons (2 from 4s and 1 from 3d orbitals) to form the Fe+3 ion with an electron configuration of [Ar]3d^5.

c) [Kr] 5s^2 4d^10

To find ions with this electron configuration, we need to look for elements that can either gain or lose electrons to achieve the given configuration. Since [Kr] represents 36 electrons, 5s^2 adds 2 more electrons, and 4d^10 adds 10 more electrons, we are looking for ions with a total of 48 electrons. Two such ions are: 1. Cd+2 (Cadmium ion): A neutral cadmium atom, with 48 electrons (\([Kr]5s^2 4d^10\)), can lose 2 electrons from its 5s orbital to form the Cd+2 ion with an electron configuration of [Kr] 5s^2 4d^10. 2. In+3 (Indium ion): A neutral indium atom, with 49 electrons (\([Kr]5s^2 4d^10 5p^1\)), can lose 3 electrons (2 from 5s and 1 from 5p orbitals) to form the In+3 ion with an electron configuration of [Kr] 5s^2 4d^10. These are the ions corresponding to each specific ground-state electron configuration.

Key concepts

Ions

Using these configurations is particularly useful in predicting and explaining the chemical behavior of elements. Noble gases are nonreactive due to their full outer electron shells, providing a stable configuration that other elements strive to achieve through ion formation. By gaining or losing electrons to emulate noble gas electron configurations, elements attain chemically stable, energetically favored configurations. This is a key factor in the formation of ions.

Periodic Table Trends

Understanding the periodic table trends is crucial in predicting the chemical and physical properties of elements, including their tendencies to form ions.

Elements exhibit periodic trends in electronegativity, atomic size, and ionization energy, which influence their ability to gain or lose electrons. For example, moving from left to right across a period, atomic size decreases and electronegativity increases. This means elements towards the right are more likely to gain electrons and form anions, whereas elements on the left tend to lose electrons and form cations.

Going down a group, atoms typically increase in size due to the addition of electron shells. This makes it easier for the outermost electrons to be removed, facilitating the formation of cations. Thus, the position of an element on the periodic table can often help predict whether it will more readily form a cation or an anion, and what the resulting electron configuration might be.