Question

a neutral li atom has three electrons; an li ion is formed when a li atom loses one electron. therefore, li has two electrons, which are located in the 1s orbital. the resulting electron configuration is $1s^{2}$. part d $al^{3+}$ express your answer in complete form, in order of increasing orbital. for example, $1s^{2}2s^{2}$ would be entered as 1s^22s^2. part e what do all the electron configurations have in common? all of the obtained electron configurations are isoelectronic with group 1a elements. group 6a elements. group 7a elements. noble gases. group 2a elements.

Explanation:

Part D

Step1: Determine the number of electrons in Al atom

Al (aluminum) has an atomic number of 13. So, a neutral Al atom has 13 electrons.

Step2: Calculate the number of electrons in \(Al^{3 +}\)

Since \(Al^{3+}\) is formed by losing 3 electrons, the number of electrons in \(Al^{3+}\) is \(13 - 3=10\) electrons.

Step3: Write the electron configuration

The electron configuration of 10 - electron species is \(1s^{2}2s^{2}2p^{6}\). In the required format, it is \(1s^{\wedge}22s^{\wedge}22p^{\wedge}6\)

Part E

Isoelectronic species have the same number of electrons. Noble - gas atoms have a complete valence - shell electron configuration (\(ns^{2}np^{6}\) for \(n\geq2\) or \(1s^{2}\) for \(n = 1\)). The electron configurations of the ions (e.g., \(Li^{+}\) with \(1s^{2}\) and \(Al^{3+}\) with \(1s^{2}2s^{2}2p^{6}\)) are similar to the electron configurations of noble - gas atoms (\(He:1s^{2}\), \(Ne:1s^{2}2s^{2}2p^{6}\)). Group 1A elements have \(ns^{1}\) valence - shell configuration, group 6A have \(ns^{2}np^{4}\), group 7A have \(ns^{2}np^{5}\), and group 2A have \(ns^{2}\) as their valence - shell electron configurations.

Answer:

• Part D: \(1s^{\wedge}22s^{\wedge}22p^{\wedge}6\)
• Part E: noble gases