Two atoms have the electron configurations 1s2 2s2 sp6 and 1s2 2s2 2p6 3s1. The first ionization energy of one is 2080 kJ/mol and that of the other is 496 kJ/mol Match each ionization energy with one of the given electron configurations. Justify your choice.
The Correct Answer and Explanation is:
To determine which atom corresponds to each ionization energy, we first need to analyze the given electron configurations and their implications regarding the stability of the outermost electrons.
- Electron Configurations:
- Atom A: (1s^2 \, 2s^2 \, 2p^6)
- Atom B: (1s^2 \, 2s^2 \, 2p^6 \, 3s^1)
Interpretation of Electron Configurations:
- Atom A has a total of 10 electrons in the first two energy levels, filling the (1s), (2s), and (2p) subshells completely, making it a noble gas configuration (neon). This configuration is very stable because all subshells are filled.
- Atom B has 11 electrons, with the additional (3s^1) electron. This configuration suggests that Atom B is sodium (Na), which is in Group 1 of the periodic table, and is less stable due to the presence of a single valence electron in the (3s) subshell.
- Ionization Energies:
- Ionization Energy of 2080 kJ/mol: This high value is indicative of an atom with a stable electron configuration. Atoms with noble gas configurations require significantly more energy to remove an electron because they are energetically favorable.
- Ionization Energy of 496 kJ/mol: This lower value suggests that the electron being removed is relatively easier to ionize, which is characteristic of alkali metals like sodium, where the outer electron is not tightly bound due to its higher energy level and less nuclear charge felt by the outermost electron.
Matching Ionization Energies with Electron Configurations:
- The higher ionization energy (2080 kJ/mol) corresponds to Atom A ((1s^2 \, 2s^2 \, 2p^6)) because it reflects the stability of the filled electron configuration.
- The lower ionization energy (496 kJ/mol) corresponds to Atom B ((1s^2 \, 2s^2 \, 2p^6 \, 3s^1)) as the presence of a single outer electron in a higher energy level makes it easier to remove.
In conclusion, Atom A (noble gas configuration) has a higher ionization energy due to its stable electron arrangement, while Atom B (sodium) has a lower ionization energy due to its single valence electron in the (3s) subshell.