Which of the following electron configurations represent an excited state? [He] 2s^1 2p^5 [Kr] 4d^10 5s^2 5p^1 [Ar] 3d^10 4s^2 4p^5 [Ne] 3s^2 3p^2 4s^1
The Correct Answer and Explanation is:
The correct electron configuration that represents an excited state is [Ne] 3s² 3p² 4s¹.
Explanation:
In an atom, electrons fill orbitals following the Aufbau principle, which dictates that electrons should fill the lowest energy orbitals first. For example, in the ground state configuration, the 3p orbital would fully fill with two electrons, and the 4s orbital would fill before the 3d orbital.
However, when an atom absorbs energy, one or more electrons may jump to a higher energy level or an excited state. In this case, [Ne] 3s² 3p² 4s¹, one of the electrons from the 3p orbital is promoted to the 4s orbital, which is an excited state because the 4s orbital is typically filled after the 3p orbital.
Let’s break down the configurations:
- [He] 2s¹ 2p⁵: This is a configuration that could be for a fluorine atom (F). The 2p orbital should have 6 electrons in the ground state. Therefore, this configuration is incorrect because it has one fewer electron than expected.
- [Kr] 4d¹⁰ 5s² 5p¹: This is the configuration of a neutral element like indium (In). The 5p orbital is only partially filled, and this configuration is normal, not excited.
- [Ar] 3d¹⁰ 4s² 4p⁵: This corresponds to an atom like bromine (Br), which is in its ground state, with a full 4s and 4p subshell. No electron has been promoted.
- [Ne] 3s² 3p² 4s¹: This is the excited state of an atom like silicon (Si). The 3p orbital is partially filled in the ground state with two electrons, but here one electron has been promoted to the 4s orbital, which is not the normal filling order, indicating an excited state.
Thus, the configuration [Ne] 3s² 3p² 4s¹ represents an excited state because it shows an electron in an orbital (4s) that would typically not be filled before the 3p orbital.
