Give the noble gas core ground state electron configuration for Mo3+
The Correct Answer and Explanation is :
The noble gas core ground state electron configuration for the molybdenum ion Mo³⁺ is:
[ \text{[Kr]} \, 4d^5 ]
Explanation:
To understand the electron configuration for Mo³⁺, we first need to look at the neutral molybdenum atom (Mo). Molybdenum has an atomic number of 42, meaning it has 42 electrons in its neutral state. The electron configuration of neutral molybdenum is:
[ \text{[Kr]} \, 5s^2 \, 4d^4 ]
This configuration shows that molybdenum has 2 electrons in the 5s subshell and 4 electrons in the 4d subshell, following the order of filling based on the Aufbau principle, which prioritizes lower energy levels before filling higher ones.
When molybdenum is ionized to form Mo³⁺, it loses three electrons. The typical order of electron loss follows the principle that electrons are removed from the outermost shell first, which is the 5s subshell in this case. Therefore, Mo³⁺ loses the 2 electrons from the 5s subshell first:
[ 5s^2 \to 0 ]
Next, one electron is removed from the 4d subshell. This results in the following electron configuration for Mo³⁺:
- Starting configuration: [ \text{[Kr]} \, 5s^2 \, 4d^4 ]
- After losing 2 from 5s and 1 from 4d: [ \text{[Kr]} \, 4d^5 ]
Thus, Mo³⁺ has a noble gas core electron configuration of [ \text{[Kr]} \, 4d^5 ]. This ionization leads to a more stable electron configuration, particularly for transition metals, where half-filled or fully filled d subshells are energetically favorable due to exchange energy and symmetry considerations. Molybdenum in the +3 oxidation state is often encountered in various chemical reactions and compounds, particularly in catalysis and materials science.