Question

write the complete ground-state electron configuration of cr²⁺. for multi-digit superscripts or coefficients, use each number in succession.

Explanation:

Step1: Determine the electron number of Cr atom

Chromium (Cr) has an atomic number of 24. So, a neutral Cr atom has 24 electrons. The electron configuration of neutral Cr is \([Ar]3d^54s^1\) due to the stability of half - filled \(3d\) subshell.

Step2: Consider the ion formation

\(Cr^{2+}\) is formed by losing 2 electrons. Electrons are lost from the outermost shell first. The \(4s\) orbital is considered the outermost (higher - energy orbital in the context of electron loss for transition metals). So, we lose 1 electron from \(4s\) and 1 electron from \(3d\).

Step3: Write the electron configuration of \(Cr^{2+}\)

After losing 2 electrons, the electron configuration of \(Cr^{2+}\) is \([Ar]3d^4\). Expanding \([Ar]\) (which has 18 electrons), we get \(1s^22s^22p^63s^23p^63d^4\)

Answer:

\(1s^22s^22p^63s^23p^63d^4\)