The electron configuration of \(_{22}X^{2+}\) ion is
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Correct Answer: Option D
Explanation:
\(_{22}X^{2+}\) has 22 electrons but has been ionized to 2+ by giving off 2 electrons. Hence, the ion \(_{22}X^{2+}\) has 20 electrons.
The electronic configuration = 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 4s\(^2\).
\(_{22}X^{2+}\) has 22 electrons but has been ionized to 2+ by giving off 2 electrons. Hence, the ion \(_{22}X^{2+}\) has 20 electrons.
The electronic configuration = 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 4s\(^2\).