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The electron configuration of \(_{22}X^{2+}\) ion is

Chemistry
JAMB 2000

The electron configuration of \(_{22}X^{2+}\) ion is

  • A. 1s2 2s22p6 3s2 3p6 4s2 3d2
  • B. 1s2 2s22p6 3s2 3p6 4s2 3d1
  • C. 1s2 2s22p6 3s2 3p6
  • D. 1s2 2s22p6 3s2 3p6 3d2
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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\(^6\) 3d\(^2\) 

the general trend for the first-row transition metals is that they lose electrons from their 4s orbitals first before 3d when forming ions. but this is not true for chromium and copper due to stability related to half-filled or completely-filled 3d orbitals


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