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 4s2
jamb 2000chemistry

The correct answer is D. 1s2 2s22p6 3s2 3p6 4s2

\(_{22}X^{2+}\) 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\(^2\).

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