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
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).