A) s-electrons in \[Mg(Z=12)\]
B) p-electrons in \[Cl(Z=17)\]
C) d-electrons in \[Fe(Z=26)\]
D) p-electrons in \[Ne(Z=10)\]
Correct Answer: B
Solution :
Electronic configuration of \[F{{e}^{2+}}\]is \[[Ar]3{{d}^{6}}4{{s}^{0}}.\] \[\therefore \]Number of electrons = 6 \[Mg-1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}\](6d electrons) It matches with the 6d electrons of \[F{{e}^{2+}}\] \[Cl-1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}\,3{{s}^{2}}3{{p}^{5}}\] (11p elections) It does not match with the 6d electrons of\[F{{e}^{2+}}\]. \[Fe-[Ar]3{{d}^{6}}4{{s}^{2}}\] (6d electrons) It matches with the 6d electrons of \[F{{e}^{2+}}.\] \[Ne-1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}\](6p electrons) It matches with the 6d electrons of \[F{{e}^{2+}}.\] Hence, \[Cl\] has 11p electrons which does not matches in number with 6d electrons of \[F{{e}^{2+}}.\]
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