A) \[N,\,O,\,F,\,Ne\]
B) \[N{{a}^{+}},M{{n}^{2+}},A{{l}^{3+}},S{{i}^{4+}}\]
C) \[C{{l}^{-}},Ar,C{{a}^{2+}},T{{i}^{4+}}\]
D) \[Be,M{{g}^{2+}},Ca,S{{i}^{2+}}\]
Correct Answer: C
Solution :
\[Cl\] (at.no. = 17), \[C{{l}^{-}}\] (no. of electrons =18) \[Ar\](at.no. = 18), no. of electrons = 18 \[Ca\](at.no.= 20), \[C{{a}^{2+}}\] no. of electrons= 18 \[Ti\](at.no. = 22), \[T{{i}^{4+}}\] no. of electrons =18 \[\therefore \]all these are isoelectronic
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