A) \[N{{a}^{+}}>S{{i}^{4+}}>A{{l}^{3+}}>M{{g}^{2+}}\]
B) \[M{{g}^{2+}}>N{{a}^{+}}>A{{l}^{3+}}>S{{i}^{4+}}\]
C) \[N{{a}^{+}}>M{{g}^{2+}}>A{{l}^{3+}}>S{{i}^{4+}}\]
D) \[S{{i}^{4+}}>A{{l}^{3+}}>M{{g}^{2+}}>N{{a}^{+}}\]
Correct Answer: C
Solution :
\[N{{a}^{+}},M{{g}^{2+}},A{{l}^{3+}}\] and\[S{{i}^{4+}}\]are isoelectronic but nuclear charge per electron is greatest for\[S{{i}^{4+}},\]hence it has smallest size and nuclear charge per electron for\[N{{a}^{+}}\]is smallest, therefore it has largest size.
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