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