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Assertion (A) : $\mathrm{Na}^{+}$and $\mathrm{Mg}^{2+}$ ions are isoelectronic but the ionic radius of $\mathrm{Na}^{+}$is greater than that of $\mathrm{Mg}^{2+}$.
Reason (R) : The effective nuclear charge of $\mathrm{Na}^{+}$ion is less than that of $\mathrm{Mg}^{2+}$ ion.
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Reason (R) : The effective nuclear charge of $\mathrm{Na}^{+}$ion is less than that of $\mathrm{Mg}^{2+}$ ion.
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Verified Answer
The correct answer is:
Both (A) and (R) are correct and (R) is correct
explanation of (A).
explanation of (A).
$\mathrm{Na}^{+}$and $\mathrm{Mg}^{2+}$ are isoelectrionic species having $10 e^{-}$ each. But $\mathrm{Mg}^{2+}$ has 12 protons while $\mathrm{Na}^{+}$has 11 protons. Hence effective nuclear charge is greater in case of $\mathrm{Mg}^{2+}$.
ionic radius $\propto \frac{1}{\text { effective nuclear charge }}$
ionic radius $\propto \frac{1}{\text { effective nuclear charge }}$
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