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Q. Assertion $Na^{+}$ and $Al^{3+}$are isoelectronic but the magnitude of ionic radius of $Al^{3+}$ is less than that of $Na^{+}$.
Reason The magnitude of effective nuclear charge of the outer shell electrons in $Al^{3+}$ is greater than that in $Na^{+}$.

AIIMSAIIMS 2013

Solution:

In case of isoelectronic species, effective nuclear charge increases with increase in atomic number, $Z$ (with increase in protons) This results in decrease in size Hence. ionic radius of $Al ^{3+} < Na ^{+}$