Question

Arrange the following isoelectronic species in order of increasing ionization enthalpy (IE).

$O^{2-},F^{-},{Na}^{+},{Mg}^{2+}$

• A. $O^{2-}<F^{-}<{Na}^{+}<{Mg}^{2+}$
• B. $F^{-}<O^{2-}<{Na}^{+}<{Mg}^{2+}$
• C. $O^{2-}<F^{-}<{Mg}^{2+}<{Na}^{+}$
• D. None of these

Answer 1

The correct option is A $O^{2-}<F^{-}<{Na}^{+}<{Mg}^{2+}$

$O^{2-},F^{-},{Na}^{+},{Mg}^{2+}$ have the same number of electrons $\left(10\right)$. Their radii would be different because of their different nuclear charges. The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Anion with the greater negative charge will have a larger radius. In this case, the net repulsion of the electrons will outweigh the nuclear charge and the ion will expand in size. The order of ionization enthalpy (IE) will be opposite to that of ionic radii as smaller the size, the control of nucleus on outermost electron will be more and hence more energy will be required to remove an electron.

Hence the correct order is A i.e $O^{2-}<F^{-}<{Na}^{+}<{Mg}^{2+}$