Periodic Classification and Periodic Properties 2 Question 48
48. Arrange the following in order of their
(i) decreasing ionic size $\mathrm{Mg}^{2+}, \mathrm{O}^{2-}, \mathrm{Na}^{+}, \mathrm{F}^{-}$
(ii) increasing first ionisation energy $\mathrm{Mg}, \mathrm{Al}, \mathrm{Si}, \mathrm{Na}$
(iii) increasing bond length $\mathrm{F}{2}, \mathrm{~N}{2}, \mathrm{Cl}{2}, \mathrm{O}{2}$
$(1985,3 \mathrm{M})$
Show Answer
Solution:
- (i) $\mathrm{Mg}^{2+}, \mathrm{O}^{2-}, \mathrm{Na}^{+}$and $\mathrm{F}^{-}$are all isoelectronic, has 10 electrons each. Among isoelectronic species, the order of size is cation $<$ neutral $<$ anion.
Also, between cations, higher the charge, smaller the size and between anions, greater the negative charge, larger the size. Therefore, the decreasing order of ionic radii : $\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}$
(ii) First ionisation energy increases from left to right in a period. However, exception occur between group 2 and 13 and group 15 and 16 where trend is reversed on the grounds of stability of completely filled and completely half-filled orbitals. Therefore, Ionisation energy (1st) : $\mathrm{Na}<\mathrm{Al}<\mathrm{Mg}<\mathrm{Si}$
(iii) If the atoms are from same period, bond length is inversely proportional to bond order. In a group, bond length is related directly to atomic radius. Therefore,
bond length $\mathrm{N}{2}<\mathrm{O}{2}<\mathrm{F}{2}<\mathrm{Cl}{2}$
