Periodic Trends and Chemical Reactivity:
The physical and chemical properties of elements depend on the valence shell electronic configuration as discussed earlier. The elements on the left side of the periodic table have less ionisation energy and readily loose their valence electrons. On the other hand, the elements on right side of the periodic table have high electron affinity and readily accept electrons. As a consequence of this, elements of these extreme ends show high reactivity when compared to the elements present in the middle. The noble gases having completely filled electronic configuration neither accept nor lose their electron readily and hence they are chemically inert in nature.
The ionisation energy is directly related to the metallic character and the elements located in the lower left portion of the periodic table have less ionisation energy and therefore show metallic character. On the other hand the elements located in the top right portion have very high ionisation energy and are non-metallic in nature.
Let us analyse the nature of the compounds formed by elements from both sides of the periodic table. Consider the reaction of alkali metals and halogens with oxygen to give the corresponding oxides.
4 Na + O2 → 2 Na2O
2 Cl2 + 7 O2 → 2 Cl2O7
Since sodium oxide reacts with water to give strong base sodium hydroxide, it is a basic oxide. Conversely Cl2O7 gives strong acid called perchloric acid upon reaction with water So, it is an acidic oxide.
Na2O + H2O → 2NaOH
Cl2O7 + H2O → 2 HClO4
Thus, the elements from the two extreme ends of the periodic table behave differently as expected.
As we move down the group, the ionisation energy decreases and the electropositive character of elements increases. Hence, the hydroxides of these elements become more basic. For example,let us consider the nature of the second group hydroxides:
Be(OH)2 amphoteric; Mg(OH)2 weakly basic; Ba(OH)2 strongly basic Beryllium hydroxide reacts with both acid and base as it is amphoteric in nature.
Be(OH)2 + HCl → BeCl2 + 2H2O
Be(OH)2 + 2 NaOH → Na2BeO2 + 2H2O
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