The heat of atomization decreases in order ________.
Li > Na > K > Rb > Cs
The heat of atomization is the energy required to convert one mole of an element from its standard state (usually solid for metals) into individual gaseous atoms. For metallic elements like alkali metals, this energy is primarily needed to overcome the metallic bonds holding the atoms together in the solid lattice.
A higher heat of atomization indicates stronger metallic bonding, meaning more energy is needed to break the bonds and separate the atoms into the gaseous state. Conversely, a lower heat of atomization means weaker metallic bonding.
The alkali metals are Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), and Cesium (Cs), found in Group 1 of the periodic table. As we move down this group, several properties change, which affects the strength of the metallic bond:
Since the strength of the metallic bond decreases down the group from Li to Cs, the energy required to break these bonds (the heat of atomization) also decreases in the same order.
Based on the decreasing strength of metallic bonding down the alkali metal group, the heat of atomization decreases in the following order:
Lithium > Sodium > Potassium > Rubidium > Cesium
$$ \text{Li} > \text{Na} > \text{K} > \text{Rb} > \text{Cs} $$
This means Lithium has the highest heat of atomization among the alkali metals due to its relatively strong metallic bonding (smallest size), while Cesium has the lowest heat of atomization due to its weakest metallic bonding (largest size).