An alkali metal in row 4 has lower ionization energy than another alkali metal in row two because of the greater size of its atom.
Alkali metals belong to the first column of the periodic table. They readily lose their one valence electron in the outermost shell to form ionic compounds with the nonmetals. Lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr) are alkali metals.
Ionization energy means the measure of the capability of a neutral atom to lose or donate its electrons for a chemical reaction. It is an endothermic process that requires energy to lose electrons. Ionization energy in a periodic table increases from left to right while decreases when we move from top to bottom. This is all because of the size of an atom.
The valence electron of an alkali metal in row 4 is far away from the nucleus of the atom and experiences the least forces of attraction. It can readily lose its valence electron in the outermost shell without the requirement of greater energy. Consequently, the alkali metal in row 4 has low ionization energy.
On the other hand, the alkali metal in row 2 has a smaller size. Its valence electron experiences a larger force of attraction from the nucleus. It, therefore, requires greater ionization energy to lose the electron. This means that the alkali metal in row 2 has high ionization energy.
To know more about ionization energy, refer to the following link:
brainly.com/question/20658080
#SPJ4