Which element requires the least amount of energy to remove the loosely held electron from a gaseous atom in the ground state?
a. bromine
b. calcium
c. sodium
d. silver
How do you know?
To determine which element requires the least amount of energy to remove a loosely held electron from a gaseous atom in the ground state, we need to consider the concept of ionization energy. Ionization energy refers to the energy required to remove an electron from an atom.
The general trend for ionization energy is that it increases across a period from left to right on the periodic table, and decreases down a group from top to bottom. This is due to the change in atomic radius and the increasing positive charge of the nucleus.
In the given options, bromine (Br) is the third element in Group 17 (also known as the halogens), calcium (Ca) is the second element in Group 2, sodium (Na) is the first element in Group 1 (also known as the alkali metals), and silver (Ag) is in the transition metals category.
Based on the periodic trend mentioned earlier, we can conclude that:
- The halogens have higher ionization energies compared to the alkali metals.
- The alkali metals have relatively low ionization energies.
Therefore, among the given options, sodium (Na) would require the least energy to remove the loosely held electron from a gaseous atom in the ground state. Hence, the correct answer is c. sodium.