How do the trends in ionization energy correlate with the trends in atomic radii?
The correct answer and explanation is:
Correct Answer: Ionization energy increases as atomic radius decreases.
Explanation:
Ionization energy refers to the amount of energy required to remove an electron from a gaseous atom or ion. Atomic radius, on the other hand, is the distance from the nucleus of an atom to the outermost electron shell. These two properties are inversely related across the periodic table due to the influence of nuclear charge and electron shielding.
Across a period (from left to right on the periodic table), atomic radius decreases. This is because the number of protons in the nucleus increases, which strengthens the electrostatic attraction between the positively charged nucleus and the negatively charged electrons. Although electrons are being added across the period, they enter the same energy level and do not effectively shield each other from the increasing nuclear charge. As a result, the outer electrons are pulled closer to the nucleus, leading to a smaller atomic radius. With the electrons more tightly held, more energy is required to remove one, so ionization energy increases.
Down a group (from top to bottom), atomic radius increases. This occurs because new electron shells are added, increasing the distance between the nucleus and the outermost electrons. Additionally, inner electrons shield the outer electrons from the full attractive force of the nucleus, weakening the pull on the outermost electrons. Therefore, it becomes easier to remove an electron, and ionization energy decreases.
This inverse relationship means that elements with a large atomic radius tend to have low ionization energy, and those with a small atomic radius tend to have high ionization energy. For example, alkali metals like sodium have large radii and low ionization energies, while noble gases like neon have small radii and high ionization energies. This trend reflects the balance between nuclear attraction and electron distance from the nucleus.