Choose the element with the higher first ionization energy from each pair. d. S or Sn

Verified step by step guidance

1. Understand the concept of first ionization energy, which is the energy required to remove the most loosely bound electron from a neutral atom in its gaseous state.
2. Recall the periodic trend for ionization energy: it generally increases across a period (from left to right) and decreases down a group (from top to bottom) in the periodic table.
3. Locate sulfur (S) and tin (Sn) on the periodic table. Sulfur is in the third period and Group 16 (chalcogens), while tin is in the fifth period and Group 14 (carbon group).
4. Compare their positions: Sulfur is higher up and to the right of tin in the periodic table.
5. Conclude that sulfur, being higher and further to the right than tin, generally has a higher first ionization energy due to its smaller atomic radius and greater effective nuclear charge.

Key Concepts

Ionization Energy

Ionization energy is the energy required to remove an electron from a gaseous atom or ion. It is a key factor in determining an element's reactivity and is influenced by the atomic size and the effective nuclear charge. Generally, ionization energy increases across a period and decreases down a group in the periodic table.

Periodic Trends

Periodic trends refer to the predictable patterns observed in the properties of elements as you move across or down the periodic table. For ionization energy, elements on the right side of the table tend to have higher ionization energies than those on the left, while elements higher up in a group have higher ionization energies than those lower down due to decreased electron shielding.

Comparison of Elements

When comparing elements like sulfur (S) and tin (Sn), it is essential to consider their positions in the periodic table. Sulfur is located in period 3 and group 16, while tin is in period 5 and group 14. This difference in period and group indicates that sulfur will generally have a higher ionization energy than tin due to its smaller atomic radius and higher effective nuclear charge.