Sulfur (S) is in Group 16. When it forms an ion, what charge would you predict, and why?
AS²⁺, because sulfur can lose 2 electrons from its outer shell
BS²⁻, because sulfur needs 2 more electrons to complete its outer shell and reach the argon configuration
CS⁶⁺, because sulfur has 6 valence electrons that can all be removed
DS¹⁻, because sulfur needs only 1 electron to fill the next available orbital
Sulfur has 6 valence electrons and needs 2 more to reach 8 and achieve the electron configuration of argon (the next noble gas). The driving force is stability — gaining 2 electrons fills the outer shell. While sulfur could theoretically lose electrons (forming S²⁺ or higher), doing so would require breaking the noble gas core, which requires far more energy than is gained. Main-group nonmetals on the right side of the table characteristically gain electrons rather than lose them. Option D is wrong — sulfur has 6 valence electrons and needs 2, not 1, to complete the shell.
Question 2 Multiple Choice
Why does sodium form Na⁺ and not Na²⁺, even though a higher positive charge might seem to mean more electron loss?
ASodium only has one valence electron, so it physically cannot lose a second electron
BLosing the first electron gives sodium the stable neon configuration; losing the second would require breaking into a filled, stable inner shell at enormously greater energy cost
CNa²⁺ would be too small to form stable ionic bonds with anions
DSodium's ionization energy drops to zero after the first electron is removed
Sodium's electron configuration is [Ne]3s¹. Removing the single 3s electron gives Na⁺ the neon configuration — a stable, filled outer shell. Removing a second electron would require pulling from the 2p subshell, which is part of the filled noble-gas core. The second ionization energy is enormously higher than the first because of this stability. Ion formation stops when the next noble gas configuration is reached; further removal dismantles a stable shell rather than shedding a lone outer electron. Option A is incorrect — sodium has more electrons, but removing them is energetically prohibitive.
Question 3 True / False
The charge of a main-group ion directly tells you how many electrons were gained or lost, and this number is determined by how far the neutral atom is from a noble gas configuration.
TTrue
FFalse
Answer: True
This is the core principle of ion formation for main-group elements. The number of electrons transferred is dictated by the shortest path to a filled outer shell — losing a few electrons (metals, left side) or gaining a few (nonmetals, right side). This directly predicts ion charges: Group 1 → 1+, Group 2 → 2+, Group 16 → 2−, Group 17 → 1−. The charge is not arbitrary; it is the count of electrons transferred, and those electrons are transferred because doing so delivers the atom to a more stable configuration.
Question 4 True / False
Atoms form ions because they are attracted to the opposite charges of nearby ions in an ionic compound.
TTrue
FFalse
Answer: False
Ion formation is driven by the energetic stability of reaching a noble-gas electron configuration — an intrinsic property of the atom's electron structure, not a response to nearby ions. Sodium has the same tendency to lose one electron whether or not a chloride ion is nearby. The attraction between oppositely charged ions (ionic bonding) stabilizes the compound after ions form, but it is not the cause of electron transfer. Confusing the stability of the resulting bond with the driving force for electron loss or gain is a common error.
Question 5 Short Answer
Why does aluminum form Al³⁺ rather than Al²⁺, and what general principle does this illustrate about main-group ion charges?
Think about your answer, then reveal below.
Model answer: Aluminum has 3 valence electrons (configuration [Ne]3s²3p¹). Removing all three gives it the stable neon configuration. Stopping at Al²⁺ would leave it with one 3s electron still in the outer shell — less stable than a fully stripped outer shell reaching neon. The general principle: main-group ions form by losing or gaining exactly as many electrons as needed to reach the electron configuration of the nearest noble gas. The charge equals the number of electrons transferred, which is determined by the most energetically favorable path to a stable (filled-shell) configuration.
The ionization energy profile tells the story: the first three ionization energies of aluminum increase gradually (removing 3p¹, then 3s²), but the fourth jumps enormously because it starts removing from the neon core. This energy cliff explains why Al³⁺ is strongly preferred over Al⁴⁺ — further removal would dismantle a stable noble-gas configuration.