an atom x has three electrons in its outermost shell. which ion will most likely be formed by x a. x2– b. x3– c. x2 d. x3

An atom X with three electrons in its outermost shell most likely forms the ion X³⁺, written as option d (x3 in the shorthand). Atoms lose or gain electrons to reach a stable outer shell, and X sits three electrons past a stable configuration. Losing those three valence electrons is energetically cheaper than gaining five to fill the shell. So X tends to shed three electrons, leaving a positive charge of +3. The two negative options (x2– and x3–) would require a metal to gain electrons, which is not its tendency.

This answer rests on electron counting, not on a single rule. Read the mechanism below to see why the count of three valence electrons points to a +3 ion, why charge is written after the number in real notation, and where the reasoning can bend.

Valence electron count drives ion charge

Electrons occupy shells in order of increasing energy, and the outermost shell is where chemistry happens. For a main-group atom, the number of electrons in that outer shell equals its group number or that group minus ten. A count of three places X in group 13 of the periodic table, among boron, aluminum, gallium, indium, and thallium.

The driving force is the stability of a filled or empty outer shell. Atoms do not "want" electrons in a psychological sense. They settle into configurations that lower their overall energy, and a complete shell does that. An atom with three valence electrons faces two routes to a closed configuration. It can lose three electrons to expose the filled shell beneath, or gain five to complete the current one.

Losing three costs less than gaining five for two reasons. First, fewer electron transfers mean less energy spent. Second, a positive ion that forms is stabilized by the surrounding atoms or ions in most real environments. The empty outer shell left behind is itself a stable arrangement, matching the noble-gas configuration of the period above. That is why the +3 outcome dominates.

Charge notation matters here. Chemists write the number first and the sign second, so the three-plus ion is X³⁺. The answer choice "x3" is shorthand for that, while "x3–" is a three-minus ion. The sign flips the meaning, so read the negative options as anions and the positive one as the cation.

The electron count alone does not guarantee the ion. A thin explainer stops at "three valence electrons, so +3." That is the right first approximation, and it answers the exam question. But the same count predicts a different behavior in some elements, and knowing where the shortcut fails is what separates real understanding from memorized rules.

Quantified context: charges, sizes, and energies

Three is a specific number, and it maps to real, measurable properties. Lose three electrons and the ion's radius shrinks sharply because the remaining electrons feel a higher effective nuclear charge. A neutral aluminum atom has a radius of roughly 143 picometers. The Al³⁺ ion collapses to about 54 picometers, a drop of roughly 60 percent, because three positive charges now pull on fewer electrons.

Ionization energy quantifies the cost of removal. For aluminum, the first ionization energy is about 578 kilojoules per mole. The second is roughly 1,817, and the third is about 2,745. The rising values show that each successive electron is harder to remove. The fourth would jump dramatically higher because it would break into the filled inner shell, which is why aluminum stops at +3 and does not reach +4 under normal chemistry.

Charge density scales with the ratio of charge to radius. A +3 ion packed into a small radius produces a dense pocket of positive charge. That high charge density lets X³⁺ polarize nearby molecules and bind tightly to negative ions or electron-rich atoms. This is why group 13 metals form strong bonds and why their compounds often behave as acids rather than bases.

A rough energy balance explains the preference. Removing three electrons costs the sum of three ionization energies. Gaining five electrons would require five electron affinities, and the last few would be strongly endothermic because you would be forcing electrons onto an already negative ion. The arithmetic favors the +3 route in nearly every case. For perspective on shell structure, three valence electrons sit one short of a half-filled subshell, which is part of why group 13 tends toward +3 rather than +1 in many environments.

One counterintuitive detail: in group 13, the lighter and heavier members behave differently. Boron, at the top, is a metalloid that forms mostly covalent compounds and rarely exists as a free B³⁺ ion in solution. The heavier members like thallium can show a +1 state because of an effect called the inert pair, where the outer s electrons resist removal. So the "three valence electrons means +3 ion" rule is strongest for aluminum and gallium and weakest at the extremes of the group.

Why the +3 state usually wins over +1

Two competing stable states exist for many group 13 elements. The +1 state removes the single outermost p electron and leaves two s electrons behind. The +3 state removes all three valence electrons. Which one forms depends on how tightly the s electrons are held.

For light group 13 elements, the +3 state dominates because the energy gap between the s and p electrons is modest and removing all three reaches a clean noble-gas configuration. For heavy elements like thallium, relativistic effects pull the s electrons close to the nucleus and make them hard to remove. The extra stability of that inert pair makes the +1 state competitive and sometimes preferred. Lead and bismuth, their heavier neighbors, show the same effect more strongly.

The environment also steers the outcome. In a solid crystal or a liquid solution, ions are surrounded by other charges and solvent molecules. A highly charged +3 ion attracts a tight shell of solvent or anions, which lowers its energy and stabilizes the state. In a vacuum, the bare +3 ion is expensive to create, but in condensed phases the surrounding stabilization pays most of the cost. This is why the exam's plain "which ion forms" expects the +3 answer, even though a gas-phase experiment at high energy could produce other states.

Temperature and phase add another layer. At high temperatures, the first electron comes off easily, and the ion can lose more electrons than it would at room temperature. This is why plasmas and high-temperature vapors show multiple charge states at once. A single answer like X³⁺ assumes ordinary chemistry near room temperature, not an extreme environment. State the assumption when you reason about it.

Comparison of possible charges

The four answer choices describe four different ions with different stability. The table ranks them for a typical group 13 metal under ordinary conditions.

The two anions (a and b) are essentially ruled out for a metallic X because metals have low electron affinity and low electronegativity. Forcing extra electrons onto the atom costs energy and does not reach a closed shell. The X²⁺ option is real for some heavy elements under specific conditions, but it gives up the clean closure that the +3 state achieves. The X³⁺ answer wins because it matches both the electron count and the energy arithmetic.

Notice that option c can confuse students who see "2" and think "second shell" or "two electrons left." The correct reading is two electrons lost, leaving a +2 charge. The sign and magnitude must be read together. This is a common trap in multiple-choice electron counting.

Red flags that signal a wrong answer

Watch for these mistakes when working the problem. Each one hides a unit or sign error that flips the answer.

Choosing an anion (x2– or x3–) because the atom "gains" something. Metals lose electrons; nonmetals gain them. Dropping the sign on the charge and writing "x3" as if it means the same as x3–. The sign is the whole answer. Counting only the shell number instead of the valence electrons inside it. Forgetting that the +3 state assumes a group 13 metal, not a nonmetal. Treating a single valence count as a universal rule and ignoring inert-pair elements like thallium. Inverting the notation and putting the number after the sign, which produces a different ion.

Practical actions for checking any ion problem

Apply a fixed sequence when a question asks which ion forms. Start by locating the element in the periodic table and reading off its group. Count the valence electrons from that group number. Decide whether reaching a closed shell needs losing or gaining electrons, and count the fewer moves. Write the charge with the number first and the sign second. Verify the sign against the element's metal or nonmetal character.

Use a threshold check. If the valence count is three or fewer, losing the valence electrons usually reaches a stable configuration first, so expect a positive ion. If the count is five or more, gaining electrons to fill the shell tends to dominate, so expect a negative ion. Counts of four sit at a balance point and often depend on the specific element. For three valence electrons, move toward a +3 cation unless the element is boron or thallium, which have special behavior.

Write the answer in both styles. The shorthand "x3" and the formal "X³⁺" mean the same thing only if the sign is positive. State the charge explicitly and note the electrons lost. Then check the state. Are you assuming room-temperature condensed-phase chemistry? If not, qualify the answer because high temperatures can change the charge state.

Practice distinguishing the four options by drawing them. Mark a circle for the nucleus, add three valence dots for the neutral atom, and then show the loss of three to reach an empty outer shell. The picture makes the +3 result obvious and prevents sign confusion. Do this once and the rule sticks.

Limitations and what the count cannot tell you

The electron-count rule answers the exam question, but it carries real limits. It predicts the most likely ion under ordinary conditions, not the only possible ion. Real elements show multiple oxidation states, and the dominant one depends on the compound, the solvent, and the energy available. Aluminum forms +3 almost always, but gallium can show +1 in some compounds, and thallium often prefers +1 in solution.

The rule also says nothing about bonding. The +3 ion in a formula does not guarantee that a free, isolated X³⁺ ion exists. In many compounds, the element shares electrons covalently rather than sitting as a bare ion. Boron and aluminum form mostly covalent bonds in many molecules, so the "+3 state" is a bookkeeping charge, not always a free cation. Treat the charge as a formal assignment and check the structure when it matters.

Finally, conditions change outcomes. High temperature, high pressure, or a strong ligand can shift the preferred charge. The answer X³⁺ is the most probable result for a group 13 metal near room temperature in a typical chemical environment. State that assumption when you explain the answer, and you avoid overclaiming. The count of three valence electrons is the starting point, and the energy balance plus environment decide the final state.