SCIENCE · INVESTIGATE AND EXPLAIN

Infer a shell boundary from successive ionization evidence

Keep individual ion-removal energies separate from their cumulative cost.

Learning outcome: Compare all three packets and mark the largest adjacent ratio. Capture a step just before and just after each jump, retaining both individual and cumulative energies.

Starting recap

Review the atomic electron ledger, signed energy changes and dimensional conversions. Distinguish a mathematical parameter from a unique material identification.

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Try

For packet1 (selector1), removing two electrons costs 600+1300=1900 kJ/mol; the third costs another7000, for cumulative8900.

Open the experiment

Predict

Does a large third ionization mean the third electron alone costs the sum of all three steps?

Your prediction and reason

Experiment

Compare all three packets and mark the largest adjacent ratio. Capture a step just before and just after each jump, retaining both individual and cumulative energies.

Locate a large successive removal jump

Worked teaching model · synthetic values · Three fictional successive ionization sequences in kJ/mol: [500,4500,7000,9500,12000], [600,1300,7000,9500,12500], [700,1500,2400,12000,15000]. Each term removes one electron from the next gaseous ion. No packet is a real element identification.

Synthetic packet: 1 choice Electrons removed: 2 count

Reset keeps your prediction, captures and practice history.

Axes rescale with settings. Compare numerical values, not screen slopes. Discrete stems show steps, contributors or partial sums; connected samples show the declared parameter relationship, not a measured time trajectory.

Numerical diagram coordinates

successive electron removed	step energy (kJ/mol)
1	600
2	1300
3	7000
4	9500
5	12500

Calculated values under declared assumptions
Quantity	Value	Unit
Current step energy	1300	kJ/mol
Cumulative removal energy	1900	kJ/mol
Largest adjacent ratio	5.384615	ratio
Removal count before largest jump	2	electrons

Successive and cumulative energies answer different questions.

A large increase after a few removals supports a shell-boundary interpretation within the stipulated main-group picture. It does not uniquely establish a real element or prove that an ion with that charge forms in every environment.

Capture contrasting evidence and retain the interpretation limits. These supplied models do not identify an actual material or certify practical laboratory skill.

Successive and cumulative energies answer different questions.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Specify the gaseous process

First ionization removes an electron from a neutral gaseous atom. Second ionization starts with the singly charged gaseous ion, not with another neutral atom. Each successive term therefore refers to a different species. The cumulative energy to make a triply charged ion from a neutral atom is the sum of three terms, not merely the third entry.

Read an original packet

Our fictional packet has sequential energies 600, 1300, 7000, 9500 and 12500 kJ/mol. Two removals cost 1900 kJ/mol; three cost 8900 kJ/mol. Adjacent ratios are about 2.17, 5.38, 1.36 and 1.32. The most pronounced jump occurs after two removals. Under a main-group shell interpretation, that supports two relatively accessible outer electrons followed by removal from a more tightly bound shell.

An inference has assumptions

The packet is not a real element’s spectrum, and the largest ratio is a diagnostic rather than a universal algorithm for every atom. Electron-electron repulsion, penetration and subshell structure matter. Gas-phase ionization costs also do not determine compound stability alone: electron attachment, lattice or solvation energies and chemical environment enter the complete energy account.

Predict before changing the packet

For the packet beginning 500, 4500, 7000, identify the jump before opening the ledger. The largest ratio is nine between the first and second steps; the first removal is relatively inexpensive compared with the next. For the packet beginning 700, 1500, 2400, 12000, the largest jump follows the third removal. Explain the mechanism rather than assigning actual element names to invented numbers.

Transfer to a partial record

A new supplied sequence begins 800, 1700, 2600, 14000 kJ/mol. The cumulative cost for three electrons is 5100 kJ/mol, while the fourth step alone is 14000. A shell break after three removals is consistent with the large jump, but there is no basis here to infer a fifth energy or a particular reaction yield. State both what the packet supports and what it leaves unidentified.

Evidence and safety boundary

Conceptual sources checked 30 October 2026: Periodic ionization trends. All prose, numerical packets and diagrams are original; no textbook exercises or images are reproduced. The activity is supplied mathematical or synthetic evidence, not a performed materials experiment. No unknown substances, heating, high voltage or chemical synthesis are requested. Retain prior predictions and a changed-case explanation; numerical agreement alone does not certify independent understanding or practical competence.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Explain why a shell-boundary inference does not prove a unique element or the charge of every stable compound.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Explain periodic size through competing model parameters Next: Distinguish bond depth, equilibrium and restoring slope

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