The periodic table becomes more useful when you can explain a position or trend from an atom’s electrons. Counting occupied shells helps you identify a period. Counting outer-shell electrons helps you recognise a main-group family. Explaining a size or reactivity comparison requires one more step: deciding how nuclear attraction, distance and shielding change.
This lesson uses four neutral atoms labelled A, B, C and D. These letters are exercise labels, not element symbols. Their proton numbers identify real main-group elements, but the anonymised table makes you show the reasoning. The comparisons are theoretical; no reaction or handling of reactive metals is required.
Read the supplied atom table
| Exercise label | Proton number | Total electrons | Supplied shell arrangement |
|---|---|---|---|
| A | 11 | 11 | 2, 8, 1 |
| B | 12 | 12 | 2, 8, 2 |
| C | 17 | 17 | 2, 8, 7 |
| D | 19 | 19 | 2, 8, 8, 1 |
For a neutral atom, the total electron number equals the proton number. Check each supplied arrangement by addition: A has 2 + 8 + 1 = 11 electrons, while D has 2 + 8 + 8 + 1 = 19. This check catches a copied digit before it affects the rest of the answer.
The shell notation records electron counts by occupied shell. It is a simplified model for these selected atoms, not a claim that all shells always hold a maximum of eight electrons. Use the supplied arrangement rather than extending an “eight everywhere” shortcut to unfamiliar elements.
Worked analysis 1: locate each atom
A has three occupied shells, so it is in period 3. Its single outer-shell electron places it in group 1. B also has three occupied shells but two outer electrons, giving period 3, group 2. C has three shells and seven outer electrons, giving period 3, group 17 in the modern 1–18 numbering system.
D has four occupied shells and one outer electron. It belongs in period 4, group 1. A and D therefore share a group, whereas A, B and C share a period. The same-group pair need not have the same total number of electrons; it shares the relevant outer-shell pattern.
Complete partial-table layout: The diagram shows selected columns labelled group 1, group 2 and group 17, with an explicitly marked gap for omitted columns. In period 3, A occupies group 1, B group 2 and C group 17. In period 4, D occupies group 1; other cells are omitted rather than claimed empty. A: 2,8,1. B: 2,8,2. C: 2,8,7. D: 2,8,8,1. No spacing in this simplified diagram represents an atomic radius.
Some older school resources call the main-group halogen column group VII or VIIA. This lesson uses group 17 and states the convention to prevent confusion with modern group 7, a different column. A number written as a group label is not automatically an ionic charge.
For instance, C has seven valence electrons; it is not thereby a +7 ion. The table describes neutral atoms. Predicting a likely simple ion is a separate task about electron gain or loss, and the periodic position of an element does not move when that element forms an ion.
Build a cause chain rather than naming a trend
A useful comparison names the atoms, identifies what stays similar, then explains what changes. Across the selected period-3 atoms, the occupied outer shell is the third shell. Their inner-shell arrangement is 2,8. Nuclear charge increases from A to B to C, while the added outer-shell electrons do not shield one another as effectively as inner electrons do.
Down the selected group-1 pair, D has an additional occupied shell. Its outer electron is generally farther from the nucleus and more shielded by inner electrons than A’s outer electron. The larger proton number alone cannot decide the resulting attraction; distance and shielding must also be considered.
These are qualitative explanations. The simplified shell drawings do not supply measured radii or exact effective nuclear charges. Do not assign numerical attractions by subtracting every inner electron from the proton number and presenting the result as a precise measurement.
Worked analysis 2: compare two atoms in one period
Which is expected to have the smaller atomic radius, A or B? A is the neutral atom with 11 protons; B has 12. Both have three occupied shells, and both have ten inner-shell electrons in the supplied arrangement.
B’s nucleus has a greater positive charge. The extra electron is added to the same outer shell, so it does not introduce a new occupied shell or an equally strong additional inner-shell shield. The outer electrons experience a stronger net nuclear attraction in B. The expected comparison is therefore radius of B less than radius of A.
The explanation is not “B is smaller because it has fewer electrons”: B actually has more electrons. Nor is it enough to write “size decreases across a period” when the task asks why. The same-shell comparison and increased effective attraction explain the direction of the trend for this pair.
Both quantities must be atomic radii defined consistently. Comparing the radius of neutral A with that of a B ion would introduce a different question. If a data table mixes atomic and ionic radii, identify the column before applying any trend.
Worked analysis 3: compare a same-group reactivity pattern
A and D are sodium and potassium respectively. For the familiar comparison of these metals reacting with water under comparable conditions, potassium is more reactive. This is a stated classroom reaction context, not a general ranking for every chemical process.
Both atoms have one outer electron. D’s outer electron occupies the fourth shell rather than the third and is more shielded by inner electrons. Despite D’s greater nuclear charge, that electron is more readily removed. This supports the school-level explanation of increasing reactivity from sodium to potassium in this context.
A complete short answer could read: “D is more reactive in the stated water-reaction comparison. Its outer electron is in a more distant shell with greater shielding, so it is less strongly held and is lost more readily.” The atom labels make the conclusion checkable against the supplied table.
This reasoning does not predict an exact reaction time, flame size or temperature. Those observations depend on quantities, surfaces, conditions and other factors. Do not invent such measurements from the electron arrangements, and do not attempt the reaction to complete this paper exercise.
Worked analysis 4: repair an overgeneralisation
A learner argues: “Every element farther down the periodic table is more reactive because its atoms are bigger.” Identify the problem before replacing the claim.
The statement has no defined family or reaction context. Our result concerns two group-1 metals and their reaction with water. It does not establish a universal rule for nonmetals, noble gases or transition metals. Larger radius and easier electron removal are relevant to this chosen metal comparison, but a different reaction may involve electron gain and other energetic contributions.
A bounded repair is: “For sodium and potassium in the stated comparison, potassium’s more shielded, more distant valence electron helps explain its greater reactivity.” This revised answer says enough to solve the problem without claiming knowledge that the small dataset cannot support.
Three practice tasks with solved answers
Practice 1: verify a position
Neutral atom E has proton number 13 and supplied arrangement 2,8,3. Determine its period and modern group, checking the electron count first.
Answer: The sum is 13, consistent with neutrality. Three occupied shells give period 3. Three outer electrons place this main-group atom in group 13. It is not in modern group 3. It shares a period with A, B and C, but not a group with A or B.
Practice 2: explain another radius comparison
Compare neutral B, with arrangement 2,8,2, and neutral C, with arrangement 2,8,7. Which has the smaller expected atomic radius, and what two structural facts support your answer?
Answer: C is expected to be smaller. Both have three occupied shells and the same inner-shell arrangement, but C has the greater nuclear charge. The additional electrons occupy the same outer shell and do not fully cancel that increased attraction. This comparison concerns neutral atoms, not their common ions.
Practice 3: detect a faulty explanation
A response says: “D is smaller than A because 19 protons attract more strongly than 11; therefore D is less reactive with water.” Correct both conclusions using the supplied arrangements.
Answer: D has an extra occupied shell, greater distance to the outer electron and greater inner-electron shielding. Its atomic radius is larger, and its outer electron is more readily removed. In the sodium–potassium water-reaction comparison, D is more reactive. The original answer considered nuclear charge but left out the structural changes that matter down this group.
Frequently asked questions
How do I find the period from these arrangements?
Count occupied shells in the neutral ground-state arrangements supplied. Three occupied shells place A, B and C in period 3; four place D in period 4.
Do seven outer electrons mean modern group 7?
No. For the main-group atom C, seven outer electrons correspond to group 17. Some older main-group notation uses VII or VIIA, so identify the numbering system.
Does a higher proton number always mean a smaller atom?
No. Across the selected same-period pair, stronger effective attraction explains contraction. Down a group, additional shells and shielding can produce a larger radius despite more protons.
Are atomic and ionic radius interchangeable?
No. An ion has a different electron count from its parent atom. The worked radius comparisons here use neutral atoms consistently.
Does the group number give the charge on an atom?
No. All atoms in the supplied table are neutral. Group position and outer-electron count help interpret bonding patterns, but they are not an automatic ionic charge.
Does reactivity always increase down every group?
No. The worked claim concerns sodium and potassium reacting with water under comparable conditions. It does not establish a universal trend for every family or reaction.
Can the shell model predict an exact reaction rate?
No. It supports a qualitative explanation for this comparison. Exact rates also depend on conditions and cannot be calculated from these arrangements alone.
Sources and further reading
The references below support the scientific background. The cases, calculations and illustrations in this lesson are original. External teaching guidance does not establish current KCSE marking rules or grade placement.
- OpenStax Chemistry 2e: the periodic table
- OpenStax Chemistry 2e: electron configurations
- OpenStax Chemistry 2e: periodic variations
- University of Washington Chemistry: alkali-metal reactivity
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