Gr 11 · Atomic Structure · Deep Study

Understanding Atoms

Everything in chemistry follows from atomic structure. This guide builds from subatomic particles through electron configuration, periodic trends, and ion formation — with the reasoning behind every pattern.

5 sections Worked examples with steps Checkpoint questions Intuition-first explanations
1
Subatomic Particles
The building blocks that define every element

All matter is made of atoms. Every atom has a tiny, dense nucleus containing protons and neutrons, surrounded by electrons in shells. The number of protons is not just a label — it is the element. Swap the proton count and you have a completely different element with different chemistry.

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Why protons define the element:
The number of protons determines the nuclear charge, which sets how many electrons the neutral atom has, which determines every chemical property. Change one proton and the entire electron configuration restructures — you have a new element. Neutrons add mass and can affect nuclear stability (isotopes), but they barely change chemical behaviour because chemistry happens in the electron shells.
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Core definitions:
Z = atomic number = number of protons (identifies the element)
A = mass number = protons + neutrons
neutrons = A − Z
Isotopes = same Z, different A (different number of neutrons)
Atomic number
Z = protons
Mass number
A = p + n
Neutrons
n = A − Z
Electrons (neutral)
e− = Z
p+ p+ p+ n0 Nucleus Z=3 A=4 n=1 e- e- e- Electron shells (orbitals outside nucleus) Protons (p+) Neutrons (n0)
Atom with Z=3 (Li-like): nucleus contains protons (red, charge +1) and neutrons (grey, charge 0). Electrons (blue) occupy shells outside the nucleus.
★ Easy
Identifying an atom from Z and A
Element X has mass number A=35 and atomic number Z=17. Identify the element and find neutrons and electrons.
Show solution
1
Identify the element from Z
Z = 17 → Chlorine (Cl)
2
Find neutrons using A − Z
neutrons = A − Z = 35 − 17 = 18
3
Electrons for a neutral atom
electrons = Z = 17
Answer: Cl-35. Neutrons = 18. Electrons = 17.
★★ Intermediate
Average atomic mass from isotopes
Cl-35 (abundance 75.76%) and Cl-37 (abundance 24.24%) are chlorine's two stable isotopes. Calculate the average atomic mass.
Show solution
1
Both are isotopes of chlorine
Z = 17 for both.
Cl-35: neutrons = 18. Cl-37: neutrons = 20.
2
Weighted average formula
avg mass = (mass × abundance) + (mass × abundance)
3
Calculate
= (35 × 0.7576) + (37 × 0.2424)
= 26.52 + 8.97
= 35.48 u
4
Interpret: this matches the periodic table

The periodic table shows Cl = 35.45 u — a decimal, never a whole number, because it is a weighted average of isotopes.

Answer: Average atomic mass ≈ 35.48 u
Checkpoint 1
a) An atom has Z=8 and A=16. What element is it? How many neutrons and electrons does it have?
b) Two atoms belong to the same element but have different mass numbers. What specifically differs between them?
c) Why do isotopes have nearly identical chemical properties?

a) Z=8 is Oxygen (O). Neutrons = A − Z = 16 − 8 = 8. Electrons = Z = 8.

b) They have different numbers of neutrons (and therefore different mass numbers A). These are isotopes — same element, same proton count, different neutron count.

c) Chemical properties are determined by electron configuration, which depends on the number of electrons (= number of protons = Z). Isotopes have the same Z and therefore the same electron arrangement and virtually identical chemical behaviour. Neutrons affect nuclear stability and mass, not chemistry.

2
Electron Configuration
Why electrons fill orbitals in a specific order — and what that means for bonding

Electrons do not randomly occupy space around the nucleus. They fill orbitals in a precise order governed by energy. The outermost electrons (valence electrons) determine how an atom interacts with other atoms — bonding, reactivity, and ion formation all stem from the valence electron count.

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Why fill lowest energy first?
Systems naturally minimise energy. An electron in a higher-energy orbital is less stable. Left to find its own level, it will drop to the lowest available orbital. The rules (Aufbau, Pauli, Hund) are just the systematic description of electrons doing exactly this. The one exception — 4s filling before 3d — happens because 4s is actually lower in energy than 3d for most atoms.
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The three rules:
Aufbau: fill lowest energy orbitals first. Order: 1s 2s 2p 3s 3p 4s 3d 4p 5s...
Pauli exclusion: max 2 electrons per orbital; they must have opposite spins.
Hund's rule: when orbitals of equal energy exist (e.g. three 2p orbitals), place one electron in each before pairing any.
Energy Level Diagram — Electron Filling Order Energy 1s 2s 2p Hund: 1 e- per box first 3s 3p 4s 4s fills before 3d (lower energy) 3d
Energy level diagram: each box is one orbital (max 2e−). Arrows show electrons with spin up (↑) or spin down (↓). Note 4s is lower energy than 3d for most elements, so it fills first.
★ Easy
Writing the configuration for Mg
Write the full and shorthand electron configuration for magnesium (Z=12).
Show solution
1
Fill orbitals in order for 12 electrons
1s: 2e− → 2s: 2e− → 2p: 6e− → 3s: 2e−
Total so far: 2+2+6+2 = 12 ✓
2
Full configuration
1s² 2s² 2p&sup6; 3s²
3
Noble gas shorthand (replace 1s²2s²2p&sup6; with [Ne])
[Ne]3s²
4
Identify valence electrons

Outermost shell is n=3. There are 2 electrons in 3s. Mg has 2 valence electrons — hence it is in Group 2.

Answer: 1s²2s²2p&sup6;3s²   or   [Ne]3s²
★★ Intermediate
Iron and its ions
Write configurations for Fe (Z=26), Fe²+, and Fe³+. Explain which ion is more stable and why.
Show solution
1
Fe (Z=26) — neutral
[Ar]3d&sup6;4s²    (4s fills before 3d per Aufbau)
2
Fe²+ — loses 4s electrons first
Remove 2 electrons from 4s (outermost shell loses first)
[Ar]3d&sup6;
3
Fe³+ — loses one more from 3d
Remove 1 more electron from 3d
[Ar]3d&sup5;
4
Which is more stable?

[Ar]3d&sup5; is a half-filled d subshell (one electron in each of 5 orbitals). Half-filled subshells have extra stability because electrons are evenly spread, minimising repulsion. Fe³+ with [Ar]3d&sup5; is actually more stable than Fe²+ with [Ar]3d&sup6;.

Answer: Fe: [Ar]3d&sup6;4s²  |  Fe²+: [Ar]3d&sup6;  |  Fe³+: [Ar]3d&sup5; (more stable, half-filled d)
Checkpoint 2
a) Write the full electron configuration for phosphorus (Z=15).
b) Write the shorthand (noble gas) configuration for calcium (Z=20).
c) Fe³+ has configuration [Ar]3d&sup5; rather than [Ar]3d&sup6;. Why does Fe lose its 4s electrons before losing 3d electrons?

a) P (Z=15): 1s² 2s² 2p&sup6; 3s² 3p³ — total = 2+2+6+2+3 = 15 ✓. Note: the three 3p electrons each occupy a separate orbital (Hund's rule).

b) Ca (Z=20): [Ar]4s². Ar has 18 electrons, then 4s fills with 2 more.

c) When forming ions, electrons are removed from the outermost shell first (highest principal quantum number). For Fe, the 4s electrons (n=4) are in the outermost shell even though 4s filled before 3d. Once you are removing electrons, the 4s is the most exposed and least tightly held, so they go first.

3
The Periodic Table
Why properties repeat and how the table is organised

Mendeleev noticed that when elements are arranged by properties, patterns repeat at regular intervals — this is periodicity. Today we know the repeating pattern exists because elements in the same group have the same number of valence electrons, and valence electrons dictate chemical behaviour.

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Why does the table work?
Every element in Group 1 has 1 valence electron. Every element in Group 17 has 7 valence electrons. They all behave similarly because their outermost electron count is the same. As you move down a group, you add electron shells, but the valence count stays constant — so properties stay similar. Periodicity is just electron configuration made visual.
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Key layout rules:
Period (row) = all atoms have the same number of electron shells (n = period number)
Group (column) = all atoms have the same number of valence electrons
Group number = valence electron count (for main-group elements)
Periodic Table — Key Regions (Periods 1–4) P1 P2 P3 P4 1 2 3-12 13 14 15 16 17 18 transition metals H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Br Kr Alkali metals (Gr 1) Alkaline earth (Gr 2) Halogens (Gr 17) Noble gases (Gr 18) Nonmetals Metals
Simplified periodic table (periods 1–4, main groups). Colour-coding shows the key families. Groups 3–12 (transition metals) sit between Groups 2 and 13.
★ Easy
Predicting properties from position
Na is in Period 3, Group 1. What can you predict about its structure and reactivity?
Show solution
1
Period 3 → 3 electron shells
n = 1, 2, 3 shells filled. Config: [Ne]3s¹
2
Group 1 → 1 valence electron
1 valence electron in outermost shell (3s¹)
3
Predict behaviour

Alkali metal: very reactive (only 1 e− to lose), forms +1 ion (Na⁺), similar properties to Li (above) and K (below).

Answer: 3 shells, 1 valence e−, highly reactive, forms Na+, similar to Li and K
★★ Intermediate
Predicting an unknown element
Predict the properties of the element with Z=33 (arsenic, As) using its periodic table position alone.
Show solution
1
Find period and group
Z=33 is in Period 4, Group 15
2
Period 4 → 4 shells. Group 15 → 5 valence electrons
Config: [Ar]3d¹⁰4s²4p³
3
Region of table

As sits on the metalloid staircase — it is a metalloid (semiconductor-like properties). Group 15 with 5 valence electrons: can form −3 ion or share 3 electrons in covalent bonds.

4
Similar to group neighbours N and P

As shares chemical properties with nitrogen (Z=7) and phosphorus (Z=15) — same group, same valence count.

Answer: Period 4, Group 15, metalloid, 5 valence e−, can form −3 ion, similar to N and P
Checkpoint 3
a) An element has 2 electron shells and 7 valence electrons. Identify it by name and give its group and period.
b) Which group of the periodic table always has exactly 2 valence electrons?
c) Why do elements in the same group have similar chemical properties?

a) 2 shells = Period 2. 7 valence electrons = Group 17. That is Fluorine (F), the most electronegative element.

b) Group 2 (alkaline earth metals) — always 2 valence electrons in ns² configuration.

c) Elements in the same group have the same number of valence electrons. Since chemical reactions involve gaining, losing, or sharing valence electrons, elements with identical valence counts react in the same ways and form the same types of compounds. The inner electron shells differ (more shells as you go down), but they are shielded from chemistry by the outer shell.

4
Periodic Trends
Nuclear charge vs shielding — the two forces behind every trend

Every periodic trend can be explained by the same two competing forces: nuclear charge (the positive protons pulling electrons inward) and electron shielding (inner electrons blocking some of that pull). The direction of each trend depends on which force wins in a given direction.

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Why nuclear charge wins across a period:
Going from Na to Cl across Period 3, the number of protons increases from 11 to 17. All those electrons remain in the same shell (n=3). The extra protons pull the electron cloud inward with more force, but the shielding barely changes because no new inner shells are added. Result: the atom shrinks, electrons are held tighter (higher IE), and the atom attracts bonding electrons more strongly (higher EN).

Why shielding wins down a group:
Going from F to Cl to Br in Group 17, each step adds a completely new electron shell between the nucleus and the valence electrons. Those inner shells shield the valence electrons from the nuclear charge. Even though the nucleus gets more protons, the valence electrons “feel” less pull. Result: the atom grows, electrons are easier to remove, and EN decreases.
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All three trends in one rule:
Atomic radius: increases down, decreases across (opposite of IE and EN)
Ionization energy: decreases down, increases across
Electronegativity: decreases down, increases across
Highest EN: F = 4.0   Lowest EN: Fr ≈ 0.7
Direction of Increasing Trends Group 1 Group 18 P1 P3 P7 Atomic radius increases radius ↑ IE & EN increase → IE & EN increase ↑ F EN=4.0 Fr EN~0.7
Trend directions on the periodic table. Green arrows show direction of increasing atomic radius (down and left). Blue arrows show direction of increasing ionization energy and electronegativity (up and right, toward F).
★ Easy
Comparing Na and Cl across Period 3
Compare the atomic radius, ionization energy, and electronegativity of Na (Z=11) and Cl (Z=17).
Show solution
1
Both are in Period 3 (same shell count). Cl is further right.
Na: Z=11, Group 1. Cl: Z=17, Group 17.
2
Atomic radius: decreases across a period
Na has larger atomic radius than Cl.
(More protons in Cl pull electrons in tighter)
3
IE and EN: increase across a period
Cl has higher IE and higher EN than Na.
(Cl holds its electrons tighter; Cl is a much stronger electron attractor)
Answer: Na: larger radius, lower IE, lower EN. Cl: smaller radius, higher IE, higher EN.
★★ Intermediate
Arranging elements by ionization energy
Arrange K, Ca, and Br in order of increasing ionization energy. Justify each step.
Show solution
1
Locate each element
K: Period 4, Group 1
Ca: Period 4, Group 2
Br: Period 4, Group 17
2
All in Period 4 → IE increases going right
K (Gr1) < Ca (Gr2) < Br (Gr17)
3
Reasoning for each

K has only 1 valence electron, loosely held by 19 protons — easiest to remove. Ca has 2 valence electrons, tighter grip. Br has 17 protons pulling on 7 valence electrons in the same shell — hardest to remove of the three.

Answer: K < Ca < Br (increasing IE across Period 4)
Checkpoint 4
a) Which has a larger atomic radius: Na (Period 3, Group 1) or K (Period 4, Group 1)? Explain why.
b) Arrange F, Cl, and Br in order of decreasing electronegativity.
c) Why does ionization energy generally increase across a period?

a) K has a larger atomic radius. Both are in Group 1 (same number of valence electrons), but K is in Period 4 and has 4 electron shells compared to Na's 3. The extra shell pushes the outermost electrons farther from the nucleus, more than offsetting the greater nuclear charge.

b) F > Cl > Br (decreasing EN down Group 17). F is the most electronegative element on the table (4.0). Going down the group, more electron shells shield the nucleus from bonding electrons, so each element attracts shared electrons less strongly.

c) Across a period, the number of protons increases while electrons are added to the same shell (no new inner shells, so shielding barely changes). The increased nuclear charge pulls valence electrons more tightly, making them harder to remove — hence higher ionization energy.

5
Ions
Why atoms gain or lose electrons — and what happens to their size

Atoms form ions to achieve the electron configuration of the nearest noble gas — a full outer shell. This is the most stable arrangement because filled shells have symmetrical electron distribution and strong shielding. Metals lose electrons to reach the configuration of the noble gas before them; nonmetals gain electrons to reach the configuration of the noble gas after them.

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Why do metals lose electrons and nonmetals gain them?
Metals (left side) have 1–3 valence electrons. It costs relatively little energy to remove a small number of electrons. Once lost, the atom reaches a full inner shell — stable. Nonmetals (right side) have 5–7 valence electrons. They are so close to a full shell that gaining 1–3 electrons is energetically favourable — the nucleus pulls additional electrons in readily. The driving force is always: reach the nearest noble gas configuration.
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Core rules:
Cation = atom loses e− → positive charge (metals, Groups 1–13)
Anion = atom gains e− → negative charge (nonmetals, Groups 15–17)
Cations are smaller than the parent atom (fewer electrons, same nuclear charge)
Anions are larger than the parent atom (more electrons, same nuclear charge — more repulsion)
Ion Formation — Size Change Na Z=11, 11e− [Ne]3s¹ 11 protons, 3 shells loses 1e− Na⁺ 10e− [Ne] Smaller: fewer e− Cl Z=17, 17e− [Ne]3s²3p&sup5; 17 protons, 3 shells gains 1e− Cl⁻ 18e− [Ar] Larger: more e−, repulsion
Na loses 1 electron to form Na+ (smaller — same 11 protons, only 10 electrons, tighter pull). Cl gains 1 electron to form Cl− (larger — same 17 protons, but 18 electrons spread over a larger volume).
★ Easy
Na and Cl ion formation
Write the electron configurations of Na (Z=11) and Cl (Z=17) after they form ions. Which noble gas do each resemble?
Show solution
1
Na loses 1 electron to reach 10e−
Na: [Ne]3s¹ → Na⁺: [Ne]    (10 electrons, like Ne)
2
Cl gains 1 electron to reach 18e−
Cl: [Ne]3s²3p&sup5; → Cl−: [Ar]    (18 electrons, like Ar)
3
Are they isoelectronic?

No. Na+ has 10 electrons (like Ne). Cl− has 18 electrons (like Ar). They have different electron counts, so they are NOT isoelectronic with each other — though each is isoelectronic with its respective noble gas.

Answer: Na+: [Ne] (10e−).   Cl−: [Ar] (18e−). Not isoelectronic with each other.
★★ Intermediate
Fe²+ vs Fe³+ stability
Iron forms both Fe²+ and Fe³+. Write configurations for each and explain why Fe³+ is the more stable ion.
Show solution
1
Fe neutral: [Ar]3d&sup6;4s²
26 electrons total
2
Fe²+: remove 4s electrons first (outermost shell)
[Ar]3d&sup6;    (24 electrons)
3
Fe³+: remove one more from 3d
[Ar]3d&sup5;    (23 electrons)
4
Why [Ar]3d&sup5; is more stable

A half-filled d subshell ([Ar]3d&sup5;) has one electron in each of the five d orbitals. This symmetric, evenly-spread arrangement minimises electron-electron repulsion and has extra stability (similar to why half-filled p subshells are slightly more stable). Fe³+ is therefore unusually stable for a +3 ion.

Answer: Fe²+: [Ar]3d&sup6;.   Fe³+: [Ar]3d&sup5; (half-filled d — extra stability).
Checkpoint 5
a) Write the electron configuration of Ca²+ (Z=20).
b) Na+ and Mg²+ both have 10 electrons. Which ion is smaller, and why?
c) Why do Group 17 elements form −1 ions but Group 16 elements form −2 ions?

a) Ca (Z=20): [Ar]4s². Ca²+ loses both 4s electrons: [Ar] — same configuration as argon, 18 electrons.

b) Mg²+ is smaller. Both ions have 10 electrons. However, Mg²+ has 12 protons and Na+ has only 11. More protons pulling the same number of electrons inward means the electron cloud is contracted more tightly in Mg²+. In isoelectronic series, the ion with more protons is always the smallest.

c) Group 17 elements have 7 valence electrons — they need to gain just 1 more electron to complete the octet (full outer shell). Group 16 elements have 6 valence electrons — they need to gain 2 more electrons to complete the octet. The charge on the ion reflects exactly how many electrons were gained to achieve the nearest noble gas configuration.