Atomic Structure and Electron Configuration

Part 1: Atomic Structure Review

Part 1 of 7 — Quantum Numbers & Orbitals

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Quick Reference

Particle	Charge	Location	How to Find Count
Proton $(p^{+})$	+1	Nucleus	= Atomic number (Z)
Neutron $(n^{0})$	0	Nucleus	= Mass number − Z
Electron $(e^{-})$	−1	Electron cloud	= Z (neutral atom)

🔑 Why this matters: The atomic number defines the element, and the electron count determines all chemical behavior — bonding, reactivity, and periodic trends.

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What You'll Master in Part 1

• Identifying protons, neutrons, and electrons from atomic/mass numbers
• Using isotope notation to describe different forms of an element
• Calculating particle counts in ions (cations and anions)

📌 The Three Subatomic Particles

Particle	Symbol	Charge	Location	Relative Mass
Proton	$p^{+}$	+1	Nucleus	1 amu
Neutron	$n^{0}$	0	Nucleus	1 amu
Electron	$e^{-}$	−1	Electron cloud	≈ 0 amu (1/1836 amu)

Key relationships:

• The atomic number (Z) = number of protons = number of electrons (in a neutral atom)
• The mass number (A) = protons + neutrons
• Number of neutrons = A − Z

The identity of an element is determined entirely by its number of protons. Change the proton count and you change the element.

🔑 Key Point: The atomic number (protons) defines the element. Everything else — neutrons, electrons — can vary.

📝 Isotope Notation

Atoms of the same element can have different numbers of neutrons. These variants are called isotopes.

We write isotope notation as:

$\boxed{^{A}_{Z}X}$

where A is the mass number (top), Z is the atomic number (bottom), and X is the element symbol.

Example: $^{14}_{6}\text{C}$ — Carbon-14

• Protons = 6
• Electrons = 6 (neutral atom)
• Neutrons = 14 − 6 = 8

Isotopes of an element have identical chemical behavior because they have the same number of electrons. Their physical properties (mass, nuclear stability) differ.

Quick Check: Identifying Particles

How many protons are in an atom of phosphorus (P, atomic number 15)?

Calculating Neutrons

Chlorine-37 ($^{37}_{17}\text{Cl}$) has a mass number of 37 and an atomic number of 17. How many neutrons does it have?

Remember: neutrons = mass number − atomic number

🔋 Ions: Gaining and Losing Electrons

When an atom gains or loses electrons, it becomes an ion:

• Cation (positive ion): atom loses electrons → fewer electrons than protons
  • Na → $Na^{+}$ (11 protons, 10 electrons)
• Anion (negative ion): atom gains electrons → more electrons than protons
  • Cl → $Cl^{-}$ (17 protons, 18 electrons)

⚠️ Important: Gaining or losing electrons does NOT change the atomic number or the identity of the element. Only changing protons does that.

Ion Particle Counts

How many electrons does the ion $Ca^{2+}$ have? (Calcium has atomic number 20.)

Comprehensive Review

An atom of $^{56}_{26}\text{Fe}$ — let's verify you can identify all its particles.

Part 2: Energy Levels and Subshells

Part 2 of 7 — Orbital Filling Order

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Energy Level Overview

Level (n)	Subshells Available	Max Electrons ($2n^2$)
1	1s	2
2	2s, 2p	8
3	3s, 3p, 3d	18
4	4s, 4p, 4d, 4f	32

The filling order does not follow simple numerical order — 4s fills before 3d!

🔑 Why this matters: The Aufbau filling order determines where every electron goes, and it explains why the periodic table is shaped the way it is.

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What You'll Master in Part 2

• Understanding principal energy levels and the $2n^2$ formula
• Knowing the four subshell types (s, p, d, f) and their capacities
• Memorizing the Aufbau filling order with the diagonal rule

📌 Principal Energy Levels

The principal quantum number (n) describes the main energy level of an electron:

n	Name	Max Electrons $(2n^{2})$
1	First shell	2
2	Second shell	8
3	Third shell	18
4	Fourth shell	32

The formula for the maximum number of electrons in a principal energy level is:

$\boxed{\text{Max electrons} = 2n^2}$

As n increases, the energy level is farther from the nucleus on average and the electrons have higher energy.

📌 Subshells: s, p, d, f

Each principal energy level is divided into subshells, labeled s, p, d, and f.

Subshell	Number of Orbitals	Max Electrons
s	1	2
p	3	6
d	5	10
f	7	14

Each orbital holds a maximum of 2 electrons (with opposite spins — the Pauli exclusion principle).

Which subshells exist in each level?

• n = 1: 1s only
• n = 2: 2s, 2p
• n = 3: 3s, 3p, 3d
• n = 4: 4s, 4p, 4d, 4f

In general, level n contains subshells s through the (n − 1)th letter in the sequence s, p, d, f.

Subshell Capacity Check

How many electrons can the 3d subshell hold at maximum?

📌 The Aufbau Filling Order

Electrons fill subshells in order of increasing energy, not simply by principal quantum number. The filling order is:

1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p → 7s → 5f → 6d → 7p

Notice that 4s fills before 3d — this is because 4s is lower in energy than 3d for most elements.

⚠️ AP Watch Out: The filling order is NOT the same as the shell order. 4s fills before 3d, 5s before 4d, 6s before 4f. This catches many students on the exam.

The diagonal rule is a visual trick to remember this order:

Write the subshells in a grid and draw diagonal arrows from upper-right to lower-left:

1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f
6s 6p 6d
7s 7p

Following the diagonals gives the correct filling order.

Filling Order Practice

Determine which subshell fills next in the Aufbau order.

Maximum Electron Calculations

Use the formula $2n^2$ to determine the maximum number of electrons in a principal energy level.

� Part 2 Summary: Energy Levels & Subshells

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🧰 Quick Reference

Subshell	l value	# Orbitals	Max Electrons
s	0	1	2
p	1	3	6
d	2	5	10
f	3	7	14

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📌 Key Concepts

Concept	Rule	Example
Energy level capacity	Max electrons = $2n^2$	n = 3 → 18 electrons max
Aufbau filling order	Fill lowest energy first	4s fills before 3d
Orbital capacity	Max 2 electrons per orbital	Opposite spins (↑↓)

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✅ Your Checklist Before Moving On

• ☐ I know the four subshell types and how many electrons each holds
• ☐ I can use the diagonal rule to determine filling order
• ☐ I understand that energy order ≠ numerical order (4s < 3d)
• ☐ I know that each orbital holds at most 2 electrons with opposite spins

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🔮 What's Next

In Part 3, we will use these rules to write complete electron configurations for real elements — from hydrogen all the way through the transition metals.

Part 3: Writing Electron Configurations

Part 3 of 7 — Writing Electron Configurations

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The Three Rules at a Glance

Rule	What It Controls	Key Idea
Aufbau Principle	Filling order	Lowest energy subshell fills first
Pauli Exclusion	Orbital capacity	Max 2 electrons per orbital (opposite spins)
Hund's Rule	Degenerate orbitals	Fill singly before pairing

🔑 Why this matters: These three rules are the complete recipe for writing any electron configuration — and they're tested heavily on the AP exam.

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What You'll Master in Part 3

• Applying all three rules to write configurations for any element
• Building configurations step-by-step with running electron counts
• Verifying configurations by checking total electrons match Z

📏 The Three Rules

1. Aufbau Principle

Electrons fill the lowest energy subshell available first.

Filling order: 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → ...

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2. Pauli Exclusion Principle

Each orbital can hold a maximum of 2 electrons, and those 2 electrons must have opposite spins (↑↓).

No two electrons in the same atom can have the same set of four quantum numbers.

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3. Hund's Rule

When filling orbitals of equal energy (degenerate orbitals, such as the three 2p orbitals), electrons fill each orbital singly first with parallel spins before any orbital gets a second electron.

Think of it like a bus: passengers sit in empty seats before doubling up.

🔑 The Three Rules: Aufbau (lowest energy first) + Pauli (max 2 per orbital, opposite spins) + Hund’s (fill degenerate orbitals singly before pairing) = the complete rules for electron configuration.

🧪 Step-by-Step Examples

Let's build electron configurations from scratch, starting simple and working up to transition metals.

Example 1: Hydrogen (H, Z = 1)

Total electrons: 1

Step	Subshell	Electrons Added	Running Total
1	1s	1	1 ✓

Configuration: $1s^{1}$

Only one electron — it goes into the lowest energy subshell, 1s.

Example 2: Carbon (C, Z = 6)

Total electrons: 6

Step	Subshell	Electrons Added	Running Total
1	1s	2	2
2	2s	2	4
3	2p	2	6 ✓

Configuration: $1s^{2}$ $2s^{2}$ $2p^{2}$

By Hund's rule, the two 2p electrons occupy two separate p orbitals with parallel spins — they don't pair up in the same orbital.

Example 3: Sodium (Na, Z = 11)

Total electrons: 11

Step	Subshell	Electrons Added	Running Total
1	1s	2	2
2	2s	2	4
3	2p	6	10
4	3s	1	11 ✓

Configuration: $1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{1}$

The first 10 electrons fill the n = 1 and n = 2 levels completely. The 11th electron starts a new shell.

Example 4: Iron (Fe, Z = 26)

Total electrons: 26

Step	Subshell	Electrons Added	Running Total
1	1s	2	2
2	2s	2	4
3	2p	6	10
4	3s	2	12
5	3p	6	18
6	4s	2	20
7	3d	6	26 ✓

Configuration: $1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{2}$ $3p^{6}$ $4s^{2}$ $3d^{6}$

Notice: 4s fills before 3d in the Aufbau order, so $4s^{2}$ appears before $3d^{6}$.

💡 Tip: Always verify your total by adding the superscripts: 2 + 2 + 6 + 2 + 6 + 2 + 6 = 26 ✓

Identify the Element

Which element has the electron configuration $1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{2}$ $3p^{4}$?

Write the Configuration

Write the full electron configuration for the following elements (e.g., 1s2 2s2 2p6). A formatted preview will appear as you type.

More Configuration Practice

Write the full electron configuration. Remember: 4s fills before 3d!

Spot the Error

Which of the following electron configurations is INCORRECT?

� Part 3 Summary: Writing Electron Configurations

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🧰 The Three Rules

Rule	What It Says	Common Mistake
Aufbau Principle	Fill the lowest energy subshell first	Putting electrons in 3d before 4s
Pauli Exclusion	Max 2 electrons per orbital (opposite spins ↑↓)	Putting 3 electrons in one orbital
Hund's Rule	Fill degenerate orbitals singly before pairing	Pairing 2p electrons before all three 2p orbitals have one

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✅ Your Checklist Before Moving On

• ☐ I can write the full configuration for any element up to Z = 36
• ☐ I know that 4s fills before 3d in the Aufbau order
• ☐ I always verify my total electron count matches the atomic number
• ☐ I understand why Hund's rule leads to unpaired electrons in partially filled subshells

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🔮 What's Next

In Part 4, you'll learn noble gas shorthand notation — a way to simplify long configurations like:

$1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{2}$ $3p^{6}$ $4s^{2}$ $3d^{6}$ → [Ar] $4s^{2}$ $3d^{6}$

This will save you time on the AP exam and make it easier to focus on the valence electrons that matter most for chemistry.

📦 Noble Gas (Shorthand) Notation

Part 4 of 7 — Simplifying Electron Configurations

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The Problem

Writing out full configurations gets long fast:

Element	Z	Full Configuration	That's a lot...
Na	11	$1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{1}$	4 subshells
Fe	26	$1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{2}$ $3p^{6}$ $4s^{2}$ $3d^{6}$	7 subshells
Br	35	$1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{2}$ $3p^{6}$ $4s^{2}$ $3d^{10}$ $4p^{5}$	8 subshells

The Solution

Replace the inner-shell electrons with the preceding noble gas in brackets:

Element	Full Configuration	→	Shorthand
Na	$1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{1}$	→	[Ne] $3s^{1}$
Fe	$1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{2}$ $3p^{6}$ $4s^{2}$ $3d^{6}$	→	[Ar] $4s^{2}$ $3d^{6}$
Br	$1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{2}$ $3p^{6}$ $4s^{2}$ $3d^{10}$ $4p^{5}$	→	[Ar] $4s^{2}$ $3d^{10}$ $4p^{5}$

🔑 Why this matters: Noble gas shorthand lets you focus on the valence electrons — the ones that actually determine chemical behavior and bonding.

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What You'll Master in Part 4

• Identifying the correct noble gas core for any element
• Converting between full and shorthand notation
• Recognizing that shorthand highlights the chemically important electrons

📌 The Noble Gases

Noble Gas	Symbol	Atomic Number	Full Configuration
Helium	He	2	$1s^{2}$
Neon	Ne	10	$1s^{2}$ $2s^{2}$ $2p^{6}$
Argon	Ar	18	$1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{2}$ $3p^{6}$
Krypton	Kr	36	$1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{2}$ $3p^{6}$ $4s^{2}$ $3d^{10}$ $4p^{6}$
Xenon	Xe	54	[Kr] $4d^{10}$ $5s^{2}$ $5p^{6}$
Radon	Rn	86	[Xe] $4f^{14}$ $5d^{10}$ $6s^{2}$ $6p^{6}$

To use shorthand notation:

1. Find the noble gas that comes just before your element in the periodic table.
2. Write that noble gas symbol in brackets.
3. Continue the configuration from where the noble gas left off.

🧪 Step-by-Step Examples

Example 1: Sodium (Na, Z = 11)

Total electrons: 11

Step	Action	Result
1	Write the full configuration	$1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{1}$
2	Identify the preceding noble gas	Neon (Ne, Z = 10)
3	Ne accounts for:	$1s^{2}$ $2s^{2}$ $2p^{6}$ (10 electrons)
4	Remaining electrons: 11 − 10 = 1	$3s^{1}$

Shorthand: [Ne] $3s^{1}$

🔑 The single $3s^{1}$ electron is sodium's valence electron — the one it loses to form $Na^{+}$.

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Example 2: Iron (Fe, Z = 26)

Total electrons: 26

Step	Action	Result
1	Write the full configuration	$1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{2}$ $3p^{6}$ $4s^{2}$ $3d^{6}$
2	Identify the preceding noble gas	Argon (Ar, Z = 18)
3	Ar accounts for:	$1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{2}$ $3p^{6}$ (18 electrons)
4	Remaining electrons: 26 − 18 = 8	$4s^{2}$ $3d^{6}$

Shorthand: [Ar] $4s^{2}$ $3d^{6}$

⚠️ Why not [Kr]? Krypton has Z = 36, which is more than 26. Always use the noble gas that comes before your element.

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Example 3: Bromine (Br, Z = 35)

Total electrons: 35

Step	Action	Result
1	Write the full configuration	$1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{2}$ $3p^{6}$ $4s^{2}$ $3d^{10}$ $4p^{5}$
2	Identify the preceding noble gas	Argon (Ar, Z = 18)
3	Ar accounts for:	$1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{2}$ $3p^{6}$ (18 electrons)
4	Remaining electrons: 35 − 18 = 17	$4s^{2}$ $3d^{10}$ $4p^{5}$

Shorthand: [Ar] $4s^{2}$ $3d^{10}$ $4p^{5}$

💡 Notice how the shorthand cuts a 8-subshell configuration down to just 3 subshells — and immediately shows the 7 valence electrons $(4s^{2} + 4p^{5})$ that determine bromine's chemistry.

Identify the Noble Gas Core

Which noble gas core would you use for the shorthand notation of Selenium (Se, Z = 34)?

Shorthand Notation

What is the correct noble gas shorthand notation for phosphorus (P, Z = 15)?

Convert to Shorthand

Convert the following full electron configurations to noble gas shorthand (e.g., [Ne] 3s2 3p5). A formatted preview will appear as you type.

Noble Gas Shorthand Identification

Match each element with its correct shorthand electron configuration.

� Part 4 Summary: Noble Gas Shorthand

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🧰 The Method

Step	Action	Example (Bromine, Z = 35)
1	Write the full configuration	$1s^{2}$ $2s^{2}$ $2p^{6}$ $3s^{2}$ $3p^{6}$ $4s^{2}$ $3d^{10}$ $4p^{5}$
2	Find the preceding noble gas	Argon (Ar, Z = 18)
3	Replace the noble gas core with brackets	[Ar]
4	Write the remaining subshells	$4s^{2}$ $3d^{10}$ $4p^{5}$
Result	Noble gas shorthand	[Ar] $4s^{2}$ $3d^{10}$ $4p^{5}$

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📌 Noble Gas Reference

Noble Gas	Z	Use for elements with Z =
He	2	3–10
Ne	10	11–18
Ar	18	19–36
Kr	36	37–54
Xe	54	55–86
Rn	86	87+

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✅ Your Checklist Before Moving On

• ☐ I can identify the correct noble gas core for any element
• ☐ I can convert a full configuration to noble gas shorthand
• ☐ I can convert noble gas shorthand back to the full configuration
• ☐ I understand that shorthand highlights the valence electrons — the ones that determine chemical behavior

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🔮 What's Next

In Part 5, we tackle the important exceptions to the Aufbau filling order (Chromium and Copper) and learn how to write electron configurations for ions — including the critical rule that 4s electrons are removed before 3d when forming cations.

Part 5: Exceptions and Ion Configurations

Part 5 of 7 — Exceptions & Ion Configurations

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The Two Must-Know Exceptions

Element	Expected Config	Actual Config	Why?
Cr (Z=24)	[Ar] $4s^{2}$ $3d^{4}$	[Ar] $3d^{5}$ $4s^{1}$	Half-filled $d^{5}$ is extra stable
Cu (Z=29)	[Ar] $4s^{2}$ $3d^{9}$	[Ar] $3d^{10}$ $4s^{1}$	Fully filled $d^{10}$ is extra stable

And the critical ion rule: remove electrons from the highest n first (4s before 3d).

🔑 Why this matters: These exceptions and the ion formation rule are among the most frequently tested topics on the AP Chemistry exam.

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What You'll Master in Part 5

• Recognizing and writing the Cr and Cu exceptions
• Forming cation configurations by removing from the highest n first
• Writing anion configurations by adding electrons
• Identifying isoelectronic species

⚠️ The Two Critical Exceptions

Chromium (Cr, Z = 24)

Expected: [Ar] $4s^{2}$ $3d^{4}$
Actual: [Ar] $3d^{5}$ $4s^{1}$

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Copper (Cu, Z = 29)

Expected: [Ar] $4s^{2}$ $3d^{9}$
Actual: [Ar] $3d^{10}$ $4s^{1}$

Why? Half-filled $(d^{5})$ and fully filled $(d^{10})$ subshells have extra stability due to:

• Exchange energy: More favorable electron-electron interactions when orbitals are symmetrically occupied.
• Electrons in the 4s and 3d subshells are very close in energy, so the stabilization from a half-filled or fully filled d subshell outweighs the cost of promoting one electron from 4s.

Other elements in the same columns (Mo, Ag, etc.) show similar exceptions, but Cr and Cu are the ones you must know for the AP exam.

⚠️ AP Must-Know: Cr is [Ar] $3d^{5}$ $4s^{1}$ $(not 4s^{2} 3d^{4})$ and Cu is [Ar] $3d^{10}$ $4s^{1}$ $(not 4s^{2} 3d^{9})$. Half-filled and fully filled d subshells have extra stability.

Exception Check

What is the correct electron configuration for chromium (Cr, Z = 24)?

🔋 Electron Configurations of Ions

Cations (Positive Ions)

When forming cations, electrons are removed from the subshell with the highest principal quantum number (n) first.

⚠️ Critical Rule for Transition Metals: Remove electrons from 4s before 3d, even though 4s filled first!

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Example 1: $Fe^{2+}$ (Z = 26, 24 electrons)

Step	Action	Result
1	Write neutral Fe configuration	[Ar] $4s^{2}$ $3d^{6}$
2	Identify highest n to remove from	4s (n = 4)
3	Remove 2 electrons from 4s	$4s^{0}$ (both 4s electrons gone)

$Fe^{2+}$: [Ar] $3d^{6}$

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Example 2: $Fe^{3+}$ (Z = 26, 23 electrons)

Step	Action	Result
1	Start from $Fe^{2+}$	[Ar] $3d^{6}$
2	4s is already empty — remove from 3d	Remove 1 electron from 3d
3	Final configuration	[Ar] $3d^{5}$

$Fe^{3+}$: [Ar] $3d^{5}$

🔑 Notice: $Fe^{3+}$ has a half-filled $3d^{5}$ subshell, giving it extra stability — this is why $Fe^{3+}$ is a very common ion.

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Anions (Negative Ions)

When forming anions, electrons are added to the next available subshell.

Example 3: $Cl^{-}$ (Z = 17, 18 electrons)

Step	Action	Result
1	Write neutral Cl configuration	[Ne] $3s^{2}$ $3p^{5}$
2	Add 1 electron to 3p	$3p^{5}$ → $3p^{6}$
3	Final configuration	[Ne] $3s^{2}$ $3p^{6}$

$Cl^{-}$: [Ne] $3s^{2}$ $3p^{6}$

💡 $Cl^{-}$ has 18 electrons — the same as Argon. They are isoelectronic!

Ion Configuration Practice

What is the electron configuration of $Cu^{2+}$?

Recall: Cu (Z = 29) has the configuration [Ar] $3d^{10}$ $4s^{1}$ (exception).

Write Ion Configurations

Write the noble gas shorthand electron configuration for each ion (e.g., [Ar] 3d6). A formatted preview will appear as you type.

Remember: Remove electrons from the highest n first!

Tricky Ion Problems

These require careful attention to exception rules and ion formation rules.

� Part 5 Summary: Exceptions & Ion Configurations

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🧰 The Two Critical Exceptions

Element	Expected	Actual	Why
Cr (Z = 24)	[Ar] $4s^{2}$ $3d^{4}$	[Ar] $3d^{5}$ $4s^{1}$	Half-filled $d^{5}$ = extra stability
Cu (Z = 29)	[Ar] $4s^{2}$ $3d^{9}$	[Ar] $3d^{10}$ $4s^{1}$	Fully filled $d^{10}$ = extra stability

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📌 Ion Configuration Rules

Ion Type	Rule	Example
Cations (+)	Remove from highest n first	$Fe^{2+}$: remove $4s^{2}$ → [Ar] $3d^{6}$
Anions (−)	Add to next available subshell	$Cl^{-}$: add to 3p → [Ne] $3s^{2}$ $3p^{6}$
Isoelectronic	Same $e^{-}$ count = same configuration	$Na^{+}$, $F^{-}$, Ne all have 10 $e^{-}$

⚠️ AP Trap: For transition metal cations, always remove 4s electrons before 3d — even though 4s filled first!

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✅ Your Checklist Before Moving On

• ☐ I know the configurations of Cr and Cu (and why they are exceptions)
• ☐ I can write ion configurations by removing from the highest n first
• ☐ I can identify isoelectronic species
• ☐ I will not mistakenly remove 3d electrons before 4s when forming cations

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🔮 What's Next

In Part 6, we explore orbital diagrams (box-arrow notation) and the four quantum numbers that uniquely identify every electron in an atom.

Part 6: Orbital Diagrams and Quantum Numbers

Part 6 of 7 — Problem-Solving Workshop

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From Configuration to Quantum Address

Level of Detail	What It Tells You	Example (for a 2p electron)
Configuration	Which subshells are occupied	$2p^{4}$
Orbital diagram	Spin of each electron	↑↓ ↑ ↑
Quantum numbers	Exact "address" of one electron	n=2, l=1, $m_l$=−1, $m_s$=+½

🔑 Why this matters: Quantum numbers give every electron a unique identity — like a GPS coordinate inside the atom.

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What You'll Master in Part 6

• Drawing orbital diagrams with correct Hund's rule application
• Assigning all four quantum numbers (n, l, $m_l$, $m_s$) to any electron
• Calculating the number of orbitals in a subshell using 2l + 1

📝 Orbital Diagrams (Box-Arrow Notation)

An orbital diagram represents each orbital as a box (or line) and each electron as an arrow:

• ↑ represents spin-up ($m_s = +\frac{1}{2}$)
• ↓ represents spin-down ($m_s = -\frac{1}{2}$)

Example: Nitrogen (N, Z = 7)

1s	2s	2p
↑↓	↑↓	↑ ↑ ↑

Each of the three 2p orbitals gets one electron first (Hund's rule) before any pairing occurs. All three unpaired electrons have the same spin direction.

🔑 Key Rule: Hund’s Rule in action — fill each orbital singly with parallel spins before any pairing. This minimizes electron-electron repulsion.

Example: Oxygen (O, Z = 8)

1s	2s	2p
↑↓	↑↓	↑↓ ↑ ↑

Oxygen has 8 electrons. After filling the three 2p orbitals singly (like nitrogen), the 8th electron pairs up in the first 2p orbital.

Hund's Rule Application

How many unpaired electrons does nitrogen (N, Z = 7) have?

📌 The Four Quantum Numbers

Every electron in an atom is described by a unique set of four quantum numbers — like a full mailing address.

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1. Principal Quantum Number (n)

• Allowed values: 1, 2, 3, 4, ...
• Describes the main energy level (shell)
• Higher n = higher energy, larger orbital

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2. Angular Momentum (Azimuthal) Quantum Number (l)

• Allowed values: 0 to (n − 1)
• Describes the subshell shape
• l = 0 → s, l = 1 → p, l = 2 → d, l = 3 → f

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3. Magnetic Quantum Number ($m_l$)

• Allowed values: −l to +l (including 0)
• Describes the orientation of the orbital in space
• For p orbitals (l = 1): $m_l$ = −1, 0, +1 → three orientations

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4. Spin Quantum Number ($m_s$)

• Allowed values: $+\frac{1}{2}$ or $-\frac{1}{2}$
• Describes the spin direction of the electron
• Two electrons in the same orbital must have opposite spins (Pauli exclusion)

📋 Allowed Values Summary

Subshell	n (example)	l	$m_l$ values	# orbitals	Max $e^{-}$
1s	1	0	0	1	2
2p	2	1	−1, 0, +1	3	6
3d	3	2	−2, −1, 0, +1, +2	5	10
4f	4	3	−3, −2, −1, 0, +1, +2, +3	7	14

Key relationship: The number of orbitals in a subshell = 2l + 1

Quantum Number Practice

Determine the quantum numbers for specified electrons.

Quantum Number Calculations

Use the relationships between quantum numbers to answer.

� Part 6 Summary: Orbital Diagrams & Quantum Numbers

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🧰 The Four Quantum Numbers

Quantum Number	Symbol	Determines	Allowed Values
Principal	n	Energy level / shell	1, 2, 3, ...
Angular momentum	l	Subshell shape	0 to n − 1
Magnetic	$m_l$	Orbital orientation	−l to +l
Spin	$m_s$	Electron spin	+½ or −½

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📌 Quick Formulas

Formula	Meaning	Example (l = 2, d subshell)
# orbitals = 2l + 1	Orbitals in a subshell	2(2) + 1 = 5 orbitals
max $e^{-}$ = 2(2l + 1)	Electrons in a subshell	2(5) = 10 electrons

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✅ Your Checklist Before Moving On

• ☐ I can draw orbital diagrams using boxes and arrows (↑↓)
• ☐ I apply Hund's rule: fill all orbitals singly before pairing
• ☐ I know all four quantum numbers and their allowed values
• ☐ I can determine the quantum numbers for any electron in an atom

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🔮 What's Next

Part 7 brings it all together with AP-style synthesis problems that combine electron configuration with periodic trends, ionization energy, and other core concepts.

Part 7: Synthesis & AP Review

Part 7 of 7 — Synthesis & AP Review

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Concepts You'll Integrate

Concept	From Part	How It Connects
Subatomic particles	Part 1	Identify elements from configurations
Aufbau filling order	Part 2	Write any configuration correctly
Three rules	Part 3	Avoid common errors
Noble gas shorthand	Part 4	Simplify and focus on valence electrons
Exceptions & ions	Part 5	Handle Cr, Cu, and transition metal ions
Quantum numbers	Part 6	Full electron "addresses"

🔑 Why this matters: The AP exam tests electron configuration in multiple-choice, free-response, AND as background knowledge for bonding, periodicity, and spectroscopy questions.

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What You'll Master in Part 7

• Solving multi-concept problems that combine configurations with periodic trends
• Connecting ionization energy exceptions to electron configuration
• Identifying elements from configurations and predicting ion behavior

📈 Electron Configuration and Periodic Trends

Electron configurations explain why periodic trends exist:

Trend	Across a Period (→)	Down a Group (↓)	Why
Atomic Radius	Decreases	Increases	More protons pull $e^{-}$ closer (→); higher n = farther from nucleus (↓)
Ionization Energy	Generally increases	Decreases	Greater $Z_{eff}$ holds $e^{-}$ tighter (→); valence $e^{-}$ farther out (↓)
Electronegativity	Increases	Decreases	Higher $Z_{eff}$ attracts bonding $e^{-}$ (→); distance reduces pull (↓)

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⚠️ AP Exceptions

Exception	What Happens	Why
IE: Be → B	IE drops	B removes a 2p $e^{-}$ (higher energy) vs. Be's 2s $e^{-}$
IE: N → O	IE drops	O has a paired 2p $e^{-}$; N's half-filled $2p^{3}$ has extra stability

⚠️ These exceptions are frequently tested on the AP exam. Always connect your explanation back to electron configuration and subshell occupancy.

AP-Style Question 1: Ionization Energy Exception

The first ionization energy of oxygen (Z = 8) is lower than that of nitrogen (Z = 7), even though oxygen has a higher atomic number. Which explanation best accounts for this?

AP-Style Question 2: Identifying an Element

An element has the electron configuration [Kr] $4d^{10}$ $5s^{2}$ $5p^{4}$. Which statement about this element is correct?

AP-Style Question 3: Transition Metal Ion

The ion $Ti^{2+}$ is used in some catalytic processes. What is the ground-state electron configuration of $Ti^{2+}$?

📌 Common AP Mistakes to Avoid

Mistake 1: Ion Configurations

❌ Removing electrons from the last-filled subshell (3d)
✅ Remove from the highest n first (4s before 3d for transition metals)

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Mistake 2: Forgetting Exceptions

❌ Cr: [Ar] $4s^{2}$ $3d^{4}$
✅ Cr: [Ar] $3d^{5}$ $4s^{1}$ (half-filled d subshell)

❌ Cu: [Ar] $4s^{2}$ $3d^{9}$
✅ Cu: [Ar] $3d^{10}$ $4s^{1}$ (fully filled d subshell)

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Mistake 3: Wrong Noble Gas Core

❌ Using [Kr] for elements with Z < 36
✅ Always use the noble gas that comes immediately before the element

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Mistake 4: Violating Hund's Rule

❌ Pairing electrons in a 2p orbital before all three 2p orbitals have one electron
✅ Fill all degenerate orbitals singly (with parallel spins) before pairing

Challenge Problems

Write the electron configuration in noble gas shorthand (e.g., [Ar] 3d5 4s1). A formatted preview will appear as you type.

Synthesis Questions

These questions connect electron configuration to other chemistry concepts.

📋 Final Summary: Electron Configuration Mastery

Congratulations on completing all 7 parts! Here is everything you need to know:

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🧰 Master Reference Table

Topic	Key Fact	AP Must-Know
Subshells	s, p, d, f hold 2, 6, 10, 14 $e^{-}$	Know max electrons per subshell
Three Rules	Aufbau → Pauli → Hund's	Apply in this order
Noble Gas Shorthand	Replace core with [noble gas]	Focus on valence electrons
Exceptions	Cr: [Ar] $3d^{5}$ $4s^{1}$, Cu: [Ar] $3d^{10}$ $4s^{1}$	Half-filled/full d = extra stability
Ion Configs	Remove from highest n first	4s before 3d for TM cations
Quantum Numbers	n, l, $m_l$, $m_s$	Uniquely identify every electron
Periodic Trends	Radius, IE, EN from $e^{-}$ config	Know the exceptions (Be→B, N→O)

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✅ Final Checklist

• ☐ I can write the full and shorthand configuration for any element
• ☐ I know the Cr and Cu exceptions and can explain why
• ☐ I can write ion configurations (removing 4s before 3d)
• ☐ I can assign all four quantum numbers to any electron
• ☐ I can explain periodic trends using electron configuration
• ☐ I can identify IE exceptions and explain them

You are now prepared for any electron configuration question on the AP Chemistry exam. Good luck! 🎯