1.5 - Atomic Structure and Electron Configuration

Overview

Almost everything in chemistry comes down to how electrons are arranged around a nucleus. This topic covers what an atom is made of, how Coulomb's law describes the pull between charged particles, and how to write the electron configuration of an atom or ion with the Aufbau principle. Those configurations are the key to photoelectron spectra (1.6), periodic trends (1.7), and the charges ions form (1.8).

What an atom is made of

An atom has a tiny, dense nucleus made of protons (positive) and neutrons (no charge). Electrons (negative) occupy the space around the nucleus.

Particle	Charge	Relative mass	Where it is
Proton	$+1$	1	Nucleus
Neutron	$0$	1	Nucleus
Electron	$-1$	about $\frac{1}{1836}$ (almost nothing)	Around the nucleus

Atomic number ( $Z$ ): the number of protons. It identifies the element. Every sodium atom has 11 protons.

Mass number: protons plus neutrons. Sodium-23 has $23 - 11 = 12$ neutrons.

In a neutral atom, the number of electrons equals the number of protons. An ion has gained or lost electrons: a cation lost electrons (positive), and an anion gained electrons (negative). $\mathrm{Na^+}$ has 11 protons but only 10 electrons.

Almost all of the mass is in the nucleus, but almost all of the volume belongs to the electrons: a nucleus is roughly 100,000 times smaller across than its atom. The positive nucleus attracts the negative electrons, and the electrons repel each other. Those two pushes and pulls explain the rest of this unit.

Coulomb's law

Coulomb's law gives the force between two charged particles:

F_{\text{coulombic}} \propto \frac{q_1 q_2}{r^2}

Here

q_1

and

q_2

are the charges and

r

is the distance between them. The symbol

\propto

means “is proportional to”.

Opposite charges attract and like charges repel.

Bigger charges mean a stronger force. Doubling one charge doubles the force.

Greater distance means a weaker force, and it drops off fast because $r$ is squared. Doubling the distance cuts the force to $\frac{1}{4}$ .

In an atom, the nucleus (charge

+Z

) pulls on every electron. Electrons closer to the nucleus are held more tightly, and a nucleus with more protons pulls harder. That one idea is behind ionization energy, atomic size, and every other trend in this unit. The AP exam uses Coulomb's law qualitatively or with ratios, which is why the reference sheet gives it as a proportion with no constant.

Example 1: Using Coulomb's law with ratios

Two charged particles attract each other with a force

F

. Find the new force, in terms of

F

, if (a) the distance between them is cut in half, (b) both charges are doubled, (c) one charge is tripled and the distance is doubled.

Compare the new force to the old one by putting the changes into the formula:

\frac{F_{\text{new}}}{F_{\text{old}}} = \frac{q_1' q_2'}{q_1 q_2} \times \left(\frac{r}{r'}\right)^2

(a) The charges don't change and

r' = \frac{1}{2}r

\frac{F_{\text{new}}}{F_{\text{old}}} = 1 \times \left(\frac{1}{1/2}\right)^2 = 4

(b) The distance doesn't change, and the charge product grows by

2 \times 2

\frac{F_{\text{new}}}{F_{\text{old}}} = 2 \times 2 = 4

(c) The charge product triples and the distance doubles:

\frac{F_{\text{new}}}{F_{\text{old}}} = 3 \times \left(\frac{1}{2}\right)^2 = \frac{3}{4}

Answer: (a)

4F

, (b)

4F

, (c)

0.75F

. In (c) the bigger charge helps, but the doubled distance hurts more because distance is squared.

Topic 2.2 uses the same ratio method on ions, to explain why

\mathrm{MgO}

holds together far more strongly than

\mathrm{NaCl}

Shells and subshells

Electrons in an atom are organized into shells (energy levels), numbered

n = 1, 2, 3, \ldots

A larger

n

means the electron is, on average, farther from the nucleus and higher in energy.

Each shell is split into subshells (sublevels) labeled

s

p

d

, and

f

. Shell

n

has

n

subshells: shell 1 has only

1s

, shell 2 has

2s

and

2p

, shell 3 has

3s

3p

, and

3d

, and so on. Each subshell is made of orbitals, and each orbital holds at most 2 electrons.

Subshell	Number of orbitals	Maximum electrons
$s$	1	2
$p$	3	6
$d$	5	10
$f$	7	14

So shell

n

holds up to

2n^2

electrons: 2 in shell 1, 8 in shell 2, 18 in shell 3. You don't need to assign quantum numbers to electrons; that isn't tested on the AP exam.

Core and valence electrons

Valence electrons are the electrons in the outermost shell (the highest $n$ ). They are the ones involved in bonding and reactions.

Core electrons are all the inner electrons. For a main-group element, the core matches the noble gas at the end of the previous row.

1.5 - Atomic Structure and Electron Configuration

Sodium has 11 electrons: 2 in shell 1 and 8 in shell 2 (10 core electrons), plus 1 in shell 3 (its single valence electron). The circles are a bookkeeping picture, not real orbits. For main-group elements, the number of valence electrons matches the last digit of the group number: group 1 has 1, group 2 has 2, and groups 13 to 18 have 3 to 8 (helium has 2).

Electron configurations and the Aufbau principle

An electron configuration lists how many electrons are in each subshell. Sodium is

1s^2\,2s^2\,2p^6\,3s^1

: the big number is the shell, the letter is the subshell, and the superscript is the number of electrons in it. The superscripts add up to the total number of electrons.

The Aufbau principle (German for “building up”) says that in the ground state, electrons fill the lowest-energy subshells first. The approximate 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

. In potassium and calcium, the

4s

subshell is lower in energy, so the 19th and 20th electrons go there. The

3d

subshell starts filling at scandium.

Two more rules matter when you draw orbital diagrams (one box per orbital, one arrow per electron):

Pauli exclusion principle: an orbital holds at most two electrons, and they must have opposite spins (one up arrow, one down arrow).

Hund's rule: within a subshell, electrons spread out one per orbital before any of them pair up.

Oxygen's four

2p

electrons fill all three

2p

orbitals singly, and then the fourth pairs up. That paired electron matters in 1.7, where it explains a dip in ionization energy.

Reading configurations from the periodic table

You don't have to memorize the filling order. The periodic table is built from it: each row fills one shell, and each block tells you which subshell is filling.

The $s$ block is 2 columns wide, the $p$ block 6, the $d$ block 10, and the $f$ block 14, matching the subshell capacities.

The $d$ block in period 4 is the $3d$ subshell (one less than the period number). The $f$ block in period 6 is $4f$ (two less).

Helium sits above the $p$ block, but its electrons are $1s^2$ .

To write a configuration, start at hydrogen and read across the rows until you reach your element, writing each block you pass through. Noble-gas notation saves time: write the previous noble gas in brackets for the core, then add the rest. Sodium becomes

[\mathrm{Ne}]\,3s^1

A few elements, such as chromium and copper, don't follow the Aufbau order exactly. Writing configurations for those exceptions isn't assessed on the AP exam.

Example 2: Writing electron configurations

Write the full and noble-gas configurations for (a) phosphorus and (b) selenium, and give the number of valence electrons in each.

(a) Phosphorus,

Z = 15

. Fill in order until 15 electrons are placed:

\mathrm{P}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^3 = [\mathrm{Ne}]\,3s^2\,3p^3

Check:

2 + 2 + 6 + 2 + 3 = 15

. The outer shell is

n = 3

, with

2 + 3 = 5

valence electrons, which fits group 15.

(b) Selenium,

Z = 34

. Selenium is in period 4, so pass through

4s

and

3d

before reaching

4p

\mathrm{Se}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^4 = [\mathrm{Ar}]\,4s^2\,3d^{10}\,4p^4

Check:

18 + 2 + 10 + 4 = 34

. The outer shell is

n = 4

, so selenium has

2 + 4 = 6

valence electrons (group 16). The filled

3d^{10}

subshell belongs to shell 3, inside the outer shell, so it isn't counted as valence.

Electron configurations of ions

Anions: add electrons in the next open spots. Chlorine, $[\mathrm{Ne}]\,3s^2\,3p^5$ , gains one electron to become $\mathrm{Cl^-}$ , $[\mathrm{Ne}]\,3s^2\,3p^6$ .

Cations: remove electrons from the outermost shell (highest $n$ ) first. Sodium, $[\mathrm{Ne}]\,3s^1$ , loses its $3s$ electron to become $\mathrm{Na^+}$ , $[\mathrm{Ne}]$ .

Transition metals lose their $4s$ electrons before their $3d$ electrons, even though $4s$ filled first. The $4s$ electrons are in shell 4, the outermost shell. Iron, $[\mathrm{Ar}]\,4s^2\,3d^6$ , forms $\mathrm{Fe^{2+}}$ , $[\mathrm{Ar}]\,3d^6$ , and $\mathrm{Fe^{3+}}$ , $[\mathrm{Ar}]\,3d^5$ .

Species with the same electron configuration are isoelectronic.

\mathrm{S^{2-}}

\mathrm{Cl^-}

, Ar,

\mathrm{K^+}

, and

\mathrm{Ca^{2+}}

all have 18 electrons arranged as

1s^2\,2s^2\,2p^6\,3s^2\,3p^6

Example 3: Configurations of ions

Write the ground-state configurations of (a)

\mathrm{S^{2-}}

, (b)

\mathrm{Ca^{2+}}

, and (c)

\mathrm{Zn^{2+}}

(a) Sulfur (

Z = 16

) is

[\mathrm{Ne}]\,3s^2\,3p^4

. The

2-

charge means two extra electrons, which finish the

3p

subshell:

\mathrm{S^{2-}}

[\mathrm{Ne}]\,3s^2\,3p^6

, the same as

[\mathrm{Ar}]

(18 electrons).

(b) Calcium (

Z = 20

) is

[\mathrm{Ar}]\,4s^2

. Remove both

4s

electrons:

\mathrm{Ca^{2+}}

[\mathrm{Ar}]

(18 electrons).

(c) Zinc (

Z = 30

) is

[\mathrm{Ar}]\,4s^2\,3d^{10}

. The two electrons come out of

4s

, the outermost shell, not

3d

\mathrm{Zn^{2+}}:\ [\mathrm{Ar}]\,3d^{10} \quad (\text{not } [\mathrm{Ar}]\,4s^2\,3d^8)

Answer:

\mathrm{S^{2-}}

and

\mathrm{Ca^{2+}}

are both

[\mathrm{Ar}]

, so they are isoelectronic with argon, and

\mathrm{Zn^{2+}}

[\mathrm{Ar}]\,3d^{10}

(28 electrons).

Ionization energy and Coulomb's law

Ionization energy is the energy needed to remove an electron from an atom or ion in the gas phase. Coulomb's law lets you predict which electrons take more energy to remove, without any numbers. Two things matter:

Distance from the nucleus. Electrons in a higher shell (larger $n$ ) are farther from the nucleus, so they are attracted less strongly and are easier to remove.

Effective nuclear charge ( $Z_{\text{eff}}$ ). Core electrons sit between the nucleus and the outer electrons and cancel part of the nuclear charge. This is called shielding. The net charge an outer electron actually feels is $Z_{\text{eff}}$ . A simple estimate is protons minus core electrons. For sodium's $3s$ electron, $Z_{\text{eff}} \approx 11 - 10 = +1$ .

The CED asks you to compare two kinds of electrons:

Different subshells in the same atom. In lithium, removing a $1s$ electron takes about 6.26 MJ/mol, but removing the $2s$ electron takes only about 0.52 MJ/mol, roughly 12 times less. The $1s$ electrons are much closer to the nucleus and have no inner electrons shielding them.

The same subshell in different atoms. Sodium and magnesium both lose a $3s$ electron first, and both have 10 core electrons shielding it. Magnesium has 12 protons instead of 11, so its $3s$ electrons feel $Z_{\text{eff}} \approx +2$ instead of $+1$ , and it takes more energy to remove one: 738 kJ/mol for Mg compared with 496 kJ/mol for Na.

(1 MJ/mol = 1000 kJ/mol. Photoelectron spectra usually use MJ/mol, and tables of ionization energies usually use kJ/mol.)

Example 4: Explaining ionization energy with Coulomb's law (AP-style)

Which takes more energy: removing a

2p

electron from a neon atom or removing the

3s

electron from a sodium atom? Justify your answer using Coulomb's law.

Step 1: Write the configurations. Ne is

1s^2\,2s^2\,2p^6

and Na is

1s^2\,2s^2\,2p^6\,3s^1

Step 2: Compare distance. Neon's

2p

electron is in shell 2. Sodium's

3s

electron is in shell 3, farther from the nucleus.

Step 3: Compare the charge each electron feels. Neon's

2p

electrons are shielded only by the two

1s

electrons, so

Z_{\text{eff}} \approx 10 - 2 = +8

. Sodium's

3s

electron is shielded by 10 core electrons, so

Z_{\text{eff}} \approx 11 - 10 = +1

Answer: Removing a

2p

electron from neon takes more energy. By Coulomb's law, that electron is attracted more strongly because it feels a much larger effective nuclear charge and is closer to the nucleus. (The measured values are about 2.08 MJ/mol for Ne's

2p

and 0.50 MJ/mol for Na's

3s

.) A full-credit answer names both the charge and the distance, not just one.

Common mistakes

Filling $3d$ before $4s$ , or removing $3d$ electrons before $4s$ when a transition metal forms a cation. The $4s$ subshell fills first and empties first.

Forgetting to add or remove electrons for an ion. The superscripts must add up to the number of electrons, not the number of protons.

Counting a filled $d$ subshell as valence electrons for a main-group element. Selenium has 6 valence electrons, not 16.

Treating Coulomb's law as linear in distance. Doubling the distance gives one fourth of the force, not half.

Explaining ionization energy with only one factor. Say how the nuclear charge (protons and shielding) and the distance (which shell) each affect the attraction.

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