ChemGem

How to Find Valence Electrons from Ionization Energies (Step-by-Step, with Practice)

Why do atoms shrink across a row and hold their electrons more tightly? The answer is nuclear pull and shielding. This lesson uses those ideas to explain ionization energy, electron affinity and the big jump that reveals valence electrons.

Spot it, solve it, practice it: 4 steps, 1 worked example, 9 practice problems. Free, no account.

ChemGemUpdated Section 3.5

The route · 4 steps

  1. Use the group as a first clue to the valence electrons.
  2. Compare each successive ionization energy with the one before it.
  3. Find the jump much larger than the others.
  4. Electrons removed before that jump are the valence electrons; the next is a core electron.
Try one now See a worked example
Spot it

How to recognize ionization energy questions

Look for successive ionization energies (a list in kJ/mol), core electrons, shielding, or electron affinity compared with ionization.

Example wording, with the clue words marked:

An atom has successive ionization energies 590, 1145, and 4912 kJ/mol. How many valence electrons does the large jump suggest?

Not sure it’s this type? Compare the neighbors:

Step-by-step

Worked example: ionization energy

The idea

First ionization energy: harder to remove an electron. First ionization energy generally increases UP and RIGHT; noble gases ARE included. It is the energy needed to remove one electron from a neutral gaseous atom. Compare Na and Cl: Cl generally holds its outer electron more tightly. There are exceptions, such as Mg/Al.
First ionization energy: the energy to remove one electron from a neutral gaseous atom. It generally rises up and to the right; noble gases are included. Cl holds its outer electron more tightly than Na.
Complete counts: lithium 2 plus 1 electrons, sodium 2 plus 8 plus 1, chlorine 2 plus 8 plus 7, magnesium 2 plus 8 plus 2. Nuclei label their proton counts. Magnesium has a large jump between second and third ionization energies.
Li 2 + 1, Na 2 + 8 + 1, Cl 2 + 8 + 7, Mg 2 + 8 + 2. In neutral atoms, electrons equal protons. The rings are schematic, not tracks.
Two forces behind the trends
Down a group
An occupied shell is added.
Inner electrons shield the outer ones from part of the nuclear pull.
The atom usually gets larger.
Across a period
Nuclear charge rises; electrons enter the same main shell.
Attraction generally strengthens.
Radius shrinks; first ionization energy generally rises.
First ionization energy across Period 2
0500100015002000LiBeBCNOFNeB below BeO below NElement, left to rightkJ/mol
The overall rise has two dips. B’s electron sits in a higher-energy 2p subshell; O’s first paired 2p electrons repel. Mg/Al in Period 3 is the same kind of exception as Be/B. Trend arrows are not exception-free laws.
Three energies, three processes
QuantityWhat happensPattern
First ionization energyRemove one electron from a gaseous neutral atomGenerally up and right
Successive ionization energiesRemove more electrons from ever more positive ionsEach is larger than the last; a very large jump means a core electron
Electron affinityAdd an electron to a gaseous atomFavorable attachment releases energy; sign depends on convention. Less regular than radius
Use your course’s sign convention for electron affinity.

On the periodic table: Group 2 atoms, such as magnesium (Mg), have 2 valence electrons, so they often show the jump between the second and third ionizations.

On the periodic table: Beryllium (Be) and boron (B) sit side by side in Period 2, as do nitrogen (N) and oxygen (O): the two first-ionization exceptions.

On the periodic table: The atomic number gives the proton count (N 7, O 8, Mg 12). The decimal mass (14.01, 16.00, 24.31) is an average, not an isotope’s mass number; use supplied rounded masses when given.

Worked example · 4 steps

Magnesium is in Group 2. Where should its large ionization-energy jump fall, and why?

  1. 1
    Read the group

    On the periodic table: Mg is Group 2, Period 3, atomic number 12.

    SymbolAtomic numberGroupShells
    Mg1222 + 8 + 2
  2. 2
    Count valence electrons

    The outer shell holds 2 electrons; the 10 inside are core electrons.

    Mg: [Ne] 3s2 → 12 − 10 = 2 valence electrons
  3. 3
    Predict the jump

    Removing a third electron would reach the core: closer to the nucleus and less shielded, from an already positive ion.

    IonizationElectron removedExpected size
    1st3s (valence)Small
    2nd3s (valence)Somewhat larger
    3rd2p (core)Very large jump
  4. 4
    Check against data

    The group is a starting clue, not a substitute for supplied data. Measured values show the jump.

    Magnesium’s successive ionization energies
    1st738 kJ/mol2nd1451 kJ/mol3rd7733 kJ/mol
    1451738 ≈ 2.0     77331451 ≈ 5.3
  5. ✓
    Answer

    The jump falls between the 2nd and 3rd ionizations: Mg has 2 valence electrons, and the third electron comes from the core.

Your turn

Ionization energy practice problems

Answer first, then check. Hints open one at a time; the answer and full solution open after you try.

Problem 1 of 9 · warm-up

An atom has successive ionization energies 590, 1145, and 4912 kJ/mol. How many valence electrons does the large jump suggest?

Hint 1 · the clue

Find the unusually large increase, not the largest absolute value alone.

Hint 2 · the next move

"Successive ionization energies" and "the large jump" mean find where the energy suddenly leaps; electrons before it are valence.

Your path:

  1. Compare each step to the one before
  2. Find the unusually large jump
  3. Electrons removed before the jump are valence
Answer

Answer: 2

Full worked solution

What you know: Successive ionization energies: 590, 1145, and 4912 kJ/mol. What you want: The number of valence electrons.

  1. Compare each step, not the raw numbers. 590 → 1145 is about 1.9 times larger.
  2. 1145 → 4912 is about 4.3 times larger. That is the unusually large jump.
  3. The jump comes right before the third electron. That third electron must be a core electron, much closer to the nucleus.
  4. So the first two removed were valence electrons: 2.

Check: These are the measured values for calcium (Ca), a Group 2 atom, which shows exactly this pattern: a big jump between the second and third ionizations.

Watch out: Each ionization energy is always higher than the last, so a rise alone means nothing. Look for the rise that is much bigger than the others.

Answer: 2

Problem 2 of 9

An atom has successive ionization energies 550, 1064, and 4138 kJ/mol. How many valence electrons does the large jump suggest?

Hint 1 · the clue

Find the unusually large increase, not the largest absolute value alone.

Hint 2 · the next move

"Successive ionization energies" and "the large jump" mean find where the energy suddenly leaps; electrons before it are valence.

Your path:

  1. Compare each step to the one before
  2. Find the unusually large jump
  3. Electrons removed before the jump are valence
Answer

Answer: 2

Full worked solution

What you know: Successive ionization energies: 550, 1064, and 4138 kJ/mol. What you want: The number of valence electrons.

  1. Compare each step, not the raw numbers. 550 → 1064 is about 1.9 times larger.
  2. 1064 → 4138 is about 3.9 times larger. That is the unusually large jump.
  3. The jump comes right before the third electron. That third electron must be a core electron, much closer to the nucleus.
  4. So the first two removed were valence electrons: 2.

Check: These are the measured values for strontium (Sr), a Group 2 atom, which shows exactly this pattern: a big jump between the second and third ionizations.

Watch out: Each ionization energy is always higher than the last, so a rise alone means nothing. Look for the rise that is much bigger than the others.

Answer: 2

Problem 3 of 9

An atom has successive ionization energies 503, 965, and 3458 kJ/mol. How many valence electrons does the large jump suggest?

Hint 1 · the clue

Find the unusually large increase, not the largest absolute value alone.

Hint 2 · the next move

"Successive ionization energies" and "the large jump" mean find where the energy suddenly leaps; electrons before it are valence.

Your path:

  1. Compare each step to the one before
  2. Find the unusually large jump
  3. Electrons removed before the jump are valence
Answer

Answer: 2

Full worked solution

What you know: Successive ionization energies: 503, 965, and 3458 kJ/mol. What you want: The number of valence electrons.

  1. Compare each step, not the raw numbers. 503 → 965 is about 1.9 times larger.
  2. 965 → 3458 is about 3.6 times larger. That is the unusually large jump.
  3. The jump comes right before the third electron. That third electron must be a core electron, much closer to the nucleus.
  4. So the first two removed were valence electrons: 2.

Check: These are the measured values for barium (Ba), a Group 2 atom, which shows exactly this pattern: a big jump between the second and third ionizations.

Watch out: Each ionization energy is always higher than the last, so a rise alone means nothing. Look for the rise that is much bigger than the others.

Answer: 2

Problem 4 of 9

An atom has successive ionization energies 900, 1757, and 14,849 kJ/mol. How many valence electrons does the large jump suggest?

Hint 1 · the clue

Find the unusually large increase, not the largest absolute value alone.

Hint 2 · the next move

"Successive ionization energies" and "the large jump" mean find where the energy suddenly leaps; electrons before it are valence.

Your path:

  1. Compare each step to the one before
  2. Find the unusually large jump
  3. Electrons removed before the jump are valence
Answer

Answer: 2

Full worked solution

What you know: Successive ionization energies: 900, 1757, and 14,849 kJ/mol. What you want: The number of valence electrons.

  1. Compare each step, not the raw numbers. 900 → 1757 is about 2.0 times larger.
  2. 1757 → 14,849 is about 8.5 times larger. That is the unusually large jump.
  3. The jump comes right before the third electron. That third electron must be a core electron, much closer to the nucleus.
  4. So the first two removed were valence electrons: 2.

Check: These are the measured values for beryllium (Be), a Group 2 atom, which shows exactly this pattern: a big jump between the second and third ionizations.

Watch out: Each ionization energy is always higher than the last, so a rise alone means nothing. Look for the rise that is much bigger than the others.

Answer: 2

Problem 5 of 9

An atom has successive ionization energies 738, 1451, and 7733 kJ/mol. How many valence electrons does the large jump suggest?

Hint 1 · the clue

Find the unusually large increase, not the largest absolute value alone.

Hint 2 · the next move

"Successive ionization energies" and "the large jump" mean find where the energy suddenly leaps; electrons before it are valence.

Your path:

  1. Compare each step to the one before
  2. Find the unusually large jump
  3. Electrons removed before the jump are valence
Answer

Answer: 2

Full worked solution

What you know: Successive ionization energies: 738, 1451, and 7733 kJ/mol. What you want: The number of valence electrons.

  1. Compare each step, not the raw numbers. 738 → 1451 is about 2.0 times larger.
  2. 1451 → 7733 is about 5.3 times larger. That is the unusually large jump.
  3. The jump comes right before the third electron. That third electron must be a core electron, much closer to the nucleus.
  4. So the first two removed were valence electrons: 2.

Check: These are the measured values for magnesium (Mg), a Group 2 atom, which shows exactly this pattern: a big jump between the second and third ionizations.

Watch out: Each ionization energy is always higher than the last, so a rise alone means nothing. Look for the rise that is much bigger than the others.

Answer: 2

Problem 6 of 9

Why can removing a core electron require much more energy?

Hint 1 · the clue

Core electrons sit closer to the nucleus and are less shielded than valence electrons.

Hint 2 · the next move

"Removing a core electron" and "more energy" ask why inner electrons are held so tightly.

Your path:

  1. Core electrons sit closer to the nucleus
  2. Less shielding → stronger pull
  3. Ion is already positive, pulls harder
Answer

Answer: It is more tightly bound

Full worked solution

What you know: Core electrons are the inner electrons, below the valence shell. What you want: Why removing one takes much more energy.

  1. Core electrons sit in a lower shell, closer to the nucleus.
  2. They are shielded by fewer electrons, so they feel more of the nuclear pull.
  3. By then the ion is already positive, which pulls even harder. In short: It is more tightly bound.

Why it matters: This is why ionization-energy data show a huge jump right after the last valence electron is removed.

Answer: It is more tightly bound

Problem 7 of 9

Is electron affinity the same as ionization energy?

Hint 1 · the clue

Compare what each one measures: adding an electron or removing one.

Hint 2 · the next move

"Electron affinity" versus "ionization energy" asks you to compare adding an electron with removing one.

Your path:

  1. Ionization energy: remove an electron
  2. Electron affinity: add an electron
  3. Different processes → not the same
Answer

Answer: No

Full worked solution

What you know: Two quantities: electron affinity and ionization energy. What you want: Whether they are the same.

  1. Ionization energy is the energy needed to remove an electron from a gaseous atom.
  2. Electron affinity concerns adding an electron to a gaseous atom.
  3. Removing and adding are different processes, so the answer is No.

Why it matters: The energy changes differ too. Removing an electron always takes energy in, while adding an electron to many atoms, such as Cl, releases energy.

Watch out: The two trends look similar on the table, but that does not make them the same thing.

Answer: No

Problem 8 of 9

Is first ionization energy always increasing across a row?

Hint 1 · the clue

On the periodic table: Look across Period 2: beryllium (Be) to boron (B), and nitrogen (N) to oxygen (O).

The trend is general. Check the pairs Be → B and N → O.

Hint 2 · the next move

The word "always" about a trend is the clue: trends are general patterns with known exceptions.

Your path:

  1. Trend: generally rises across a row
  2. Recall exceptions: B below Be, O below N
  3. One exception means not always
Answer

Answer: No

Full worked solution

What you know: First ionization energy generally increases across a row. What you want: Whether it always does.

  1. "Generally" is the clue: trends are patterns, not exception-free laws.
  2. Period 2 has two known exceptions: B is lower than Be, and O is lower than N.
  3. These come from subshell energy (B's electron is in 2p) and paired-electron repulsion (O's first paired 2p electron). So the answer is No.

Why it matters: One exception is enough to make "always" false.

On the periodic table: Beryllium (Be) and boron (B) sit side by side in Period 2, as do nitrogen (N) and oxygen (O). Each pair is one step to the right across the same row.

Answer: No

YOUR PERIODIC-TABLE REFERENCE

7NNitrogen14.01Group 15 · Period 2
8OOxygen16.00Group 16 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Problem 9 of 9

Magnesium’s first three ionization energies are 738, 1451 and 7733 kJ/mol. How many valence electrons does Mg have, and why is the third value so much larger than the second?

Hint 1 · the clue

Divide each ionization energy by the one before it. The step that is several times larger marks where the core begins.

Hint 2 · the next move

"First three ionization energies" and "so much larger" mean find the big jump; electrons before it are valence.

Your path:

  1. Compare each step to the one before
  2. Big jump → next electron is core
  3. Count electrons before the jump
  4. Core electron: closer, less shielded
Answer

On the periodic table: Magnesium (Mg) is in Group 2, so it has 2 valence electrons.

Mg has 2 valence electrons. The third electron would come from the inner (core) shell, which is closer to the nucleus and less shielded.

Full worked solution

What you know: Mg’s first three ionization energies: 738, 1451 and 7733 kJ/mol. What you want: the number of valence electrons, and why the third value is so large.

  1. Compare each step with the one before: 1451 ÷ 738 ≈ 2.0, and 7733 ÷ 1451 ≈ 5.3.
  2. The big jump comes before the 3rd electron, so the first 2 electrons come from the valence shell (3s).
  3. The 3rd electron would come from shell 2, the core. It is closer to the nucleus and shielded by fewer inner electrons, so it is held much more tightly. So: Mg has 2 valence electrons. The third electron would come from the inner (core) shell, which is closer to the nucleus and less shielded.

Check: Mg is in Group 2, which also predicts 2 valence electrons. Mg2+ has the neon core, [Ne].

Watch out: Every ionization energy is bigger than the one before. Look for the step that is several times larger, not just larger.

On the periodic table: Magnesium (Mg) is in Group 2, so it has 2 valence electrons.

Answer: Mg has 2 valence electrons. The third electron would come from the inner (core) shell, which is closer to the nucleus and less shielded.

YOUR PERIODIC-TABLE REFERENCE

12MgMagnesium24.31Group 2 · Period 3

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Watch out

Common ionization energy mistakes