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8.4

Atomic radius and ionization energy

Your guide: Miss SanaCounts atoms the way you count rupees: carefully.

Size and grip

An atom's electron cloud has no sharp edge, so its size is measured when it is bonded to an identical atom. Atomic radius is half the distance between the two nuclei. Two bonded carbon atoms are 154 pm apart, so the radius of carbon is 77 pm (1 pm = 10−12 m).

Across period 2, atoms get smaller: from Li (152 pm) to Ne (69 pm). Each step adds a proton to the nucleus and an electron to the same outer shell, so the bigger nuclear charge pulls the shell in.

Down a group, atoms get bigger: Li 152, Na 186, K 227, Rb 248, Cs 265 pm. Each step adds a new shell, and the inner electrons shield the outer ones from the nucleus.

Ionization energy is the energy needed to remove the most loosely held electron from an isolated gaseous atom, in kJ mol−1. Sodium needs +496 kJ mol−1. When there are more outer electrons, each one is harder to remove than the last: magnesium needs +737 for the first and +1450 for the second.

Ionization energy follows size. In a smaller atom the nucleus holds the outer electrons more tightly, so more energy is needed. It therefore generally increases across a period (Li 520 to Ne 2081, with small dips at B and O) and decreases down a group (Li 520 to Cs 377).

Smaller atom, tighter grip, higher ionization energy.

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Notes, short questions and MCQs

Read the full notes: key terms, model answers and MCQs with answers

Size and grip

An atom's electron cloud has no sharp edge, so its size is measured when it is bonded to an identical atom. Atomic radius is half the distance between the two nuclei. Two bonded carbon atoms are 154 pm apart, so the radius of carbon is 77 pm (1 pm = 10−12 m).

Across period 2, atoms get smaller: from Li (152 pm) to Ne (69 pm). Each step adds a proton to the nucleus and an electron to the same outer shell, so the bigger nuclear charge pulls the shell in.

Down a group, atoms get bigger: Li 152, Na 186, K 227, Rb 248, Cs 265 pm. Each step adds a new shell, and the inner electrons shield the outer ones from the nucleus.

Ionization energy is the energy needed to remove the most loosely held electron from an isolated gaseous atom, in kJ mol−1. Sodium needs +496 kJ mol−1. When there are more outer electrons, each one is harder to remove than the last: magnesium needs +737 for the first and +1450 for the second.

Ionization energy follows size. In a smaller atom the nucleus holds the outer electrons more tightly, so more energy is needed. It therefore generally increases across a period (Li 520 to Ne 2081, with small dips at B and O) and decreases down a group (Li 520 to Cs 377).

Smaller atom, tighter grip, higher ionization energy.

Key terms

Atomic radius
Comparing atom sizes.
Picometre (pm)
Unit for atomic radius.
Shielding
Why atoms grow down a group.
Ionization energy
How easily a metal forms ions.
Second ionization energy
Always larger than the first.

Short questions with model answers

  1. Q1. The first ionization energy of magnesium is less than the second. Give the reason. (Short question viii, p. 117)

    • After the first
    • Stronger pull

    The second electron is removed from a positive ion, so it needs more energy: 1450 versus 737 kJ mol−1.

  2. Q2. Which element has the largest and which the smallest atomic radius in period 3? (Short question vi, p. 117)

    • Trend
    • Ends of period 3

    Largest: sodium. Smallest: argon (at the right-hand end).

Common mistakes

  • ✗ More electrons across a period make atoms bigger.

    ✓ They enter the same shell while nuclear charge rises, so atoms shrink (p. 111).

  • ✗ Ionization energy increases down a group.

    ✓ It decreases: bigger atoms hold outer electrons less tightly (Table 8.7).

  • ✗ Atomic radius is the full distance between two nuclei.

    ✓ It is half that distance: 154 pm gives 77 pm for carbon (p. 111).

MCQs

  1. 1. Which order of atomic radii is correct for O, F and N? (Exercise ix, p. 117)

    1. (a) O < F < N
    2. (b) N < F < O
    3. (c) F < O < N
    4. (d) O < N < F
    Show answer

    (c) Table 8.4: F 71, O 73, N 75 pm.

  2. 2. Two bonded chlorine nuclei are 198 pm apart. The atomic radius of chlorine is:

    1. (a) 198 pm
    2. (b) 99 pm
    3. (c) 396 pm
    4. (d) 49.5 pm
    Show answer

    (b) Half the distance, as for carbon on p. 111.

  3. 3. Which group 1 element has the lowest ionization energy in Table 8.7?

    1. (a) Li
    2. (b) Na
    3. (c) K
    4. (d) Cs
    Show answer

    (d) Cs 377 kJ mol−1, the biggest atom.

  4. 4. Across a period, ionization energy generally:

    1. (a) Increases
    2. (b) Decreases
    3. (c) Stays the same
    4. (d) Becomes negative
    Show answer

    (a) p. 112.

  5. 5. Atoms get bigger down a group mainly because:

    1. (a) A new shell is added each period
    2. (b) Protons are lost
    3. (c) Nuclear charge falls
    4. (d) Electrons are lost
    Show answer

    (a) p. 111.

Quick revision

  • Atomic radius is half the distance between two identical bonded nuclei, in pm.
  • Radius decreases across a period and increases down a group.
  • Ionization energy increases across a period and decreases down a group.
  • Each extra electron removed needs more energy than the last.

Before you start

What is Class 9 Chemistry Atomic radius and ionization energy?
Atomic radius and ionization energy is the open Class 9 Chemistry lesson for Matric Part 1, Punjab board. It is a step-by-step lesson, not a PDF of the book.
Which boards is this Class 9 Chemistry lesson for?
The same Punjab textbook: Lahore, Rawalpindi, Gujranwala, Faisalabad, Multan, Sargodha, Bahawalpur, Dera Ghazi Khan, Sahiwal.
Can I study this in Urdu?
Yes. English, Urdu or Roman Urdu. Science terms stay as the exam writes them.
Is this a notes PDF?
No. The book stays with you. This page is the lesson and the check.

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