Atomic Structure & the Periodic Table — Science

  1. Grade 8 — Atomic Structure

    ↑ Deepens: matter is now explained by the atom.

  2. Grade 8 — The Periodic Table
  3. Grade 11 — Atomic Models & Subatomic Particles

    ↑ Deepens: the atomic model, now down to electron configuration.

  4. Grade 11 — Isotopes & Average Atomic Mass
  5. Grade 11 — Electron Configuration
  6. Grade 11 — The Periodic Table
  7. Grade 11 — Periodic Trends
  8. Grade 11 — Nuclear Chemistry

Apply it in a new setting

Problem 1

Element X occurs as two isotopes: X-10 (mass 10.013 amu, 20.0% abundance) and X-11 (mass 11.009 amu, 80.0% abundance). What is the average atomic mass of element X, in amu?
Hints

Average atomic mass is a weighted average: multiply each isotope's mass by its fractional abundance, then add.

Convert the percents to fractions: 20.0% → 0.200 and 80.0% → 0.800.

avg = (0.200 × 10.013) + (0.800 × 11.009).

Answer and explanation

10.81 amu

Weighted average = Σ (fractional abundance × isotope mass).

= (0.200 × 10.013) + (0.800 × 11.009) = 2.003 + 8.807.

= 10.81 amu.

Problem 2

Chlorine has two stable isotopes: Cl-35 (mass 34.969 amu) and Cl-37 (mass 36.966 amu). The average atomic mass of chlorine is 35.45 amu. What is the percent abundance of Cl-35? Give the percent (just the number).
Hints

Let the fraction of Cl-35 be x; then the fraction of Cl-37 is (1 − x).

Set up the weighted average: 34.969x + 36.966(1 − x) = 35.45.

Solve for x, then multiply by 100 to get a percent.

Answer and explanation

75.9 %

Let x = fraction of Cl-35, so (1 − x) = fraction of Cl-37: 34.969x + 36.966(1 − x) = 35.45.

34.969x + 36.966 − 36.966x = 35.45 → −1.997x = −1.516 → x = 0.759.

x = 0.759 = 75.9% Cl-35.

Problem 3

A radioactive isotope used in medicine has a half-life of 8.0 days. A hospital receives an 80. mg sample. How many milligrams of the isotope remain after 24 days?
Hints

First find how many half-lives fit into the elapsed time: 24 ÷ 8.0.

Each half-life leaves half of what you had: multiply by ½ once per half-life.

80 → 40 → 20 → …

Answer and explanation

10 mg

Number of half-lives = 24 days ÷ 8.0 days = 3.

Halve the sample three times: 80 → 40 → 20 → 10.

10 mg remain.

Problem 4

A sulfide ion, S²⁻, forms when a sulfur atom (atomic number 16) gains electrons to complete its octet. How many electrons does one S²⁻ ion have?
Hints

A neutral atom has as many electrons as its atomic number.

A negative charge means the atom has GAINED that many electrons.

Start with 16 electrons in neutral sulfur, then add 2 for the 2− charge.

Answer and explanation

18 electrons

Neutral sulfur (Z = 16) has 16 electrons.

A 2− charge means 2 extra electrons were gained: 16 + 2.

= 18 electrons.

Problem 5

A ground-state phosphorus atom (atomic number 15) fills its orbitals following Hund's rule. How many unpaired electrons does it have?
Hints

Write the configuration: 1s² 2s² 2p⁶ 3s² 3p³ — then focus on the outermost (valence) subshell.

Hund's rule: electrons spread out one per orbital across a subshell before any pairing begins.

The 3p subshell has three orbitals and three electrons to place.

Answer and explanation

3 electrons

Phosphorus: 1s² 2s² 2p⁶ 3s² 3p³; the 1s, 2s, 2p, and 3s subshells are completely filled (all paired).

By Hund's rule the three 3p electrons each occupy a separate 3p orbital before pairing.

That leaves 3 unpaired electrons.

Problem 6

Among the Period 3 main-group elements, one has the highest electronegativity (setting aside the noble gas argon). Which element is it, and what is the correct reason?
  1. Sodium (Na), because with only one valence electron it attracts bonding electrons most strongly.
  2. Argon (Ar), because a completely full outer shell lets it pull on electrons the hardest.
  3. Chlorine (Cl), because electronegativity rises across a period as the nuclear charge increases and the atomic radius shrinks.
  4. Magnesium (Mg), because middle-of-the-period metals attract shared electrons best.
Hints

Electronegativity is the pull an atom exerts on the electrons it shares in a bond.

Across a period (left → right), nuclear charge rises and atoms get smaller — what does that do to the pull on bonding electrons?

Set aside the noble gas and look at the element farthest right that still forms ordinary bonds.

Answer and explanation

Chlorine (Cl), because electronegativity rises across a period as the nuclear charge increases and the atomic radius shrinks.

Across Period 3, increasing nuclear charge and shrinking radius pull bonding electrons more tightly, so electronegativity rises from left to right.

Ignoring argon (a noble gas that essentially does not form ordinary bonds), the rightmost element is chlorine.

So the most electronegative is chlorine (choice C), because of the across-a-period trend — not a full shell or a 'metals attract best' idea.

Problem 7

The first four ionization energies of an element are, in order, 738, 1451, 7733, and 10540 kJ/mol. The huge jump between the 2nd and 3rd values tells you the element most likely has which valence structure?
  1. 1 valence electron (Group 1), because only one electron is easy to remove.
  2. 3 valence electrons (Group 13), because the big jump comes after three removals.
  3. 4 valence electrons (Group 14), because there are four listed ionization energies.
  4. 2 valence electrons (Group 2), because after 2 valence electrons are gone the 3rd must come from a stable inner shell.
Hints

A sudden large jump in ionization energy appears when you start pulling electrons out of a full inner (core) shell.

Count how many electrons come off relatively 'easily' before the big jump.

Two low values, then a spike at the 3rd removal — how many valence electrons is that?

Answer and explanation

2 valence electrons (Group 2), because after 2 valence electrons are gone the 3rd must come from a stable inner shell.

Valence electrons come off relatively easily; removing a core electron takes far more energy, which shows up as a sudden jump.

Here the jump is between the 2nd (1451) and 3rd (7733) ionization energies, so exactly 2 electrons were easy to remove.

That means 2 valence electrons — a Group 2 element (choice D).

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