Atoms, Electrons & Periodicity — Chapter quiz · Physical Science: Chemistry, Grade 11
Connect atomic evidence, isotopes, electron arrangements, spectra, and periodic patterns.
Problem 1
Answer and explanation
20
Neutrons = mass number − atomic number = 39 − 19 = 20.
Problem 2
- Dalton's ratios of combining masses
- Thomson's cathode-ray tube
- Rutherford's gold-foil experiment
- Bohr's study of hydrogen light
Answer and explanation
Rutherford's gold-foil experiment
The sharp backward deflection of a few alpha particles in Rutherford's gold-foil experiment revealed the nucleus.
Problem 3
- Neutrons weigh slightly less than 1 amu
- Instruments cannot measure whole numbers
- Electrons add fractional mass
- It is a weighted average of the element's isotopes
Answer and explanation
It is a weighted average of the element's isotopes
The listed mass is the abundance-weighted average of all the element's naturally occurring isotopes.
Problem 4
Answer and explanation
69.8 amu
69.0 × 0.600 + 71.0 × 0.400 = 41.4 + 28.4 = 69.8 amu.
Problem 5
- 1 s² 2 s² 2p⁴
- 1 s² 2 s² 2p⁶
- 1 s² 2 s² 2p²
- 1 s² 2p⁶
Answer and explanation
1 s² 2 s² 2p⁴
Filling in order until 8 electrons are placed gives 1 s² 2 s² 2p⁴ (2 + 2 + 4 = 8).
Problem 6
Answer and explanation
8
2n² = 2 × 2² = 2 × 4 = 8 electrons.
Problem 7
- 3d before 4 s
- 4 s before 3d
- 4p before 4 s
- 3d before 3p
Answer and explanation
4 s before 3d
The 4 s sublevel is slightly lower in energy than 3d, so 4 s fills first.
Problem 8
- shell
- period
- group
- block
Answer and explanation
group
Vertical columns are groups (families); their shared valence-electron count drives similar chemistry.
Problem 9
Answer and explanation
5
For main-group elements, valence electrons = group number − 10 = 15 − 10 = 5.
Problem 10
Answer and explanation
Elements in the same group have the same number of valence electrons in their outermost energy level. Because valence electrons determine how an atom bonds and reacts, group-mates form similar bonds and behave alike chemically.
Problem 11
- decreases
- stays constant
- drops to zero
- increases
Answer and explanation
increases
Rising effective nuclear charge across a period holds the outer electron more tightly, so ionization energy increases.
Problem 12
- Potassium
- They are equal
- It cannot be determined
- Sodium
Answer and explanation
Potassium
Potassium sits below sodium in Group 1; atomic radius increases down a group because a new energy level is added, so potassium is larger.
Problem 13
Answer and explanation
Charge=17−18=−1; atomic number 17 identifies chlorine, so Cl⁻. A shell diagram is an organizing model, not a literal path for an electron.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 14
Answer and explanation
Z=17, A=35, chlorine with charge −1: ³⁵Cl⁻. Changing neutron number while keeping 17 protons makes another chlorine isotope. Changing only electron number changes its ionization state.
Z=17, A=35, chlorine with charge −1: ³⁵Cl⁻. Changing neutron number while keeping 17 protons makes another chlorine isotope. Changing only electron number changes its ionization state.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 15
Answer and explanation
10(.20)+11(.80)=10.8u. The result lies between the two masses and closer to the more abundant one. Mass number counts nucleons; average atomic mass is a weighted value, not a nucleon count.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 16
Answer and explanation
24.6=24.0x+26.0(1−x)=26.0−2.0x, so x=0.70. Fractions are 70% lighter and 30% heavier; 0.70(24.0)+0.30(26.0)=24.6 u.
24.6=24.0x+26.0(1−x)=26.0−2.0x, so x=0.70. Fractions are 70% lighter and 30% heavier; 0.70(24.0)+0.30(26.0)=24.6 u.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 17
Answer and explanation
1 s²2 s²2p⁶3s¹; eleven electrons total and one outer-shell electron. The 2n² shell capacity is not the actual filling order of sublevels.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 18
Answer and explanation
Neutral sulfur: 1s²2s²2p⁶3s²3p⁴, totaling 16 electrons. S²⁻ has 18 electrons, with 3p⁶. Its 16 protons retain sulfur identity.
Neutral sulfur: 1s²2s²2p⁶3s²3p⁴, totaling 16 electrons. S²⁻ has 18 electrons, with 3p⁶. Its 16 protons retain sulfur identity.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 19
Answer and explanation
Outer n=3 gives period 3; five valence electrons give main-group 15; this is phosphorus. The simple group-number valence rule applies to main-group patterns, not all transition elements.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 20
Answer and explanation
Both fit Group 17. Similar valence patterns support related bonding tendencies, while B has an additional occupied shell and different nuclear charge/shielding. Exact radius, energy and reaction behavior need the specified species and conditions.
Both fit Group 17. Similar valence patterns support related bonding tendencies, while B has an additional occupied shell and different nuclear charge/shielding. Exact radius, energy and reaction behavior need the specified species and conditions.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 21
Answer and explanation
K has an occupied outer shell farther from the nucleus with more inner shielding, outweighing its greater proton count in this comparison. General trends have exceptions; do not use a trend arrow as an exact numerical table.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 22
Answer and explanation
All have ten electrons; proton counts are F=9, Ne=10, Mg=12. The usual order is F⁻ > Ne > Mg²⁺ under a compatible radius comparison. Defined atomic/ionic radii and environments must be matched before treating reference numbers as directly interchangeable.
All have ten electrons; proton counts are F=9, Ne=10, Mg=12. The usual order is F⁻ > Ne > Mg²⁺ under a compatible radius comparison. Defined atomic/ionic radii and environments must be matched before treating reference numbers as directly interchangeable.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 23
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
5 energy units
9−4 = 5.
Problem 24
Answer and explanation
No. It supports discrete energy states; modern orbitals are probability distributions.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 25
Answer and explanation
The first photon has 3.0 eV, or about 4.81×10⁻¹⁹ J, giving λ≈4.14×10⁻⁷ m =414 nm. A later allowed 1.5-to-0 transition could emit a 1.5 eV photon; the selected model does not supply transition probabilities.
The first photon has 3.0 eV, or about 4.81×10⁻¹⁹ J, giving λ≈4.14×10⁻⁷ m =414 nm. A later allowed 1.5-to-0 transition could emit a 1.5 eV photon; the selected model does not supply transition probabilities.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.