Unit 1 Practice: Hard
Atomic Structure and Properties · 20 questions. Try each question before revealing the answer and worked explanation.
Question 1
An element has two main isotopes with masses 62.93 amu and 64.93 amu. Its average atomic mass is 63.55 amu. Determine the approximate abundance of the lighter isotope.
Show answer and explanation
Answer: About 69%
Work:
Let x = fraction of lighter isotope.
Then heavier isotope fraction = 1−x.
62.93x + 64.93(1−x) = 63.55
Expand:
62.93x + 64.93 − 64.93x = 63.55
−2.00x = −1.38
x = 0.690
Therefore:
69.0%
Explanation: Because the average is closer to 62.93 than 64.93, the lighter isotope must be more abundant.
Question 2
A compound containing only C, H, and O produces 8.80 g CO₂ and 3.60 g H₂O during combustion. Determine the moles of C atoms and H atoms originally present.
Show answer and explanation
Answer: 0.200 mol C and 0.400 mol H
Work:
CO₂:
8.80 g / 44.01 g/mol ≈ 0.200 mol CO₂
Every CO₂ contains one C:
0.200 mol C
H₂O:
3.60 g / 18.02 g/mol ≈ 0.200 mol H₂O
Every H₂O contains two H:
0.200 × 2 = 0.400 mol H
Explanation: Combustion analysis uses the products to determine how much C and H were present in the original material.
Question 3
A compound is 52.2% C, 13.0% H, and 34.8% O. Determine its empirical formula.
Show answer and explanation
Answer: C₂H₆O
Work:
Assume 100 g.
C:
52.2/12.01 ≈ 4.35 mol
H:
13.0/1.008 ≈ 12.90 mol
O:
34.8/16.00 = 2.175 mol
Divide by 2.175:
C ≈ 2.00 H ≈ 5.93 ≈ 6 O = 1
Formula:
C₂H₆O
Question 4
A compound has empirical formula C₂H₆O and molar mass about 92.1 g/mol. Determine its molecular formula.
Show answer and explanation
Answer: C₄H₁₂O₂
Work:
Empirical formula mass:
2(12.01)+6(1.008)+16.00
≈ 46.07 g/mol
Multiplier:
92.1/46.07 ≈ 2
Multiply subscripts:
C₄H₁₂O₂
Question 5
Explain why atomic radius decreases from Si to Cl even though electrons are being added.
Show answer and explanation
Answer: The increase in nuclear charge is stronger than the increase in shielding.
Explanation: Si through Cl all add electrons to the same principal energy level.
At the same time, each new element gains another proton.
Because electrons in the same shell do not perfectly shield one another, effective nuclear charge increases.
The nucleus therefore pulls the valence shell inward.
Question 6
Why does first ionization energy generally increase across a period but contain exceptions?
Show answer and explanation
Answer: Effective nuclear charge generally increases, but subshell energy and electron pairing can make particular electrons easier to remove.
Explanation: The overall trend is upward because electrons are held more strongly.
However, moving from an s to a p subshell can lower ionization energy because p electrons are higher energy.
Electron pairing in a p orbital can also increase repulsion and make one electron easier to remove.
Question 7
Which has the greater second ionization energy: Na or Mg? Explain.
Show answer and explanation
Answer: Na
Explanation: After losing one electron:
Na⁺ = [Ne]
Mg⁺ = [Ne]3s¹
A second ionization of Na would remove a core electron from the stable neon configuration.
A second ionization of Mg simply removes its remaining 3s valence electron.
Therefore Na's second ionization energy is much larger.
Question 8
Rank O²⁻, F⁻, Na⁺, and Mg²⁺ from largest to smallest ionic radius.
Show answer and explanation
Answer: O²⁻ > F⁻ > Na⁺ > Mg²⁺
Explanation: All four species are isoelectronic, meaning they each have 10 electrons.
Their proton counts increase:
O = 8 F = 9 Na = 11 Mg = 12
More protons pull the same number of electrons more strongly.
Therefore radius becomes smaller as nuclear charge increases.
Question 9
Why is O²⁻ larger than Mg²⁺ even though both contain 10 electrons?
Show answer and explanation
Answer: Mg²⁺ has four more protons pulling on the same number of electrons.
Explanation: O²⁻ has 8 protons.
Mg²⁺ has 12.
The much greater nuclear charge of Mg²⁺ contracts its electron cloud.
Question 10
A PES spectrum corresponds to 1s²2s²2p⁶3s²3p³. Identify the element.
Show answer and explanation
Answer: Phosphorus, P
Explanation: Count the electrons:
2 + 2 + 6 + 2 + 3 = 15 electrons
Atomic number 15 is phosphorus.
Question 11
Which subshell in Question 10 would generally produce the peak representing the greatest number of electrons?
Show answer and explanation
Answer: 2p⁶
Explanation: PES peak intensity is related to the number of electrons occupying a subshell.
The 2p subshell contains six electrons, more than any other listed subshell.
Question 12
Explain one way PES could distinguish Mg from Al.
Show answer and explanation
Answer: Al has a 3p electron while Mg does not.
Explanation: Mg:
[Ne]3s²
Al:
[Ne]3s²3p¹
Al's PES spectrum should contain an additional lower-binding-energy peak corresponding to the 3p electron.
That extra peak identifies Al.
Question 13
An element shows a huge increase between its third and fourth ionization energies. What group is it likely in?
Show answer and explanation
Answer: Group 13
Explanation: A large jump after the third electron suggests the atom has three valence electrons.
Removing the fourth electron would break into a stable core.
Main-group elements with three valence electrons are in Group 13.
Question 14
An element shows a very large increase between its second and third ionization energies. What common ionic charge would you predict?
Show answer and explanation
Answer: +2
Explanation: The first two electrons are relatively easy to remove.
The third would be a core electron.
Therefore the atom tends to lose two valence electrons and form a 2+ ion.
Question 15
Why is chromium commonly written [Ar]3d⁵4s¹ instead of [Ar]3d⁴4s²?
Show answer and explanation
Answer: The half-filled 3d⁵ arrangement has extra stability.
Explanation: The 4s and 3d orbitals are very close in energy.
Moving one 4s electron into 3d creates a half-filled d subshell with one electron in each d orbital.
This arrangement is somewhat lower in energy.
Question 16
Why is copper commonly written [Ar]3d¹⁰4s¹ instead of [Ar]3d⁹4s²?
Show answer and explanation
Answer: A completely filled 3d¹⁰ subshell provides extra stability.
Explanation: Because 4s and 3d are close in energy, moving one 4s electron into 3d creates the favorable filled d subshell.
Question 17
A 5.00 g mixture contains 2.00 g CaCO₃ and 3.00 g another substance. Determine the mass percent of each component.
Show answer and explanation
Answer: 40.0% CaCO₃ and 60.0% other substance
Work:
CaCO₃:
(2.00/5.00)(100) = 40.0%
Other:
(3.00/5.00)(100) = 60.0%
Explanation: The component percentages should add to 100%.
Question 18
Why do isotopes of the same element usually have almost identical chemical behavior?
Show answer and explanation
Answer: They have the same number and arrangement of electrons.
Explanation: Chemical reactions mostly involve valence electrons.
Isotopes differ in their number of neutrons, which changes mass but usually does not significantly change the electron configuration responsible for bonding.
Question 19
A student says, “Cl has more protons than Na, so Cl should be the larger atom.” Explain why that reasoning is incorrect.
Show answer and explanation
Answer: More protons actually pull the electrons closer when the electrons are in the same principal shell.
Explanation: Na and Cl are both in Period 3.
Cl has greater effective nuclear charge.
Instead of making the atom larger, the additional nuclear attraction contracts the electron cloud.
Therefore:
Na is larger than Cl.
Question 20
Explain the relationship among effective nuclear charge, atomic radius, and ionization energy across a period.
Show answer and explanation
Answer: As effective nuclear charge increases, atomic radius generally decreases and ionization energy generally increases.
Explanation: Moving across a period adds protons while valence electrons remain in the same main energy level.
The stronger effective nuclear charge pulls electrons closer:
Zeff ↑ → radius ↓
Because the electrons are held more strongly, more energy is needed to remove one:
radius ↓ / attraction ↑ → ionization energy ↑
These trends come from the same underlying electrostatic idea rather than being unrelated facts.