Atoms, Electrons & Periodicity — Chapter Practice, Grade 11

An atom's mass and positive charge are concentrated in a tiny nucleus

Most particles passed through (mostly empty space), but the rare sharp deflections showed a small, dense, positively charged nucleus.

electron

Electrons are almost weightless compared with protons and neutrons, so they contribute almost nothing to atomic mass.

46

Neutrons = A − Z = 80 − 34 = 46.

18

A 2− ion has gained 2 electrons: electrons = protons + 2 = 16 + 2 = 18.

Thomson spread the positive charge thinly throughout the whole atom, but Rutherford's sharp alpha-particle deflections showed the positive charge and nearly all the mass are packed into a tiny, dense central nucleus instead of being spread out.

15

The atomic number is the proton count, so protons = 15.

Electrons affect ion charge; protons identify the element.

Magnesium, charge 2+; positive charges exceed negative by two.

State what the supplied observations or conditions establish, and separate that from any assumption.

Quantum models describe orbital probability distributions rather than definite classical trajectories; simple drawings omit that distinction.

It does not specify a particular electron position or trajectory. Its shading must be interpreted as a probability representation under the stated model.

An orbital describes an electron state with a spatial probability distribution. A cloud or boundary drawing does not show a solid shell, exact electron location or a classical path around the nucleus. Allowed energy levels account for discrete transitions; different orbitals can have different shapes and energies. Bohr-style shells remain useful bookkeeping for simple comparisons, but a multi-electron atom requires a quantum model and electron interactions rather than planets moving on fixed circular roads.

It does not specify a particular electron position or trajectory. Its shading must be interpreted as a probability representation under the stated model.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the chemistry relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

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.

It is the weighted average of all the element's naturally occurring isotopes

The listed value averages every natural isotope, weighted by abundance, so it is rarely a whole number.

protons; neutrons

Isotopes share the proton count (same element) but differ in neutrons, which changes the mass number.

20.2 amu

Average = 20.0 × 0.900 + 22.0 × 0.100 = 18.0 + 2.2 = 20.2 amu.

63.55 amu

62.93 × 0.6915 = 43.52 amu; 64.93 × 0.3085 = 20.03 amu; sum = 63.55 amu.

Chlorine-35 must be far more abundant than chlorine-37 (roughly 75% versus 25%), because a weighted average is pulled toward the mass of the most common isotope.

80 %

Let the mass-10 fraction be x: 10x + 11(1 − x) = 10.8 → 11 − x = 10.8 → x = 0.20. So the mass-11 fraction is 0.80 = 80%.

An arithmetic average would incorrectly assume equal abundance.

10.0×0.20+11.0×0.80=10.8 u.

State what the supplied observations or conditions establish, and separate that from any assumption.

Yes, if their isotope abundances differ; proton count remains the same.

The X-11 fraction would increase, while the X-10 fraction x would decrease.

To infer an unknown abundance in a two-isotope model, define one fraction x and use one minus x for the other. The measured mean supplies a weighted-balance equation. Solving it is justified only if the two isotope masses and the sample model are appropriate; an unrecognized third isotope, mass bias or uncertainty changes the inference. A solution outside zero to one is a warning that the inputs or assumptions are inconsistent, not a physically negative abundance.

The X-11 fraction would increase, while the X-10 fraction x would decrease.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the chemistry relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

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.

2

By the Pauli exclusion principle, an orbital holds at most two electrons, and they must have opposite spins.

d

The d sublevel has 5 orbitals × 2 electrons = 10 electrons.

18

2n² = 2 × 3² = 2 × 9 = 18 electrons.

1 s² 2 s² 2p⁶ 3 s² 3p³

Filling in Aufbau order and stopping at 15 electrons gives 1 s² 2 s² 2p⁶ 3 s² 3p³ (2+2+6+2+3 = 15).

Electrons fill in order of increasing energy, not simply increasing n. The 4 s sublevel happens to be slightly lower in energy than 3d, so by the Aufbau principle electrons occupy 4 s first.

7

Valence electrons are those in the highest energy level (n = 4): 4 s² + 4p⁵ = 7. The inner 3d¹⁰ electrons are core, not valence.

Check electron total and distinguish an orbital from a shell.

1s² 2s² 2p⁶; exponents total 10.

State what the supplied observations or conditions establish, and separate that from any assumption.

It loses the two outer 3s electrons, giving [Ne]; the nucleus remains magnesium.

Removing an electron from an isolated atom requires energy. Other reaction steps and interactions must be included in the full favorable or unfavorable account.

Valence electrons are the electrons most involved in a chemical bonding model. Main-group outer configurations help explain recurring groups and common ion charges: sodium often loses its 3s electron and chlorine often gains one electron into its 3p subshell. The octet rule is a useful main-group pattern, not a cause or universal command that atoms want eight electrons. Whether a reaction occurs depends on the complete energy balance, interaction partners and conditions, including lattice or solution stabilization.

Removing an electron from an isolated atom requires energy. Other reaction steps and interactions must be included in the full favorable or unfavorable account.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the chemistry relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

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.

group (family)

Vertical columns are groups, or families; horizontal rows are periods.

valence electrons

Matching valence-electron counts give group-mates similar bonding behavior.

6

For main-group elements, valence electrons = group number − 10 = 16 − 10 = 6.

The d-block, groups 3–12

The transition metals make up the d-block in the center of the table, groups 3–12.

A period is a horizontal row; every element in it has its outermost electrons in the same highest energy level (same n). A group is a vertical column; its elements all have the same number of valence electrons, which is why they share similar chemistry.

metalloid

Elements along the staircase, like silicon, are metalloids with in-between properties that make them useful semiconductors.

Group labels 1–18 differ from valence-electron counts.

Period 4, group 2, for the corresponding neutral ground-state main-group atom.

State what the supplied observations or conditions establish, and separate that from any assumption.

d-electron involvement and variable oxidation states complicate those patterns.

Its first shell is filled by two electrons. The relevant pattern is a filled available valence shell, not a requirement for eight in every element.

Main-group columns often share an outer-shell electron pattern. Group 1 typically has ns¹ and Group 17 ns²np⁵; Group 18 has a filled outer shell, with helium’s 1s² as a key exception to an eight-electron slogan. Transition elements require more attention to nearby d states and variable oxidation numbers. The pattern helps predict likely bonding behavior, but it does not prescribe an atom’s objective or supply a complete reaction energy calculation.

Its first shell is filled by two electrons. The relevant pattern is a filled available valence shell, not a requirement for eight in every element.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the chemistry relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

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.

decreases

Rising effective nuclear charge across a period pulls the electron cloud inward, so atoms get smaller.

fluorine (F)

Fluorine, at the top right of the table (excluding noble gases), is the most electronegative element.

decreases

Down a group, the outer electron is farther out and better shielded, so it is easier to remove — ionization energy decreases.

7

Zeff ≈ 9 − 2 = 7. The 7 valence electrons feel a strong pull, which is why fluorine is so small and so electronegative.

Across a period, protons are added to the nucleus while the new electrons enter the same energy level, adding little shielding. Effective nuclear charge rises, so the stronger pull draws the whole electron cloud inward and the atom shrinks.

high ionization energy

A small radius means the outer electron is close to the nucleus and held tightly, so it takes a lot of energy to remove — a high ionization energy.

Trend statements require a specified property and comparable definition.

Na⁺ is smaller; losing the outer electron removes the occupied outer level and changes electron interactions.

State what the supplied observations or conditions establish, and separate that from any assumption.

Orbital structure, electron pairing and measurement definitions produce exceptions; consult data for precise comparisons.

Na⁺ is smallest in the usual comparison; all three have ten electrons but Na⁺ has the largest proton count, 11.

Isoelectronic species have the same electron count under the compared state. O²⁻, F⁻ and Na⁺ each have ten electrons, but their proton counts rise from eight to eleven. Within this selected series, stronger nuclear attraction generally contracts the electron distribution, giving O²⁻ the larger and Na⁺ the smaller radius under compatible definitions. This is different from saying every positive ion of every element is smaller than every negative ion. Match the electron count and comparison conditions first.

Na⁺ is smallest in the usual comparison; all three have ten electrons but Na⁺ has the largest proton count, 11.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the chemistry relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

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.

Write the relationship first; substitute with consistent units.

4 energy units

7−3 = 4.

Only allowed state differences occur; each gives a photon with a specific frequency.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

List the given quantities, select a relationship, and check units before substituting.

6 energy units

10−4=6.

Match additional lines, check resolution and calibration, and evaluate whether a mixture could give the same subset.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

Emission lowers atomic energy; absorption raises it.

E=hf=3.315×10⁻¹⁹ J, about 3.3×10⁻¹⁹ J.

State what the supplied observations or conditions establish, and separate that from any assumption.

Different allowed transitions have different energy differences, producing different photon frequencies.

Several positions constrain the species pattern and reduce ambiguity from overlap, intensity differences and visual perception.

A low-density excited atomic gas can emit discrete spectral lines associated with electronic transitions. A continuous spectrum spans a broad range rather than only isolated ideal lines. Line positions help identify species when compared with calibrated reference patterns; intensity also depends on population, excitation, detection and other conditions. Overlapping species or broadening can complicate an observed spectrum. A colored stripe diagram is a schematic key to positions, not an actual camera record or a guarantee of equal line intensities.

Several positions constrain the species pattern and reduce ambiguity from overlap, intensity differences and visual perception.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the chemistry relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

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.