Periodic Trends practice questions

Question 1

A student heats a sample of solid strontium ($\text{Sr}$) in pure nitrogen gas ($\text{N}_2$) until a reaction occurs to form a binary ionic compound. Based on the positions of strontium and nitrogen on the periodic table, what is the predicted empirical formula of the product?

Accepted Answer

$\text{Sr}_3\text{N}_2$

Answer

$\text{SrN}$

Answer

$\text{Sr}_2\text{N}_3$

Answer

$\text{SrN}_2$

Question 2

A student analyzes a pure solid substance composed of two elements with an electronegativity difference of $2.1$. Which of the following correctly identifies the predominant type of bonding in the substance and its electrical conductivity?

Accepted Answer

Ionic | Conducts when molten

Answer

Covalent | Conducts in the solid state

Answer

Covalent | Conducts when molten

Answer

Ionic | Conducts in the solid state

Question 3

The valence electron configurations for three consecutive Period 3 elements are provided in the table below.

| Element | Atomic number | Valence electron configuration |
|---|---|---|
| $\text{Mg}$ | 12 | $3s^2$ |
| $\text{Al}$ | 13 | $3s^2 3p^1$ |
| $\text{Si}$ | 14 | $3s^2 3p^2$ |

Which of the following correctly ranks the three elements in order of increasing first ionization energy?

Accepted Answer

$\text{Al} < \text{Mg} < \text{Si}$

Answer

$\text{Mg} < \text{Al} < \text{Si}$

Answer

$\text{Mg} < \text{Si} < \text{Al}$

Answer

$\text{Al} < \text{Si} < \text{Mg}$

Question 4

A student subjects a crystal of sodium chloride, $\text{NaCl(s)}$, to a mechanical shearing force, causing the crystal to cleave cleanly along a plane as represented in the diagram below. Which of the following claims best explains why ionic solids such as $\text{NaCl(s)}$ are brittle and fracture when a stress is applied?

Accepted Answer

The crystal fractures because the applied force shifts layers of ions, aligning ions of like charge adjacent to one another and producing strong electrostatic repulsions.

Answer

The crystal fractures because the applied force disrupts a sea of delocalized valence electrons, preventing them from cushioning the movement of cations.

Answer

The crystal fractures because directional covalent bonds between the sodium and chlorine atoms are rigid and cannot bend under mechanical pressure.

Answer

The crystal fractures because the applied force pushes oppositely charged ions closer together, increasing attractive forces until the lattice collapses.

Question 5

A student studying electron-capture species for a gas-phase detector compares isolated gaseous fluorine and chlorine atoms. Electron affinity is reported as the magnitude of the energy released when one mole of gaseous atoms each gains an electron.

| Atom | Valence configuration before electron addition | Energy released $\text{(kJ/mol)}$ |
|------|-----------------------------------------------|--------------------------------------|
| $\text{F(g)}$ | $2s^2 2p^5$ | $328$ |
| $\text{Cl(g)}$ | $3s^2 3p^5$ | $349$ |

Which statement best explains the relative electron affinities of the two atoms?

Accepted Answer

The electron affinity of $\text{Cl(g)}$ is greater because the added electron experiences less electron-electron repulsion in chlorine's more spatially extended $3p$ subshell than in fluorine's compact $2p$ subshell.

Answer

The electron affinity of $\text{F(g)}$ is greater because its smaller atomic radius results in a stronger attraction between the nucleus and the added electron.

Answer

The electron affinity of $\text{F(g)}$ is greater because its $2p$ subshell contains fewer electrons before the addition than the $3p$ subshell of $\text{Cl(g)}$.

Answer

The electron affinity of $\text{Cl(g)}$ is greater because chlorine has fewer occupied energy levels and therefore less electron shielding than fluorine.

Question 6

A student analyzes the set of isoelectronic species shown in the table below.

| Species | Atomic Number (Z) | Number of Electrons |
| :--- | :---: | :---: |
| $\text{O}^{2-}$ | 8 | 10 |
| $\text{F}^-$| 9 | 10 |
| $\text{Na}^+$ | 11 | 10 |
| $\text{Mg}^{2+}$ | 12 | 10 |

Which of the following claims correctly identifies the species with the smallest ionic radius and provides the correct justification?

Accepted Answer

$\text{Mg}^{2+}$ has the smallest ionic radius because its greater nuclear charge exerts a stronger attractive force on the same number of electrons.

Answer

$\text{O}^{2-}$ has the smallest ionic radius because it has fewer electron shells, resulting in reduced electron-electron repulsion.

Answer

$\text{Mg}^{2+}$ has the smallest ionic radius because its electrons occupy fewer energy levels than those of the other species.

Answer

$\text{O}^{2-}$ has the smallest ionic radius because its lower nuclear charge exerts a stronger electrostatic attraction per electron.

Question 7

A student is asked to analyze the relative melting points of three binary ionic solids: $\text{MgO}$, $\text{NaF}$, and $\text{NaCl}$. Which of the following correctly ranks the melting points of these solids from highest to lowest and provides the correct justification?

Accepted Answer

$\text{MgO} > \text{NaF} > \text{NaCl}$, because $\text{MgO}$ has higher ionic charges ($+2/-2$) than $\text{NaF}$ and $\text{NaCl}$ ($+1/-1$), and $\text{F}^-$ has a smaller ionic radius than $\text{Cl}^-$, resulting in a shorter internuclear distance.

Answer

$\text{NaCl} > \text{NaF} > \text{MgO}$, because larger ions create stronger lattice energies, and lower ion charges result in weaker repulsive forces within the crystal lattice.

Answer

$\text{MgO} > \text{NaCl} > \text{NaF}$, because $\text{Cl}^-$

Answer

$\text{MgO} > \text{NaF} > \text{NaCl}$, because the internuclear distance in $\text{MgO}$ is shorter than in $\text{NaF}$, which has a greater effect on lattice energy than the charges of the constituent ions.

Question 8

The complete photoelectron spectra of two neutral, second-period elements, X and Y, are shown below.

Which of the following correctly identifies Element Y and explains why its $\text{1s}$ peak is located at a higher binding energy than the $\text{1s}$ peak of Element X?

Accepted Answer

Element Y is $\text{O}$; its nucleus contains a greater number of protons, exerting a stronger coulombic attraction on the core electrons.

Answer

Element Y is $\text{C}$; its $\text{1s}$ electrons experience less shielding from valence electrons, causing them to be held more tightly by the nucleus.

Answer

Element Y is $\text{O}$; its $\text{1s}$ electrons experience greater electron-electron repulsion within the core subshell, which increases their binding energy.

Answer

Element Y is $\text{N}$; its $\text{2p}$ subshell is exactly half-filled, which provides extra stability that shifts all binding energy peaks to higher values.

Question 9

A student hypothesizes that inner-shell (core) electrons shield outer-shell (valence) electrons from the full attractive force of the nucleus. Which of the following observations from photoelectron spectroscopy (PES) data provides the best evidence to support this hypothesis?

Accepted Answer

The observation that the binding energy of the valence electron drops sharply from neon ($Z = 10$) to sodium ($Z = 11$), despite the increase in nuclear charge.

Answer

The observation that the binding energy of $1s$ electrons increases continuously across Period 2 elements from lithium ($Z = 3$) to neon ($Z = 10$).

Answer

The observation that the relative peak height for the $2p$ subshell in neon is three times greater than the relative peak height for its $2s$ subshell.

Answer

The observation that the binding energy of $2s$ electrons is slightly higher than the binding energy of $2p$ electrons within the same shell of a neon atom.

Question 10

A student compares the lattice energies of four alkali halide ionic compounds: $\text{LiF}$, $\text{NaF}$, $\text{NaCl}$, and $\text{KCl}$. All four compounds crystallize in identical crystal lattice structures where each ion has a charge of magnitude 1. Which of the following correctly ranks the compounds in order of decreasing lattice energy (greatest to least magnitude) and provides the correct justification?

Accepted Answer

$\text{LiF} > \text{NaF} > \text{NaCl} > \text{KCl}$, because smaller ionic radii result in a shorter internuclear distance between ions, which increases the magnitude of Coulombic attraction.

Answer

$\text{LiF} > \text{NaF} > \text{NaCl} > \text{KCl}$, because an increase in core electron shielding down a group reduces the attraction between cations and anions.

Answer

$\text{KCl} > \text{NaCl} > \text{NaF} > \text{LiF}$, because larger ionic radii lead to greater electron cloud polarizability, which strengthens the lattice interactions.

Answer

$\text{KCl} > \text{NaCl} > \text{NaF} > \text{LiF}$, because ions with a larger number of total electron shells exert a stronger electrostatic attraction on neighboring counterions.

Question 11

In an ionic crystal with a rock salt ($\text{NaCl}$) lattice, anions ($\text{X}^-$) of radius $r_-$

Accepted Answer

$0.414$

Answer

$0.732$

Answer

$1.000$

Answer

$1.414$

Question 12

A student compares the atomic structures and periodic trends of nitrogen ($\text{N}$) and oxygen ($\text{O}$). The atomic numbers and ground-state electron configurations for both elements are given in the table below.

| Element | Atomic Number | Electron Configuration |
|---|---|---|
| $\text{N}$ | 7 | $1s^2 2s^2 2p^3$ |
| $\text{O}$ | 8 | $1s^2 2s^2 2p^4$ |

Which of the following claims correctly compares the first ionization energies of nitrogen and oxygen and provides the correct reasoning?

Accepted Answer

$\text{N}$ has a higher first ionization energy than $\text{O}$ because removing an electron from $\text{O}$ involves a paired electron in a $2p$ orbital, where electron-electron repulsion makes the electron easier to remove.

Answer

$\text{N}$ has a higher first ionization energy than $\text{O}$ because the valence electrons in $\text{N}$ experience a greater effective nuclear charge than the valence electrons in $\text{O}$.

Answer

$\text{O}$ has a higher first ionization energy than $\text{N}$ because $\text{O}$ has a greater nuclear charge, attracting its valence electrons more strongly toward the nucleus.

Answer

$\text{O}$ has a higher first ionization energy than $\text{N}$ because the valence electrons in $\text{O}$ occupy a lower principal energy level than those in $\text{N}$.

Question 13

A student compares the first ionization energies and valence-electron configurations of two pairs of Period 3 elements.

| Element | Valence-electron configuration | First ionization energy $\text{(kJ/mol)}$ |
|---|---|---:|
| $\text{Mg}$ | $[\text{Ne}]3s^2$ | $738$ |
| $\text{Al}$ | $[\text{Ne}]3s^2 3p^1$ | $578$ |
| $\text{P}$ | $[\text{Ne}]3s^2 3p^3$ | $1{,}012$ |
| $\text{S}$ | $[\text{Ne}]3s^2 3p^4$ | $1{,}000$ |

The student observes that the first ionization energy decreases from $\text{Mg}$ to $\text{Al}$ and from $\text{P}$ to $\text{S}$. Which row correctly predicts the electronegativity comparison for each pair?

Accepted Answer

$\chi(\text{Al})>\chi(\text{Mg})$ ; $\chi(\text{S})>\chi(\text{P})$

Answer

\(\chi(\text{Al})

Answer

$\chi(\text{Al})\chi(\text{P})$

Answer

$\chi(\text{Al})>\chi(\text{Mg})$ ; \(\chi(\text{S})

Question 14

To estimate steric interactions in a porous material, a student compares the electron-cloud radii of the species shown in the table.

| Species | Number of protons | Number of electrons |
|---|---:|---:|
| $\text{S}^{2-}$ | 16 | 18 |
| $\text{Ar}$ | 18 | 18 |
| $\text{Ca}^{2+}$ | 20 | 18 |

Which statement gives the correct ordering of electron-cloud radii from greatest to least and the best justification?

Accepted Answer

$\text{S}^{2-} > \text{Ar} > \text{Ca}^{2+}$, because all three species have the same electron configuration, and increasing nuclear charge pulls the electron cloud closer to the nucleus.

Answer

$\text{Ca}^{2+} > \text{Ar} > \text{S}^{2-}$, because the additional protons provide greater shielding of the outer electrons, allowing the electron cloud to expand.

Answer

$\text{Ca}^{2+} > \text{Ar} > \text{S}^{2-}$, because for the same electron configuration, a greater nuclear charge attracts the electrons more strongly.

Answer

$\text{S}^{2-} > \text{Ar} > \text{Ca}^{2+}$, because forming $\text{S}^{2-}$ adds a new occupied principal energy level, whereas forming $\text{Ca}^{2+}$ removes one.

Question 15

In a spectroscopy experiment, a student supplies energy to dissociate gaseous $\text{H}_2$. Some of the resulting hydrogen atoms later recombine. To isolate the energy associated with the bond, the student represents the changes below and compares all species at the same temperature.

| Process | Isolated bond change |
|---------|----------------------|
| 1 | $\text{H}_2\text{(g)} \rightarrow 2\text{H(g)}$ |
| 2 | $2\text{H(g)} \rightarrow \text{H}_2\text{(g)}$ |

No other bonds are broken or formed. Which statement correctly identifies the signs of $\Delta H_1$ and $\Delta H_2$ and explains them using a particle-attraction model?

Accepted Answer

$\Delta H_1>0$ and \(\Delta H_2

Answer

$\Delta H_10$, because eliminating the attraction between the nuclei and the shared electron density places the separated atoms at lower potential energy than the bonded molecule.

Answer

$\Delta H_10$, because the $\text{H-H}$ bond stores energy that is released when the bond breaks and must be supplied again when the bond forms.

Question 16

A student compares the photoelectron spectra of gaseous phosphorus and sulfur atoms. The displayed portion of each spectrum contains the peak produced by ejecting a $3s$ electron, and the peak intensities have been normalized.

Which statement best interprets the relative positions of the peaks?

Accepted Answer

The $3s$ electrons in $\text{S}$ experience a greater effective nuclear charge because $\text{S}$ has one more proton, while its additional same-shell electron provides only partial additional shielding.

Answer

The $3s$ electrons in $\text{S}$ experience a greater effective nuclear charge because $\text{S}$ has more neutrons, which increase the attraction between the nucleus and the electrons.

Answer

The $3s$ electrons in $\text{P}$ experience a greater effective nuclear charge because the lower binding-energy peak indicates that more energy is required to remove an electron.

Answer

The $3s$ electrons in $\text{P}$ experience a greater effective nuclear charge because the additional $3p$ electron in $\text{S}$ completely shields the charge of its additional proton.

Question 17

A student examines idealized, equal-volume microscopic regions of two iron alloys. The atomic radii and relative atomic masses used in the models are shown in the table.

| Atom | Atomic radius (pm) | Relative atomic mass |
|------|--------------------|----------------------|
| $\text{Fe}$ | 126 | 55.8 |
| $\text{C}$ | 70 | 12.0 |
| $\text{Ni}$ | 124 | 58.7 |

Each particle shown represents one atom, and each rectangular region has the same volume and depth. Based on the representations and the data, which statement correctly identifies the two alloy types and compares their densities and expected malleabilities?

Accepted Answer

Alloy I is interstitial, and alloy II is substitutional. Alloy I is denser because its four added carbon atoms contribute more mass than the two nickel substitutions add; alloy II is more malleable because its similarly sized atoms distort the lattice less.

Answer

Alloy I is interstitial, and alloy II is substitutional. The alloys have the same density because each has nine atoms at lattice positions; alloy II is more malleable.

Answer

Alloy I is substitutional, and alloy II is interstitial. Alloy I is denser, and alloy II is more malleable; the alloy type is determined by which sample contains more solute atoms.

Answer

Alloy I is interstitial, and alloy II is substitutional. Alloy I is denser and more malleable because the smaller carbon atoms fit into gaps without interfering with the sliding of iron-atom layers.

Question 18

A student compares four halogen-containing oxyacids that could be used in an antimicrobial rinse.

| Acid | Structural formula |
|------|--------------------|
| Chlorous acid, $\text{HClO}_2$ | $\text{H}-\text{O}-\text{Cl}=\text{O}$ |
| Bromous acid, $\text{HBrO}_2$ | $\text{H}-\text{O}-\text{Br}=\text{O}$ |
| Hypochlorous acid, $\text{HClO}$ | $\text{H}-\text{O}-\text{Cl}$ |
| Hypobromous acid, $\text{HBrO}$ | $\text{H}-\text{O}-\text{Br}$ |

Which choice ranks the acids from greatest to least acid strength and gives the best justification?

Accepted Answer

$\text{HClO}_2 > \text{HBrO}_2 > \text{HClO} > \text{HBrO}$, because an additional oxygen atom stabilizes the conjugate base, and Cl withdraws electron density more strongly than Br.

Answer

$\text{HBrO}_2 > \text{HClO}_2 > \text{HBrO} > \text{HClO}$, because an additional oxygen atom stabilizes the conjugate base and Br is more electronegative than Cl.

Answer

$\text{HClO}_2 > \text{HClO} > \text{HBrO}_2 > \text{HBrO}$, because the electronegativity of the halogen is the dominant factor, so every Cl-containing acid is stronger than every Br-containing acid.

Answer

$\text{HClO} > \text{HBrO} > \text{HClO}_2 > \text{HBrO}_2$, because an additional oxygen atom localizes the negative charge in the conjugate base, while Cl is more electronegative than Br.

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