Chemical Bonding & Bond Properties 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

A student draws a Lewis structure for the carbonate ion, $\text{CO}_3^{2-}$, showing one carbon-oxygen double bond and two carbon-oxygen single bonds. Experimental measurements show that all three carbon-oxygen bond lengths in $\text{CO}_3^{2-}$ are identical ($128\text{ pm}$). Which of the following statements correctly identifies the bond order and justifies the equal bond lengths in $\text{CO}_3^{2-}$?

Accepted Answer

All three $\text{C}-\text{O}$ bonds have equal length and a bond order of $1.33$ because the $\pi$ electrons are delocalized equally across all three carbon-oxygen bonds.

Answer

One $\text{C}-\text{O}$ bond is shorter than the other two because double bonds have higher bond orders and shorter bond lengths than single bonds.

Answer

All three $\text{C}-\text{O}$ bonds have equal length and a bond order of $1.33$ because the carbon atom rapidly alternates forming a double bond with each of the three oxygen atoms.

Answer

All three $\text{C}-\text{O}$ bonds have equal length and a bond order of $1.00$ because single bonds form between carbon and each oxygen atom to minimize electron-electron repulsion.

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 is asked to select the Lewis diagram for sulfur dioxide, $\text{SO}_2$, that represents the most favorable formal-charge arrangement. Which of the following diagrams correctly represents this structure?

Accepted Answer

Structure C

Answer

Structure A

Answer

Structure B

Answer

Structure D

Question 6

A laboratory inventory includes three solid materials: silicon carbide ($\text{SiC}$), potassium bromide ($\text{KBr}$), and brass (an alloy of $\text{Cu}$ and $\text{Zn}$). Which of the following options correctly identifies the primary type of bonding present in each material?

Accepted Answer

$\text{SiC}$: Network covalent; $\text{KBr}$: Ionic; Brass: Metallic

Answer

$\text{SiC}$: Molecular covalent; $\text{KBr}$: Ionic; Brass: Metallic

Answer

$\text{SiC}$: Network covalent; $\text{KBr}$: Metallic; Brass: Ionic

Answer

$\text{SiC}$: Ionic; $\text{KBr}$: Network covalent; Brass: Metallic

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

A student plots the potential energy as a function of internuclear distance for a molecule of $\text{O}_2(g)$, shown as the solid curve in the diagram. The student then wants to plot the curve for a molecule of $\text{N}_2(g)$. Which of the labeled dashed curves best represents the potential energy profile for $\text{N}_2(g)$?

Accepted Answer

Curve B, because $\text{N}_2$ has a higher bond order than $\text{O}_2$, which leads to a shorter equilibrium bond length and a greater bond energy.

Answer

Curve A, because $\text{N}_2$ has a higher bond order than $\text{O}_2$, which leads to a longer equilibrium bond length and a greater bond energy.

Answer

Curve C, because $\text{N}_2$ has a higher bond order than $\text{O}_2$, which leads to a shorter equilibrium bond length and a smaller bond energy.

Answer

Curve D, because $\text{N}_2$ has a higher bond order than $\text{O}_2$, which leads to a longer equilibrium bond length and a smaller bond energy.

Question 9

A student compares the physical properties of silicon dioxide, $\text{SiO}_2\text{(s)}$, and carbon dioxide, $\text{CO}_2\text{(s)}$. The student notes that $\text{SiO}_2\text{(s)}$ has a melting point of $1710^\circ\text{C}$, whereas $\text{CO}_2\text{(s)}$ sublimes at $-78^\circ\text{C}$ at atmospheric pressure. Which of the following statements best accounts for the large difference in phase-change temperatures between the two substances?

Accepted Answer

$\text{SiO}_2\text{(s)}$ has a much higher melting point because phase change requires breaking strong covalent bonds within a continuous network lattice, whereas phase change of $\text{CO}_2\text{(s)}$ requires overcoming much weaker intermolecular forces between discrete molecules.

Answer

$\text{SiO}_2\text{(s)}$ has a much higher melting point because melting $\text{SiO}_2\text{(s)}$ breaks the strong $\text{Si}-\text{O}$ covalent bonds within discrete $\text{SiO}_2$ molecules, whereas melting $\text{CO}_2\text{(s)}$ breaks the weaker $\text{C}=\text{O}$ double bonds within discrete $\text{CO}_2$ molecules.

Answer

$\text{CO}_2\text{(s)}$ has a much lower sublimation temperature because the polar $\text{C}=\text{O}$ bonds create repulsive forces between adjacent $\text{CO}_2$ molecules, whereas nonpolar $\text{Si}-\text{O}$ bonds in $\text{SiO}_2$ attract each other.

Answer

$\text{CO}_2\text{(s)}$ has a much lower sublimation temperature because phase change requires breaking covalent bonds in $\text{CO}_2\text{(s)}$, whereas phase change of $\text{SiO}_2\text{(s)}$ requires overcoming only London dispersion forces.

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

A student compares two structural models for benzene, $\text{C}_6\text{H}_6$, as shown in the diagram: Model 1 with localized alternating single and double carbon-carbon bonds, and Model 2 with a delocalized resonance hybrid. Which of the following experimental observations provides the best evidence supporting Model 2 over Model 1?

Accepted Answer

All six carbon-carbon bonds have an identical measured length of $139\text{ pm}$, which is intermediate between a standard $\text{C}-\text{C}$ single bond ($154\text{ pm}$) and a standard $\text{C}=\text{C}$ double bond ($134\text{ pm}$).

Answer

Benzene reacts with hydrogen gas in the presence of a metal catalyst to form cyclohexane, $\text{C}_6\text{H}_{12}$, consuming three moles of $\text{H}_2$ for every one mole of benzene reacted.

Answer

The six carbon atoms in the ring lie in a single geometric plane with bond angles of $120^\circ$, consistent with $sp^2$ hybridization at each carbon atom.

Answer

Elemental analysis shows that benzene consists of $92.3\%$ carbon and $7.7\%$ hydrogen by mass, establishing an empirical formula of $\text{CH}$.

Question 12

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 13

Sulfur hexafluoride, $\text{SF}_6$, is a dense, colorless gas widely used as an electrical insulator in high-voltage equipment. Based on VSEPR theory, which of the following best predicts the molecular geometry and ideal bond angles around the central sulfur atom in an $\text{SF}_6$ molecule?

Accepted Answer

Octahedral geometry with bond angles of $90^\circ$ and $180^\circ$

Answer

Seesaw geometry with bond angles of less than $90^\circ$ and less than $120^\circ$

Answer

Trigonal bipyramidal geometry with bond angles of $90^\circ$ and $120^\circ$

Answer

Square planar geometry with bond angles of $90^\circ$ and $180^\circ$

Question 14

A student compares the properties of two crystalline ionic solids, $\text{MgO}$ and $\text{NaCl}$. Which of the following statements correctly identifies the compound with the higher melting point and provides the correct Coulombic justification?

Accepted Answer

$\text{MgO}$ has a higher melting point than $\text{NaCl}$ because the magnitude of the charges on $\text{Mg}^{2+}$ and $\text{O}^{2-}$ is greater than on $\text{Na}^+$ and $\text{Cl}^-$, producing stronger Coulombic attractions.

Answer

$\text{NaCl}$ has a higher melting point than $\text{MgO}$ because $\text{Cl}^-$ is larger than $\text{O}^{2-}$, creating stronger ion-ion attractive forces.

Answer

$\text{MgO}$ has a higher melting point than $\text{NaCl}$ because $\text{MgO}$ has a greater molar mass, resulting in stronger London dispersion forces.

Answer

$\text{NaCl}$ has a higher melting point than $\text{MgO}$ because single charges permit closer packing in the crystal lattice, leading to greater interionic attraction.

Question 15

A student applies a sharp mechanical force to a crystal of sodium chloride, $\text{NaCl(s)}$. The particulate representation below shows the arrangement of ions in a section of the crystal lattice before and after the force causes the top layers of ions to shift relative to the bottom layers. Based on the representation, which of the following best explains why the ionic crystal cleaves along the plane of force rather than deforming malleably?

Accepted Answer

The applied force shifts the layers so that ions of like charge align, creating strong repulsive electrostatic forces that cause the crystal to cleave.

Answer

The applied force causes like-charged ions to align, producing attractive electrostatic forces that hold the shifted layers together.

Answer

The applied force breaks nonpolar covalent bonds between adjacent ions, causing the crystal lattice to fracture.

Answer

The applied force causes valence electrons to become delocalized, reducing electrostatic attraction and allowing layers to slide past one another.

Question 16

A student compares the scratch hardness and cleavage behavior of three crystalline solids. Cleavage is observed when a sharp blow causes a solid to split along flat, repeating surfaces.

| Solid | Approximate Mohs hardness | Observation after a sharp blow | Structural description |
|---|---:|---|---|
| Diamond, $\text{C(s)}$ | 10 | Flat surfaces form along certain orientations | Each $\text{C}$ atom is covalently bonded to four other $\text{C}$ atoms in a three-dimensional network. |
| Quartz, $\text{SiO}_2\text{(s)}$ | 7 | Irregular surfaces form without cleavage | Covalent $\text{Si}-\text{O}$ bonds extend throughout a three-dimensional framework without distinct layers. |
| Graphite, $\text{C(s)}$ | 1–2 | Thin, flat sheets separate readily | Strong covalent bonds connect $\text{C}$ atoms within each sheet, and London dispersion forces act between sheets. |

Which of the following best explains the hardness and cleavage results?

Accepted Answer

Hardness and cleavage test different responses: diamond's strong three-dimensional network resists scratching but can split along planes where fewer bonds oppose separation; quartz lacks layered weak regions, and graphite separates between sheets held by dispersion forces.

Answer

Diamond's high hardness means that its cleavage cannot involve breaking covalent bonds; therefore, diamond must contain unrepresented molecular layers held together by weak intermolecular forces.

Answer

Quartz's lack of cleavage means that quartz should be harder than diamond, because a solid that fractures irregularly always requires more energy to scratch than one that cleaves.

Answer

Graphite is soft and cleaves readily because the $\text{C}-\text{C}$ covalent bonds within each sheet are much weaker than those in diamond, so both tests separate atoms within a sheet.

Question 17

A materials scientist compares two samples at the same temperature.

| Sample | Composition and structure |
|--------|---------------------------|
| P | Pure iron containing only $\text{Fe}$ atoms at regular lattice positions |
| S | Steel containing mostly $\text{Fe}$ atoms and a small number of $\text{C}$ atoms in interstitial spaces |

Which choice correctly ranks the samples from greatest to least hardness and explains the ranking?

Accepted Answer

Sample S $>$ Sample P, because interstitial carbon deforms the iron lattice and makes layers of atoms less able to slide past one another.

Answer

Sample P $>$ Sample S, because the regular lattice of pure iron prevents layers of atoms from sliding past one another.

Answer

Sample P $>$ Sample S, because interstitial carbon increases the separation between some iron atoms and therefore allows layers of atoms to slide more easily.

Answer

Sample S $>$ Sample P, because the carbon atoms form a continuous covalent network throughout the steel.

Question 18

A student compares the bonding around each carbon atom in ethene, $\text{H}_2\text{C}=\text{CH}_2$, with the bonding around each carbon atom in ethyne, $\text{HC}\equiv\text{CH}$.

Which row correctly identifies the hybridization of each carbon atom in the two molecules?

Accepted Answer

Ethene carbon: $sp^2$; ethyne carbon: $sp$

Answer

Ethene carbon: $sp^2$; ethyne carbon: $sp^3$

Answer

Ethene carbon: $sp^3$; ethyne carbon: $sp$

Answer

Ethene carbon: $sp^3$; ethyne carbon: $sp^3$

Chemical Bonding & Bond Properties MCQ Practice & Quizzes

Where This Topic Shows Up

Find More Questions On Related Topics

Practice Questions

Quizzes