Atomic Structure and Chemical Bonding Practice Questions — JAMB UTME — Page 3

Question 41Question →

Match each observed physical property or phenomenon of ionic (electrovalent) compounds on the left with its correct underlying thermodynamic or structural explanation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

1 matches with the explanation of charge product dependence on lattice energy; 2 matches with the explanation of Fajans' rules of polarization; 3 matches with the explanation of hydration enthalpy exceeding lattice enthalpy; 4 matches with the explanation of ion mobility in molten versus solid states.

Each physical property directly corresponds to its underlying quantum mechanical or thermodynamic principle: lattice energy scales with charge product (

MgO

vs

NaCl

), polarization of anion electron clouds by small high-charge cations creates covalent character (

AlI_3

), exothermic dissolution occurs when hydration energy exceeds lattice energy (

NaOH

), and electrical conduction requires mobile ions that are locked in solids but liberated upon melting (

CaF_2

).

Step-by-Step Solution

1

Analyze the high melting point of

MgO

versus

NaCl

Lattice energy is governed by Coulomb's law:

E \propto \frac{|q_1 q_2|}{r}

.

MgO

consists of

Mg^{2+}

and

O^{2-}

(product = 4), while

NaCl

consists of

Na^+

and

Cl^-

(product = 1). Higher charge product leads to stronger lattice attraction and a higher melting point.

Identify the primary thermodynamic factor controlling lattice strength in ionic crystals.

2

Analyze the anomalous covalent behavior of

AlI_3

Apply Fajans' rules: Covalency increases with high cation charge density and large anion size.

Al^{3+}

has high charge density and

I^-

is large and easily polarized, leading to electron cloud sharing (covalent character).

Explain deviations from purely electrovalent behavior using polarization principles.

3

Analyze the thermochemistry of dissolution of

NaOH

Dissolution enthalpy

\Delta H_{soln} = \Delta H_{lat} + \Delta H_{hyd}

. If hydration enthalpy released is greater in magnitude than the lattice enthalpy required to separate ions, the net process is exothermic.

Relate lattice energy and hydration energy to dissolution energetics.

4

Analyze electrical conductivity in solid versus molten

CaF_2

Solid ionic compounds contain ions held rigidly in a lattice structure. When melted, thermal energy breaks the lattice, producing free-moving ions capable of carrying electrical current.

Distinguish between mobile charge carriers (molten state) and immobile lattice positions (solid state).

Key Concept

Thermodynamic and structural factors governing ionic lattice stability, polarization (Fajans' rules), solution energetics, and state-dependent conductivity.

Question 42Question →

During the formation of the hydronium ion ( $H_3O^+$ ), a water molecule reacts with a proton ( $H^+$ ). Which condition must be satisfied by the oxygen atom in water to enable this coordinate (dative) bond formation?

It must possess at least one unshared lone pair of electrons available for donation.

It must completely transfer one of its valence electrons to the proton.

It must receive one electron from the proton to complete a shared electron pair.

It must expand its valence shell beyond eight electrons to accept the proton.

Show answer & explanation

Answer: It must possess at least one unshared lone pair of electrons available for donation.

Answer

It must possess at least one unshared lone pair of electrons available for donation.

A coordinate (dative) covalent bond is formed when a single donor atom provides both electrons of a shared pair to an acceptor species with an empty orbital. In the reaction of

H_2O

with

H^+

, the oxygen atom acts as the donor because it possesses unshared lone pairs of valence electrons.

Step-by-Step Solution

1

Analyze the electronic configuration of the reacting species.

The water molecule (

H_2O

) has two single covalent bonds and two unshared lone pairs on the oxygen atom, while the hydrogen ion (

H^+

) has an empty valence shell with no electrons.

Identifying the electron distribution helps determine how the bond is formed between the two species.

2

Apply the definition of coordinate (dative) covalent bonding.

Because

H^+

carries no electrons, the shared pair forming the new

O-H

bond must come entirely from one of the lone pairs on the oxygen atom.

A coordinate bond is formed when one atom provides both electrons of the shared bonding pair to an electron-deficient species.

Key Concept

Coordinate (Dative) Covalent Bonding

Question 43Question →

Consider four main group elements $P$ , $Q$ , $R$ , and $S$ with atomic numbers $11$ , $12$ , $16$ , and $17$ respectively. Which combination of these elements forms an electrovalent compound with the highest melting point?

The combination of

Q

and

S

, because the

+2

and

-2

ionic charges maximize the electrostatic lattice attraction.

The combination of

P

and

R

, because single electron transfer minimizes interionic repulsion within the crystal lattice.

The combination of

Q

and

R

, because a

1:2

stoichiometric ratio produces the highest crystal packing density.

The combination of

P

and

S

, because two

P^+

cations double the total electrostatic pull on a single

S^{2-}

anion.

Show answer & explanation

Answer: The combination of $Q$ and $S$ , because the $+2$ and $-2$ ionic charges maximize the electrostatic lattice attraction.

Answer

The combination of element Q (atomic number 12) and element S (atomic number 16) forms QS, which has the highest melting point due to the +2 and -2 ionic charges maximizing electrostatic lattice attraction.

Element Q (atomic number 12) has electronic configuration 2,8,2 and loses two electrons to form Q2+. Element S (atomic number 16) has electronic configuration 2,8,6 and gains two electrons to form S2-. The compound formed between Q and S (QS) consists of divalent ions. By Coulomb's Law, lattice energy is proportional to the product of ionic charges (|q1 * q2|). The charge product for QS is 4, which is double that of QR2 or P2S (charge product 2) and four times that of PR (charge product 1). Consequently, QS possesses the highest lattice energy and highest melting point.

Step-by-Step Solution

1

Determine the identity and valency of each element from its atomic number

P

(

Z=11

, Sodium) forms

P^+

cations;

Q

(

Z=12

, Magnesium) forms

Q^{2+}

cations;

S

(

Z=16

, Sulfur) forms

S^{2-}

anions;

R

(

Z=17

, Chlorine) forms

R^-

anions.

Electronic configurations determine the number of valence electrons lost or gained to achieve stable octet structures.

2

Write the chemical formulas for the electrovalent compounds formed by valid metal-nonmetal pairs

Possible ionic compounds are

PR

(

P^+ R^-

),

P_2S

(

(P^+)_2 S^{2-}

),

QR_2

(

Q^{2+} (R^-)_2

), and

QS

(

Q^{2+} S^{2-}

).

Electrovalent compounds form when metals transfer electrons to non-metals to achieve electrical neutrality.

3

Compare the electrostatic lattice energies of the resulting crystal lattices

Lattice energy is directly proportional to the product of ionic charges (

E_{\text{lattice}} \propto |q_1 q_2|

). For

QS

,

|q_1 q_2| = |(+2)(-2)| = 4

. For

QR_2

and

P_2S

,

|q_1 q_2| = 2

. For

PR

,

|q_1 q_2| = 1

.

According to Coulomb's Law, higher ionic charges create substantially stronger electrostatic forces of attraction between ions in the solid lattice.

4

Relate lattice energy to the physical property of melting point

Higher lattice energy requires significantly more thermal energy to break the ionic bonds, making

QS

the compound with the highest melting point.

The melting point of an electrovalent compound increases as the strength of the lattice attraction increases.

Key Concept

Lattice energy dependence on ionic charge magnitude (Coulomb's Law in ionic crystals)

Question 44Question →

An electrovalent compound is insoluble in water when its lattice enthalpy is smaller in magnitude than the total hydration enthalpy of its constituent gaseous ions.

True

False

Show answer & explanation

Answer: False

Answer

False. An electrovalent compound is soluble in water when the magnitude of its hydration enthalpy exceeds its lattice enthalpy, allowing ion-water electrostatic attractions to overcome ionic crystal lattice forces.

The statement is false because for an electrovalent compound to dissolve in water, the hydration energy released when ions interact with water molecules must overcome the lattice energy holding the crystal together. If hydration enthalpy is greater in magnitude than lattice enthalpy, the compound is soluble rather than insoluble.

Step-by-Step Solution

1

Identify the enthalpy changes during the dissolution of an ionic solid.

Dissolution depends on two key thermodynamic quantities: lattice enthalpy (energy required to separate solid ions into gaseous ions) and hydration enthalpy (energy released when gaseous ions are solvated by water).

The overall enthalpy of solution is approximated by

\Delta H_{\text{solution}} = \Delta H_{\text{lattice}} + \Delta H_{\text{hydration}}

.

2

Evaluate the condition where

|\Delta H_{\text{lattice}}| < |\Delta H_{\text{hydration}}|

.

The energy released during ion hydration is greater than the energy required to break the ionic lattice.

This leads to an exothermic dissolution process (

\Delta H_{\text{solution}} < 0

), which strongly favors the compound dissolving in water.

3

Determine the truth value of the statement.

The statement asserts that such a compound is insoluble, which contradicts chemical thermodynamic principles.

Therefore, the given statement is false.

Key Concept

Lattice Enthalpy vs Hydration Enthalpy in Ionic Compound Solubility

Question 45Question →

When a mechanical stress is applied to a solid metal, the material deforms without shattering. Which of the following structural features of metallic bonding is directly responsible for this malleability?

The ability of layers of metal cations to slide past each other without disrupting the electrostatic attraction to the delocalized electron sea

The rapid breaking and re-formation of rigid directional covalent bonds between adjacent metal atoms

The rigid positioning of delocalized electrons that locks positive metal cations into fixed lattice positions

The expansion of localized electron orbitals which absorbs mechanical force without shifting metal cations

Show answer & explanation

Answer: The ability of layers of metal cations to slide past each other without disrupting the electrostatic attraction to the delocalized electron sea

Answer

The ability of layers of metal cations to slide past each other without disrupting the electrostatic attraction to the delocalized electron sea

In metallic lattices, delocalized valence electrons move freely throughout the array of positive metal cations. When a mechanical force is applied, layers of cations slide past one another. The mobile electron sea adapts immediately to the shifted cations, maintaining the non-directional electrostatic attraction throughout the lattice so that the metal deforms (malleability) instead of fracturing.

Step-by-Step Solution

1

Identify the atomic-scale structure of a metallic lattice.

Solid metals consist of a giant lattice of positive metal cations immersed in a fluid sea of delocalized valence electrons.

Understanding the non-directional nature of metallic bonds is necessary to explain physical properties.

2

Analyze how applied mechanical force alters the lattice structure.

Under mechanical stress, planes of positive cations slide past one another.

Applied mechanical forces induce shear stress across crystal lattice planes.

3

Determine why the metallic structure deforms rather than breaking.

Because the delocalized electrons are mobile and non-directional, they adjust immediately to the shifted cation layers, maintaining attractive electrostatic forces throughout the lattice and preventing repulsive cleavage.

Non-directional electrostatic attraction preserves structural cohesion during deformation.

Key Concept

Metallic Bonding and Malleability

Question 46Question →

Ammonium chloride ( $NH_4Cl$ ) is synthesized by reacting ammonia gas with hydrogen chloride gas. Which types of chemical bonding exist within solid ammonium chloride?

Covalent, coordinate (dative), and ionic bonds

Covalent, ionic, and metallic bonds

Ionic, coordinate, and hydrogen bonds

Covalent, metallic, and coordinate bonds

Show answer & explanation

Answer: Covalent, coordinate (dative), and ionic bonds

Answer

Covalent, coordinate (dative), and ionic bonds

Ammonium chloride (

NH_4Cl

) exhibits three distinct types of chemical bonds: (1) Covalent bonds between nitrogen and three hydrogen atoms in the original ammonia molecule, (2) A coordinate (dative) bond formed when nitrogen donates its lone pair to a hydrogen ion (

H^+

) to form the ammonium ion (

NH_4^+

), and (3) An ionic (electrovalent) bond between the

NH_4^+

cation and the

Cl^-

anion.

Step-by-Step Solution

1

Analyze the structure of the ammonia molecule (

NH_3

).

Nitrogen shares electron pairs with three hydrogen atoms, forming three polar covalent bonds and retaining one unshared lone pair of electrons.

Covalent bonding occurs when non-metal atoms share pairs of valence electrons.

2

Analyze the formation of the ammonium ion (

NH_4^+

).

Nitrogen donates its lone pair of electrons to an electron-deficient hydrogen ion (

H^+

), forming one coordinate (dative) covalent bond.

A coordinate bond is formed when one atom provides both electrons for the shared pair.

3

Analyze the interaction between the ammonium ion (

NH_4^+

) and the chloride ion (

Cl^-

).

Electrostatic attraction between the positively charged

NH_4^+

cation and negatively charged

Cl^-

anion forms an ionic bond.

Oppositely charged ions attract each other to form a stable crystal lattice.

Key Concept

Coexistence of covalent, dative, and ionic bonding in polyatomic salts

Estimated Time:1m 0s

Question 47Question →

Match each structural feature of the electron sea model on the left with the macroscopic metal property it directly accounts for on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

1. Movement of delocalized electrons toward a positive terminal matches High electrical conductivity. 2. Layers of cations sliding past each other matches Malleability and ductility. 3. Absorption and re-emission of light by free electrons matches Lustrous (shiny) appearance. 4. Strong non-directional electrostatic attraction matches High melting and boiling points.

Each structural feature in the electron sea model directly dictates a specific macroscopic behavior: delocalized electron motion provides electrical conductivity, cation layer flexibility allows deformation (malleability/ductility), light oscillation by surface electrons causes shiny luster, and extensive electrostatic forces produce high thermal melting thresholds.

Step-by-Step Solution

1

Relate electric charge transport to metallic conduction

Free electrons moving toward a positive potential corresponds to high electrical conductivity.

Electric current in solid metals consists of a net flow of delocalized valence electrons.

2

Analyze deformation behavior of metallic lattices under pressure

Sliding layers of cations buffered by the electron sea corresponds to malleability and ductility.

Metals deform without shattering because metallic bonding is non-directional.

3

Connect light interaction with free electron oscillations

Free surface electrons absorbing and re-emitting light photons corresponds to luster.

Unbound electrons respond dynamically to electromagnetic waves, reflecting light.

4

Examine thermal stability of the lattice bonding

Strong omnidirectional electrostatic attraction corresponds to high melting and boiling points.

Separating metallic particles requires inputting significant thermal energy to overcome electrostatic bonds.

Key Concept

Electron Sea Model of Metallic Bonding

Question 48Question →

Match each physical or chemical behavior of metallic substances on the left with its corresponding atomic-scale mechanism on the right.

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Show answer & explanation

Answer

High thermal conductivity corresponds to rapid kinetic energy transfer by delocalized electrons; the decrease in electrical conductivity at higher temperatures corresponds to increased scattering from vibrating metal cations; metallic lustre corresponds to photon absorption and re-emission by surface delocalized electrons; and the higher melting points of transition metals correspond to combined

s

-electron delocalization and

d

-orbital overlap.

Each property is accurately matched with its fundamental physical cause: thermal conduction is driven by kinetic energy transfer by mobile electrons; thermal reduction of electrical conductivity stems from enhanced cation scattering; lustre arises from rapid light re-emission by surface electrons; and high transition metal melting points are due to combined

s

-electron delocalization and

d

-orbital bonding.

Step-by-Step Solution

1

Analyze the mechanism for heat conduction in metals.

Thermal conduction occurs because delocalized valence electrons move freely and quickly pass kinetic energy down the temperature gradient.

Free electrons carry kinetic energy much faster than localized lattice atom collisions alone.

2

Analyze how temperature affects electrical resistance/conductivity in metals.

Heating increases the vibrational amplitude of positive cations in the lattice, creating greater resistance (scattering) for moving electron streams.

Impeding the mean free path of drift electrons reduces electrical conductivity.

3

Analyze the optical reflection property of metals.

Incident light causes surface delocalized electrons to oscillate and instantly re-radiate light photons across continuous energy levels.

The sea of mobile electrons acts as a reflective barrier to light waves.

4

Compare cohesive energy differences between alkali metals and transition metals.

Transition elements utilize both outer

s

valence electrons and partially filled inner

d

subshells to form additional covalent bonds, significantly increasing lattice strength and melting point.

Greater electrostatic attraction and inter-atomic orbital overlap increase the energy required to break the lattice.

Key Concept

Metallic Bonding mechanisms relating atomic-scale electron sea and lattice structures to macroscopic physical properties

Question 49Question →

An element $M$ has the ground-state electronic configuration $1s^2 2s^2 2p^6 3s^2$ , while element $X$ has the configuration $1s^2 2s^2 2p^5$ . What is the chemical formula of the compound formed between $M$ and $X$ , and what type of bonding holds the compound together?

MX_2

; electrovalent (ionic) bonding

MX

; electrovalent (ionic) bonding

M_2X

; covalent bonding

MX_2

; covalent bonding

Show answer & explanation

Answer: $MX_2$ ; electrovalent (ionic) bonding

Answer

The correct compound formula is

MX_2

and the bonding is electrovalent (ionic).

The correct answer states that the compound formula is

MX_2

formed by electrovalent (ionic) bonding. Element

M

has 2 valence electrons (

3s^2

) which it transfers to two atoms of element

X

(each having 7 valence electrons,

2s^2 2p^5

), forming

M^{2+}

and two

X^-

ions. Bonding via complete electron transfer between metal and non-metal is electrovalent.

Step-by-Step Solution

1

Analyze the electronic configuration of element

M

to determine its valency and ion charge.

Element

M

(

1s^2 2s^2 2p^6 3s^2

) has 2 valence electrons in its outermost shell (

3s

). It donates 2 electrons to achieve a stable octet, forming the cation

M^{2+}

.

Elements with 1 or 2 outer electrons readily lose them to achieve stable noble gas electron configurations.

2

Analyze the electronic configuration of element

X

to determine its valency and ion charge.

Element

X

(

1s^2 2s^2 2p^5

) has 7 valence electrons in its outermost shell (

2s^2 2p^5

). It accepts 1 electron to complete its octet, forming the anion

X^-

.

Non-metals with 7 valence electrons require 1 additional electron for octet stability.

3

Balance the ionic charges to establish the formula unit and identify the bond type.

To maintain electrical neutrality, one

M^{2+}

ion combines with two

X^-

ions, resulting in the chemical formula

MX_2

. The complete transfer of electrons from a metal to a non-metal yields electrostatic attraction, which constitutes electrovalent (ionic) bonding.

Ionic compounds must be electrically neutral, and bonding formed by complete electron transfer is electrovalent.

Key Concept

Ionic (Electrovalent) Bond Formation and Formula Determination

Question 50Question →

Ionic compounds are generally brittle because applying a mechanical stress causes layers of ions to shift, bringing ions of identical charge into alignment and causing strong electrostatic repulsion.

True

False

Show answer & explanation

Answer: True

Answer

True. Applying mechanical stress causes layers of an ionic lattice to shift, bringing like-charged ions into direct alignment; the immediate electrostatic repulsion forces the crystal planes apart, rendering ionic solids brittle.

The statement is true because mechanical stress displaces adjacent rows in an ionic lattice, shifting like-charged ions into alignment. The powerful electrostatic repulsion generated between identical charges forces the crystal layers apart, causing the ionic solid to shatter.

Step-by-Step Solution

1

Identify the arrangement of particles in a solid ionic lattice.

An ionic crystal consists of alternating positive cations and negative anions organized in a regular three-dimensional array held by electrostatic attraction.

Determining the initial alternating charge pattern is necessary to understand how movement alters interionic forces.

2

Analyze the structural shift caused by an applied mechanical force.

The force causes one layer of ions to slide past another by one atomic position.

Mechanical impact displaces crystal planes relative to each other.

3

Evaluate the net electrostatic forces after displacement.

Ions of identical charge (

+

and

+

, or

-

and

-

) are brought into direct alignment, creating powerful repulsive forces that shatter the lattice along cleavage planes.

Electrostatic repulsion between like charges overcomes binding attraction, explaining the characteristic brittleness of ionic solids.

Key Concept

Brittleness and Mechanical Cleavage of Ionic Lattice Structures

Question 51Question →

Pair each of the given chemical molecules with its corresponding molecular geometry as predicted by Valence Shell Electron Pair Repulsion (VSEPR) theory.

Click a left item, then click its matching right item

Show answer & explanation

Answer

BF_3

matches Trigonal planar,

CH_4

matches Tetrahedral,

BeCl_2

matches Linear, and

H_2O

matches Bent (V-shaped).

According to VSEPR theory, molecular shape depends on the total number of bonding pairs and lone pairs surrounding the central atom.

BeCl_2

has 2 bonding pairs with no lone pairs, yielding a linear geometry.

BF_3

has 3 bonding pairs with no lone pairs, producing a trigonal planar geometry.

CH_4

has 4 bonding pairs with no lone pairs, resulting in a tetrahedral geometry.

H_2O

has 2 bonding pairs and 2 non-bonding lone pairs, creating a bent (V-shaped) geometry.

Step-by-Step Solution

1

Determine the number of valence electron pairs (bonding pairs and lone pairs) surrounding the central atom for each chemical species.

BF_3

has 3 bonding pairs and 0 lone pairs;

CH_4

has 4 bonding pairs and 0 lone pairs;

BeCl_2

has 2 bonding pairs and 0 lone pairs;

H_2O

has 2 bonding pairs and 2 lone pairs.

VSEPR theory states that electron pairs around a central atom arrange themselves to minimize electrostatic repulsion.

2

Deduce the resulting molecular geometry for each molecule based on the arrangement of bonding and lone pairs.

3 bond pairs (0 lone pairs) = Trigonal planar; 4 bond pairs (0 lone pairs) = Tetrahedral; 2 bond pairs (0 lone pairs) = Linear; 2 bond pairs + 2 lone pairs = Bent.

Lone pairs exert greater repulsive force than bonding pairs, bending the molecular framework accordingly.

Key Concept

VSEPR Theory and Molecular Geometries

Question 52Question →

Sulfur tetrafluoride ( $SF_4$ ) and xenon tetrafluoride ( $XeF_4$ ) are both covalent fluorides containing four fluorine atoms bonded to a central atom. Based on VSEPR theory and hybridization models, which of the following correctly pairs the molecular geometry and central atom hybridization state for $SF_4$ and $XeF_4$ , respectively?

SF_4

: Seesaw geometry with

sp^3d

hybridization;

XeF_4

: Square planar geometry with

sp^3d^2

hybridization

SF_4

: Trigonal bipyramidal geometry with

sp^3d

hybridization;

XeF_4

: Octahedral geometry with

sp^3d^2

hybridization

SF_4

: Tetrahedral geometry with

sp^3

hybridization;

XeF_4

: Tetrahedral geometry with

sp^3

hybridization

SF_4

: Square planar geometry with

sp^3d^2

hybridization;

XeF_4

: Seesaw geometry with

sp^3d

hybridization

Show answer & explanation

Answer: $SF_4$ : Seesaw geometry with $sp^3d$ hybridization; $XeF_4$ : Square planar geometry with $sp^3d^2$ hybridization

Answer

SF_4

has a seesaw molecular geometry with

sp^3d

hybridization on the sulfur atom, while

XeF_4

has a square planar molecular geometry with

sp^3d^2

hybridization on the xenon atom.

For

SF_4

, sulfur has 6 valence electrons forming 4 single bonds and retaining 1 lone pair, giving 5 electron domains (

sp^3d

hybridization). The lone pair occupies an equatorial site of the trigonal bipyramid, yielding a seesaw shape. For

XeF_4

, xenon has 8 valence electrons forming 4 single bonds and retaining 2 lone pairs, giving 6 electron domains (

sp^3d^2

hybridization). The two lone pairs lie opposite each other along the axial axis, resulting in a square planar geometry.

Step-by-Step Solution

1

Determine valence electron count and electron domains for

SF_4

Sulfur has 6 valence electrons. Bonding 4 fluorine atoms consumes 4 electrons, leaving 2 non-bonding electrons (1 lone pair). Total steric number = 4 bonding pairs + 1 lone pair = 5 electron domains.

VSEPR theory requires calculating the total number of electron pairs around the central atom.

2

Assign hybridization and molecular geometry for

SF_4

A steric number of 5 corresponds to

sp^3d

hybridization and trigonal bipyramidal electron geometry. With 1 equatorial lone pair, the molecular shape is seesaw.

Lone pairs prefer equatorial positions in trigonal bipyramidal systems to minimize electron repulsion, leading to a seesaw geometry.

3

Determine valence electron count and electron domains for

XeF_4

Xenon has 8 valence electrons. Bonding 4 fluorine atoms consumes 4 electrons, leaving 4 non-bonding electrons (2 lone pairs). Total steric number = 4 bonding pairs + 2 lone pairs = 6 electron domains.

Steric number governs hybridization and electron arrangement for the central xenon atom.

4

Assign hybridization and molecular geometry for

XeF_4

A steric number of 6 corresponds to

sp^3d^2

hybridization and octahedral electron geometry. With 2 axial lone pairs opposing each other (180° apart), the molecular geometry is square planar.

Axial placement of two lone pairs in an octahedral arrangement maximizes separation and produces a planar square arrangement of bonded fluorine atoms.

Key Concept

Molecular Shapes, VSEPR Theory, and Hybridization

Estimated Time:2m 0s

Question 53Question →

Match each of the following chemical species or systems with the predominant type of intermolecular force or interaction present between its units in the liquid or solid state.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Water matches with Hydrogen bonding; Trichloromethane matches with Permanent dipole-dipole interaction; Argon matches with London dispersion forces; Hydrated sodium ion matches with Ion-dipole interaction.

Water undergoes hydrogen bonding due to the presence of highly polar

\text{O-H}

bonds. Trichloromethane exhibits permanent dipole-dipole attractions because it is a polar molecule without hydrogen attached to

\text{N, O, F}

. Argon is a nonpolar noble gas that relies solely on temporary induced dipoles (London dispersion forces). The hydrated sodium ion experiences electrostatic ion-dipole interactions between the positive ion and polar water molecules.

Step-by-Step Solution

1

Analyze the chemical composition and molecular polarity of each species.

Water (

\text{H}_2\text{O}

) has

\text{O-H}

bonds; Trichloromethane (

\text{CHCl}_3

) is a polar molecule lacking

\text{N-H}

,

\text{O-H}

, or

\text{F-H}

bonds; Argon (

\text{Ar}

) is a nonpolar noble gas; Hydrated sodium ion (

\text{Na}^+_{(aq)}

) features an ionic species surrounded by polar solvent molecules.

Determining polarity, presence of electronegative elements bonded to hydrogen, and ionic charge is necessary to classify intermolecular forces.

2

Assign the predominant intermolecular force to each species.

\text{H}_2\text{O}

forms hydrogen bonds;

\text{CHCl}_3

experiences permanent dipole-dipole attractions;

\text{Ar}

relies on London dispersion forces;

\text{Na}^+_{(aq)}

displays ion-dipole attractions.

Hydrogen bonding requires

\text{H}

attached directly to

\text{F, O, N}

; dipole-dipole forces operate between permanent dipoles; dispersion forces exist in all species but predominate in nonpolar units; ion-dipole interactions occur between ions and polar molecules.

Key Concept

Intermolecular Forces and Hydrogen Bonding

Estimated Time:1m 30s

Question 54Question →

According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, what is the molecular shape and central atom hybridization of sulfur dioxide ( $SO_2$ )?

Bent (V-shaped) with

sp^2

hybridization

Trigonal planar with

sp^2

hybridization

Linear with

sp

hybridization

Bent (V-shaped) with

sp^3

hybridization

Show answer & explanation

Answer: Bent (V-shaped) with $sp^2$ hybridization

Answer

Bent (V-shaped) geometry with

sp^2

hybridization of the central sulfur atom.

The central sulfur atom in

SO_2

has three electron regions around it (two bonding regions to oxygen atoms and one lone pair). Three electron regions dictate a trigonal planar electron geometry and

sp^2

hybridization. Because one of these regions is a lone pair, the observable molecular geometry formed by the atoms is bent (V-shaped).

Step-by-Step Solution

1

Determine the valence electron count and electron domains around the central sulfur atom in

SO_2

.

Sulfur (Group 16) brings 6 valence electrons. It forms 2 double bonds with 2 oxygen atoms (2 bonding domains) and retains 1 lone pair of electrons. Total electron domains = 3.

VSEPR theory uses the total number of electron domains (bonding regions + lone pairs) to determine spatial arrangement.

2

Determine the hybridization of the central atom.

3 electron domains correspond to an

sp^2

hybridization state.

Mixing one s orbital and two p orbitals yields three equivalent

sp^2

hybrid orbitals directed towards the corners of an equilateral triangle.

3

Determine the molecular geometry based on bonding domains and lone pairs.

With 3 electron domains (2 bonding domains and 1 lone pair), the electron geometry is trigonal planar, but the molecular geometry (shape formed by atoms) is Bent (V-shaped) with a bond angle of slightly less than

120^\circ

.

Lone pair-bonding pair repulsion compresses the

O-S-O

bond angle slightly below the ideal

120^\circ

trigonal planar angle.

Key Concept

VSEPR Theory, Molecular Geometry, and Hybridization

Estimated Time:1m 0s

Question 55Question →

In solid-state chemistry, the distinct physical behaviors of metals arise directly from the structural characteristics of metallic bonds. Match each observable metallic property or behavior on the left with its corresponding atomic-scale explanation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

High electrical conductivity corresponds to unconfined valence electrons drifting directionally under an applied potential difference. High malleability and ductility correspond to non-directional electrostatic attractions allowing cation layers to slide past each other while maintaining cohesive forces. Higher melting point of iron compared to sodium corresponds to the contribution of delocalized d-orbital electrons alongside s-electrons. Lustrous reflective appearance corresponds to the oscillation of free valence electrons absorbing and rapidly re-emitting incident photons.

High electrical conductivity is explained by the movement of unconfined valence electrons drifting directionally when a potential difference is applied. High malleability and ductility stem from non-directional electrostatic forces allowing metal cation planes to slide over each other without breaking cohesive bonds. The higher melting point of transition metals like iron compared to alkali metals like sodium is caused by extra binding strength provided by delocalized d-orbital electrons in addition to s-electrons. Metallic luster is caused by mobile valence electrons absorbing incident light energy and immediately re-emitting it.

Step-by-Step Solution

1

Analyze the microscopic origin of electrical conduction in metallic crystals.

Electrical conduction requires mobile charge carriers. In metals, delocalized valence electrons move freely across the lattice under an electric potential.

Relates macroscopic electric current to electron mobility.

2

Analyze how mechanical force affects metal cation layers.

Deformation causes layers of cations to slip over each other. Because metallic bonds are non-directional, the electron sea adjusts instantly to keep the lattice bound without brittle cleavage.

Explains malleability and ductility via non-directional bonding.

3

Compare the bonding strength of alkali metals versus transition metals.

Sodium donates only one s-electron per atom into the sea, whereas iron donates both s and unpaired inner d-electrons, greatly increasing the electrostatic cohesive energy and melting point.

Explains variation in thermal resistance and hardness across different metals.

4

Analyze the interaction between light waves and delocalized electron clouds.

Mobile surface electrons readily absorb light energy and oscillate, promptly re-radiating light photons to generate a high spectral reflectance (luster).

Connects optical reflectivity to electron sea excitation.

Key Concept

Electron Sea Model and Metal Property Mechanisms

Question 56Question →

Match each chemical species with its corresponding central atom hybridization state and molecular geometry as predicted by Valence Shell Electron Pair Repulsion (VSEPR) theory.

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Answer

Chlorate ion (

ClO_3^-

) matches

sp^3

hybridization with a trigonal pyramidal shape; Xenon difluoride (

XeF_2

) matches

sp^3d

hybridization with a linear shape; Tetrachloroiodate ion (

ICl_4^-

) matches

sp^3d^2

hybridization with a square planar shape; Sulfur dioxide (

SO_2

) matches

sp^2

hybridization with a bent shape.

Each chemical species is accurately paired by identifying the total number of electron domains around its central atom:

ClO_3^-

has 4 domains (

sp^3

, trigonal pyramidal shape),

XeF_2

has 5 domains (

sp^3d

, linear shape),

ICl_4^-

has 6 domains (

sp^3d^2

, square planar shape), and

SO_2

has 3 domains (

sp^2

, bent shape).

Step-by-Step Solution

1

Determine the valence electron count and steric number (bonding domains + lone pairs) for the central atom of each species.

Chlorate ion (

ClO_3^-

): 3 bonds + 1 lone pair = steric number 4; Xenon difluoride (

XeF_2

): 2 bonds + 3 lone pairs = steric number 5; Tetrachloroiodate ion (

ICl_4^-

): 4 bonds + 2 lone pairs = steric number 6; Sulfur dioxide (

SO_2

): 2 double-bond domains + 1 lone pair = steric number 3.

The total number of electron domains determines both the hybridization of atomic orbitals and the underlying electron pair geometry.

2

Assign the hybridization corresponding to each steric number.

Steric number 4 corresponds to

sp^3

; steric number 5 corresponds to

sp^3d

; steric number 6 corresponds to

sp^3d^2

; steric number 3 corresponds to

sp^2

.

Hybrid orbital set size equals the number of electron domains.

3

Apply VSEPR theory rules to deduce the molecular geometry (accounting only for atomic positions).

ClO_3^-

is trigonal pyramidal (

sp^3

);

XeF_2

is linear (

sp^3d

);

ICl_4^-

is square planar (

sp^3d^2

);

SO_2

is bent (

sp^2

).

Non-bonding lone pairs cause greater repulsion and define the non-spherical molecular geometry relative to electron geometry.

Key Concept

VSEPR Theory and Hybridization of Molecular Species and Polyatomic Ions

Question 57Question →

When ammonia ( $NH_3$ ) reacts with an acid to form the ammonium ion ( $NH_4^+$ ), the central nitrogen atom shares its lone pair of electrons to form a dative covalent bond with a proton ( $H^+$ ). According to Valence Shell Electron Pair Repulsion (VSEPR) theory, what is the molecular geometry and hybridization state of the nitrogen atom in the $NH_4^+$ ion?

Tetrahedral geometry with

sp^3

hybridization

Trigonal pyramidal geometry with

sp^3

hybridization

Trigonal planar geometry with

sp^2

hybridization

Square planar geometry with

sp^3d^2

hybridization

Show answer & explanation

Answer: Tetrahedral geometry with $sp^3$ hybridization

Answer

Tetrahedral geometry with

sp^3

hybridization

In the ammonium ion (

NH_4^+

), the central nitrogen atom is surrounded by four single covalent bonds (including one dative covalent bond formed with

H^+

) and zero lone pairs. Four electron domains arrange themselves to minimize repulsion into a symmetric tetrahedral arrangement with bond angles of

109.5^\circ

. Mixing one s orbital and three p orbitals gives

sp^3

hybridization.

Step-by-Step Solution

1

Determine the valence electron count and electron domain total around the central nitrogen atom in

NH_4^+

.

Nitrogen brings 5 valence electrons, four hydrogen atoms contribute 1 electron each, and the +1 charge indicates the loss of 1 electron. Total valence electrons =

5 + 4 - 1 = 8

electrons (4 pairs).

VSEPR theory uses the total number of electron pairs around the central atom to predict spatial geometry.

2

Identify the number of bonding pairs and lone pairs on the nitrogen atom.

All 4 electron pairs are involved in single covalent bonds (3 covalent, 1 dative covalent) with hydrogen atoms, leaving 0 lone pairs.

Molecular geometry depends specifically on the arrangement of bonding pairs when 4 electron domains are present.

3

Determine the molecular geometry and hybridization from the electron pair arrangement.

Four bonding domains with zero lone pairs yield a symmetrical regular tetrahedral shape with a bond angle of

109.5^\circ

and

sp^3

hybridization.

Four equivalent steric domains require one s orbital and three p orbitals to hybridize into four

sp^3

hybrid orbitals.

Key Concept

Molecular geometry and hybridization of polyatomic ions using VSEPR theory

Question 58Question →

Match each of the following chemical species on the left with its corresponding central atom hybridization state and molecular geometry on the right, as predicted by Valence Shell Electron Pair Repulsion (VSEPR) theory. Which set of pairs accurately connects each species to its geometry and hybridization?

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Show answer & explanation

Answer

Iodine trifluoride (

IF_3

) matches with

sp^3d

hybridization and T-shaped geometry; Pentafluoroxenate ion (

XeF_5^-

) matches with

sp^3d^3

hybridization and pentagonal planar geometry; Sulfite ion (

SO_3^{2-}

) matches with

sp^3

hybridization and trigonal pyramidal geometry; Nitronium ion (

NO_2^+

) matches with

sp

hybridization and linear geometry.

Each chemical species is correctly paired according to its total steric number (sum of bonding electron domains and non-bonding lone pairs).

IF_3

has steric number 5 (

sp^3d

, T-shaped),

XeF_5^-

has steric number 7 (

sp^3d^3

, pentagonal planar),

SO_3^{2-}

has steric number 4 (

sp^3

, trigonal pyramidal), and

NO_2^+

has steric number 2 (

sp

, linear).

Step-by-Step Solution

1

Determine valence electron count and steric number for

IF_3

Iodine has 7 valence electrons + 3 from fluorine = 10 electrons (5 pairs). Steric number = 5 (3 bonding pairs, 2 lone pairs).

Steric number 5 corresponds to

sp^3d

hybridization. Two equatorial lone pairs force the 3 terminal fluorines into a T-shaped arrangement.

2

Determine valence electron count and steric number for

XeF_5^-

Xenon has 8 valence electrons + 5 from fluorine + 1 from overall negative charge = 14 electrons (7 pairs). Steric number = 7 (5 bonding pairs, 2 lone pairs).

Steric number 7 corresponds to

sp^3d^3

hybridization. The two lone pairs occupy axial positions above and below the equatorial plane, creating a pentagonal planar molecular shape.

3

Determine valence electron count and steric number for

SO_3^{2-}

Sulfur has 6 valence electrons + 2 from charge = 8 valence shell electrons. It forms 3 sigma bonds with oxygen and retains 1 lone pair. Steric number = 4.

Steric number 4 corresponds to

sp^3

hybridization. Three bonding domains and 1 lone pair produce a trigonal pyramidal molecular shape.

4

Determine valence electron count and steric number for

NO_2^+

Nitrogen has 5 valence electrons - 1 from positive charge = 4 electrons. It forms two double bonds with oxygen atoms and has 0 lone pairs. Steric number = 2.

Steric number 2 corresponds to

sp

hybridization, resulting in a linear geometry with a

180^\circ

bond angle.

Key Concept

VSEPR Theory, Steric Numbers, and Central Atom Hybridization States

Estimated Time:2m 30s

Question 59Question →

Match each of the following chemical species with the predominant intermolecular force operating between its molecules in the liquid or solid state.

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Answer

Hydrogen fluoride matches with Hydrogen bonding; Trichloromethane matches with Permanent dipole-dipole interactions; Solid iodine matches with London dispersion forces (in a non-polar crystalline lattice); Methane matches with Weak London dispersion forces (in a small non-polar molecule).

Hydrogen fluoride forms hydrogen bonds due to the extreme electronegativity difference between H and F. Trichloromethane exhibits permanent dipole-dipole attractions because of its permanent net molecular dipole. Solid iodine is non-polar but has a large polarizable electron cloud leading to substantial London dispersion forces in its solid crystal. Methane is non-polar and small, possessing only weak London dispersion forces.

Step-by-Step Solution

1

Analyze the polarity and chemical structure of each given substance.

HF

is highly polar with

H-F

bonds;

CHCl_3

is a polar asymmetrical molecule;

I_2

is a non-polar diatomic solid;

CH_4

is a non-polar tetrahedral gas.

Intermolecular forces depend strictly on molecular polarity, presence of

N-H

,

O-H

, or

F-H

bonds, and molecular size/polarizability.

2

Identify specific conditions for hydrogen bonding.

HF

satisfies the requirement of hydrogen attached to highly electronegative fluorine, giving rise to intermolecular hydrogen bonds.

Hydrogen bonding requires a hydrogen atom covalently bonded to

N

,

O

, or

F

interacting with a lone pair on a neighbouring electronegative atom.

3

Differentiate dipole-dipole forces from dispersion forces in neutral covalent compounds.

CHCl_3

possesses a permanent dipole moment giving dipole-dipole forces, while

I_2

and

CH_4

are non-polar and rely on London dispersion forces, with

I_2

having larger dispersion forces due to greater electron cloud polarizability.

Dispersion forces scale with molecular size and electron count, while dipole-dipole forces require permanent polar bonds in asymmetrical shapes.

Key Concept

Classification and Origin of Intermolecular Forces

Question 60Question →

Which of the following species features a central atom with $sp^2$ hybridization and a trigonal planar molecular geometry?

CO_3^{2-}

H_3O^+

PCl_3

SO_3^{2-}

Show answer & explanation

Answer: $CO_3^{2-}$

Answer

The trioxocarbonate(IV) ion (

CO_3^{2-}

) possesses a central carbon atom with

sp^2

hybridization and a trigonal planar molecular shape.

In the trioxocarbonate(IV) ion (

CO_3^{2-}

), the central carbon atom forms three sigma bonds with three oxygen atoms and contains zero non-bonding lone pairs. According to VSEPR theory, three electron charge clouds position themselves as far apart as possible (

120^\circ

bond angles), adopting a trigonal planar geometry which requires

sp^2

hybridization of the central carbon atom.

Step-by-Step Solution

1

Determine the valence electron count and steric number for the central carbon in

CO_3^{2-}

Carbon brings 4 valence electrons, plus 2 electrons from the net charge (-2), total 6 valence electrons available for bonding. Carbon forms 3 sigma bonds with three oxygen atoms and has 0 lone pairs. Steric number = 3.

VSEPR theory uses the number of electron domains (steric number) around the central atom to determine orbital hybridization and electron geometry.

2

Determine the hybridization and molecular geometry based on VSEPR theory

Steric number 3 with 0 lone pairs corresponds to

sp^2

hybridization and a trigonal planar molecular shape.

Three electron domains orient themselves at

120^\circ

bond angles to minimize electron pair repulsion.

3

Compare with the distractor species (

H_3O^+

,

PCl_3

, and

SO_3^{2-}

)

Each of

H_3O^+

,

PCl_3

, and

SO_3^{2-}

has 3 bonding pairs and 1 non-bonding lone pair on its central atom (steric number 4).

A steric number of 4 corresponds to

sp^3

hybridization, and 3 bonding pairs with 1 lone pair produces a trigonal pyramidal geometry rather than trigonal planar.

Key Concept

VSEPR Theory and Hybridization of Species with Three Attached Atoms

Atomic Structure and Chemical Bonding

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