Atomic Structure and Chemical Bonding Practice Questions — JAMB UTME — Page 4

Question 61Question →

The sulfite ion ( $SO_3^{2-}$ ) is formed when sulfur dioxide dissolves in aqueous basic media. Based on Valence Shell Electron Pair Repulsion (VSEPR) theory, which of the following correctly describes the hybridization of the central sulfur atom and the molecular geometry of the $SO_3^{2-}$ ion?

sp^3

hybridization with trigonal pyramidal molecular geometry

sp^3

hybridization with tetrahedral molecular geometry

sp^2

hybridization with trigonal planar molecular geometry

sp^2

hybridization with bent molecular geometry

Show answer & explanation

Answer: $sp^3$ hybridization with trigonal pyramidal molecular geometry

Answer

The central sulfur atom in the sulfite ion (

SO_3^{2-}

) exhibits

sp^3

hybridization and has a trigonal pyramidal molecular geometry.

The central sulfur atom in

SO_3^{2-}

possesses 4 electron domains (3 bonding pairs and 1 non-bonding lone pair), yielding

sp^3

hybridization. According to VSEPR theory, an

AX_3E

electron arrangement produces a trigonal pyramidal molecular shape.

Step-by-Step Solution

1

Determine total valence electrons for

SO_3^{2-}

Sulfur provides 6, each of the three oxygens provides 6, plus 2 electrons from the

2-

charge:

6 + (3 \times 6) + 2 = 26

valence electrons (13 pairs).

Accurate valence electron counting is necessary to determine bonding and non-bonding electron distribution.

2

Determine electron domain count and steric number on the central atom

Sulfur forms 3 single sigma bonds to oxygen atoms and retains 1 lone pair of non-bonding electrons (

3 \text{ bonding pairs} + 1 \text{ lone pair} = 4 \text{ total electron domains}

).

Steric number determines the orbital hybridization of the central atom.

3

Determine hybridization and molecular geometry using VSEPR theory

A steric number of 4 corresponds to

sp^3

hybridization. With 3 bonding pairs and 1 lone pair (an

AX_3E

system), the electron geometry is tetrahedral, but the actual molecular geometry is trigonal pyramidal.

Molecular geometry is named based only on the spatial arrangement of the bonded atoms, not the non-bonding lone pair.

Key Concept

VSEPR theory and hybridization determination for polyatomic ions with lone pairs

Estimated Time:1m 30s

Question 62Question →

Arrange the following chemical substances in order of increasing boiling point, starting from the substance with the lowest boiling point to the one with the highest boiling point based on the nature and relative strength of their intermolecular forces.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order of increasing boiling point is Methane (

CH_4

) < Hydrogen sulfide (

H_2S

) < Ammonia (

NH_3

) < Water (

H_2O

).

The sequence reflects the increasing magnitude of intermolecular forces: non-polar Methane (

CH_4

) relies solely on weak London dispersion forces (lowest boiling point). Polar Hydrogen sulfide (

H_2S

) has dipole-dipole interactions but lacks hydrogen bonding because sulfur is not electronegative enough. Ammonia (

NH_3

) undergoes hydrogen bonding due to nitrogen's high electronegativity. Water (

H_2O

) forms an extensive network of strong hydrogen bonds, resulting in the highest boiling point.

Step-by-Step Solution

1

Identify the type of intermolecular forces operating in each substance

CH_4

is non-polar (London dispersion forces only);

H_2S

is polar (dipole-dipole and dispersion forces);

NH_3

is polar with hydrogen bonding;

H_2O

is polar with strong, extensive hydrogen bonding.

Boiling point depends directly on the total magnitude of attraction between molecules in the liquid state.

2

Compare non-hydrogen-bonding substances (

CH_4

vs

H_2S

)

CH_4

has the weakest intermolecular forces (dispersion only), while

H_2S

has additional permanent dipole-dipole attractions.

Permanent dipole-dipole interactions in polar molecules generally create stronger attraction than non-polar dispersion forces of comparable size.

3

Compare hydrogen-bonding substances (

NH_3

vs

H_2O

)

Both

NH_3

and

H_2O

form hydrogen bonds, but

H_2O

forms up to four hydrogen bonds per molecule in a 3D network, whereas

NH_3

is limited by its single lone pair to fewer hydrogen bonds per molecule.

Oxygen is more electronegative than nitrogen, and water has an optimal 1:1 ratio of lone pairs to hydrogen atoms for maximum hydrogen-bonding capacity.

4

Synthesize the complete sequence from lowest to highest boiling point

CH_4

<

H_2S

<

NH_3

<

H_2O

Intermolecular attraction strength increases in the order: London dispersion forces < dipole-dipole interactions < moderate hydrogen bonding < extensive hydrogen bonding.

Key Concept

Relative strengths of intermolecular forces (London dispersion, dipole-dipole, and hydrogen bonding) and their effect on physical properties like boiling point.

Estimated Time:2m 0s

Question 63Question →

Complete the statements regarding the neutron activation of sodium and its subsequent radioactive decay by filling in the appropriate values.

Fill in the blanks below

When Sodium-23 (

^{23}_{11}\text{Na}

) is bombarded with a neutron, it forms radioactive Sodium-24 (

^{24}_{11}\text{Na}

), which subsequently undergoes beta decay with a half-life of

15\text{ hours}

. Starting with an initial mass of

0.80\text{ g}

of pure

^{24}_{11}\text{Na}

, after an elapsed time of

45\text{ hours}

, the mass of

^{24}_{11}\text{Na}

remaining is grams, and the element formed as the stable daughter product is .

Show answer & explanation

Answer

The remaining mass of Sodium-24 after 45 hours is 0.10 grams, and the stable daughter element formed is Magnesium.

In 45 hours, exactly three 15-hour half-lives elapse, reducing the initial 0.80 g sample to 0.10 g. During beta decay, the emission of a beta particle (

^{0}_{-1}\beta

) increases the atomic number of the nuclide from 11 to 12, converting Sodium to Magnesium.

Step-by-Step Solution

1

Calculate the total number of half-lives elapsed (

n

)

n = \frac{t}{t_{1/2}} = \frac{45\text{ hours}}{15\text{ hours}} = 3\text{ half-lives}

The total elapsed time divided by the half-life period yields the number of decay cycles.

2

Determine the remaining mass of Sodium-24 using exponential decay

N_t = N_0 \times \left(\frac{1}{2}\right)^n = 0.80\text{ g} \times \left(\frac{1}{2}\right)^3 = 0.80\text{ g} \times 0.125 = 0.10\text{ g}

With each half-life cycle, the active radioactive mass decreases by half.

3

Write the balanced nuclear reaction for the beta decay of Sodium-24 to identify the daughter element

^{24}_{11}\text{Na} \rightarrow ^{24}_{12}\text{Mg} + ^{0}_{-1}\beta

In beta particle emission (

^{0}_{-1}\beta

), a neutron converts to a proton, increasing the atomic number

Z

from 11 to 12 while leaving the mass number

A = 24

unchanged. Element 12 on the periodic table is Magnesium (Mg).

Key Concept

Radioactive decay law and nuclear equation balancing for beta decay

Question 64Question →

Although fluorine and chlorine are both Group 17 halogens, hydrogen fluoride ( $HF$ ) exhibits intermolecular hydrogen bonding in the liquid state while hydrogen chloride ( $HCl$ ) does not. Which of the following statements best accounts for this difference?

Fluorine has a higher electronegativity and smaller atomic radius than chlorine, creating a highly polar

H-F

bond.

The intramolecular covalent bond in an

HF

molecule is weaker than the covalent bond in an

HCl

molecule.

Hydrogen fluoride has a greater molar mass than hydrogen chloride, generating stronger dispersion forces.

Chlorine is more electronegative than fluorine, causing

HCl

molecules to experience non-polar interactions.

Show answer & explanation

Answer: Fluorine has a higher electronegativity and smaller atomic radius than chlorine, creating a highly polar $H-F$ bond.

Answer

Fluorine's high electronegativity combined with its small atomic size allows for strong electrostatic attraction between the hydrogen atom of one molecule and the lone pair on the fluorine atom of an adjacent molecule.

Hydrogen bonding requires hydrogen to be covalently attached to a very small, highly electronegative atom (N, O, or F). Fluorine fulfills both conditions, giving the

H-F

bond a large dipole moment and high localized charge density that attracts adjacent

HF

molecules strongly.

Step-by-Step Solution

1

Identify the structural requirements for hydrogen bonding.

Hydrogen bonding occurs specifically when hydrogen is bonded directly to small, highly electronegative atoms (N, O, or F).

High electronegativity creates a strong dipole with high positive charge density on the hydrogen atom.

2

Compare fluorine and chlorine properties.

Fluorine has an electronegativity of 4.0 and a small atomic radius, whereas chlorine has an electronegativity of 3.0 and a larger atomic radius.

Although chlorine is electronegative, its larger atomic volume diffuses electron charge density, preventing effective hydrogen bond formation.

3

Select the statement that correctly attributes hydrogen bond formation to electronegativity and atomic size.

The statement emphasizing fluorine's higher electronegativity and smaller atomic size relative to chlorine provides the correct rationale.

It fulfills the fundamental chemical criteria required for hydrogen bond formation.

Key Concept

Intermolecular Forces and Hydrogen Bonding

Estimated Time:1m 0s

Question 65Question →

Match each physical property or characteristic of ionic (electrovalent) bonding on the left with its correct microscopic or structural explanation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

High melting and boiling points match with the need for extensive thermal energy to disrupt the giant 3D lattice; Electrical conductivity in molten/aqueous state matches with lattice destruction freeing mobile charge carriers; Solubility in water matches with hydration energy overcoming lattice energy; Non-directional nature matches with the electrostatic field acting uniformly in all directions.

Each property of ionic compounds directly stems from its underlying electrostatic structure: High melting points are caused by the strong 3D electrostatic attractions requiring high thermal energy to break; Electrical conductivity in molten/dissolved states occurs because ions are set free as mobile charge carriers; Solubility in water occurs when hydration energy exceeds lattice energy; Non-directionality arises because an ion's electrostatic field attracts opposite charges equally in all spatial directions.

Step-by-Step Solution

1

Analyze the high melting/boiling points of ionic compounds.

Recognize that ions are held in a giant lattice by strong electrostatic forces in all dimensions, requiring high heat energy to overcome.

Relates macro property (melting point) to micro structure (lattice binding energy).

2

Examine the electrical conduction mechanism in ionic substances.

In solid state, ions are fixed in lattice positions. Melting or dissolving releases these ions as mobile charge carriers.

Conduction requires free charge carriers, which are absent in solid ionic crystals.

3

Evaluate the dissolution of ionic compounds in polar solvents.

Polar water molecules surround separated ions (solvation/hydration), releasing energy that overcomes the lattice energy holding the crystal together.

Solubility depends on the thermodynamic balance between hydration enthalpy and lattice enthalpy.

4

Assess the directional nature of ionic bonding.

Because electrostatic attraction operates spherically in space, ionic bonds have no preferred angle or directional vector.

Charges attract equally in all directions, unlike localized shared electron pairs in covalent bonds.

Key Concept

Physical Properties and Structural Basis of Electrovalent (Ionic) Bonding

Question 66Question →

In the reaction between phosphine ( $PH_3$ ) and a hydrogen ion ( $H^+$ ) to form a phosphonium ion ( $PH_4^+$ ), a dative covalent bond is formed. Which of the following statements correctly describes the electronic transfer mechanism during this bond formation?

PH_3

donates an unshared electron pair to

H^+

, which provides an empty orbital.

H^+

donates an electron pair to

PH_3

to complete the phosphorus valence shell.

Both

PH_3

and

H^+

contribute one valence electron each to form the shared pair.

PH_3

completely transfers one electron to

H^+

, forming an electrovalent bond.

Show answer & explanation

Answer: $PH_3$ donates an unshared electron pair to $H^+$ , which provides an empty orbital.

Answer

Phosphine (

PH_3

) donates an unshared electron pair to the hydrogen ion (

H^+

), which provides an empty orbital.

The phosphorus atom in phosphine (

PH_3

) has a non-bonding lone pair of valence electrons. The hydrogen ion (

H^+

) possesses an empty 1s orbital. When

PH_3

reacts with

H^+

, the lone pair on phosphorus is donated into the empty orbital of

H^+

, forming a dative (coordinate) covalent bond where both shared electrons originate from phosphine.

Step-by-Step Solution

1

Identify the valence shell electron structure of the reactants

Phosphorus (Group 15) in

PH_3

forms three single covalent bonds with hydrogen atoms and retains one unshared lone pair of electrons. The hydrogen ion (

H^+

) has lost its electron, leaving an empty 1s orbital.

Determining available lone pairs and empty orbitals is necessary to identify electron donor and acceptor species.

2

Apply the concept of coordinate (dative) covalent bonding

A dative covalent bond occurs when one atom/molecule provides both electrons of the shared pair (the Lewis base/donor), and another atom/ion accepts the pair into an empty orbital (the Lewis acid/acceptor).

This distinguishes dative bonding from normal covalent bonding where each participating atom contributes one electron.

3

Deduce the specific roles of

PH_3

and

H^+

PH_3

serves as the lone pair donor and

H^+

serves as the electron pair acceptor, forming the

[PH_4]^+

ion.

Matching the electronic structures to donor/acceptor definitions confirms the correct mechanism.

Key Concept

Coordinate (Dative) Covalent Bonding and Electron Pair Donation

Estimated Time:1m 0s

Question 67Question →

Complete the following statement regarding the electronic structure, hybridization, and molecular geometry of phosphorus trichloride ( $PCl_3$ ).

Fill in the blanks below

In a molecule of phosphorus trichloride (

PCl_3

), the central phosphorus atom undergoes hybridization and exhibits a molecular geometry.

Show answer & explanation

Answer

The central phosphorus atom in

PCl_3

undergoes

sp^3

hybridization and adopts a trigonal pyramidal molecular geometry.

The central phosphorus atom in

PCl_3

has 5 valence electrons. It uses 3 valence electrons to form single covalent bonds with three chlorine atoms, leaving 2 non-bonding valence electrons as a single lone pair. The total steric number is 4 (3 bonding pairs + 1 lone pair), which corresponds to

sp^3

hybridization. According to VSEPR theory, four electron pairs arrange tetrahedrally, but the presence of three bonded atoms and one lone pair produces a trigonal pyramidal molecular shape.

Step-by-Step Solution

1

Determine the valence electron count and steric number for the central atom.

Phosphorus (Group 15) has 5 valence electrons. It forms 3 single covalent bonds with chlorine atoms and retains 1 lone pair of electrons. Steric number = 3 bonding pairs + 1 lone pair = 4.

The steric number dictates the set of hybridized orbitals used by the central atom.

2

Determine the hybridization of the central atom.

A steric number of 4 corresponds to

sp^3

hybridization.

Four electron domains require four degenerate hybridized orbitals formed from one s orbital and three p orbitals.

3

Differentiate between electron pair geometry and molecular geometry.

The electron pair geometry is tetrahedral, but because one position is occupied by a non-bonding lone pair, the molecular geometry is trigonal pyramidal.

VSEPR theory specifies that molecular geometry considers only the spatial arrangement of the atomic nuclei.

Key Concept

VSEPR Theory, Steric Number, and Hybridization

Estimated Time:1m 0s

Question 68Question →

Match each of the following physical phenomena or chemical systems on the left with the predominant type of intermolecular force or interaction responsible for it on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Sodium chloride dissolution matches Ion-dipole interactions; Liquefaction of argon gas matches London dispersion forces; Lower density of ice matches Extensive three-dimensional hydrogen bonding; Higher boiling point of polar

\text{HCl}

over

\text{Ar}

matches Permanent dipole-dipole interactions.

Each system exhibits physical behaviors dictated by its specific intermolecular interaction: ion-dipole attractions enable ionic solvation; temporary induced dipoles (dispersion forces) allow nonpolar noble gases to condense; directional 3D hydrogen bonding creates an expanded lattice in ice; permanent dipole-dipole forces provide extra attraction in polar compounds like

\text{HCl}

.

Step-by-Step Solution

1

Analyze the nature of the chemical species involved in each phenomenon (ionic, polar, nonpolar, or hydrogen-bonded).

\text{NaCl}

in water involves ions and polar molecules; argon gas consists of isolated nonpolar atoms; ice involves water molecules forming a rigid lattice;

\text{HCl}

consists of polar molecules.

Identifying molecular polarity and ionic state determines which category of intermolecular interaction dominates.

2

Pair each phenomenon with the correct fundamental intermolecular force.

Ions + polar solvent

\rightarrow

Ion-dipole; Nonpolar atoms

\rightarrow

London dispersion forces; Water lattice expansion

\rightarrow

Extensive 3D hydrogen bonding; Permanent molecular dipoles

\rightarrow

Permanent dipole-dipole forces.

Connecting microscopic force definitions to observed physical properties yields the correct matches.

Key Concept

Intermolecular Forces and Hydrogen Bonding

Question 69Question →

The hydronium ion ( $H_3O^+$ ) is formed when a hydrogen ion bonds to a water molecule. Based on Valence Shell Electron Pair Repulsion (VSEPR) theory, what is the molecular geometry and the hybridization state of the central oxygen atom in $H_3O^+$ ?

Trigonal pyramidal geometry with

sp^3

hybridization

Tetrahedral geometry with

sp^3

hybridization

Trigonal planar geometry with

sp^2

hybridization

Bent geometry with

sp^2

hybridization

Show answer & explanation

Answer: Trigonal pyramidal geometry with $sp^3$ hybridization

Answer

The central oxygen atom in

H_3O^+

is

sp^3

hybridized, and the ion exhibits a trigonal pyramidal molecular geometry.

In

H_3O^+

, the central oxygen atom forms three single covalent bonds with hydrogen atoms and retains one lone pair. The total of four electron pairs creates a tetrahedral electron arrangement, which corresponds to

sp^3

hybridization. Because one of the four domains is a non-bonding lone pair, the positions of the atomic nuclei define a trigonal pyramidal molecular geometry.

Step-by-Step Solution

1

Calculate valence electron pairs around the central oxygen atom in

H_3O^+

Oxygen supplies 6 valence electrons, three hydrogens supply 3, and subtracting 1 for the positive charge leaves 8 valence electrons (4 electron pairs).

Determining steric number requires counting both bonding and non-bonding electron pairs.

2

Determine the electron-pair geometry and hybridization

Four electron pairs arrange tetrahedrally around oxygen, requiring

sp^3

hybridization.

Four steric domains around a central atom always correspond to an

sp^3

set of hybrid orbitals.

3

Deduce the molecular shape

With 3 bonding pairs and 1 non-bonding lone pair, the molecular shape is trigonal pyramidal.

Molecular geometry reflects only the arrangement of the atomic nuclei around the central atom.

Key Concept

VSEPR Theory and Hybridization of Polyatomic Ions

Question 70Question →

Despite hydrogen fluoride ( $HF$ ) possessing stronger individual hydrogen bonds than water ( $H_2O$ ), liquid water has a significantly higher boiling point ( $100^\circ\text{C}$ ) than liquid hydrogen fluoride ( $19.5^\circ\text{C}$ ). What is the primary structural reason for this higher boiling point in water?

Each water molecule can form an average of four hydrogen bonds in an extensive three-dimensional network, whereas each hydrogen fluoride molecule forms an average of only two hydrogen bonds.

The intramolecular covalent bond between hydrogen and oxygen in water is significantly stronger than the intramolecular covalent bond in hydrogen fluoride.

Oxygen has a higher electronegativity than fluorine, resulting in a larger permanent dipole moment in water molecules.

Hydrogen fluoride molecules in the liquid state are predominantly attracted by weak London dispersion forces rather than hydrogen bonds.

Show answer & explanation

Answer: Each water molecule can form an average of four hydrogen bonds in an extensive three-dimensional network, whereas each hydrogen fluoride molecule forms an average of only two hydrogen bonds.

Answer

Each water molecule can form an average of four hydrogen bonds in an extensive three-dimensional network, whereas each hydrogen fluoride molecule forms an average of only two hydrogen bonds.

The correct answer correctly identifies that the number of hydrogen bonds per molecule determines the bulk physical property. Water possesses 2 hydrogen atoms and 2 lone pairs on oxygen, creating an optimal ratio that allows 4 hydrogen bonds per molecule in a 3D network. Hydrogen fluoride has 3 lone pairs but only 1 hydrogen atom, creating a hydrogen deficit that limits the system to an average of 2 hydrogen bonds per molecule.

Step-by-Step Solution

1

Analyze the structural capacity for hydrogen bonding in hydrogen fluoride (

HF

).

An

HF

molecule has

1

hydrogen atom and

3

lone pairs on the fluorine atom. Because hydrogen atoms are the limiting factor, each molecule can only participate in an average of

2

hydrogen bonds (

1

donated,

1

accepted).

Hydrogen bonding requires both a hydrogen atom bonded to a highly electronegative atom and an available unshared electron pair.

2

Analyze the structural capacity for hydrogen bonding in water (

H_2O

).

An

H_2O

molecule has

2

hydrogen atoms and

2

lone pairs on the central oxygen atom. This 1:1 stoichiometry of hydrogens to lone pairs allows each water molecule to form an average of

4

hydrogen bonds in a 3D network.

The equal number of hydrogen atoms and lone pairs maximizes the overall density of the hydrogen-bonding network.

3

Compare the total intermolecular energy required to separate the molecules during boiling.

Even though a single

F-H\cdots F

bond is stronger than a single

O-H\cdots O

bond, twice as many hydrogen bonds must be broken per mole of water, requiring significantly more thermal energy to vaporize liquid water.

Boiling point depends on the total energy required to overcome all intermolecular attractions in the liquid bulk.

Key Concept

Hydrogen bonding capacity and network density

Estimated Time:1m 0s

Question 71Question →

What are the molecular shape and the hybridization state of the central atom in a molecule of xenon trioxide ( $XeO_3$ )?

Trigonal pyramidal shape with

sp^3

hybridization

Trigonal planar shape with

sp^2

hybridization

Tetrahedral shape with

sp^3

hybridization

T-shaped shape with

sp^3d

hybridization

Show answer & explanation

Answer: Trigonal pyramidal shape with $sp^3$ hybridization

Answer

Trigonal pyramidal shape with

sp^3

hybridization

In xenon trioxide (

XeO_3

), the central xenon atom possesses 8 valence electrons. It forms 3 double bonds with oxygen atoms (using 6 electrons for bonding) and retains 1 non-bonding lone pair. The steric number is 4 (3

\sigma

-bonds + 1 lone pair), which directs the hybridisation state to

sp^3

. While the four electron domains adopt a tetrahedral spatial distribution, the presence of one lone pair results in a trigonal pyramidal molecular geometry.

Step-by-Step Solution

1

Determine the valence electron count of the central atom and total electron domains.

Xenon (Group 18) has 8 valence electrons. It forms three double bonds with oxygen atoms (utilizing 6 valence electrons), leaving 2 unshared electrons (1 lone pair).

VSEPR theory requires calculating the total number of electron domains (

\sigma

-bonds + lone pairs) on the central atom.

2

Calculate the steric number and determine the hybridization state.

Steric number = 3

\sigma

-bonds + 1 lone pair = 4 electron domains, which corresponds to

sp^3

hybridization.

Four electron domains arrange in a tetrahedral arrangement requiring four

sp^3

hybrid orbitals.

3

Deduce the molecular shape from the electron domain geometry.

With 4 electron domains (3 bonding, 1 non-bonding), the electron geometry is tetrahedral while the molecular shape (atomic arrangement) is trigonal pyramidal.

Molecular shape describes only the arrangement of atomic nuclei around the central atom, excluding lone pairs.

Key Concept

VSEPR Theory, Steric Number, and Hybridization

Question 72Question →

Complete the following statement regarding the molecular geometry and central atom hybridization of phosphorus pentachloride ( $PCl_5$ ).

Fill in the blanks below

In a molecule of phosphorus pentachloride (

PCl_5

), the central phosphorus atom forms five bonding pairs with no lone pairs, resulting in a molecular shape and an hybridization state.

Show answer & explanation

Answer

The molecular shape of phosphorus pentachloride (

PCl_5

) is trigonal bipyramidal, and the hybridization state of the central phosphorus atom is

sp^3d

.

Phosphorus in

PCl_5

shares its 5 valence electrons with 5 chlorine atoms, forming 5 single covalent bonds with zero lone pairs. According to VSEPR theory, five valence electron pairs position themselves as far apart as possible in a trigonal bipyramidal arrangement (with 90° axial-equatorial and 120° equatorial-equatorial bond angles). The central phosphorus atom undergoes

sp^3d

hybridization by mixing one 3s, three 3p, and one 3d orbital to produce five

sp^3d

hybrid orbitals.

Step-by-Step Solution

1

Determine valence electron count and electron domains around the central phosphorus atom

Phosphorus (Group 15) has 5 valence electrons. In

PCl_5

, phosphorus forms 5 single covalent bonds with 5 chlorine atoms, yielding 5 bonding pairs and 0 lone pairs (total 5 electron domains).

VSEPR theory uses the total number of valence electron pairs (bonding pairs plus lone pairs) around the central atom to determine spatial arrangement.

2

Deduce the molecular shape and hybridization scheme for 5 electron domains

To minimize electrostatic repulsion, 5 electron domains adopt a trigonal bipyramidal geometry. Creating 5 equivalent hybrid orbitals requires mixing one s orbital, three p orbitals, and one d orbital, resulting in

sp^3d

hybridization.

Elements in Period 3 (like phosphorus) have accessible 3d orbitals allowing an expanded octet of 10 valence electrons.

Key Concept

VSEPR theory predictions and orbital hybridization for molecules with five valence electron pairs (

sp^3d

trigonal bipyramidal).

Question 73Question →

Match each chemical species to its correct molecular geometry and central atom hybridization state.

Click a left item, then click its matching right item

Show answer & explanation

Answer

BeCl_2

matches Linear shape,

sp

hybridization;

BF_3

matches Trigonal planar shape,

sp^2

hybridization;

CH_4

matches Tetrahedral shape,

sp^3

hybridization;

SF_6

matches Octahedral shape,

sp^3d^2

hybridization.

Each chemical species is matched to its corresponding molecular geometry and central atom hybridization based on the number of sigma bonds and lone pairs present on the central atom.

Step-by-Step Solution

1

Determine steric number for

BeCl_2

Be

forms 2 single bonds with 0 lone pairs, giving a steric number of 2 (

sp

hybridization, linear shape).

Two electron domains arrange at 180° to minimize electron pair repulsion.

2

Determine steric number for

BF_3

B

forms 3 single bonds with 0 lone pairs, giving a steric number of 3 (

sp^2

hybridization, trigonal planar shape).

Three electron domains arrange at 120° in a single plane.

3

Determine steric number for

CH_4

C

forms 4 single bonds with 0 lone pairs, giving a steric number of 4 (

sp^3

hybridization, tetrahedral shape).

Four electron domains arrange symmetrically in three-dimensional space at 109.5°.

4

Determine steric number for

SF_6

S

forms 6 single bonds with 0 lone pairs, giving a steric number of 6 (

sp^3d^2

hybridization, octahedral shape).

Six electron domains arrange symmetrically at 90° axial/equatorial positions.

Key Concept

Valence Shell Electron Pair Repulsion (VSEPR) Theory and Orbital Hybridization

Estimated Time:1m 0s

Question 74Question →

Arrange the following hydrogen halides in order of increasing boiling point, starting from the compound with the lowest boiling point to the compound with the highest boiling point.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence in order of increasing boiling point is Hydrogen chloride (

HCl

), Hydrogen bromide (

HBr

), Hydrogen iodide (

HI

), and Hydrogen fluoride (

HF

).

The correct sequence ranks the Group 17 hydrides by increasing boiling point:

HCl < HBr < HI < HF

. For

HCl

,

HBr

, and

HI

, boiling points rise systematically with increasing atomic size and molar mass because larger electron clouds enhance polarizability and London dispersion forces.

HF

breaks this trend and has the highest boiling point in the group due to the presence of strong intermolecular hydrogen bonding.

Step-by-Step Solution

1

Identify the primary intermolecular forces present in each hydrogen halide.

HCl

,

HBr

, and

HI

interact mainly via London dispersion forces and dipole-dipole interactions, while

HF

forms strong intermolecular hydrogen bonds.

Fluorine is extremely small and electronegative, fulfilling the requirements for hydrogen bonding.

2

Compare van der Waals forces among

HCl

,

HBr

, and

HI

.

Boiling point increases progressively from

HCl

to

HBr

to

HI

.

As molecular size and electron cloud volume increase down Group 17, polarizability increases, leading to stronger temporary dipoles and stronger London dispersion forces.

3

Determine the position of

HF

within the series.

HF

has an anomalously high boiling point compared to the rest of the group.

Intermolecular hydrogen bonding is considerably stronger than van der Waals dispersion forces, making

HF

the least volatile of the four hydrogen halides.

Key Concept

Boiling point trends in hydrides are governed by the interplay of London dispersion forces (which scale with molar mass and polarizability) and hydrogen bonding.

Estimated Time:1m 0s

Question 75Question →

Complete the statements regarding the artificial nuclear transmutation of aluminium-27 by filling in the blanks with the correct numerical values.

Fill in the blanks below

When aluminium-27 (

^{27}_{13}\text{Al}

) is bombarded with an alpha particle (

^{4}_{2}\alpha

), it produces phosphorus-30 (

^{30}_{15}\text{P}

) and a neutron (

^{A}_{Z}\text{n}

). According to the law of conservation of mass number and atomic number, the mass number

A

of the emitted neutron is and its atomic number

Z

is .

Show answer & explanation

Answer

The mass number A is 1 and the atomic number Z is 0.

In any balanced nuclear reaction, the sum of the superscripts (mass numbers) and the sum of the subscripts (atomic numbers) must be equal on both sides of the equation. For the reactants (

^{27}_{13}\text{Al} + ^{4}_{2}\alpha

), the total mass number is

27 + 4 = 31

and the total atomic number is

13 + 2 = 15

. For the products (

^{30}_{15}\text{P} + ^{A}_{Z}\text{n}

), setting

30 + A = 31

yields

A = 1

, and setting

15 + Z = 15

yields

Z = 0

. Thus, the emitted particle is a neutron (

^{1}_{0}\text{n}

).

Step-by-Step Solution

1

Balance the total mass numbers on both sides of the nuclear equation.

Total mass number on reactants side = 27 + 4 = 31. Mass number of phosphorus-30 = 30. Therefore, mass number of the neutron A = 31 - 30 = 1.

The total sum of mass numbers must remain equal before and after a nuclear reaction.

2

Balance the total atomic numbers (nuclear charge) on both sides of the nuclear equation.

Total atomic number on reactants side = 13 + 2 = 15. Atomic number of phosphorus-30 = 15. Therefore, atomic number of the neutron Z = 15 - 15 = 0.

The total sum of atomic numbers (positive charges) must be conserved in nuclear reactions.

Key Concept

Conservation of mass number and atomic number in nuclear transmutation equations.

Estimated Time:1m 0s

Question 76Question →

A sample of iodine-131 ( $^{131}\text{I}$ ), a radioactive isotope used in medical diagnosis, has a half-life of $8\text{ days}$ . If only $2.5\text{ g}$ of the sample remains active after an elapsed time of $24\text{ days}$ , what was the initial mass of the sample?

7.5\text{ g}

10\text{ g}

20\text{ g}

40\text{ g}

Show answer & explanation

Answer: $20\text{ g}$

Answer

The initial mass of the iodine-131 sample was

20\text{ g}

.

Over an elapsed time of

24\text{ days}

with a half-life of

8\text{ days}

, exactly

3

half-lives pass (

24 \div 8 = 3

). Since the remaining mass is

2.5\text{ g}

, working backward requires doubling the mass three times:

2.5\text{ g} \rightarrow 5.0\text{ g} \rightarrow 10.0\text{ g} \rightarrow 20.0\text{ g}

, giving an initial mass of

20\text{ g}

.

Step-by-Step Solution

1

Determine the number of half-lives (

n

) that have elapsed.

n = \frac{\text{Total Time}}{\text{Half-life}} = \frac{24\text{ days}}{8\text{ days}} = 3\text{ half-lives}

Dividing total elapsed time by the half-life period gives the total count of half-life cycles.

2

Apply the radioactive decay relationship to calculate initial mass (

N_0

).

N = N_0 \left(\frac{1}{2}\right)^n \implies 2.5\text{ g} = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8}

Radioactive decay follows an exponential model where remaining amount is reduced by half each cycle.

3

Solve for initial mass (

N_0

).

N_0 = 2.5\text{ g} \times 8 = 20\text{ g}

Multiplying the remaining mass by

2^3

reverses the exponential decay process.

Key Concept

Radioactive Half-Life and Exponential Decay Calculations

Question 77Question →

An atom of an element $X$ forms a stable monoatomic anion $X^{2-}$ containing 18 electrons and 18 neutrons. What is the mass number of element $X$ ?

16

34

36

38

Show answer & explanation

Answer: 34

Answer

34

The monoatomic anion

X^{2-}

carries a negative charge of 2, indicating that it has gained 2 electrons compared to its neutral atomic state. Since the ion has 18 electrons, the neutral atom of element

X

has

18 - 2 = 16

electrons, which corresponds to an atomic number of 16 (16 protons). The mass number (

A

) is defined as the total number of protons and neutrons in the nucleus. Thus, mass number

A = 16\text{ protons} + 18\text{ neutrons} = 34

.

Step-by-Step Solution

1

Determine the atomic number (number of protons) of element

X

from its anion

X^{2-}

.

Number of protons

Z = 18 - 2 = 16

.

An anion with a

2-

charge has gained 2 electrons. Therefore, the neutral atom has 2 fewer electrons than the ion, which equals its atomic number.

2

Calculate the mass number (

A

) of element

X

by summing the number of protons and neutrons.

Mass number

A = 16 + 18 = 34

.

Mass number is the total sum of protons and neutrons present in the nucleus of an atom (

A = Z + N

).

Key Concept

Relationship between ion charge, subatomic particles, atomic number, and mass number

Question 78Question →

A neutral copper atom undergoes oxidation to form a copper(II) ion, $Cu^{2+}$ . Given that the atomic number of copper ( $Cu$ ) is 29, which of the following represents the correct ground-state electronic configuration of $Cu^{2+}$ ?

1s^2 2s^2 2p^6 3s^2 3p^6 3d^9

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7

1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^8

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^9

Show answer & explanation

Answer: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^9$

Answer

The correct ground-state electronic configuration of the copper(II) ion (

Cu^{2+}

) is

1s^2 2s^2 2p^6 3s^2 3p^6 3d^9

.

The correct option correctly applies the rules of electronic configuration for transition metals. Neutral copper (

Z=29

) has an ground state of

1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1

. When ionized to

Cu^{2+}

, two electrons are removed: the outermost

4s^1

electron is lost first, followed by one electron from the

3d^{10}

subshell, leaving

1s^2 2s^2 2p^6 3s^2 3p^6 3d^9

.

Step-by-Step Solution

1

Determine the total number of electrons in neutral copper (

Cu

) and write its ground-state configuration.

Atomic number

Z = 29

, so neutral

Cu

has 29 electrons. Due to the extra stability of a completely filled

d

-subshell, the configuration is

1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1

.

Full subshell stability causes one electron from the

4s

orbital to promote to the

3d

subshell in neutral copper.

2

Calculate the number of electrons lost to form the

Cu^{2+}

ion.

Cu^{2+}

loses 2 electrons, leaving

29 - 2 = 27

electrons.

A

+2

charge indicates the loss of two valence electrons.

3

Remove two electrons following the rule for transition metal cation formation.

Remove the 1 electron from the outermost

4s

orbital first, then 1 electron from the

3d

orbital, leaving

3d^9

.

Electrons in the highest principal energy level (

n=4

) are lost before inner

(n-1)d

electrons (

n=3

).

Key Concept

Electronic configuration of d-block cations

Estimated Time:1m 0s

Question 79Question →

Match each specific electronic structure scenario or electron assignment with the corresponding quantum principle or thermodynamic factor that governs it.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings match copper's anomalous configuration to full-subshell exchange energy stability, identical quantum set restriction to the Pauli exclusion principle, unpaired degenerate orbital filling in nitrogen to Hund's rule, and

4s

filling prior to

3d

to the Aufbau

(n+l)

principle.

Each scenario directly aligns with its foundational quantum principle: copper's ground-state configuration (

[Ar]3d^{10}4s^1

) is stabilized by high exchange energy of the full

d

-subshell; identical quantum numbers are forbidden by the Pauli exclusion principle; single filling of degenerate

2p

orbitals in nitrogen obeys Hund's rule of maximum multiplicity; and subshell filling order (

4s

before

3d

) is governed by the Aufbau

(n+l)

rule.

Step-by-Step Solution

1

Analyze the copper configuration anomaly (

[Ar]3d^{10}4s^1

).

Identified as a deviation due to exchange energy stabilization of a completely filled

d

subshell.

Completely filled subshells provide enhanced stability via maximum exchange interactions and spherical symmetry.

2

Examine the prohibition of identical four-quantum-number sets

(n, l, m_l, m_s)

.

Identified as a direct statement of the Pauli exclusion principle.

Two electrons in an orbital must have opposite spin quantum numbers (

+\frac{1}{2}

and

-\frac{1}{2}

).

3

Evaluate nitrogen's

2p^3

configuration (

2p_x^1 2p_y^1 2p_z^1

).

Identified as an application of Hund's rule of maximum multiplicity.

Electrons occupy degenerate subshell orbitals singly with parallel spins to minimize inter-electronic coulomb repulsion.

4

Assess the sequence of filling

4s

prior to

3d

.

Identified as following the Aufbau principle via the

(n+l)

rule.

For

4s

,

n+l = 4+0 = 4

; for

3d

,

n+l = 3+2 = 5

. Lower

(n+l)

subshells fill first.

Key Concept

Quantum Rules, Subshell Energies, and Electronic Configuration Anomalies

Question 80Question →

Element $E$ has a relative atomic mass of $12.011$ and exists as two naturally occurring isotopes: $^{12}\text{E}$ with a natural abundance of $98.9\%$ and $^{A}\text{E}$ with a natural abundance of $1.1\%$ . What is the mass number ( $A$ ) of the second isotope?

Show answer & explanation

Answer: 13

Answer

The mass number of the second isotope is 13.

Relative atomic mass is calculated as the weighted average of the mass numbers of all naturally occurring isotopes. Setting up the equation

12.011 = \frac{(98.9 \times 12) + (1.1 \times A)}{100}

yields

1201.1 = 1186.8 + 1.1A

, which simplifies to

1.1A = 14.3

and gives

A = 13

.

Step-by-Step Solution

1

State the relationship for relative atomic mass based on isotopic abundance.

\text{RAM} = \frac{(\text{Abundance}_1 \times \text{Mass}_1) + (\text{Abundance}_2 \times \text{Mass}_2)}{100}

Relative atomic mass is the weighted average of the atomic masses of naturally occurring isotopes of an element.

2

Substitute the given numerical values into the formula.

12.011 = \frac{(98.9 \times 12) + (1.1 \times A)}{100}

Inserting the known abundances (

98.9\%

and

1.1\%

) and mass number (

12

) sets up an algebraic equation for the unknown mass number

A

.

3

Solve the algebraic equation for

A

.

1201.1 = 1186.8 + 1.1A \implies 1.1A = 14.3 \implies A = 13

Subtracting the contribution of the first isotope and dividing by the abundance of the second isotope gives the integer mass number

13

.

Key Concept

Calculating isotopic mass number from relative atomic mass and fractional abundances

Atomic Structure and Chemical Bonding

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