Atomic Structure and Chemical Bonding Practice Questions — JAMB UTME

Question 1Question →

Which of the following physical properties of a metal is directly attributed to the presence of a mobile 'sea' of delocalized valence electrons?

High electrical conductivity in the solid state

High solubility in polar solvents such as water

Low melting and boiling points due to weak intermolecular forces

Conduction of electricity only when molten or dissolved in aqueous solution

Show answer & explanation

Answer: High electrical conductivity in the solid state

Answer

High electrical conductivity in the solid state

High electrical conductivity in the solid state is a direct consequence of delocalized valence electrons moving freely throughout the metallic crystal lattice when an electric potential difference is applied.

Step-by-Step Solution

1

Identify the essential structural feature of metallic bonding.

Metallic bonding consists of a giant lattice of positive metal ions surrounded by a sea of mobile, delocalized valence electrons.

Understanding the presence of free-moving valence electrons explains the physical behavior of metals.

2

Relate delocalized electrons to electrical conduction.

When an electric field is applied, the mobile electrons drift towards the positive potential, carrying electric charge through the solid metal.

Electrical conduction requires mobile charge carriers, which in metals are the delocalized valence electrons present even in the solid state.

Key Concept

Metallic bonding and electrical conductivity due to delocalized electrons

Estimated Time:1m 0s

Question 2Question →

A piece of copper wire can be hammered into a thin sheet without shattering, demonstrating malleability. Which of the following structural features best explains this physical property at the submicroscopic level?

The layers of positive metal cations can slide over one another while maintaining non-directional electrostatic attraction with the sea of delocalized electrons.

Rigid, highly directional covalent bonds between adjacent copper atoms easily break and reform along new planes of pressure.

Valence electrons are permanently transferred between copper atoms to form cation-anion pairs that easily slip past each other.

The metallic lattice is held together primarily by weak intermolecular dispersion forces that readily yield under mechanical stress.

Show answer & explanation

Answer: The layers of positive metal cations can slide over one another while maintaining non-directional electrostatic attraction with the sea of delocalized electrons.

Answer

The layers of positive metal cations can slide over one another while maintaining non-directional electrostatic attraction with the sea of delocalized electrons.

In metallic bonding, valence electrons are delocalized and free to move throughout the metallic crystal lattice. When stress is applied to a metal like copper, layers of positive metal cations slide past one another. The sea of delocalized electrons adjusts to the new position of the cations, maintaining the non-directional electrostatic attraction and preventing the metal from fracturing.

Step-by-Step Solution

1

Identify the chemical bond type and structural model for copper metal.

Copper is a metal with a metallic lattice consisting of positive metal ions (cations) surrounded by a sea of mobile, delocalized electrons.

Understanding the nature of metallic bonding is essential to explaining physical properties such as malleability and ductility.

2

Analyze how mechanical force affects the metallic lattice.

When hammered, layers of cations shift relative to each other, but the delocalized electron cloud adapts instantaneously to the new shape.

Because electrostatic attraction in metallic bonding is non-directional, moving cation layers does not result in strong repulsive forces or broken bonds.

Key Concept

Malleability of metals in the delocalized electron sea model

Estimated Time:1m 0s

Question 3Question →

Which of the following best describes the fundamental attraction responsible for metallic bonding in solid metals?

Electrostatic attraction between positive metal cations and a sea of delocalized valence electrons

Mutual sharing of localized electron pairs between adjacent atoms

Electrostatic attraction between oppositely charged discrete ions formed by electron transfer

Weak intermolecular forces existing between neutral metal atoms

Show answer & explanation

Answer: Electrostatic attraction between positive metal cations and a sea of delocalized valence electrons

Answer

The metallic bond is defined as the strong electrostatic attraction between positively charged metal cations fixed in a lattice and a surrounding sea of mobile, delocalized valence electrons.

The correct option accurately defines metallic bonding as the electrostatic force of attraction binding positive metal ions to a fluid sea of delocalized valence electrons, which accounts for characteristic metallic properties like electrical conductivity and malleability.

Step-by-Step Solution

1

Identify the valence electron behavior in metals

Metal atoms lose their outer valence electrons to form positive cations, producing a shared pool of delocalized electrons free to move throughout the giant structure.

Low ionization energies in metals allow valence electrons to become detached easily from individual atoms.

2

Determine the nature of the attractive force holding the lattice together

Strong electrostatic attraction acts non-directionally between the positive ions and the mobile electron sea.

Opposite charges attract each other, forming a stable metallic lattice.

Key Concept

Nature of Metallic Bonding

Estimated Time:45s

Question 4Question →

Calculate the mass number and atomic number of the final daughter nuclide in the given radioactive decay series.

Fill in the blanks below

When a nucleus of Thorium-232 (

^{232}_{90}\text{Th}

) undergoes a decay series emitting

6

alpha particles (

\alpha

) and

4

beta particles (

\beta^-

), the resulting stable daughter nuclide has a mass number of and an atomic number of .

Show answer & explanation

Answer

The final stable daughter nuclide has a mass number of 208 and an atomic number of 82.

Emission of 6 alpha particles reduces the mass number by

6 \times 4 = 24

(from 232 to 208) and the atomic number by

6 \times 2 = 12

. The subsequent emission of 4 beta particles increases the atomic number by

4 \times 1 = 4

, resulting in a final atomic number of

90 - 12 + 4 = 82

.

Step-by-Step Solution

1

Calculate the total change in mass number (

A

) caused by alpha and beta particle emissions.

Each alpha particle (

^{4}_{2}\text{He}

) reduces

A

by 4, while beta particles (

^{0}_{-1}\text{e}

) do not change

A

. Total decrease in

A = 6 \times 4 = 24

.

Alpha particles carry 4 atomic mass units, whereas beta particles have virtually zero mass number.

2

Determine the final mass number.

Final mass number =

232 - 24 = 208

.

Subtracting the lost nucleon mass from the initial mass number gives the mass number of the resulting nucleus.

3

Calculate the total change in atomic number (

Z

) caused by alpha and beta particle emissions.

Each alpha particle reduces

Z

by 2, while each beta particle increases

Z

by 1. Total change in

Z = (6 \times -2) + (4 \times +1) = -12 + 4 = -8

.

Alpha decay removes 2 protons, whereas beta minus decay converts a neutron into a proton, increasing nuclear charge by 1.

4

Determine the final atomic number.

Final atomic number =

90 - 8 = 82

.

Applying the net charge change to the initial atomic number of Thorium (90) yields 82, which corresponds to Lead (Pb).

Key Concept

Conservation of mass number and atomic number in decay series

Question 5Question →

Atoms of the same chemical element that possess the same atomic number but differ in their mass number are known as which of the following?

Isotopes

Allotropes

Isomers

Isobars

Show answer & explanation

Answer: Isotopes

Answer

Atoms of the same element with identical atomic numbers but different mass numbers are called isotopes.

Isotopes are species of the same chemical element (having the same number of protons and atomic number

Z

) that contain different numbers of neutrons, resulting in different mass numbers

A

.

Step-by-Step Solution

1

Define atomic number and mass number relations in atoms of the same element

The atomic number represents the number of protons, which determines the element's identity. The mass number is the total count of protons and neutrons.

Understanding subatomic composition is essential to identifying relationship terminology.

2

Identify the relationship where atomic number remains fixed while mass number varies

Variation in mass number among atoms of the same element occurs due to a differing number of neutrons in their nuclei.

Differing neutron counts in identical chemical elements define isotopy.

Key Concept

Subatomic Particles and Isotopy

Estimated Time:45s

Question 6Question →

What is the ground-state electronic configuration of the iron(II) ion, $Fe^{2+}$ ? (Atomic number of $Fe = 26$ )

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

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

1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5

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

Show answer & explanation

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

Answer

The ground-state electronic configuration of

Fe^{2+}

is

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

(or

[Ar] 3d^6

).

The correct configuration is

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

. Neutral iron has 26 electrons with the configuration

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

. Upon forming the

Fe^{2+}

ion, two electrons are removed from the outermost shell (

n=4

, the

4s

orbital), leaving 24 electrons and a

3d^6

valence structure.

Step-by-Step Solution

1

Determine the electronic configuration of the neutral iron atom (

Fe

,

Z = 26

).

The neutral configuration is

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

.

Electrons fill subshells in order of increasing energy according to the Aufbau principle.

2

Determine which electrons are lost when forming the

Fe^{2+}

cation.

Two electrons are removed from the principal energy level with the highest quantum number (

n = 4

), which is the

4s

subshell.

Transition metals lose their outermost

4s

electrons first upon ionization because

4s

electrons experience greater shielding and are higher in energy once subshells are populated.

3

Write the resulting configuration for

Fe^{2+}

.

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

.

Removing two electrons from

4s^2

leaves an empty

4s

orbital and retains six electrons in the

3d

subshell.

Key Concept

Cation formation in d-block transition elements requires losing

ns

electrons before

(n-1)d

electrons.

Estimated Time:45s

Question 7Question →

Complete the following statement regarding the molecular geometry and atomic orbital hybridization of sulfur hexafluoride ( $SF_6$ ).

Fill in the blanks below

In a molecule of sulfur hexafluoride (

SF_6

), the central sulfur atom has six bonding electron pairs and no lone pairs, resulting in a molecular shape described as and a central atom hybridization of .

Show answer & explanation

Answer

The molecular shape of sulfur hexafluoride (

SF_6

) is octahedral and the hybridization of the central sulfur atom is

sp^3d^2

.

In

SF_6

, the central sulfur atom shares its 6 valence electrons with six fluorine atoms, creating 6 bonding pairs and 0 lone pairs. VSEPR theory predicts an octahedral geometry to minimize electron repulsions. The combination of six atomic orbitals (one s, three p, and two d) results in

sp^3d^2

hybridization.

Step-by-Step Solution

1

Count the total number of electron pairs around the central sulfur atom in

SF_6

.

Sulfur contributes 6 valence electrons and each of the six fluorine atoms contributes 1 electron for sharing, forming 6 single covalent bonds (6 bonding pairs) and 0 lone pairs.

According to VSEPR theory, steric number equals the sum of bonding pairs and lone pairs around the central atom.

2

Determine the molecular geometry based on VSEPR theory.

With 6 electron domains and 0 lone pairs, the electron pairs orient themselves toward the corners of a regular octahedron to minimize repulsion, giving an octahedral molecular geometry with bond angles of

90^\circ

.

Symmetrical arrangement of 6 identical bond pairs around a central atom produces an octahedral shape.

3

Determine the hybridization of the central sulfur atom.

Mixing one s orbital, three p orbitals, and two d orbitals yields six equivalent

sp^3d^2

hybrid orbitals directed toward the vertices of an octahedron.

Six equivalent bonding orbitals require

sp^3d^2

hybridization.

Key Concept

VSEPR prediction of molecular shape and valence orbital hybridization for hypervalent molecules.

Question 8Question →

In the hydronium ion ( $H_3O^+$ ), the central oxygen atom is bonded to three hydrogen atoms and retains one lone pair of electrons. Based on the Valence Shell Electron Pair Repulsion (VSEPR) theory, what are the molecular shape and the hybridization state of the central oxygen atom in $H_3O^+$ ?

Tetrahedral shape with

sp^3

hybridization

Trigonal pyramidal shape with

sp^3

hybridization

Trigonal planar shape with

sp^2

hybridization

Bent shape with

sp^2

hybridization

Show answer & explanation

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

Answer

Trigonal pyramidal shape with

sp^3

hybridization

The central oxygen atom in

H_3O^+

possesses four electron domains (three

\sigma

bonding pairs with hydrogen atoms and one lone pair). Four total domains result in a steric number of 4, requiring

sp^3

hybridization. Because one of the four tetrahedral positions is occupied by a lone pair rather than a bonded atom, the visual arrangement of the nuclei forms a trigonal pyramidal molecular geometry.

Step-by-Step Solution

1

Determine the total number of valence electron pairs (steric number) around the central oxygen atom in

H_3O^+

.

Oxygen contributes 6 valence electrons, each hydrogen contributes 1, and the

+1

positive charge subtracts 1 electron:

(6 + 3 - 1) = 8

valence electrons (4 electron pairs).

VSEPR theory requires calculating the steric number to determine hybridization and basic electron geometry.

2

Determine the hybridization state from the steric number.

Steric number = 4 (3 bonding pairs + 1 lone pair), which corresponds to

sp^3

hybridization.

Four electron domains require four hybrid orbitals formed by combining one

s

and three

p

atomic orbitals.

3

Differentiate between electron pair geometry and molecular geometry.

While the 4 electron pairs adopt a tetrahedral electron geometry, the arrangement of the 3 bonded atoms around the central oxygen produces a trigonal pyramidal molecular shape.

Molecular shape describes only the spatial arrangement of atomic nuclei, excluding lone pair positions.

Key Concept

VSEPR Theory and Molecular Hybridization

Estimated Time:1m 15s

Question 9Question →

Match each physical property or behavioral phenomenon of metals on the left with its corresponding microscopic structural feature of metallic bonding on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

High thermal conductivity matches rapid kinetic energy transfer via mobile valence electrons; Malleability and ductility match sliding of cation layers due to non-directional bonding; Maintenance of electrical conductivity matches uninterrupted delocalized electron sea cohesion during lattice movement; Higher melting points of transition metals match the participation of unpaired

d

-orbital electrons alongside

s

-electrons in metallic bonding.

Each physical property directly corresponds to specific structural attributes of the delocalized electron sea model: thermal conduction relies on kinetic energy transport by mobile electrons, malleability depends on non-directional layer slipping, conductivity preservation relies on continuous electron sea mobility, and melting point strength in transition metals relies on additional

d

-electron contributions to bonding.

Step-by-Step Solution

1

Analyze thermal conductivity

Identify mobile electrons as the primary mechanism for heat transfer in metals.

Kinetic energy is rapidly dispersed by mobile valence electrons colliding with lattice ions and other electrons.

2

Analyze malleability and ductility

Relate mechanical deformation to non-directional electrostatic forces.

Planes of positive metal cations can slide past each other because delocalized electrons adjust continuously to cushion repulsive cation-cation forces.

3

Analyze electrical conductivity during deformation

Connect continuous conductivity to fluid electron sea nature.

Deforming a metal does not break discrete bonds or interrupt the delocalized sea of electrons carrying charge.

4

Analyze transition metal melting points

Evaluate electron contribution to bonding strength.

Transition metals draw upon both

(n-1)d

and

ns

electrons for metallic cohesion, strengthening the bond far beyond single valence

s

-electron systems.

Key Concept

Electron sea model, non-directional bonding, and structural origins of metallic physical properties

Question 10Question →

Ethanol ( $C_2H_5OH$ ) and dimethyl ether ( $CH_3OCH_3$ ) are structural isomers with the same relative molecular mass ( $46\text{ g mol}^{-1}$ ). However, ethanol boils at $78.4^\circ\text{C}$ while dimethyl ether boils at $-24^\circ\text{C}$ . Which of the following statements correctly explains this difference in boiling points?

Ethanol molecules are held together by strong intermolecular hydrogen bonding, whereas dimethyl ether molecules experience weaker dipole-dipole interactions.

The intramolecular

O-H

covalent bond in ethanol requires more thermal energy to break than the

C-O

covalent bond in dimethyl ether during boiling.

Dimethyl ether has weaker London dispersion forces because its effective molecular weight in the liquid state is lower than that of ethanol.

Ethanol undergoes complete ionic dissociation in liquid phase, resulting in strong electrovalent attractions between ions.

Show answer & explanation

Answer: Ethanol molecules are held together by strong intermolecular hydrogen bonding, whereas dimethyl ether molecules experience weaker dipole-dipole interactions.

Answer

Ethanol molecules are held together by strong intermolecular hydrogen bonding, whereas dimethyl ether molecules experience weaker dipole-dipole interactions.

The correct answer identifies that ethanol can form strong intermolecular hydrogen bonds due to its hydroxyl group (

-OH

), while dimethyl ether lacks a hydrogen atom attached directly to an electronegative atom and relies only on weaker dipole-dipole and van der Waals forces. Consequently, ethanol requires significantly higher thermal energy to separate its molecules into the gas phase.

Step-by-Step Solution

1

Analyze the molecular structures of ethanol (

C_2H_5OH

) and dimethyl ether (

CH_3OCH_3

).

Ethanol contains a hydrogen atom directly bonded to oxygen (

-OH

), enabling hydrogen bond donation and acceptance. Dimethyl ether (

CH_3-O-CH_3

) has oxygen bonded only to carbon atoms, preventing hydrogen bond formation between its own molecules.

Hydrogen bonding requires a hydrogen atom attached to a highly electronegative atom (

N

,

O

, or

F

).

2

Compare the nature and strength of intermolecular forces in both compounds.

Ethanol exhibits strong intermolecular hydrogen bonding in addition to dipole-dipole and dispersion forces. Dimethyl ether exhibits only dipole-dipole forces and dispersion forces.

Hydrogen bonds are significantly stronger than permanent dipole-dipole attractions.

3

Relate intermolecular force strength to boiling point trends.

More thermal energy is needed to separate ethanol molecules during vaporization, giving ethanol a much higher boiling point (

78.4^\circ\text{C}

) compared to dimethyl ether (

-24^\circ\text{C}

).

Boiling point increases with stronger intermolecular forces of attraction.

Key Concept

Intermolecular Hydrogen Bonding vs. Dipole-Dipole Attractions

Question 11Question →

A radioactive sample of an isotope has a half-life of $20\text{ minutes}$ . If the initial mass of the sample is $80\text{ g}$ , what mass of the isotope, in grams, will remain undecayed after $60\text{ minutes}$ ?

Show answer & explanation

Answer: 10

Answer

10 g of the isotope remains undecayed.

The correct answer is derived from the half-life formula

N_t = N_0 \times (1/2)^n

. Since

60\text{ minutes}

contains three

20\text{ minute}

half-life periods (

n = 3

), the remaining mass is

80\text{ g} \times (1/2)^3 = 80\text{ g} / 8 = 10\text{ g}

.

Step-by-Step Solution

1

Determine the number of elapsed half-lives

3 half-lives

Divide the total time elapsed (

60\text{ minutes}

) by the duration of one half-life (

20\text{ minutes}

).

2

Calculate the remaining mass after 3 half-lives

10 g

Apply the radioactive decay relation

N_t = N_0 \times \left(\frac{1}{2}\right)^n

, yielding

80 \times \left(\frac{1}{2}\right)^3 = 10\text{ g}

.

Key Concept

Radioactive Half-Life and Exponential Decay

Question 12Question →

A sample of a radioactive isotope has a half-life of $12\text{ days}$ . If the initial mass of the sample is $48\text{ g}$ , what mass of the isotope remains undecayed after $36\text{ days}$ ?

6\text{ g}

16\text{ g}

12\text{ g}

24\text{ g}

Show answer & explanation

Answer: $6\text{ g}$

Answer

The mass of the isotope remaining undecayed after

36\text{ days}

is

6\text{ g}

.

Radioactive decay follows exponential kinetics. The total elapsed time of

36\text{ days}

represents

3

half-lives of

12\text{ days}

each. Halving the initial

48\text{ g}

sample three consecutive times (

48 \rightarrow 24 \rightarrow 12 \rightarrow 6

) yields

6\text{ g}

.

Step-by-Step Solution

1

Calculate the number of half-lives (

n

) that have elapsed.

n = \frac{\text{Total elapsed time}}{\text{Half-life}} = \frac{36\text{ days}}{12\text{ days}} = 3\text{ half-lives}

.

Determining the number of elapsed half-lives is necessary to apply exponential halving.

2

Apply the exponential decay formula

N = N_0 \times \left(\frac{1}{2}\right)^n

.

N = 48\text{ g} \times \left(\frac{1}{2}\right)^3 = 48\text{ g} \times \frac{1}{8} = 6\text{ g}

.

Radioactive decay follows exponential kinetics, where the remaining mass decreases by half for each elapsed half-life.

Key Concept

Radioactive Half-Life and Exponential Decay Kinetics

Question 13Question →

Arrange the following atomic subshells in order of increasing energy according to the Aufbau principle ( $(n+l)$ rule):

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order of increasing orbital energy is

3s \rightarrow 3p \rightarrow 4s \rightarrow 3d

.

By applying the

(n+l)

rule (Madelung's rule) derived from the Aufbau principle, subshells fill in order of increasing

(n+l)

energy sum. Calculating these values yields:

3s

(

3+0=3

),

3p

(

3+1=4

),

4s

(

4+0=4

), and

3d

(

3+2=5

). For subshells with equal sums (

3p

and

4s

), the one with the smaller principal quantum number

n

is lower in energy (

3p

with

n=3

is lower than

4s

with

n=4

). Therefore, the correct sequence from lowest to highest energy is

3s \rightarrow 3p \rightarrow 4s \rightarrow 3d

.

Step-by-Step Solution

1

Determine the

(n+l)

value for each subshell

For

3s

:

3+0 = 3

. For

3p

:

3+1 = 4

. For

4s

:

4+0 = 4

. For

3d

:

3+2 = 5

.

According to the Aufbau principle and the

(n+l)

rule, orbitals fill in order of increasing

(n+l)

energy values.

2

Compare subshells with identical

(n+l)

values

Both

3p

and

4s

have

(n+l) = 4

. Since

3p

has a smaller principal quantum number (

n=3

compared to

n=4

),

3p

is lower in energy than

4s

.

When two orbitals have the same

(n+l)

value, the orbital with the lower principal quantum number

n

has lower energy.

3

Sequence the subshells from lowest to highest energy

The final sequence is

3s < 3p < 4s < 3d

.

Sorting by

(n+l)

sum first, then by

n

value for ties, yields the correct Aufbau order.

Key Concept

Aufbau Principle and the (n+l) Rule

Question 14Question →

Match each physical property of metals on the left with the microscopic structural explanation on the right that best accounts for it.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Thermal conductivity matches with rapid transfer of thermal energy by mobile delocalized electrons through the lattice. High metallic lustre matches with re-emission of absorbed light by oscillating free electrons on the metal surface. Ductility matches with sliding of positive cation layers past each other into wires without breaking metallic bonds.

Thermal conductivity is caused by mobile delocalized electrons rapidly transferring heat throughout the metal lattice. Metallic lustre occurs because surface free electrons absorb and re-emit light photons. Ductility is enabled by positive cation layers sliding past each other within the sea of delocalized electrons without disrupting the non-directional metallic bonding.

Step-by-Step Solution

1

Identify the cause of heat transport in metals

Thermal conductivity relies on delocalized valence electrons carrying thermal energy rapidly through the crystal lattice.

Free electrons gain kinetic energy when heated and collide with surrounding ions and electrons to distribute heat.

2

Identify the cause of optical reflection (lustre)

Metallic lustre is produced by surface delocalized electrons oscillating in response to incoming light waves and reflecting them.

The un-bound nature of free valence electrons allows immediate absorption and re-radiation of visible light photons.

3

Identify the mechanical feature enabling deformation into wires

Ductility relies on layers of cations sliding past one another while held together by non-directional metallic attraction.

Because metallic bonding is non-directional, moving cation layers do not cause repulsive strain that breaks the crystal structure.

Key Concept

Properties of metals in terms of the delocalized electron sea model

Question 15Question →

Match each chemical species on the left with its correct molecular shape and central atom hybridization state on the right based on Valence Shell Electron Pair Repulsion (VSEPR) theory.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Chlorine trifluoride (

ClF_3

) matches T-shaped geometry with

sp^3d

hybridization; Xenon tetrafluoride (

XeF_4

) matches Square planar geometry with

sp^3d^2

hybridization; Boron trifluoride (

BF_3

) matches Trigonal planar geometry with

sp^2

hybridization; Triiodide ion (

I_3^-

) matches Linear geometry with

sp^3d

hybridization.

Each species is correctly paired by evaluating its total steric number (sum of sigma bonds and lone pairs) on the central atom to establish hybridization, then removing lone pair domains to find the molecular geometry. Chlorine trifluoride (

ClF_3

) has a steric number of 5 (

sp^3d

) with 2 lone pairs giving a T-shape. Xenon tetrafluoride (

XeF_4

) has a steric number of 6 (

sp^3d^2

) with 2 lone pairs giving a square planar shape. Boron trifluoride (

BF_3

) has a steric number of 3 (

sp^2

) with 0 lone pairs giving a trigonal planar shape. The triiodide ion (

I_3^-

) has a steric number of 5 (

sp^3d

) with 3 lone pairs in equatorial positions giving a linear shape.

Step-by-Step Solution

1

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

For

ClF_3

, steric number is

3 \text{ bonding} + 2 \text{ lone pairs} = 5

. For

XeF_4

, steric number is

4 \text{ bonding} + 2 \text{ lone pairs} = 6

. For

BF_3

, steric number is

3 \text{ bonding} + 0 \text{ lone pairs} = 3

. For

I_3^-

, steric number is

2 \text{ bonding} + 3 \text{ lone pairs} = 5

.

Steric number dictates the electron-pair geometry and the hybridization state of the central atom.

2

Assign the hybridization state based on the steric number.

Steric number 3 gives

sp^2

; steric number 5 gives

sp^3d

; steric number 6 gives

sp^3d^2

.

The number of hybridized orbitals equals the steric number of electron domains.

3

Determine molecular shape by accounting for lone pair positions under VSEPR principles.

ClF_3

(5 domains, 2 equatorial lone pairs) is T-shaped;

XeF_4

(6 domains, 2 axial lone pairs) is square planar;

BF_3

(3 domains, 0 lone pairs) is trigonal planar;

I_3^-

(5 domains, 3 equatorial lone pairs) is linear.

Lone pairs occupy positions that minimize electron repulsion (equatorial in trigonal bipyramidal, axial in octahedral).

Key Concept

VSEPR Theory and Central Atom Hybridization

Question 16Question →

What is the ground-state electronic configuration of the chromium(III) ion, $Cr^{3+}$ ? (Atomic number of $Cr = 24$ )

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

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

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

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

Show answer & explanation

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

Answer

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

Neutral chromium (

Z = 24

) has the ground-state electron configuration

1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1

. When forming the

Cr^{3+}

ion, three electrons must be removed. Outer shell (

4s

) electrons are lost first, followed by

(n-1)d

electrons. Removing one electron from

4s

and two from

3d

gives the configuration

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

.

Step-by-Step Solution

1

Determine the electronic configuration of neutral chromium (

Cr

)

Ground state

Cr

(

Z = 24

) is

1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1

Chromium exhibits an anomalous configuration due to the extra stability of a half-filled

3d

subshell.

2

Apply the rule for cation formation in transition metals

Electrons in the outermost shell (

n=4

,

4s

subshell) are lost before electrons in the inner subshell (

n=3

,

3d

subshell).

The

4s

electrons experience lower effective nuclear attraction in the ionized state and are situated at a higher principal energy level.

3

Remove 3 electrons to form

Cr^{3+}

Remove 1 electron from

4s

and 2 electrons from

3d

:

(3d^5 4s^1) - 3e^- = 3d^3 4s^0

Removing 1 electron from

4s

leaves

3d^5

; removing two more from

3d

leaves

3d^3

.

Key Concept

Electronic configuration of d-block transition metal cations

Estimated Time:1m 0s

Question 17Question →

Arrange the following atomic subshells in order of increasing energy according to the Aufbau principle and the $(n+l)$ rule, starting with the subshell of lowest energy:

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order of subshells from lowest to highest energy is

5s \rightarrow 4d \rightarrow 5p \rightarrow 4f

.

According to the Aufbau principle and the

(n+l)

rule, subshells fill in order of increasing

(n+l)

value.

5s

has

(n+l) = 5+0 = 5

, making it lowest in energy. Both

4d

(

(n+l) = 4+2 = 6

) and

5p

(

(n+l) = 5+1 = 6

) have a sum of

6

, but

4d

has lower energy than

5p

because its principal quantum number

n=4

is smaller.

4f

has

(n+l) = 4+3 = 7

, placing it highest in energy. Thus, the correct sequence is

5s \rightarrow 4d \rightarrow 5p \rightarrow 4f

.

Step-by-Step Solution

1

Calculate the

(n+l)

value for each atomic subshell

For

5s

:

n=5, l=0 \implies n+l = 5

.
For

4d

:

n=4, l=2 \implies n+l = 6

.
For

5p

:

n=5, l=1 \implies n+l = 6

.
For

4f

:

n=4, l=3 \implies n+l = 7

.

According to Madelung's rule, orbitals fill in order of increasing

(n+l)

values.

2

Order subshells by increasing

(n+l)

sum

5s

(

(n+l)=5

) is lowest in energy, while

4f

(

(n+l)=7

) is highest.

Subshells with smaller

(n+l)

values are filled before those with larger

(n+l)

values.

3

Break ties for subshells with identical

(n+l)

values (

4d

and

5p

)

4d

(

n=4

) has lower energy than

5p

(

n=5

).

When two subshells share the same

(n+l)

value, the subshell with the smaller principal quantum number

n

is lower in energy.

4

Assemble the complete sequence from lowest to highest energy

5s < 4d < 5p < 4f

Combines the

(n+l)

rule and the tie-breaking principal quantum number rule.

Key Concept

Aufbau Principle and the (n+l) Rule

Question 18Question →

Element $Q$ occurs naturally as two isotopes, $^{35}Q$ and $^{37}Q$ . If the relative atomic mass of element $Q$ is $35.5$ , what is the percentage abundance of the heavier isotope $^{37}Q$ ?

25%

75%

50%

35.5%

Show answer & explanation

Answer: 25%

Answer

The percentage abundance of

^{37}Q

is 25%.

The relative atomic mass (

35.5

) lies between

35

and

37

, closer to

35

. Setting up the weighted average equation

35.5 = \frac{35(100 - x) + 37x}{100}

yields

x = 25\%

for

^{37}Q

, while

^{35}Q

accounts for the remaining

75\%

.

Step-by-Step Solution

1

Set up the relative atomic mass equation using percentage abundances.

Let the percentage abundance of

^{37}Q

be

x\%

. Therefore, the abundance of

^{35}Q

is

(100 - x)\%

.

The sum of natural abundances for all isotopes of an element must equal 100%.

2

Substitute the isotopic mass numbers and relative atomic mass into the weighted average formula.

35.5 = \frac{35(100 - x) + 37x}{100}

Relative atomic mass is calculated as the weighted average of the mass numbers of naturally occurring isotopes.

3

Solve the linear equation for

x

.

3550 = 3500 - 35x + 37x \implies 50 = 2x \implies x = 25

Multiplying by 100 and rearranging terms isolates

x = 25

.

Key Concept

Calculation of isotopic abundances from relative atomic mass

Question 19Question →

A naturally occurring sample of neon gas consists of $90\%$ $^{20}\text{Ne}$ and $10\%$ $^{22}\text{Ne}$ . What is the relative atomic mass of neon in this sample?

Show answer & explanation

Answer: 20.2

Answer

The relative atomic mass of neon in the sample is 20.2.

The relative atomic mass of an element is calculated by summing the products of the mass number of each isotope and its fractional abundance:

\text{RAM} = (20 \times 0.90) + (22 \times 0.10) = 18.0 + 2.2 = 20.2

.

Step-by-Step Solution

1

Calculate the weighted contribution of

^{20}\text{Ne}

20 \times 0.90 = 18.0

The isotope

^{20}\text{Ne}

accounts for

90\%

of the sample.

2

Calculate the weighted contribution of

^{22}\text{Ne}

22 \times 0.10 = 2.2

The isotope

^{22}\text{Ne}

accounts for

10\%

of the sample.

3

Sum the weighted contributions to find the relative atomic mass

18.0 + 2.2 = 20.2

The relative atomic mass is the weighted average of the atomic masses of naturally occurring isotopes.

Key Concept

Calculation of Relative Atomic Mass from Isotopic Abundance

Question 20Question →

An element $X$ with atomic number $12$ combines with an element $Y$ with atomic number $17$ to form a solid compound. Which of the following statements correctly accounts for the electrical conductivity of this compound?

It conducts electricity in the molten state because the giant ionic lattice breaks down, allowing the ions to move freely.

It conducts electricity in the solid state because delocalized valence electrons migrate freely through the crystal lattice.

It does not conduct electricity when dissolved in water because polar solvent molecules neutralize ionic charges.

It conducts electricity in the solid state because covalent electron sharing forms continuous conductive pathways.

Show answer & explanation

Answer: It conducts electricity in the molten state because the giant ionic lattice breaks down, allowing the ions to move freely.

Answer

The compound conducts electricity in the molten state because the giant ionic lattice breaks down, allowing the ions to move freely.

In solid electrovalent compounds, ions are locked into fixed lattice coordinates by strong electrostatic attraction and cannot migrate. Heating the compound until it melts breaks down the lattice, liberating the cations and anions so they can move freely under an applied electrical potential.

Step-by-Step Solution

1

Determine the type of bonding present in the compound formed by

X

and

Y

.

Element

X

(atomic number 12) has an electronic configuration of

2,8,2

(metal). Element

Y

(atomic number 17) has an electronic configuration of

2,8,7

(non-metal). Metal

X

transfers 2 electrons to two atoms of non-metal

Y

, forming an electrovalent (ionic) compound

XY_2

consisting of

X^{2+}

and

Y^-

ions.

Electrovalent bonding occurs between electropositive metals and electronegative non-metals via full electron transfer.

2

Evaluate the state of charge carriers in the solid state versus the molten state.

In the solid state, strong electrostatic forces hold

X^{2+}

and

Y^-

ions in rigid, fixed lattice positions, so no charge carriers can move. In the molten state, heat breaks the lattice, enabling the ions to move freely toward oppositely charged electrodes.

Electrical conduction requires mobile charge carriers. In ionic substances, these carriers are mobile ions present only in liquid (molten) or aqueous states.

Key Concept

Ionic bonding and electrical conductivity of ionic compounds

Atomic Structure and Chemical Bonding

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