CHEMICAL BONDING & MOLECULAR STRUCTURE MCQs for NEET UG CHEMISTRY — Medium, Page 2

31 Medium

Understanding +4 | -1

Dipole moment is shown by

View answer and explanation

Correct option: D

Both (a) and (c )

Option D (Both (a) and (c )) is correct because cis-1,2-dichloroethane and cis-1,2-dichloro-2-pentene have a net dipole moment due to the asymmetry in the molecule.

Solution approach

Correct Answer + Core Reason: Both (a) and (c ). These molecules have a net dipole moment due to their structure.

Step-by-Step Solution:

Step 1: Analyze cis-1,2-dichloroethane. In the cis isomer, the two C-Cl bond dipoles are on the same side of the molecule and do not cancel each other out, resulting in a net dipole moment.

Step 2: Analyze trans-1,2-dichloroethane. In the trans isomer, the two C-Cl bond dipoles are on opposite sides of the molecule and cancel each other out, resulting in a zero net dipole moment.

Step 3: Analyze cis-1,2-dichloro-2-pentene. Similar to cis-1,2-dichloroethane, the cis configuration of the chlorine atoms leads to a net dipole moment.

Step 4: Conclude. Therefore, cis-1,2-dichloroethane and cis-1,2-dichloro-2-pentene show dipole moment.

Why other options are wrong:
- Option A: cis- 1, 2-dichloro ethane: This is partially correct but incomplete as option (c) is also correct.
- Option B: trans-1, 2-dichloro ethane: This is incorrect because the trans isomer has zero dipole moment due to cancellation.
- Option C: trans-1 2-dichloro-2 peptene: This is partially correct but incomplete as option (a) is also correct.

Key Concepts:
- Dipole moment
- Cis and trans isomers
- Vector addition of bond dipoles

Memory Aids:
- Cis isomers often have a dipole moment, while trans isomers may not due to cancellation.

Quick Tips:
- Draw the structures of the isomers to visualize the dipole moments.

32 Medium

32 of 262Question permalink

Remembering +4 | -1

The decreasing order of bond dissociation energies of

\mathrm{C}-\mathrm{C}, \mathrm{C}-\mathrm{H}

and

\mathrm{H}-\mathrm{H}

bonds in

View answer and explanation

Correct option: A

\mathrm{H}-\mathrm{H}>-\mathrm{C}-\mathrm{H}>-\mathrm{C}-\mathrm{C}-

Option A (

\mathrm{H}-\mathrm{H}>-\mathrm{C}-\mathrm{H}>-\mathrm{C}-\mathrm{C}-

) is correct because the H-H bond is the strongest, followed by C-H, and then C-C.

Solution approach

Correct Answer + Core Reason: $\mathrm{H}-\mathrm{H}>-\mathrm{C}-\mathrm{H}>-\mathrm{C}-\mathrm{C}-$ . This order reflects the relative strengths of the bonds.

Step-by-Step Solution:

Step 1: Consider the bond dissociation energies of the given bonds.

Step 2: The bond dissociation energy of H-H is approximately 436 kJ/mol.

Step 3: The bond dissociation energy of C-H is approximately 413 kJ/mol.

Step 4: The bond dissociation energy of C-C is approximately 347 kJ/mol.

Step 5: Therefore, the decreasing order of bond dissociation energies is H-H > C-H > C-C.

Why other options are wrong:
- Option B: $-\mathrm{C}-\mathrm{C}->\mathrm{C}-\mathrm{H}>\mathrm{H}-\mathrm{H}$ is incorrect because the C-C bond is the weakest.
- Option C: $-\mathrm{C}-\mathrm{H}>-\mathrm{C}-\mathrm{C}->\mathrm{H}-\mathrm{H}$ is incorrect because the H-H bond is the strongest.
- Option D: $-\mathrm{C}-\mathrm{C}-\mathrm{H}-\mathrm{H}>-\mathrm{C}-\mathrm{H}$ is incorrect because it is not a valid comparison.

Key Concepts:
- Bond dissociation energy
- Bond strength

Memory Aids:
- Stronger bonds require more energy to break.

Quick Tips:
- Remember the general trend of bond strengths for common bonds.

33 Medium

33 of 262Question permalink

Remembering +4 | -1

Which of the following is a polar molecule?

View answer and explanation

Correct option: C

\mathrm{NH}_3

Option C is correct because NH3 has a lone pair on the nitrogen atom, resulting in an asymmetrical distribution of electron density and thus a net dipole moment, making it a polar molecule.

Solution approach

Correct Answer + Core Reason: $\mathrm{NH}_3$ . Ammonia has a pyramidal shape and a lone pair on the nitrogen atom, resulting in a net dipole moment.

Step-by-Step Solution:

Step 1: Analyze the molecular geometry of each option.

Step 2: $\mathrm{BF}_3$ : Trigonal planar, symmetrical, nonpolar.

Step 3: $\mathrm{CH}_4$ : Tetrahedral, symmetrical, nonpolar.

Step 4: $\mathrm{NH}_3$ : Pyramidal, asymmetrical, polar.

Step 5: $\mathrm{CO}_2$ : Linear, symmetrical, nonpolar.

Step 6: Conclusion: $\mathrm{NH}_3$ is a polar molecule.

Why other options are wrong:
- Option A: $\mathrm{BF}_3$ : This is incorrect because $\mathrm{BF}_3$ is a symmetrical trigonal planar molecule, and the bond dipoles cancel out, resulting in a zero net dipole moment.
- Option B: $\mathrm{CH}_4$ : This is incorrect because $\mathrm{CH}_4$ is a symmetrical tetrahedral molecule, and the bond dipoles cancel out, resulting in a zero net dipole moment.
- Option D: $\mathrm{CO}_2$ : This is incorrect because $\mathrm{CO}_2$ is a linear molecule, and the bond dipoles cancel out, resulting in a zero net dipole moment.

Key Concepts:
- Polarity of Molecules
- Molecular Geometry
- Dipole Moment

Memory Aids:
- Remember that asymmetrical molecules with polar bonds are generally polar.

Quick Tips:
- Draw the Lewis structure and determine the molecular geometry to predict the polarity of a molecule.

34 Medium

34 of 262Question permalink

Application +4 | -1

Which one of the following constitutes a group of the isoelectronic species?

View answer and explanation

Correct option: B

\mathrm{N}{\mathrm{O}^ + },\mathrm{C}_2^{2 - },\mathrm{C}{\mathrm{N}^ - },{\mathrm{N}_2}

Option B (

\mathrm{N}{\mathrm{O}^ + },\mathrm{C}_2^{2 - },\mathrm{C}{\mathrm{N}^ - },{\mathrm{N}_2}

) is correct because all species in this group have 14 electrons.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{O}^ + },\mathrm{C}_2^{2 - },\mathrm{C}{\mathrm{N}^ - },{\mathrm{N}_2}$ . All species have 14 electrons.

Step-by-Step Solution:

Step 1: Define isoelectronic species: Isoelectronic species are atoms, ions, or molecules that have the same number of electrons.

Step 2: Calculate the number of electrons for each species:
- $\mathrm{C}_2^{2 - }$ : 2*6 + 2 = 14
- $\mathrm{O}_2^ -$ : 2*8 + 1 = 17
- $\mathrm{CO}$ : 6 + 8 = 14
- $\mathrm{NO}$ : 7 + 8 = 15
- $\mathrm{N}{\mathrm{O}^ + }$ : 7 + 8 - 1 = 14
- $\mathrm{C}{\mathrm{N}^ - }$ : 6 + 7 + 1 = 14
- ${\mathrm{N}_2}$ : 2*7 = 14
- $\mathrm{O}_2^{2 - }$ : 2*8 + 2 = 18
- $\mathrm{CO}_3^{2 - }$ : 6 + 3*8 + 2 = 32

Step 3: Identify the group with all species having the same number of electrons (14): $\mathrm{N}{\mathrm{O}^ + },\mathrm{C}_2^{2 - },\mathrm{C}{\mathrm{N}^ - },{\mathrm{N}_2}$ .

Why other options are wrong:
- Option A: $\mathrm{C}_2^{2 - },\mathrm{O}_2^ - ,\mathrm{CO},\mathrm{NO}$ : The number of electrons are 14, 17, 14, 15 respectively. Not isoelectronic.
- Option C: $\mathrm{C}{\mathrm{N}^ - },{\mathrm{N}_2},\mathrm{O}_2^{2 - },\mathrm{CO}_3^{2 - }$ : The number of electrons are 14, 14, 18, 32 respectively. Not isoelectronic.
- Option D: ${\mathrm{N}_2},\mathrm{O}_2^ - ,\mathrm{N}{\mathrm{O}^ + },\mathrm{CO}$ : The number of electrons are 14, 17, 14, 14 respectively. Not isoelectronic.

Key Concepts:
- Isoelectronic species
- Number of electrons

Memory Aids:
- Remember to account for the charge when calculating the number of electrons in ions.

Quick Tips:
- Calculate the number of electrons for each species to determine if they are isoelectronic.

35 Medium

35 of 262Question permalink

Application +4 | -1

The value of electronegativity of atoms A and B are 1.20 and 4.0 respectively. The percentage of ionic character of

\mathrm{A}-\mathrm{B}

bond is:

View answer and explanation

Correct option: D

72.24 \%

Option D (72.24 %) is correct because the percentage of ionic character can be estimated using the Pauling equation: % ionic character = (1 - exp(-0.25 * (ΔEN)²)) * 100, where ΔEN is the electronegativity difference.

Solution approach

Correct Answer + Core Reason: 72.24 %. The percentage of ionic character is calculated using the electronegativity difference.

Step-by-Step Solution:

Step 1: Calculate the electronegativity difference (ΔEN) between atoms A and B.
- ΔEN = |Electronegativity of B - Electronegativity of A| = |4.0 - 1.20| = 2.8

Step 2: Use the Pauling equation to estimate the percentage of ionic character:
- % ionic character = (1 - exp(-0.25 * (ΔEN)²)) * 100

Step 3: Substitute the value of ΔEN into the equation:
- % ionic character = (1 - exp(-0.25 * (2.8)²)) * 100
- % ionic character = (1 - exp(-0.25 * 7.84)) * 100
- % ionic character = (1 - exp(-1.96)) * 100

Step 4: Calculate the value:
- exp(-1.96) ≈ 0.141
- % ionic character = (1 - 0.141) * 100
- % ionic character = 0.859 * 100
- % ionic character = 85.9 % (Using the more accurate formula)

Step 5: Use the alternative formula:
% Ionic Character = $16 \times |X_A - X_B| + 3.5 \times |X_A - X_B|^2$
% Ionic Character = $16 \times 2.8 + 3.5 \times (2.8)^2$
% Ionic Character = $44.8 + 3.5 \times 7.84$
% Ionic Character = $44.8 + 27.44$
% Ionic Character = 72.24%

Why other options are wrong:
- Option A (43 %): This value is too low and likely results from an incorrect application of the electronegativity difference or using a wrong formula.
- Option B (50 %): This value is also too low and doesn't accurately reflect the significant electronegativity difference between A and B.
- Option C (55.3 %): This value is closer but still incorrect, possibly due to rounding errors or a slightly different empirical relationship used.

Key Concepts:
- Electronegativity
- Electronegativity difference
- Percentage of ionic character
- Pauling equation

Memory Aids:
- Remember that a larger electronegativity difference generally indicates a higher percentage of ionic character.

Quick Tips:
- Use the Pauling equation or other empirical relationships to estimate the percentage of ionic character based on the electronegativity difference.

36 Medium

36 of 262Question permalink

Understanding +4 | -1

In allene structure, three carbon atoms are joined by:

View answer and explanation

Correct option: C

Two

\mathrm{\sigma }

and two

\mathrm{\pi }

bonds

Option C (Two

\mathrm{\sigma }

and two

\mathrm{\pi }

bonds) is correct because in allene (C=C=C), the central carbon has two sigma bonds and two pi bonds.

Solution approach

Correct Answer + Core Reason: Two $\mathrm{\sigma }$ and two $\mathrm{\pi }$ bonds. The central carbon has two sigma and two pi bonds.

Step-by-Step Solution:

Step 1: Draw the structure of allene (C=C=C).

Step 2: The central carbon atom is sp hybridized and forms two sigma bonds and two pi bonds.

Step 3: Each double bond consists of one sigma and one pi bond. Since there are two double bonds, there are two sigma and two pi bonds.

Why other options are wrong:
- Option A: Incorrect number of sigma and pi bonds.
- Option B: Incorrect number of sigma and pi bonds.
- Option D: Incorrect number of sigma and pi bonds.

Key Concepts:
- Structure of allene
- Sigma and pi bonds
- Hybridization

Memory Aids:
- Remember the structure of allene and the types of bonds it contains.

Quick Tips:
- Draw the structure to visualize the sigma and pi bonds.

37 Medium

37 of 262Question permalink

Analyzing +4 | -1

Which does not use

s{p^3}

-hybrid orbitals in its bonding?

View answer and explanation

Correct option: A

\mathrm{BeF}_3^ -

Option A (

\mathrm{BeF}_3^ -

) is correct because Be in

\mathrm{BeF}_3^ -

is

sp^2

hybridized, while the others are

sp^3

hybridized.

Solution approach

Correct Answer + Core Reason: $\mathrm{BeF}_3^ -$ . Be in $\mathrm{BeF}_3^ -$ is $sp^2$ hybridized, while the others are $sp^3$ hybridized.

Step-by-Step Solution:

Step 1: Determine the hybridization of the central atom in each species.

Step 2: For $\mathrm{BeF}_3^ -$ , Be has 3 bond pairs and no lone pairs. Steric number = 3. Hybridization = $sp^2$ .

Step 3: For $\mathrm{OH}_3^ +$ , O has 3 bond pairs and 1 lone pair. Steric number = 4. Hybridization = $sp^3$ .

Step 4: For $\mathrm{NH}_4^ +$ , N has 4 bond pairs and no lone pairs. Steric number = 4. Hybridization = $sp^3$ .

Step 5: For $\mathrm{NF}_3$ , N has 3 bond pairs and 1 lone pair. Steric number = 4. Hybridization = $sp^3$ .

Step 6: Identify the species that does not have $sp^3$ hybridization.

Step 7: The species that does not have $sp^3$ hybridization is $\mathrm{BeF}_3^ -$ .

Why other options are wrong:
- Option B: $\mathrm{OH}_3^ +$ has $sp^3$ hybridization.
- Option C: $\mathrm{NH}_4^ +$ has $sp^3$ hybridization.
- Option D: $\mathrm{NF}_3$ has $sp^3$ hybridization.

Key Concepts:
- Hybridization
- Steric number
- Lone pairs
- Bond pairs

Memory Aids:
- Steric number = number of bond pairs + number of lone pairs
- Steric number 2 = sp hybridization
- Steric number 3 = $sp^2$ hybridization
- Steric number 4 = $sp^3$ hybridization

Quick Tips:
- Calculate the steric number to determine the hybridization.

38 Medium

38 of 262Question permalink

Understanding +4 | -1

In which of the following crystals of ionic compounds would you expect maximum distance between the centres of cotions and anions

View answer and explanation

Correct option: C

CsI

Option C (CsI) is correct because both Cs+ and I- are large ions, leading to a greater distance between their centers in the crystal lattice.

Solution approach

Correct Answer + Core Reason: CsI. Both Cs+ and I- are large ions.

Step-by-Step Solution:

Step 1: Consider the ionic radii of the cations and anions in each compound.

Step 2: The ionic radii increase down a group. Therefore, Cs+ is larger than Li+, and I- is larger than F-.

Step 3: The distance between the centers of the ions will be greatest when both the cation and anion are large. This occurs in CsI.

Why other options are wrong:
- Option A: LiF has small ions, so the distance will be small.
- Option B: CsF has a large cation but a small anion.
- Option D: LiI has a small cation but a large anion.

Key Concepts:
- Ionic radii
- Crystal structure

Memory Aids:
- Remember the trends in ionic radii in the periodic table.

Quick Tips:
- Larger ions will have greater distances between their centers in a crystal.

39 Medium

39 of 262Question permalink

Analyzing +4 | -1

Assertion : The

{O}-{O}

bond length in

{H}_2 {O}_2

is shorter than that of

{O}_2 {~F}_2

.

Reason:

{H}_2 {O}_2

is an ionic compound.

View answer and explanation

Correct option: D

If both the Assertion and Reason are incorrect.

Option D (If both the Assertion and Reason are incorrect) is correct because the O-O bond length in H2O2 is longer than that in O2F2, and H2O2 is a covalent compound, not ionic.

Solution approach

Correct Answer + Core Reason: If both the Assertion and Reason are incorrect. The O-O bond length in H2O2 is longer than in O2F2, and H2O2 is a covalent compound.

Step-by-Step Solution:

Step 1: Analyze the assertion: The O-O bond length in $H_2O_2$ is longer than that in $O_2F_2$ . The assertion is incorrect. The O-O bond length in $H_2O_2$ is approximately 148 pm, while in $O_2F_2$ it is approximately 122 pm. Therefore, the assertion is false.

Step 2: Analyze the reason: $H_2O_2$ is an ionic compound. The reason is incorrect. $H_2O_2$ is a covalent compound.

Step 3: Conclusion: Both the assertion and the reason are incorrect.

Why other options are wrong:
- Option A: If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion: Both are incorrect.
- Option B: If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion: Both are incorrect.
- Option C: If the Assertion is correct but Reason is incorrect: Assertion is also incorrect.

Key Concepts:
- Bond length
- Covalent compound
- Hydrogen peroxide
- Dioxygen difluoride

Memory Aids:
- Remember the bond lengths of common molecules.

Quick Tips:
- Know the difference between ionic and covalent compounds.

40 Medium

40 of 262Question permalink

Remembering +4 | -1

Bond formed due to complete transfer of electron is

View answer and explanation

Correct option: D

Ionic bond

Option D (Ionic bond) is correct because an ionic bond is formed through the complete transfer of electrons from one atom to another, resulting in the formation of ions.

Solution approach

Correct Answer + Core Reason: Ionic bond is formed due to complete transfer of electrons.

Step-by-Step Solution:

Step 1: An ionic bond is formed when one or more electrons are transferred from one atom to another.

Step 2: This transfer results in the formation of positively charged ions (cations) and negatively charged ions (anions).

Step 3: The electrostatic attraction between these oppositely charged ions constitutes the ionic bond.

Why other options are wrong:
- Option A: A covalent bond is formed by the sharing of electrons between atoms, not the complete transfer.
- Option B: A coordinate bond (also known as a dative bond) is formed when one atom donates both electrons to the bond.
- Option C: A hydrogen bond is a weak electrostatic attraction between a hydrogen atom bonded to a highly electronegative atom (such as oxygen or nitrogen) and another electronegative atom.

Key Concepts:
- Ionic bonding
- Covalent bonding
- Coordinate bonding
- Hydrogen bonding

Memory Aids:
- Ionic = Transfer of electrons

Quick Tips:
- Remember the difference between sharing (covalent) and transfer (ionic) of electrons.

41 Medium

41 of 262Question permalink

Understanding +4 | -1

Which has the largest distance between the carbon hydrogen atom?

View answer and explanation

Correct option: A

Ethane

Option A (Ethane) is correct because it has only single bonds, which are longer than the multiple bonds in ethene, ethyne, and the delocalized bonds in benzene.

Solution approach

Correct Answer + Core Reason: Ethane. Single bonds are longer than multiple bonds.

Step-by-Step Solution:

Step 1: Identify the types of bonds in each molecule.
- Ethane ( $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_3}$ ): All C-H and C-C single bonds.
- Ethene ( $\mathrm{C}{\mathrm{H}_2}=\mathrm{C}{\mathrm{H}_2}$ ): C-H single bonds and one C=C double bond.
- Ethyne ( $\mathrm{CH} \equiv \mathrm{CH}$ ): C-H single bonds and one C≡C triple bond.
- Benzene ( $\mathrm{C}_6\mathrm{H}_6$ ): C-H single bonds and delocalized pi bonds (intermediate between single and double bonds).

Step 2: Recall the order of bond lengths.
- Single bond > Double bond > Triple bond

Step 3: Determine which molecule has the largest C-H bond distance.
- Since all the molecules have C-H bonds, the molecule with only single bonds will have the largest C-H bond distance. In this case, Ethane.

Why other options are wrong:
- Option A: Ethane - Correct. Ethane has only single bonds, which are longer.
- Option B: Ethene - Incorrect. Ethene has a double bond, which shortens the C-H bond distance.
- Option C: Ethyne - Incorrect. Ethyne has a triple bond, which further shortens the C-H bond distance.
- Option D: Benzene - Incorrect. Benzene has delocalized pi bonds, which result in C-H bond distances shorter than single bonds.

Key Concepts:
- Bond length
- Single bond
- Double bond
- Triple bond
- Delocalized bonds

Memory Aids:
- Single bonds are longer than multiple bonds.

Quick Tips:
- Remember the order of bond lengths: Single > Double > Triple.

42 Medium

42 of 262Question permalink

Remembering +4 | -1

Electron deficient species are known as:

View answer and explanation

Correct option: A

Lewis acids

Option A is correct because electron-deficient species, which can accept electron pairs, are known as Lewis acids.

Solution approach

Correct Answer + Core Reason: Lewis acids. Electron deficient species are electron acceptors, which are Lewis acids.

Step-by-Step Solution:

Step 1: Define electron-deficient species. These are species that do not have a complete octet of electrons around the central atom.

Step 2: Recall the definition of Lewis acids and Lewis bases.

Step 3: Lewis acids are electron acceptors, while Lewis bases are electron donors.

Step 4: Electron-deficient species have a tendency to accept electrons to complete their octet.

Step 5: Therefore, electron-deficient species are Lewis acids.

Why other options are wrong:
- Option B: Hydrophilic means water-loving, which is not directly related to electron deficiency.
- Option C: Nucleophiles are electron-rich species that donate electrons, which is the opposite of electron deficiency.
- Option D: Lewis bases are electron donors, not electron acceptors.

Key Concepts:
- Lewis acids and bases
- Electron deficiency
- Nucleophiles

Memory Aids:
- Lewis Acid = Electron Acceptor
- Lewis Base = Electron Donor

Quick Tips:
- Remember the definitions of Lewis acids and bases in terms of electron acceptance and donation.

43 Medium

43 of 262Question permalink

Remembering +4 | -1

Covalent compounds are generally soluble in

View answer and explanation

Correct option: B

Non-polar solvents

Option B (Non-polar solvents) is correct because \"like dissolves like,\" and covalent compounds are generally non-polar.

Solution approach

Correct Answer + Core Reason: Non-polar solvents. Covalent compounds are generally non-polar and dissolve best in non-polar solvents due to similar intermolecular forces.

Step-by-Step Solution:

Step 1: Identify the type of solute and solvent.\nCovalent compounds are generally non-polar.\nWe need to determine which solvent is most similar in polarity to non-polar covalent compounds.

Step 2: Apply the \"like dissolves like\" principle.\nPolar solutes dissolve in polar solvents.\nNon-polar solutes dissolve in non-polar solvents.

Step 3: Determine the polarity of the given solvents.\nPolar solvents: Water, alcohols, etc.\nNon-polar solvents: Benzene, toluene, hexane, carbon tetrachloride, etc.

Step 4: Conclude which solvent is best for dissolving covalent compounds.\nSince covalent compounds are generally non-polar, they dissolve best in non-polar solvents.

Why other options are wrong:
- Option A: Polar solvents: Polar solvents dissolve polar compounds, not non-polar covalent compounds.
- Option C: Concentrated acids: Acids may react with some covalent compounds, but they are not generally good solvents for them. Acids are polar.
- Option D: All solvents: Covalent compounds do not dissolve in all solvents. They primarily dissolve in non-polar solvents.

Key Concepts:
- Polarity of molecules
- Intermolecular forces
- \"Like dissolves like\" principle

Memory Aids:
- \"Like dissolves like\" - substances with similar polarity tend to dissolve in each other.

Quick Tips:
- Remember that water is a polar solvent and oil is a non-polar solvent. This can help you remember the \"like dissolves like\" principle.

44 Medium

44 of 262Question permalink

Remembering +4 | -1

The half of the difference between the number of electrons in bonding molecular orbitals and antibonding molecular orbitals is known as:

View answer and explanation

Correct option: A

Bond order

Option A (Bond order) is correct because bond order is defined as half the difference between the number of electrons in bonding molecular orbitals and antibonding molecular orbitals, indicating the stability of a chemical bond.

Solution approach

Correct Answer + Core Reason: Bond order. It is defined as half the difference between bonding and antibonding electrons.

Step-by-Step Solution:

Step 1: Understand the question. We need to identify the term that describes half the difference between the number of electrons in bonding and antibonding molecular orbitals.

Step 2: Recall the definition of bond order.
- Bond order = $\frac{{1}}{{2}}$ (Number of bonding electrons - Number of antibonding electrons)

Step 3: Identify the correct term as Bond order.

Why other options are wrong:
- Option B: Proton order is not a relevant term in molecular orbital theory.
- Option C: Molecular order is not a standard term.
- Option D: Electron order is not a standard term.

Key Concepts:
- Molecular orbital theory
- Bonding and antibonding molecular orbitals
- Bond order

Memory Aids:
- Bond order = (Bonding - Antibonding) / 2

Quick Tips:
- Remember the formula for calculating bond order.

45 Medium

45 of 262Question permalink

Understanding +4 | -1

From the molecular orbital theory, one can show that the bond order in

{\mathrm{F}_2}

molecule as

View answer and explanation

Correct option: B

1

Option B (1) is correct because

{\mathrm{F}_2}

has a bond order of 1, indicating a single bond between the two fluorine atoms.

Solution approach

Correct Answer + Core Reason: 1. The bond order of F2 is 1.

Step-by-Step Solution:

Step 1: Write the electronic configuration of ${\mathrm{F}_2}$ .

Step 2: Each F atom has 9 electrons, so ${\mathrm{F}_2}$ has 18 electrons.

Step 3: The molecular orbital configuration is $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p})^2(\pi_{2p})^4(\pi_{2p}^*)^4$ .

Step 4: Calculate the bond order using the formula: Bond order = (Number of electrons in bonding orbitals - Number of electrons in antibonding orbitals) / 2.

Step 5: Bond order = (10 - 8) / 2 = 2/2 = 1.

Why other options are wrong:
- Option A (2): This is incorrect. The bond order calculation does not result in 2.
- Option C (3): This is incorrect. The bond order calculation does not result in 3.
- Option D (4): This is incorrect. The bond order calculation does not result in 4.

Key Concepts:
- Molecular Orbital Theory
- Bond Order
- Electronic Configuration

Memory Aids:
- Bond Order = (Bonding - Antibonding) / 2

Quick Tips:
- Write the correct electronic configuration before calculating the bond order.

46 Medium

46 of 262Question permalink

Analyzing +4 | -1

Which of the following species has a bond order other than 3?

View answer and explanation

Correct option: D

\mathrm{O}_2^ +

Option D (

{\mathrm{O}_2^ + }

) is correct because its bond order is 2.5, while the other options have a bond order of 3.

Solution approach

Correct Answer + Core Reason: $\mathrm{O}_2^ +$ . Its bond order is 2.5, while the others have a bond order of 3.

Step-by-Step Solution:

Step 1: Calculate the bond order for each species.

Step 2: $CO$ : Bond order = (10 - 4)/2 = 3

Step 3: $\mathrm{C}{\mathrm{N}^ - }$ : Bond order = (10 - 4)/2 = 3

Step 4: $\mathrm{N}{\mathrm{O}^ + }$ : Bond order = (10 - 4)/2 = 3

Step 5: $\mathrm{O}_2^ +$ : Bond order = (10 - 5)/2 = 2.5

Step 6: Identify the species with a bond order different from 3.

Why other options are wrong:
- Option A: $CO$ has a bond order of 3.
- Option B: $\mathrm{C}{\mathrm{N}^ - }$ has a bond order of 3.
- Option C: $\mathrm{N}{\mathrm{O}^ + }$ has a bond order of 3.

Key Concepts:
- Bond Order Calculation

Memory Aids:
- Bond Order = (Number of bonding electrons - Number of antibonding electrons) / 2

Quick Tips:
- Remember to consider the charge when calculating the number of electrons.

47 Medium

47 of 262Question permalink

Remembering +4 | -1

The observed

\mathrm{O}-\mathrm{O}

bond length in ozone molecule is

View answer and explanation

Correct option: C

128 \mathrm{~pm}

Option C (

128 \mathrm{~pm}

) is correct because the observed O-O bond length in ozone is 128 pm due to resonance.

Solution approach

Correct Answer + Core Reason: $128 \mathrm{~pm}$ . The O-O bond length in ozone is 128 pm.

Step-by-Step Solution:

Step 1: Recall the structure of ozone (O3). It exhibits resonance.

Step 2: The O-O bond length in ozone is intermediate between a single and double bond due to resonance, which is approximately 128 pm.

Why other options are wrong:
- Option A: 121 pm is closer to the length of a double bond, not the resonance hybrid in ozone.
- Option B: 148 pm is closer to the length of a single bond, not the resonance hybrid in ozone.
- Option D: 128 pm is the correct bond length.

Key Concepts:
- Resonance
- Bond length
- Structure of Ozone

Memory Aids:
- Remember that resonance leads to intermediate bond lengths.

Quick Tips:
- Know the typical bond lengths for single, double, and triple bonds.

48 Medium

48 of 262Question permalink

Application +4 | -1

The electronegativity of

{O}, {F}, {N}, {Cl}

and

{H}

are

3.5,4.0,3.2

, 3.0 and 2.1 respectively. The strongest bond will be:

View answer and explanation

Correct option: D

{O}-{H}

Option D (

{O}-{H}

) is correct because the electronegativity difference between O and H is the greatest among the given options, leading to the strongest and most polar bond.

Solution approach

Correct Answer + Core Reason: ${O}-{H}$ . It has the largest electronegativity difference.

Step-by-Step Solution:

Step 1: Calculate the electronegativity difference for ${F}-{O}$ : $|4.0 - 3.5| = 0.5$

Step 2: Calculate the electronegativity difference for ${O}-{Cl}$ : $|3.5 - 3.0| = 0.5$

Step 3: Calculate the electronegativity difference for ${N}-{H}$ : $|3.2 - 2.1| = 1.1$

Step 4: Calculate the electronegativity difference for ${O}-{H}$ : $|3.5 - 2.1| = 1.4$

Step 5: Conclusion: The ${O}-{H}$ bond has the largest electronegativity difference (1.4), making it the strongest bond.

Why other options are wrong:
- Option A: ${F}-{O}$ has an electronegativity difference of 0.5, which is smaller than that of ${O}-{H}$ .
- Option B: ${O}-{Cl}$ has an electronegativity difference of 0.5, which is smaller than that of ${O}-{H}$ .
- Option C: ${N}-{H}$ has an electronegativity difference of 1.1, which is smaller than that of ${O}-{H}$ .

Key Concepts:
- Electronegativity
- Bond Polarity
- Bond Strength

Memory Aids:
- Larger electronegativity difference = Stronger, more polar bond

Quick Tips:
- The greater the electronegativity difference between two atoms, the stronger and more polar the bond.

49 Medium

49 of 262Question permalink

Remembering +4 | -1

Which of these statements is correct according to VBT?

View answer and explanation

Correct option: B

Covalent bond is formed by partial overlap of two half-filled atomic orbitals containing electrons in opposite
spin

Option B (Covalent bond is formed by partial overlap of two half-filled atomic orbitals containing electrons in opposite spin) is correct because VBT states that a covalent bond forms from the partial overlap of two half-filled atomic orbitals with electrons of opposite spin.

Solution approach

Correct Answer + Core Reason: Covalent bond forms by partial overlap of half-filled orbitals with opposite spin.

Step-by-Step Solution:

Step 1: Valence Bond Theory (VBT) describes the formation of covalent bonds through the overlap of atomic orbitals.

Step 2: According to VBT, a covalent bond is formed when two half-filled atomic orbitals, each containing one electron with opposite spin, overlap.

Step 3: The overlapping region between the orbitals results in increased electron density, leading to the formation of a stable covalent bond.

Why other options are wrong:
- Option A: Electrons in the same spin would repel each other, hindering bond formation.
- Option C: Complete overlap of fully filled orbitals would not lead to effective bonding as it doesn't satisfy the requirement of half-filled orbitals.
- Option D: Partial overlap of fully filled orbitals is not a standard description of covalent bond formation according to VBT.

Key Concepts:
- Valence Bond Theory (VBT)
- Overlap of atomic orbitals
- Spin pairing

Memory Aids:
- VBT = Overlap = Opposite spin

Quick Tips:
- Remember that VBT emphasizes the overlap of half-filled orbitals with opposite spins for covalent bond formation.

50 Medium

50 of 262Question permalink

Application +4 | -1

Which of the following molecular species has unpaired electron (s)?

View answer and explanation

Correct option: C

\mathrm{O}_2^ -

Option C (

\mathrm{O}_2^ -

) is correct because it has an odd number of electrons, leading to unpaired electron(s).

Solution approach

Correct Answer + Core Reason: $\mathrm{O}_2^ -$ has an odd number of electrons, resulting in unpaired electrons.

Step-by-Step Solution:

Step 1: Determine the number of valence electrons for each species.

Step 2: Write the electronic configuration for each species.

Step 3: Identify the species with unpaired electrons.\nFor ${\mathrm{N}_2}$ : 14 electrons, configuration is $(\sigma_{2s})(\sigma_{2s}^{\ast})(\sigma_{2p})(\pi_{2p})^4$ . No unpaired electrons.\nFor ${\mathrm{F}_2}$ : 18 electrons, configuration is $(\sigma_{2s})^2(\sigma_{2s}^{\ast})^2(\sigma_{2p})^2(\pi_{2p})^4(\pi_{2p}^{\ast})^4$ . No unpaired electrons.\nFor $\mathrm{O}_2^ -$ : 17 electrons, configuration is $(\sigma_{2s})^2(\sigma_{2s}^{\ast})^2(\sigma_{2p})^2(\pi_{2p})^4(\pi_{2p}^{\ast})^3$ . Two unpaired electrons in $\pi_{2p}^{\ast}$ orbitals.\nFor $\mathrm{O}_2^{2 - }$ : 18 electrons, configuration is $(\sigma_{2s})^2(\sigma_{2s}^{\ast})^2(\sigma_{2p})^2(\pi_{2p})^4(\pi_{2p}^{\ast})^4$ . No unpaired electrons.

Step 4: Conclude that $\mathrm{O}_2^ -$ has unpaired electrons.

Why other options are wrong:
- Option A: ${\mathrm{N}_2}$ has all its electrons paired.
- Option B: ${\mathrm{F}_2}$ has all its electrons paired.
- Option D: $\mathrm{O}_2^{2 - }$ has all its electrons paired.

Key Concepts:
- Molecular Orbital Theory
- Electronic Configuration
- Unpaired Electrons and Paramagnetism

Memory Aids:
- Odd number of total electrons often indicates unpaired electrons.

Quick Tips:
- Draw the molecular orbital diagram to determine the number of unpaired electrons.

51 Medium

51 of 262Question permalink

Analyzing +4 | -1

Which of the following is false?

View answer and explanation

Correct option: D

Acetylene is non-linear

Option D (Acetylene is non-linear) is correct because acetylene (C₂H₂) is linear due to the sp hybridization of the carbon atoms.

Solution approach

Correct Answer + Core Reason: Acetylene is non-linear. Acetylene is linear, so this statement is false.

Step-by-Step Solution:

Step 1: Evaluate each statement.

Step 2: Methane molecule is tetrahedral in shape: Correct. Methane ( $CH_4$ ) has a tetrahedral shape due to $sp^3$ hybridization.

Step 3: Nickel tetrachloride is square planar in shape: Correct. Nickel tetrachloride ( $[NiCl_4]$

52 Medium

52 of 262Question permalink

Analyzing +4 | -1

The compound with the maximum dipole moment among the following is:

View answer and explanation

Correct option: C

o

-dichlorobenzene

Option C (

o

-dichlorobenzene) is correct because the dipole moments of the C-Cl bonds add constructively due to their orientation, resulting in a non-zero net dipole moment.

Solution approach

Correct Answer + Core Reason: $o$ -dichlorobenzene. The dipole moment is maximum due to the vector addition of bond dipoles at a smaller angle.

Step-by-Step Solution:

Step 1: Draw the structures of the isomers of dichlorobenzene: para ( $p$ ), meta ( $m$ ), and ortho ( $o$ ).

Step 2: Consider the dipole moments of the C-Cl bonds in each isomer. Each C-Cl bond has a dipole moment.

Step 3: Analyze the vector sum of the bond dipoles:
- In $p$ -dichlorobenzene, the C-Cl bond dipoles are equal in magnitude and opposite in direction, so they cancel each other out, resulting in a zero net dipole moment.
- In $m$ -dichlorobenzene, the C-Cl bond dipoles are at an angle of 120 degrees. The net dipole moment is non-zero but smaller than in the ortho isomer.
- In $o$ -dichlorobenzene, the C-Cl bond dipoles are at an angle of 60 degrees. The net dipole moment is the highest among the three isomers.
- Carbon tetrachloride ( $\mathrm{CCl}_4$ ) is a symmetrical tetrahedral molecule, so the bond dipoles cancel out, resulting in a zero net dipole moment.

Why other options are wrong:
- **Option A ( $p$ -dichlorobenzene): The dipole moments of the two C-Cl bonds cancel each other out due to symmetry.
- Option B ( $m$ -dichlorobenzene): The dipole moments do not cancel completely, but the net dipole moment is smaller than that of the ortho isomer.
- Option D (Carbon tetrachloride): It's a symmetrical tetrahedral molecule, so the bond dipoles cancel out, resulting in zero dipole moment.

Key Concepts:
- Dipole moment
- Isomers
- Vector addition of bond dipoles
- Molecular symmetry

Memory Aids:
- Ortho isomers generally have higher dipole moments than meta or para isomers.
- Symmetry leads to cancellation of dipoles.

Quick Tips:**
- Draw the structures to visualize the bond dipoles and their vector sum.
- Remember that symmetrical molecules often have zero dipole moment.

53 Medium

53 of 262Question permalink

Application +4 | -1

In

\mathrm{PO}_4^{3 - }

ion, the formal charge on each oxygen atom and

\mathrm{P} - \mathrm{O}

bond order respectively are:

View answer and explanation

Correct option: C

-0.75, 1.25

Option C (-0.75, 1.25) is correct because the formal charge on each oxygen is -0.75 and the P-O bond order is 1.25 in the phosphate ion.

Solution approach

Correct Answer + Core Reason: -0.75, 1.25. The formal charge and bond order are calculated based on the resonance structures of the phosphate ion.

Step-by-Step Solution:

Step 1: Draw the resonance structures of $\mathrm{PO}_4^{3 - }$ . Phosphorus is the central atom, and there are four oxygen atoms surrounding it. The ion has a 3- charge, and there are multiple resonance structures.

Step 2: Calculate the formal charge on each oxygen atom: Since the charge is equally distributed among the four oxygen atoms, the formal charge on each oxygen atom is -3/4 = -0.75.

Step 3: Calculate the bond order: There are a total of 5 bonds (1 double bond and 3 single bonds) distributed among 4 P-O bonds. Therefore, the bond order = (1 double bond + 3 single bonds) / 4 bonds = (2 + 3)/4 = 5/4 = 1.25.

Step 4: Determine the formal charge and bond order: Formal charge = -0.75, Bond order = 1.25.

Why other options are wrong:
- Option A: -0.75, 1.0: Incorrect bond order.
- Option B: -3, 1.25: Incorrect formal charge.
- Option D: -0.75, 0.6: Incorrect bond order.

Key Concepts:
- Formal charge
- Bond order
- Resonance structures

Memory Aids:
- Formal Charge = (Valence electrons) - (Non-bonding electrons) - (1/2 Bonding electrons).
- Bond Order = (Total number of bonds) / (Number of bond locations).

Quick Tips:
- Draw all possible resonance structures to accurately determine formal charges and bond orders.

54 Medium

54 of 262Question permalink

Application +4 | -1

The bond length of

HCl

molecule is

1.275{A^0}

and its dipole moment is

1.03 D

. The ionic character of the molecule (in per cent) ( charge of the electron=

= 4.8 \times {10^{ - 10}}esu\,)

is

View answer and explanation

Correct option: D

16.83

Option D (

16.83

) is correct because it is the calculated percentage ionic character of HCl using the given bond length and dipole moment.

Solution approach

Correct Answer + Core Reason: $16.83$ . The percentage ionic character is calculated using the formula relating dipole moment, bond length, and electronic charge.

Step-by-Step Solution:

Step 1: Convert bond length from Angstroms to cm: $d = 1.275 \text{ A}^0 = 1.275 \times 10^{-8} \text{ cm}$

Step 2: Convert dipole moment from Debye to esu cm: $\mu = 1.03 \text{ D} = 1.03 \times 10^{-18} \text{ esu cm}$

Step 3: Calculate the theoretical dipole moment assuming 100% ionic character: $\mu_{\text{theoretical}} = q \times d = (4.8 \times 10^{-10} \text{ esu}) \times (1.275 \times 10^{-8} \text{ cm}) = 6.12 \times 10^{-18} \text{ esu cm}$

Step 4: Calculate the percentage ionic character: $\% \text{ ionic character} = \frac{\mu_{\text{observed}}}{\mu_{\text{theoretical}}} \times 100 = \frac{1.03 \times 10^{-18} \text{ esu cm}}{6.12 \times 10^{-18} \text{ esu cm}} \times 100 \approx 16.83 \%$

Why other options are wrong:
- **Option A ( $100$ ): This is incorrect as it assumes the molecule is 100% ionic, which is not the case.
- Option B ( $67.3$ ): This is incorrect due to a calculation error or using incorrect conversion factors.
- Option C ( $33.66$ ): This is incorrect due to a calculation error or using incorrect conversion factors.

Key Concepts:
- Dipole moment
- Percentage ionic character
- Conversion of units (Angstroms to cm, Debye to esu cm)

Memory Aids:**
- Remember the formula: $\% \text{ ionic character} = \frac{\mu_{\text{observed}}}{\mu_{\text{theoretical}}} \times 100$

Quick Tips:
- Always ensure consistent units (esu cm) when using the given charge value.
- Remember the conversion factor for Debye to esu cm: 1 D = 10^-18 esu cm.

55 Medium

55 of 262Question permalink

Understanding +4 | -1

Which can be described as a molecule with residual bonding capacity?

View answer and explanation

Correct option: D

\mathrm{BeC}{\mathrm{l}_2}

Option D (

BeCl_2

) is correct because Beryllium in

BeCl_2

has an incomplete octet and can accept electron pairs, indicating residual bonding capacity.

Solution approach

Correct Answer + Core Reason: $BeCl_2$ has residual bonding capacity due to incomplete octet.

Step-by-Step Solution:

Step 1: A molecule with residual bonding capacity can accept more electron pairs to form additional bonds.

Step 2: $N_2$ has a triple bond and a complete octet for each nitrogen atom, so it does not have residual bonding capacity.

Step 3: $CH_4$ has four single bonds and a complete octet for the carbon atom, so it does not have residual bonding capacity.

Step 4: $NaCl$ is an ionic compound and does not form additional covalent bonds.

Step 5: $BeCl_2$ has two single bonds, and the beryllium atom has only four valence electrons, so it has an incomplete octet and can accept more electron pairs, indicating residual bonding capacity.

Why other options are wrong:
- Option A: $N_2$ has a stable triple bond and complete octet.
- Option B: $CH_4$ has a stable tetrahedral structure with a complete octet.
- Option C: $NaCl$ is an ionic compound and does not form additional covalent bonds.

Key Concepts:
- Octet rule
- Lewis acids and bases
- Residual bonding capacity

Memory Aids:
- Incomplete octet = Residual bonding capacity

Quick Tips:
- Remember that molecules with central atoms having incomplete octets can act as Lewis acids and accept electron pairs.

56 Medium

56 of 262Question permalink

Remembering +4 | -1

One would expect the elemental form of

\mathrm{Cs}

at room temperature to be:

View answer and explanation

Correct option: B

A metallic solid

Option B (A metallic solid) is correct because Cesium (Cs) is an alkali metal and exists as a metallic solid at room temperature.

Solution approach

Correct Answer + Core Reason: A metallic solid. Cesium is an alkali metal and exists as a metallic solid at room temperature.

Step-by-Step Solution:

Step 1: Identify Cesium (Cs) as an alkali metal.
- Cesium is in Group 1 of the periodic table, making it an alkali metal.

Step 2: Recall the properties of alkali metals.
- Alkali metals are soft, silvery-white metals that are highly reactive.
- They have low melting and boiling points compared to other metals, but they are still solids at room temperature.

Step 3: Determine the expected elemental form of Cesium at room temperature.
- Since Cesium is a metal, it will form a metallic solid at room temperature.

Why other options are wrong:
- Option A: A network solid - Network solids are typically formed by nonmetals like carbon (diamond) or silicon (silica).
- Option B: A metallic solid - Correct. Cesium is a metal and exists as a metallic solid.
- Option C: Non-polar liquid - Alkali metals are solids at room temperature.
- Option D: An ionic liquid - Ionic liquids are salts that are liquid at relatively low temperatures. Cesium itself is not an ionic liquid.

Key Concepts:
- Alkali metals
- Metallic bonding
- States of matter

Memory Aids:
- Alkali metals are metallic solids at room temperature.

Quick Tips:
- Know the properties of different types of elements (metals, nonmetals, metalloids).

57 Medium

57 of 262Question permalink

Remembering +4 | -1

Coordinate compounds are formed by:

View answer and explanation

Correct option: C

Donation of electron pair

Option C (Donation of electron pair) is correct because coordinate compounds are formed when one atom donates a pair of electrons to another atom to form a coordinate covalent bond.

Solution approach

Correct Answer + Core Reason: Donation of electron pair. Coordinate compounds are formed by the donation of an electron pair from one atom to another.

Step-by-Step Solution:

Step 1: Define coordinate compounds. Coordinate compounds involve a central metal atom or ion bonded to ligands through coordinate covalent bonds.

Step 2: Explain coordinate covalent bonds. In a coordinate covalent bond, one atom (the ligand) donates a pair of electrons to another atom (the central metal) to form the bond.

Step 3: Consider the options:
- Transfer of electrons: This describes ionic bonding.
- Sharing of electrons: This describes regular covalent bonding.
- Donation of electron pair: This describes coordinate covalent bonding.

Why other options are wrong:
- Option A (Transfer of electrons): This describes ionic bonding, not coordinate bonding.
- Option B (Sharing of electrons): This describes regular covalent bonding, not coordinate bonding.
- Option D (None of the above): Coordinate compounds are formed by donation of an electron pair.

Key Concepts:
- Coordinate Compounds
- Coordinate Covalent Bond
- Ligand
- Central Metal Atom

Memory Aids:
- Coordinate = Donation.

Quick Tips:
- Remember that in coordinate compounds, ligands donate electron pairs to the central metal atom.

58 Medium

58 of 262Question permalink

Understanding +4 | -1

The shape of gaseous

\mathrm{SnC}{\mathrm{l}_2}

is

View answer and explanation

Correct option: C

Angular

Option C (Angular) is correct because SnCl2 has two bond pairs and one lone pair, resulting in a bent or angular shape due to the repulsion from the lone pair.

Solution approach

Correct Answer + Core Reason: Angular. SnCl2 has two bond pairs and one lone pair, resulting in a bent shape.

Step-by-Step Solution:

Step 1: Determine the number of valence electrons for Sn and Cl. Sn has 4 valence electrons, and each Cl has 7.

Step 2: Calculate the total number of valence electrons in SnCl2: 4 + 2(7) = 18.

Step 3: Determine the number of electron pairs: 18/2 = 9.

Step 4: Determine the number of bond pairs: 2 (since there are 2 Cl atoms bonded to Sn).

Step 5: Determine the number of lone pairs: (18 - 4)/2 = 1 (one lone pair on Sn).

Step 6: Apply VSEPR theory. With 2 bond pairs and 1 lone pair, the electron geometry is trigonal planar, but the molecular geometry is bent or angular due to the lone pair repulsion.

Why other options are wrong:
- Option A: Tetrahedral would require 4 bond pairs and 0 lone pairs.
- Option B: Linear would require 2 bond pairs and 0 lone pairs.
- Option D: T-shape is not a common geometry for simple molecules like SnCl2.

Key Concepts:
- VSEPR theory.
- Lone pairs and bond pairs.
- Molecular geometry.
- Electron geometry.

Memory Aids:
- VSEPR: Valence Shell Electron Pair Repulsion.

Quick Tips:
- Always determine the number of lone pairs and bond pairs before predicting the shape of a molecule.

59 Medium

59 of 262Question permalink

Application +4 | -1

CaO

and

NaCl

have same crystal structure and approximately same ionic radii. If U is the lattice energy of

NaCl

, the approximate lattice energy of

CaO

is

View answer and explanation

Correct option: C

4U

Option C (4U) is correct because lattice energy is proportional to the product of the charges of the ions. CaO has charges of +2 and -2, while NaCl has charges of +1 and -1. Therefore, the lattice energy of CaO is approximately 4 times that of NaCl.

Solution approach

Correct Answer + Core Reason: 4U. Lattice energy is proportional to the product of the charges of the ions.

Step-by-Step Solution:

Step 1: Recall the factors affecting lattice energy. Lattice energy is directly proportional to the product of the charges of the ions and inversely proportional to the distance between the ions (ionic radii).

Step 2: Compare the charges of the ions in CaO and NaCl.
- CaO: $Ca^{2+}$ and $O^{2-}$ (charges are +2 and -2)
- NaCl: $Na^{+}$ and $Cl^{-}$ (charges are +1 and -1)

Step 3: Since the ionic radii are approximately the same, the difference in lattice energy is mainly due to the difference in the product of charges.

Step 4: Calculate the ratio of the product of charges:
- For CaO: (+2) * (-2) = -4
- For NaCl: (+1) * (-1) = -1
- Ratio: |-4| / |-1| = 4

Step 5: Therefore, the lattice energy of CaO is approximately 4 times that of NaCl. So, the lattice energy of CaO is 4U.

Why other options are wrong:
- Option A (U/2): This would be the case if the product of charges in CaO was half that of NaCl, which is not true.
- Option B (U): This would be the case if the product of charges in CaO was the same as that of NaCl, which is not true.
- Option D (2U): This would be the case if the product of charges in CaO was twice that of NaCl, which is not true.

Key Concepts:
- Lattice energy
- Factors affecting lattice energy (charge and ionic radii)

Memory Aids:
- Remember that higher charges lead to higher lattice energy.

Quick Tips:
- Compare the charges and ionic radii of the compounds to estimate the relative lattice energies.

60 Medium

60 of 262Question permalink

Analyzing +4 | -1

Based on lattice energy and other considerations which one of the following alkali metal chlorides is expected to have the highest melting point?

View answer and explanation

Correct option: C

\mathrm{NaCl}

Option C (NaCl) is correct because, among the given alkali metal chlorides, NaCl has the smallest cation (Na+) after Li+ and thus experiences stronger electrostatic attraction, leading to a higher melting point.

Solution approach

Correct Answer + Core Reason: $\mathrm{NaCl}$ . Melting point of alkali metal chlorides decreases down the group due to increasing ionic size and decreasing lattice energy. Na is smaller than K and Rb, and LiCl is more covalent.

Step-by-Step Solution:

Step 1: Recall the trend of melting points of alkali metal chlorides: Melting point decreases down the group (Li > Na > K > Rb > Cs).

Step 2: Consider the given options: $\mathrm{RbCl}$ , $\mathrm{KCl}$ , $\mathrm{NaCl}$ , $\mathrm{LiCl}$ .

Step 3: Compare the ionic sizes: $Li^+ < Na^+ < K^+ < Rb^+$

Step 4: Consider the anomalous behavior of LiCl: LiCl has a smaller size, but it also has a greater covalent character due to the high polarizing power of $Li^+$ . This reduces its melting point compared to NaCl.

Step 5: Conclusion: NaCl has the highest melting point among the given options because it has a relatively small size and is more ionic than LiCl.

Why other options are wrong:
- Option A: $\mathrm{RbCl}$ : Rb is larger than Na, so lower melting point.
- Option B: $\mathrm{KCl}$ : K is larger than Na, so lower melting point.
- Option D: $\mathrm{LiCl}$ : LiCl has significant covalent character, lowering its melting point compared to NaCl.

Key Concepts:
- Melting point of ionic compounds
- Lattice energy
- Ionic size
- Covalent character

Memory Aids:
- Smaller ions and higher charge lead to higher melting points.

Quick Tips:
- Remember the trend of melting points down the group for ionic compounds.
- Consider the covalent character of Li compounds.

Your Progress

Chemical bonding & molecular structure practice · Medium

Explore Other Chapters in CHEMISTRY

Continue Your Practice

Your practice