Delocalization of ${\rm{\pi }}$-electrons

Delocalization of ${\rm{\sigma }}$-electrons

CHEMICAL BONDING & MOLECULAR STRUCTURE MCQs for NEET UG CHEMISTRY — Easy, Page 7

181 Easy

Understanding +4 | -1

The correct increasing covalent character is:

View answer and explanation

Correct option: A

\mathrm{NaCl} < LiCl < BeC{\mathrm{l}_2}

Option A (

\mathrm{NaCl} < LiCl < BeC{\mathrm{l}_2}

) is correct because covalent character increases with increasing polarization, which is favored by smaller cations and larger anions, and higher charge.

Solution approach

Correct Answer + Core Reason: $\mathrm{NaCl} < LiCl < BeC{\mathrm{l}_2}$ . Covalent character increases with increasing polarization.

Step-by-Step Solution:

Step 1: Understand Fajans' Rules.
- Fajans' Rules describe the factors that influence the covalent character of ionic compounds.
- The rules state that covalent character is favored by:
- Small, highly charged cations
- Large, highly charged anions
- High polarizing power of the cation
- Low polarizability of the anion

Step 2: Apply Fajans' Rules to the given compounds.
- $\mathrm{NaCl}$ : Na+ is a relatively large cation with a +1 charge.
- $\mathrm{LiCl}$ : Li+ is a smaller cation than Na+ with a +1 charge. Therefore, LiCl has more covalent character than NaCl.
- $\mathrm{BeC}{\mathrm{l}_2}$ : Be2+ is a small cation with a +2 charge. The higher charge and smaller size of Be2+ make $\mathrm{BeC}{\mathrm{l}_2}$ more covalent than both NaCl and LiCl.

Step 3: Arrange the compounds in increasing order of covalent character.
- $\mathrm{NaCl} < LiCl < BeC{\mathrm{l}_2}$

Why other options are wrong:
- Option A: $\mathrm{NaCl} < LiCl < BeC{\mathrm{l}_2}$ - Correct order of increasing covalent character.
- Option B: $\mathrm{BeC}{\mathrm{l}_2} < NaCl < LiCl$ - Incorrect order.
- Option C: $\mathrm{BeC}{\mathrm{l}_2} < LiCl < NaCl$ - Incorrect order.
- Option D: $\mathrm{LiCl} < NaCl < BeC{\mathrm{l}_2}$ - Incorrect order.

Key Concepts:
- Fajans' Rules
- Covalent character
- Polarization
- Polarizing power
- Polarizability

Memory Aids:
- Fajans' Rules: Small cation, large anion, high charge = more covalent character

Quick Tips:
- Use Fajans' Rules to compare the covalent character of ionic compounds.

182 Easy

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Understanding +4 | -1

Which of the following will provide the most efficient overlap

View answer and explanation

Correct option: D

sp - sp

Option D (

sp - sp

) is correct because hybrid orbitals provide more efficient overlap than atomic orbitals, and

sp

orbitals have the highest s-character among the given options, leading to stronger and more directional bonds.

Solution approach

Correct Answer + Core Reason: $sp - sp$ . $sp$ hybridized orbitals have the highest s-character and thus provide the most efficient overlap.

Step-by-Step Solution:

Step 1: Understand the concept of orbital overlap and its relation to bond strength.

Step 2: Recognize that hybrid orbitals generally provide more efficient overlap than atomic orbitals due to their directionality.

Step 3: Compare the s-character of the given hybrid orbitals: $sp > sp^2 > sp^3$ .

Step 4: Realize that higher s-character leads to stronger and more directional bonds, resulting in more efficient overlap.

Step 5: Conclude that $sp - sp$ overlap is the most efficient.

Why other options are wrong:
- Option A (s-s): s-orbitals are non-directional and provide less efficient overlap compared to hybrid orbitals.
- Option B (s-p): p-orbitals are directional, but the overlap is less efficient than with hybrid orbitals.
- **Option C ( $sp^2 - sp^2$ ):** $sp^2$ orbitals have less s-character than $sp$ orbitals, leading to less efficient overlap.

Key Concepts:
- Orbital overlap
- Hybridization
- s-character
- Bond strength

Memory Aids:
- Higher s-character = stronger bond

Quick Tips:
- Remember that hybrid orbitals form stronger bonds than atomic orbitals.

183 Easy

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Remembering +4 | -1

Dipole moment is usually expressed in Debye Units (D). The conversion factor is

View answer and explanation

Correct option: D

1 \mathrm{~D}=3.33564 \times 10^{-30} \mathrm{~C} \mathrm{m}

Option D is correct because the conversion factor between Debye and Coulomb meter is 1 D = 3.33564 × 10^-30 C m.

Solution approach

Correct Answer + Core Reason: $1 \mathrm{~D}=3.33564 \times 10^{-30} \mathrm{~C} \mathrm{m}$ . This is the standard conversion factor between Debye and Coulomb meter.

Step-by-Step Solution:

Step 1: Recall the conversion factor between Debye and Coulomb meter. 1 Debye (D) is equal to $3.33564 \times 10^{-30}$ Coulomb meter (C m).

Step 2: Conclusion: The conversion factor is $1 \mathrm{~D}=3.33564 \times 10^{-30} \mathrm{~C} \mathrm{m}$ .

Why other options are wrong:
- Option A: $1 \mathrm{~D}=3.33564 \times 10^{-33} \mathrm{~C} \mathrm{m}$ : This is incorrect because the exponent is wrong.
- Option B: $1 \mathrm{~D}=6.53354 \times 10^{-32} \mathrm{~C} \mathrm{m}$ : This is incorrect because the value and exponent are wrong.
- Option C: $1 \mathrm{~D}=6.53554 \times 10^{-30} \mathrm{~C} \mathrm{m}$ : This is incorrect because the value is wrong.

Key Concepts:
- Dipole Moment
- Debye Unit
- Coulomb Meter

Memory Aids:
- Remember the conversion factor: 1 D = $3.33564 \times 10^{-30}$ C m.

Quick Tips:
- Pay attention to the exponent when converting between units.

184 Easy

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Remembering +4 | -1

Paramagnetism is exhibited by molecules:

View answer and explanation

Correct option: D

Containing unpaired electrons

Option D (Containing unpaired electrons) is correct because paramagnetism arises from the presence of unpaired electrons.

Solution approach

Correct Answer + Core Reason: Containing unpaired electrons. Paramagnetism is due to unpaired electrons.

Step-by-Step Solution:

Step 1: Define paramagnetism.
- Paramagnetism is the property of a substance that is weakly attracted to a magnetic field. This attraction is due to the presence of unpaired electrons.

Step 2: Explain why unpaired electrons cause paramagnetism.
- Unpaired electrons have a net magnetic moment due to their spin. When a magnetic field is applied, these magnetic moments align with the field, resulting in a weak attraction.

Step 3: Explain why paired electrons do not cause paramagnetism.
- Paired electrons have opposite spins, so their magnetic moments cancel each other out. Therefore, substances with only paired electrons are not paramagnetic.

Why other options are wrong:
- Option A: Not attracted into a magnetic field - This describes diamagnetism, not paramagnetism.
- Option B: Containing only paired electrons - Substances with only paired electrons are diamagnetic, not paramagnetic.
- Option C: Carrying a positive charge - Charge is not directly related to paramagnetism. Paramagnetism depends on unpaired electrons.
- Option D: Containing unpaired electrons - Correct. Unpaired electrons cause paramagnetism.

Key Concepts:
- Paramagnetism
- Unpaired electrons
- Magnetic moment
- Diamagnetism

Memory Aids:
- Paramagnetic: Paired electrons absent

Quick Tips:
- Look for unpaired electrons to determine if a substance is paramagnetic.

185 Easy

185 of 220Question permalink

Analyzing +4 | -1

The correct sequence of dipole moments among the chlorides of methane is

View answer and explanation

Correct option: C

C{H_3}Cl > C{H_2}C{l_2} > CHC{l_3} > CC{l_4}

Option C (

C{H_3}Cl > C{H_2}C{l_2} > CHC{l_3} > CC{l_4}

) is correct because as the symmetry decreases from

CH_3Cl

to

CHCl_3

, the dipole moment decreases, and

CCl_4

has zero dipole moment due to its tetrahedral symmetry.

Solution approach

Correct Answer + Core Reason: $C{H_3}Cl > C{H_2}C{l_2} > CHC{l_3} > CC{l_4}$ . The dipole moment decreases with increasing symmetry and cancellation of bond dipoles.

Step-by-Step Solution:

Step 1: Consider the structures of the chlorides of methane.
- $C{H_3}Cl$ : Tetrahedral, but with a net dipole moment due to the C-Cl bond.
- $C{H_2}C{l_2}$ : Tetrahedral, but with a smaller net dipole moment than $C{H_3}Cl$ because the two C-Cl bonds partially cancel each other out.
- $CHC{l_3}$ : Tetrahedral, but with an even smaller net dipole moment than $C{H_2}C{l_2}$ because the three C-Cl bonds partially cancel each other out.
- $CC{l_4}$ : Tetrahedral, symmetrical. The bond dipoles cancel each other out, resulting in a net dipole moment of zero.

Step 2: Analyze the vector sum of bond dipoles.
- In $C{H_3}Cl$ , the three C-H bonds have small dipole moments that add to the C-Cl bond dipole, resulting in a relatively large net dipole moment.
- In $C{H_2}C{l_2}$ , the two C-Cl bond dipoles partially cancel each other out, reducing the net dipole moment.
- In $CHC{l_3}$ , the three C-Cl bond dipoles partially cancel each other out even more, further reducing the net dipole moment.
- In $CC{l_4}$ , the four C-Cl bond dipoles completely cancel each other out, resulting in a zero net dipole moment.

Step 3: Arrange the molecules in decreasing order of dipole moments.
$C{H_3}Cl > C{H_2}C{l_2} > CHC{l_3} > CC{l_4}$

Why other options are wrong:
- Option A: $CHC{l_3} < C{H_2}C{l_2} < C{H_3}Cl < CC{l_4}$ is incorrect because $CCl_4$ has zero dipole moment and the dipole moment decreases from $CH_3Cl$ to $CHCl_3$ .
- Option B: $C{H_2}C{l_2} < C{H_3}Cl < CHC{l_3} < CC{l_4}$ is incorrect because $CCl_4$ has zero dipole moment and the dipole moment decreases from $CH_3Cl$ to $CHCl_3$ .
- Option D: $C{H_2}C{l_2} > CHC{l_3} > C{H_3}Cl > CC{l_4}$ is incorrect because $CH_3Cl$ has a larger dipole moment than both $CH_2Cl_2$ and $CHCl_3$ .

Key Concepts:
- Dipole moment
- Molecular geometry (VSEPR theory)
- Vector addition of bond dipoles
- Symmetry

Memory Aids:
- Remember that symmetrical molecules like $CC{l_4}$ have zero dipole moment.
- Consider the degree of cancellation of bond dipoles.

Quick Tips:
- Draw the Lewis structures and consider the molecular geometry.
- Determine the direction of bond dipoles based on electronegativity differences.
- Consider the effect of symmetry on the net dipole moment.

186 Easy

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Analyzing +4 | -1

Among

2S - 2P

mixing is not operative, the paramagnetic species among the following

View answer and explanation

Correct option: C

{C_2}

Option C (

{C_2}

) is correct because among

2S - 2P

mixing is not operative,

{C_2}

is paramagnetic.

Solution approach

Correct Answer + Core Reason: ${C_2}$ is paramagnetic when 2S-2P mixing is not operative.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each species.
$B{e_2}$ : $(\sigma_{2s})^2(\sigma_{2s}^{\ast})^2$ - Diamagnetic
${B_2}$ : $(\sigma_{2s})^2(\sigma_{2s}^{\ast})^2(\pi_{2p})^2$ - Paramagnetic (with 2s-2p mixing). Without mixing, $(\sigma_{2s})^2(\sigma_{2s}^{\ast})^2(\sigma_{2p})^2$ - Diamagnetic
${C_2}$ : $(\sigma_{2s})^2(\sigma_{2s}^{\ast})^2(\pi_{2p})^4$ - Diamagnetic (with 2s-2p mixing). Without mixing, $(\sigma_{2s})^2(\sigma_{2s}^{\ast})^2(\sigma_{2p})^2(\pi_{2p})^2$ - Paramagnetic
${N_2}$ : $(\sigma_{2s})^2(\sigma_{2s}^{\ast})^2(\pi_{2p})^4(\sigma_{2p})^2$ - Diamagnetic

Step 2: Consider the case where 2S-2P mixing is not operative.\nIn this case, the order of filling changes. For $B_2$ , it becomes diamagnetic. For $C_2$ , it becomes paramagnetic.

Step 3: Identify the paramagnetic species.
${C_2}$ is paramagnetic when 2S-2P mixing is not operative.

Why other options are wrong:
- Option A: $B{e_2}$ is diamagnetic regardless of mixing.
- Option B: ${B_2}$ is diamagnetic without mixing.
- Option D: ${N_2}$ is diamagnetic regardless of mixing.

Key Concepts:
- Molecular Orbital Theory
- 2s-2p Mixing
- Paramagnetism and Diamagnetism

Memory Aids:
- Remember the effect of 2s-2p mixing on the energy levels of molecular orbitals.

Quick Tips:
- Know the electronic configurations with and without 2s-2p mixing.

187 Easy

187 of 220Question permalink

Application +4 | -1

The phosphate of a metal has the formula

{MPO}_4

. The formula of its nitrate will be

View answer and explanation

Correct option: D

{M}\left({NO}_3\right)_3

Option D (

{M}\left({NO}_3\right)_3

) is correct because if MPO4 has a +3 and -3 charge, then M has a +3 charge. Nitrate has a -1 charge, so the formula will be M(NO3)3.

Solution approach

Correct Answer + Core Reason: ${M}\left({NO}_3\right)_3$ . If MPO4 has a +3 and -3 charge, then M has a +3 charge. Nitrate has a -1 charge, so the formula will be M(NO3)3.

Step-by-Step Solution:

Step 1: Determine the charge of the metal M. Since the phosphate is ${MPO}_4$ , and the phosphate ion ( $PO_4$ ) has a -3 charge, the metal M must have a +3 charge.

Step 2: Determine the charge of the nitrate ion. The nitrate ion is $NO_3^-$ , so it has a -1 charge.

Step 3: Balance the charges to form a neutral compound. To balance the +3 charge of M and the -1 charge of nitrate, we need 1 M ion and 3 nitrate ions. This gives a formula of ${M}\left({NO}_3\right)_3$ .

Why other options are wrong:
- **Option A ( ${MNO}_3$ ): This formula would result if M had a +1 charge and nitrate had a -1 charge, which is incorrect.
- Option B ( ${M}_2\left({NO}_3\right)_2$ ): This formula would result if M had a +1 charge and nitrate had a -1 charge, which is incorrect.
- Option C ( ${M}\left({NO}_3\right)_2$ ): This formula would result if M had a +2 charge and nitrate had a -1 charge, which is incorrect.

Key Concepts:
- Ionic compound formation
- Balancing charges

Memory Aids:
- Remember the charge of the phosphate ion is -3 and the nitrate ion is -1.

Quick Tips:**
- Determine the charges of the ions involved and then criss-cross the charges to determine the subscripts in the formula.

188 Easy

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Understanding +4 | -1

A simple example of a coordinate covalent bond is exhibited by:

View answer and explanation

Correct option: D

{H}_2 {SO}_4

Option D (

{H}_2 {SO}_4

) is correct because in sulfuric acid, one of the oxygen atoms forms a coordinate covalent bond with the sulfur atom, where sulfur donates both electrons for the bond.

Solution approach

Correct Answer + Core Reason: ${H}_2 {SO}_4$ . Sulfuric acid exhibits a coordinate covalent bond where one oxygen atom is bonded to sulfur by sharing electrons donated by sulfur only.

Step-by-Step Solution:

Step 1: Understand the definition of a coordinate covalent bond. A coordinate covalent bond is formed when both electrons in the shared pair come from one of the atoms.

Step 2: Draw the Lewis structures of the given compounds.

Step 3: Analyze each compound:
- ${HCl}$ : Hydrogen and chlorine share one electron each to form a covalent bond.
- ${NH}_3$ : Nitrogen and hydrogen share electrons to form covalent bonds. Nitrogen has a lone pair, but it's not involved in a coordinate bond within the molecule itself.
- ${C}_2 {H}_2$ : Carbon and hydrogen share electrons to form covalent bonds.
- ${H}_2 {SO}_4$ : In sulfuric acid, the sulfur atom is bonded to two oxygen atoms via single covalent bonds and to two other oxygen atoms via coordinate covalent bonds (or double bonds). The sulfur atom donates the electron pair to form the bond with these two oxygen atoms.

Step 4: Conclusion: ${H}_2 {SO}_4$ exhibits a coordinate covalent bond.

Why other options are wrong:
- **Option A ( ${HCl}$ ):** ${HCl}$ has a simple covalent bond formed by sharing of electrons between hydrogen and chlorine.
- **Option B ( ${NH}_3$ ):** ${NH}_3$ has covalent bonds between nitrogen and hydrogen. The lone pair on nitrogen can form coordinate bonds with other species (like $H^+$ to form $NH_4^+$ ), but not within the $NH_3$ molecule itself.
- **Option C ( ${C}_2 {H}_2$ ):** ${C}_2 {H}_2$ (acetylene) has covalent bonds between carbon and hydrogen and a triple bond between the two carbon atoms. No coordinate bonds are present.

Key Concepts:
- Coordinate covalent bond
- Lewis structures
- Covalent bond

Memory Aids:
- Coordinate covalent bonds are also called dative bonds.

Quick Tips:
- Look for molecules where one atom has a lone pair and can donate it to another atom.

189 Easy

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Analyzing +4 | -1

Bond energy ……..with the increase in number of lone pairs on the bonded atoms.

View answer and explanation

Correct option: C

Does not change

Option A (Decreases) is incorrect. Bond energy decreases with the increase in the number of lone pairs on the bonded atoms due to increased repulsion.

Solution approach

Correct Answer + Core Reason: Decreases. Bond energy decreases with the increase in the number of lone pairs on the bonded atoms due to increased repulsion.

Step-by-Step Solution:

Step 1: Recall the effect of lone pairs on bond energy.

Step 2: Lone pairs on bonded atoms cause repulsion between the bonding electrons and the lone pairs.

Step 3: This repulsion weakens the bond, thus decreasing the bond energy.

Why other options are wrong:
- Option B: Increases: Incorrect, lone pairs decrease bond energy due to repulsion.
- Option C: Does not change: Incorrect, lone pairs do affect bond energy.
- Option D: None of these: Incorrect, option A is correct.

Key Concepts:
- Bond energy
- Lone pair repulsion

Memory Aids:
- Lone pairs repel bonding electrons, weakening the bond.

Quick Tips:
- Consider the effect of lone pairs when comparing bond energies.

190 Easy

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Remembering +4 | -1

A bond formed by transfer of electrons between two atoms is

View answer and explanation

Correct option: A

Electrovalent bond

Option A (Electrovalent bond) is correct because an electrovalent bond (also called ionic bond) is formed by the transfer of electrons between two atoms.

Solution approach

Correct Answer + Core Reason: Electrovalent bond. Electrovalent bonds are formed by the transfer of electrons between two atoms.

Step-by-Step Solution:

Step 1: Define electrovalent bond.

Step 2: An electrovalent bond (or ionic bond) is formed by the complete transfer of one or more electrons from one atom to another, resulting in the formation of ions (cations and anions) that are held together by electrostatic attraction.

Why other options are wrong:
- Option B: Covalent bonds are formed by sharing of electrons.
- Option C: Hydrogen bonds are weak bonds between hydrogen and electronegative atoms.
- Option D: Dative bonds (coordinate covalent bonds) are formed when one atom donates both electrons to form a bond.

Key Concepts:
- Electrovalent bond (Ionic bond)
- Covalent bond
- Hydrogen bond
- Dative bond

Memory Aids:
- Electrovalent = electron transfer.

Quick Tips:
- Remember the different types of chemical bonds.

191 Easy

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Understanding +4 | -1

Resonance is due to:

View answer and explanation

Correct option: B

Delocalization of

\mathrm{\pi }

-electrons

Option B (Delocalization of

\mathrm{\pi }

-electrons) is correct because resonance involves the movement of pi electrons, not sigma electrons or atoms.

Solution approach

Correct Answer + Core Reason: Delocalization of $\mathrm{\pi }$ -electrons. Resonance involves the movement of pi electrons.

Step-by-Step Solution:

Step 1: Define resonance.
- Resonance is a phenomenon where the actual structure of a molecule is represented by a combination of two or more Lewis structures, called resonance structures or canonical forms.

Step 2: Explain the role of electrons in resonance.
- Resonance occurs due to the delocalization of electrons, usually $\mathrm{\pi }$ electrons, over multiple atoms.
- The delocalization of electrons stabilizes the molecule.

Step 3: Explain why $\mathrm{\sigma }$ electrons are not involved in resonance.
- $\mathrm{\sigma }$ electrons are involved in single bonds and are localized between two atoms. They do not participate in resonance.

Why other options are wrong:
- Option A: Delocalization of $\mathrm{\sigma }$ -electrons - Sigma electrons are localized and do not participate in resonance.
- Option B: Delocalization of $\mathrm{\pi }$ -electrons - Correct. Resonance is due to the delocalization of pi electrons.
- Option C: Migration of $H$ atoms - Migration of H atoms is tautomerism, not resonance.
- Option D: Migration of protons - Migration of protons is not resonance.

Key Concepts:
- Resonance
- Delocalization
- $\mathrm{\pi }$ electrons
- $\mathrm{\sigma }$ electrons

Memory Aids:
- Resonance = Pi electron movement

Quick Tips:
- Resonance involves the movement of electrons in pi bonds or lone pairs.

192 Easy

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Remembering +4 | -1

In ethene, the bond angle(s) is/are

View answer and explanation

Correct option: B

120^{\circ}

Option B (

120^{\circ}

) is correct because ethene (

C_2H_4

) has a trigonal planar geometry around each carbon atom due to

sp^2

hybridization, resulting in bond angles of approximately 120 degrees.

Solution approach

Correct Answer + Core Reason: $120^{\circ}$ . Ethene has $sp^2$ hybridization, leading to a trigonal planar geometry.

Step-by-Step Solution:

Step 1: Recall the structure of ethene ( $C_2H_4$ ).

Step 2: Each carbon atom in ethene is $sp^2$ hybridized.

Step 3: $sp^2$ hybridization leads to a trigonal planar geometry with bond angles of approximately 120 degrees.

Step 4: Final Answer: $120^{\circ}$

Why other options are wrong:
- Option A: $109^{\circ} 28^{\prime}$ corresponds to tetrahedral geometry ( $sp^3$ hybridization).
- Option C: $180^{\circ}$ corresponds to linear geometry ( $sp$ hybridization).
- Option D: The bond angles in ethene are approximately equal.

Key Concepts:
- Hybridization
- Ethene structure
- $sp^2$ hybridization
- Bond angle

Memory Aids:
- $sp^2$ : 120 degrees
- $sp^3$ : 109.5 degrees

Quick Tips:
- Determine the hybridization of the central atom.
- Relate the hybridization to the bond angle.

193 Easy

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Analyzing +4 | -1

Which does not have pyramidal geometry?

View answer and explanation

Correct option: B

\mathrm{NO}_3^ -

Option B (

\mathrm{NO}_3^ -

) is correct because it has a trigonal planar geometry, not pyramidal.

Solution approach

Correct Answer + Core Reason: $\mathrm{NO}_3^ -$ does not have a pyramidal geometry; it has a trigonal planar geometry.

Step-by-Step Solution:

Step 1: Determine the Lewis structure and VSEPR theory for each species.
$\mathrm{SO}_3^{2 - }$ : Sulfur has 6 valence electrons + 2 (from the charge) = 8 electrons. It forms 3 bonds with oxygen and has one lone pair. Tetrahedral electron geometry, pyramidal molecular geometry.
$\mathrm{NO}_3^ -$ : Nitrogen has 5 valence electrons + 1 (from the charge) = 6 electrons. It forms 3 bonds with oxygen (through resonance). No lone pairs. Trigonal planar electron geometry, trigonal planar molecular geometry.
$\mathrm{N}{\mathrm{H}_3}$ : Nitrogen has 5 valence electrons. It forms 3 bonds with hydrogen and has one lone pair. Tetrahedral electron geometry, pyramidal molecular geometry.
$\mathrm{C}\left( {{\mathrm{C}_6}{\mathrm{H}_5}} \right)_3^ -$ : Carbon has 4 valence electrons + 1 (from the charge) = 5 electrons. It forms 3 bonds with the phenyl groups and has one lone pair. Tetrahedral electron geometry, pyramidal molecular geometry.

Step 2: Identify the species with a non-pyramidal geometry.
$\mathrm{NO}_3^ -$ has a trigonal planar geometry.

Why other options are wrong:
- Option A: $\mathrm{SO}_3^{2 - }$ has a pyramidal geometry.
- Option C: $\mathrm{N}{\mathrm{H}_3}$ has a pyramidal geometry.
- Option D: $\mathrm{C}\left( {{\mathrm{C}_6}{\mathrm{H}_5}} \right)_3^ -$ has a pyramidal geometry.

Key Concepts:
- Lewis Structures
- VSEPR Theory
- Molecular Geometry

Memory Aids:
- Remember the common geometries and their corresponding bond angles.

Quick Tips:
- Draw the Lewis structure and apply VSEPR theory to determine the geometry.

194 Easy

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Understanding +4 | -1

Formation of

\pi

-bond:

View answer and explanation

Correct option: B

Decreases bond length

Option B is correct because the formation of a

\pi

-bond adds electron density between the atoms, increasing the attraction and decreasing the bond length.

Solution approach

Correct Answer + Core Reason: Decreases bond length. The formation of a pi-bond increases the electron density between the atoms, leading to a shorter bond length.

Step-by-Step Solution:

Step 1: Understand the relationship between bond order and bond length.

Step 2: Recall that a sigma bond is formed first, and then pi bonds are formed.

Step 3: The formation of a pi bond increases the bond order between the atoms.

Step 4: As the bond order increases, the bond length decreases due to increased attraction between the atoms.

Step 5: Therefore, the formation of a $\pi$ -bond decreases bond length.

Why other options are wrong:
- Option A: Increases bond length is incorrect. Pi bonds decrease bond length.
- Option C: Distorts the geometry of molecule is incorrect. Pi bonds can influence geometry, but they primarily decrease bond length.
- Option D: Makes homoatomic molecules more reactive is incorrect. Pi bonds can increase reactivity, but the primary effect discussed here is bond length.

Key Concepts:
- Bond order
- Bond length
- Sigma bond
- Pi bond

Memory Aids:
- Higher bond order = Shorter bond length

Quick Tips:
- Remember that multiple bonds (pi bonds) shorten the bond length.

195 Easy

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Application +4 | -1

Which of the following is a polar molecule

View answer and explanation

Correct option: A

S{F_4}

Option A (

S{F_4}

) is correct because it has a see-saw shape due to one lone pair, resulting in a net dipole moment.

Solution approach

Correct Answer + Core Reason: $S{F_4}$ . It has a see-saw shape due to one lone pair.

Step-by-Step Solution:

Step 1: Determine the geometry of each molecule.

Step 2: $S{F_4}$ : Sulfur has 4 bond pairs and 1 lone pair, so it has a see-saw shape and is polar.

Step 3: $Si{F_4}$ : Silicon has 4 bond pairs and no lone pairs, so it is tetrahedral and non-polar.

Step 4: $Xe{F_4}$ : Xenon has 4 bond pairs and 2 lone pairs, so it is square planar and non-polar.

Step 5: $B{F_3}$ : Boron has 3 bond pairs and no lone pairs, so it is trigonal planar and non-polar.

Why other options are wrong:
- Option B: $Si{F_4}$ is non-polar due to its tetrahedral geometry.
- Option C: $Xe{F_4}$ is non-polar due to its square planar geometry.
- Option D: $B{F_3}$ is non-polar due to its trigonal planar geometry.

Key Concepts:
- Polarity
- Molecular geometry
- VSEPR theory

Memory Aids:
- Lone pairs often lead to polar molecules.

Quick Tips:
- Draw the structure and consider the symmetry to determine polarity.

196 Easy

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Remembering +4 | -1

The bond angle in ethylene is

View answer and explanation

Correct option: B

120^{\circ}

Option B (

120^{\circ}

) is correct because ethylene (

C_2H_4

) has a double bond between the carbon atoms, resulting in

sp^2

hybridization and a trigonal planar geometry around each carbon with bond angles of 120 degrees.

Solution approach

Correct Answer + Core Reason: $120^{\circ}$ . Ethylene has $sp^2$ hybridized carbons with 120-degree bond angles.

Step-by-Step Solution:

Step 1: Understand the question. We need to determine the bond angle in ethylene.

Step 2: Draw the structure of ethylene ( $C_2H_4$ ). The carbon atoms are double-bonded to each other, and each carbon is bonded to two hydrogen atoms.

Step 3: Determine the hybridization of the carbon atoms. Each carbon has three sigma bonds (one C-C and two C-H), so it is $sp^2$ hybridized.

Step 4: Recall that $sp^2$ hybridization corresponds to a trigonal planar geometry with bond angles of $120^{\circ}$ .

Why other options are wrong:
- Option A: $180^{\circ}$ corresponds to linear geometry ( $sp$ hybridization).
- Option C: $109^{\circ}$ corresponds to tetrahedral geometry ( $sp^3$ hybridization).
- Option D: $90^{\circ}$ is not a typical bond angle for simple hybridization schemes.

Key Concepts:
- Hybridization
- Geometry of molecules
- Bond angles

Memory Aids:
- $sp^2$ = 3 things attached = 120 degrees

Quick Tips:
- Draw the structure and determine the hybridization to find the bond angle.

197 Easy

197 of 220Question permalink

Understanding +4 | -1

Which of the following is false?

View answer and explanation

Correct option: D

Alkyl halides have higher b.p. than corresponding alcohols

Option D (Alkyl halides have higher b.p. than corresponding alcohols) is correct because alcohols generally have higher boiling points than alkyl halides due to hydrogen bonding.

Solution approach

Correct Answer + Core Reason: Alkyl halides have higher b.p. than corresponding alcohols. Alcohols have higher boiling points due to hydrogen bonding.

Step-by-Step Solution:

Step 1: Analyze each statement:
- A: Glycerol has strong hydrogen bonding due to the presence of three -OH groups. This is true.
- B: Glycol is a poisonous alcohol. This is true (ethylene glycol is poisonous).
- C: Waxes are esters of higher alcohols with higher acids. This is true.
- D: Alkyl halides have higher b.p. than corresponding alcohols. This is generally false. Alcohols have higher boiling points due to hydrogen bonding.

Step 2: Compare the intermolecular forces in alkyl halides and alcohols. Alcohols have hydrogen bonding, while alkyl halides have dipole-dipole interactions and London dispersion forces. Hydrogen bonding is stronger.

Step 3: Conclude that statement D is false.

Why other options are wrong:
- Option A: Glycerol does indeed have strong hydrogen bonding due to its multiple -OH groups.
- Option B: Glycol (ethylene glycol) is a well-known poisonous alcohol.
- Option C: Waxes are correctly described as esters of higher alcohols and higher acids.
- Option D: Alkyl halides generally have lower boiling points than corresponding alcohols due to the absence of hydrogen bonding.

Key Concepts:
- Hydrogen bonding.
- Boiling point.
- Intermolecular forces.
- Properties of alcohols and alkyl halides.

Memory Aids:
- Alcohols have higher boiling points than alkyl halides due to hydrogen bonding.

Quick Tips:
- Remember that hydrogen bonding significantly increases the boiling point of a compound.

198 Easy

198 of 220Question permalink

Analyzing +4 | -1

Which of the following pairs of species have identical shapes

View answer and explanation

Correct option: C

Xe{F_4}

and

ICl_4^ -

Option C (

Xe{F_4}

and

ICl_4^ -

) is correct because both have a square planar shape with 4 bond pairs and 2 lone pairs around the central atom.

Solution approach

Correct Answer + Core Reason: $Xe{F_4}$ and $ICl_4^ -$ have identical shapes (square planar) because both have 4 bond pairs and 2 lone pairs around the central atom.

Step-by-Step Solution:

Step 1: Determine the shape of each species using VSEPR theory.
$NO_2^ +$ : Linear
$NO_2^ -$ : Bent
$PC{l_5}$ : Trigonal bipyramidal
$Br{F_5}$ : Square pyramidal
$Xe{F_4}$ : Square planar
$ICl_4^ -$ : Square planar
$TeC{l_4}$ : See-saw
$Xe{O_4}$ : Tetrahedral

Step 2: Identify the pair with identical shapes.
$Xe{F_4}$ and $ICl_4^ -$ both have a square planar shape.

Why other options are wrong:
- Option A: $NO_2^ + \mathrm{ }$ and $NO_2^ -$ : Linear and bent, respectively.
- Option B: $PC{l_5}$ and $Br{F_5}$ : Trigonal bipyramidal and square pyramidal, respectively.
- Option D: $TeC{l_4}$ and $Xe{O_4}$ : See-saw and tetrahedral, respectively.

Key Concepts:
- VSEPR theory
- Molecular shapes

Memory Aids:
- VSEPR theory helps predict shapes based on electron pair repulsion.

Quick Tips:
- Determine the number of bond pairs and lone pairs to predict the shape.

199 Easy

199 of 220Question permalink

Analyzing +4 | -1

In which of the following molecules / ions all the bonds are not equal

View answer and explanation

Correct option: B

S{F_4}

Option B (

S{F_4}

) is correct because it has a see-saw shape due to one lone pair, resulting in unequal bond lengths.

Solution approach

Correct Answer + Core Reason: $S{F_4}$ . Due to the presence of a lone pair, the geometry is see-saw, resulting in unequal bond lengths.

Step-by-Step Solution:

Step 1: Determine the geometry of each molecule/ion using VSEPR theory.
- $B{F_4}^ -$ : Tetrahedral, all bonds are equal.
- $S{F_4}$ : See-saw shape, due to one lone pair, resulting in unequal bond lengths.
- $Si{F_4}$ : Tetrahedral, all bonds are equal.
- $Xe{F_4}$ : Square planar, all bonds are equal.

Step 2: Identify the molecule/ion with unequal bond lengths.

Why other options are wrong:
- **Option A ( $B{F_4}^ -$ ):** $B{F_4}^ -$ has a tetrahedral geometry with four identical B-F bonds. All bond lengths are equal.
- **Option C ( $Si{F_4}$ ):** $Si{F_4}$ has a tetrahedral geometry with four identical Si-F bonds. All bond lengths are equal.
- **Option D ( $Xe{F_4}$ ):** $Xe{F_4}$ has a square planar geometry with four identical Xe-F bonds. All bond lengths are equal.

Key Concepts:
- VSEPR theory
- Molecular geometry
- Bond length

Memory Aids:
- Tetrahedral: 4 bond pairs, no lone pairs
- See-saw: 4 bond pairs, 1 lone pair
- Square planar: 4 bond pairs, 2 lone pairs

Quick Tips:
- Draw the Lewis structure to determine the number of bond pairs and lone pairs.
- Use VSEPR theory to predict the geometry.

200 Easy

200 of 220Question permalink

Application +4 | -1

A diatomic molecule has dipole moment of 1.2 D. If the bond distance is 1A, what percentage of electronic charge exists on each atom

View answer and explanation

Correct option: C

25\% \,of\,{e^ - }

Option C (

25\% \,of\,{e^ - }

) is correct because the percentage of electronic charge is calculated using the formula

\mu = q \times d

, where

\mu

is the dipole moment,

q

is the charge, and

d

is the bond distance.

Solution approach

Correct Answer + Core Reason: $25\% \,of\,{e^ - }$ . The percentage of electronic charge is calculated using the dipole moment formula.

Step-by-Step Solution:

Step 1: Convert the dipole moment from Debye to Coulomb-meters.
1 D = $3.336 \times 10^{-30}$ C m
$\mu = 1.2 \,D = 1.2 \times 3.336 \times 10^{-30} \,C \cdot m = 4.0032 \times 10^{-30} \,C \cdot m$

Step 2: Convert the bond distance from Angstroms to meters.
1 A = $10^{-10}$ m
$d = 1 \,A = 10^{-10} \,m$

Step 3: Calculate the charge (q) using the formula $\mu = q \times d$ .
$q = \frac{{\mu }}{d} = \frac{{4.0032 \times 10^{-30} \,C \cdot m}}{{10^{-10} \,m}} = 4.0032 \times 10^{-20} \,C$

Step 4: Calculate the charge of one electron.≠ = $1.602 \times 10^{-19}$ C

Step 5: Calculate the percentage of electronic charge.\nPercentage = $\frac{q}{e} \times 100 = \frac{{4.0032 \times 10^{-20} \,C}}{{1.602 \times 10^{-19} \,C}} \times 100 = 25.0\%$ \,of\, ${e^-}$

Why other options are wrong:
- **Option A ( $12\% \,of\,{e^ - }$ ): This value is incorrect due to errors in unit conversion or formula application.
- Option B ( $19\% \,of\,{e^ - }$ ): This value is incorrect due to errors in unit conversion or formula application.
- Option D ( $29\% \,of\,{e^ - }$ ): This value is incorrect due to errors in unit conversion or formula application.

Key Concepts:**
- Dipole moment calculation: $\mu = q \times d$
- Unit conversions (Debye to Coulomb-meter, Angstrom to meter)
- Electronic charge

Memory Aids:
- Remember the formula $\mu = q \times d$ .

Quick Tips:
- Pay attention to unit conversions.

201 Easy

201 of 220Question permalink

Analyzing +4 | -1

Which of the following molecules has the maximum dipole moment

View answer and explanation

Correct option: C

N{H_3}

Option C (

N{H_3}

) is correct because

NH_3

has a pyramidal structure with a lone pair, resulting in a significant dipole moment due to the additive effect of bond dipoles and the lone pair dipole.

Solution approach

Correct Answer + Core Reason: $N{H_3}$ . Ammonia has a significant dipole moment due to its geometry and lone pair.

Step-by-Step Solution:

Step 1: Analyze the structure of each molecule.
- $C{O_2}$ : Linear, symmetrical. The bond dipoles cancel each other out, resulting in a net dipole moment of zero.
- $C{H_4}$ : Tetrahedral, symmetrical. The bond dipoles cancel each other out, resulting in a net dipole moment of zero.
- $N{H_3}$ : Pyramidal. The bond dipoles and the lone pair dipole add up, resulting in a significant net dipole moment.
- $N{F_3}$ : Pyramidal. The bond dipoles oppose the lone pair dipole, resulting in a smaller net dipole moment compared to $N{H_3}$ .

Step 2: Consider the direction of bond dipoles.
- In $N{H_3}$ , both the N-H bond dipoles and the lone pair dipole point in the same direction, increasing the net dipole moment.
- In $N{F_3}$ , the N-F bond dipoles point away from the nitrogen atom, while the lone pair dipole points towards the nitrogen atom. These oppose each other, reducing the net dipole moment.

Step 3: Compare the magnitudes of dipole moments.
$N{H_3}$ has the largest dipole moment because the bond dipoles and lone pair dipole reinforce each other.
$N{F_3}$ has a smaller dipole moment because the bond dipoles and lone pair dipole oppose each other.
$C{O_2}$ and $C{H_4}$ have zero dipole moment due to their symmetrical structures.

Step 4: Therefore, $N{H_3}$ has the maximum dipole moment.

Why other options are wrong:
- Option A: $C{O_2}$ has a linear and symmetrical structure, resulting in a zero dipole moment.
- Option B: $C{H_4}$ has a tetrahedral and symmetrical structure, resulting in a zero dipole moment.
- Option D: $N{F_3}$ has a pyramidal structure, but the bond dipoles and lone pair dipole oppose each other, resulting in a smaller dipole moment compared to $N{H_3}$ .

Key Concepts:
- Dipole moment
- Molecular geometry (VSEPR theory)
- Vector addition of bond dipoles
- Lone pair dipole

Memory Aids:
- Remember that symmetrical molecules like $C{O_2}$ and $C{H_4}$ have zero dipole moment.
- Consider the direction of bond dipoles and lone pair dipoles.

Quick Tips:
- Draw the Lewis structures and consider the molecular geometry.
- Determine the direction of bond dipoles based on electronegativity differences.
- Consider the effect of lone pairs on the dipole moment.

202 Easy

202 of 220Question permalink

Analyzing +4 | -1

Which of the following contains atleast one lone pair of electrons in all of its halides?

View answer and explanation

Correct option: A

{Xe}

Option A (

{Xe}

) is correct because Xenon forms halides such as

XeF_2

,

XeF_4

, and

XeF_6

, all of which have at least one lone pair of electrons on the central Xe atom.

Solution approach

Correct Answer + Core Reason: ${Xe}$ . Xenon halides always have at least one lone pair of electrons on the central Xe atom.

Step-by-Step Solution:

Step 1: Consider the halides of Xenon.

Step 2: Xenon forms halides such as $XeF_2$ , $XeF_4$ , and $XeF_6$ .

Step 3: In $XeF_2$ , Xe has 8 valence electrons, 2 are used for bonding, leaving 3 lone pairs.

Step 4: In $XeF_4$ , Xe has 8 valence electrons, 4 are used for bonding, leaving 2 lone pairs.

Step 5: In $XeF_6$ , Xe has 8 valence electrons, 6 are used for bonding, leaving 1 lone pair.

Step 6: Consider the halides of Selenium.

Step 7: Selenium forms halides such as $SeF_6$ . In $SeF_6$ , Se has 6 valence electrons, all 6 are used for bonding, leaving 0 lone pairs.

Step 8: Consider the halides of Chromium.

Step 9: Chromium forms halides such as $CrF_6$ . In $CrF_6$ , Cr has 6 valence electrons, all 6 are used for bonding, leaving 0 lone pairs.

Step 10: Consider the halides of Nitrogen.

Step 11: Nitrogen forms halides such as $NF_3$ . In $NF_3$ , N has 5 valence electrons, 3 are used for bonding, leaving 1 lone pair. However, $NCl_5$ does not exist.

Why other options are wrong:
- Option B: Selenium can form halides like $SeF_6$ which have no lone pairs on the central atom.
- Option C: Chromium can form halides like $CrF_6$ which have no lone pairs on the central atom.
- Option D: Nitrogen does not form pentahalides.

Key Concepts:
- Lewis structures
- Lone pairs
- Halides

Memory Aids:
- Noble gases can form compounds with highly electronegative elements like F and O.

Quick Tips:
- Draw the Lewis structure of the halide to determine the number of lone pairs on the central atom.

203 Easy

203 of 220Question permalink

Understanding +4 | -1

During the formation of a chemical bond

View answer and explanation

Correct option: D

Energy decreases

Option D (Energy decreases) is correct because the formation of a chemical bond is an exothermic process, resulting in a decrease in the energy of the system.

Solution approach

Correct Answer + Core Reason: Energy decreases during chemical bond formation.

Step-by-Step Solution:

Step 1: When a chemical bond is formed between two atoms, energy is released.

Step 2: This release of energy indicates that the system (the bonded atoms) has reached a lower energy state compared to the separated atoms.

Step 3: Therefore, the formation of a chemical bond is an exothermic process, and the energy of the system decreases.

Why other options are wrong:
- Option A: Electron-electron repulsion is always present, but the net effect of bond formation is a decrease in energy due to increased nucleus-electron attraction.
- Option B: The energy of the system always changes during bond formation.
- Option C: Energy increases when a bond is broken, not when it is formed.

Key Concepts:
- Chemical bond formation
- Exothermic process
- Energy change

Memory Aids:
- Bond formation = Energy release = Lower energy

Quick Tips:
- Remember that bond formation is an exothermic process.

204 Easy

204 of 220Question permalink

Understanding +4 | -1

Which has zero dipole moment?

View answer and explanation

Correct option: C

\mathrm{Si}{\mathrm{F}_4}

Option C is correct because SiF4 has a symmetrical tetrahedral structure, causing the bond dipoles to cancel out, resulting in a zero dipole moment.

Solution approach

Correct Answer + Core Reason: $\mathrm{Si}{\mathrm{F}_4}$ . Silicon tetrafluoride has a tetrahedral shape with symmetrical arrangement of fluorine atoms, resulting in zero dipole moment.

Step-by-Step Solution:

Step 1: Analyze the molecular structure of each option.

Step 2: $\mathrm{ClF}$ : Chlorine monofluoride is a diatomic molecule with a polar bond, so it has a dipole moment.

Step 3: $\mathrm{PC}{\mathrm{l}_3}$ : Phosphorus trichloride has a pyramidal shape, so it has a dipole moment.

Step 4: $\mathrm{Si}{\mathrm{F}_4}$ : Silicon tetrafluoride is tetrahedral and symmetrical, so the bond dipoles cancel out, resulting in zero dipole moment.

Step 5: $\mathrm{CFC}{\mathrm{l}_3}$ : Trichlorofluoromethane is tetrahedral but asymmetrical, so it has a dipole moment.

Step 6: Conclusion: $\mathrm{Si}{\mathrm{F}_4}$ has zero dipole moment.

Why other options are wrong:
- Option A: $\mathrm{ClF}$ : This is incorrect because ClF has a dipole moment.
- Option B: $\mathrm{PC}{\mathrm{l}_3}$ : This is incorrect because PCl3 has a dipole moment.
- Option D: $\mathrm{CFC}{\mathrm{l}_3}$ : This is incorrect because CFCl3 has a dipole moment.

Key Concepts:
- Dipole Moment
- Molecular Geometry
- Symmetry

Memory Aids:
- Remember that symmetrical molecules with identical bonds are nonpolar.

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

205 Easy

205 of 220Question permalink

Analyzing +4 | -1

Which one species has the longest bond length?

View answer and explanation

Correct option: B

\mathrm{O}_2^ -

Option B (

\mathrm{O}_2^ -

) is correct because it has the lowest bond order among the given species, leading to the longest bond length.

Solution approach

Correct Answer + Core Reason: $\mathrm{O}_2^ -$ . It has the lowest bond order, leading to the longest bond length.

Step-by-Step Solution:

Step 1: Calculate the number of electrons in each species:
- $\mathrm{N}{\mathrm{O}^ + }$ : 7 + 8 - 1 = 14 electrons
- $\mathrm{O}_2^ -$ : 16 + 1 = 17 electrons
- $\mathrm{O}_2^ +$ : 16 - 1 = 15 electrons
- $\mathrm{N}_2^ +$ : 14 - 1 = 13 electrons

Step 2: Write the electronic configuration for each species and calculate the bond order:
- $\mathrm{N}{\mathrm{O}^ + }$ : $(\sigma_{2s})(\sigma_{2s}^{\,*})(\sigma_{2p})(\pi_{2p})$ - Bond order = (10-4)/2 = 3
- $\mathrm{O}_2^ -$ : $(\sigma_{2s})(\sigma_{2s}^{\,*})(\sigma_{2p})(\pi_{2p})(\pi_{2p}^{\,*})^3$ - Bond order = (10-7)/2 = 1.5
- $\mathrm{O}_2^ +$ : $(\sigma_{2s})(\sigma_{2s}^{\,*})(\sigma_{2p})(\pi_{2p})(\pi_{2p}^{\,*})^1$ - Bond order = (10-5)/2 = 2.5
- $\mathrm{N}_2^ +$ : $(\sigma_{2s})(\sigma_{2s}^{\,*})(\sigma_{2p})(\pi_{2p})^3$ - Bond order = (9-4)/2 = 2.5

Step 3: Recall that bond length is inversely proportional to bond order.

Step 4: Compare the bond orders and determine the species with the lowest bond order. $\mathrm{O}_2^ -$ has the lowest bond order (1.5).

Step 5: Conclude that $\mathrm{O}_2^ -$ has the longest bond length.

Why other options are wrong:
- Option A: $\mathrm{N}{\mathrm{O}^ + }$ : Higher bond order, shorter bond length.
- Option C: $\mathrm{O}_2^ +$ : Higher bond order, shorter bond length.
- Option D: $\mathrm{N}_2^ +$ : Higher bond order, shorter bond length.

Key Concepts:
- Molecular orbital theory
- Bond order calculation
- Relationship between bond order and bond length

Memory Aids:
- Higher bond order means shorter bond length

Quick Tips:
- Calculate the bond order for each species to determine the relative bond lengths.

206 Easy

206 of 220Question permalink

Remembering +4 | -1

Bond angle in

{H}_2 {O}

is:

View answer and explanation

Correct option: C

104.5^0

Option C (

104.5^0

) is correct because the bond angle in water is approximately 104.5 degrees due to the two lone pairs on the oxygen atom repelling the bonding pairs.

Solution approach

Correct Answer + Core Reason: $104.5^0$ . The bond angle in water is influenced by the repulsion of lone pairs.

Step-by-Step Solution:

Step 1: Water ( $H_2O$ ) has two bonding pairs and two lone pairs around the central oxygen atom.

Step 2: The electron pair geometry is tetrahedral, but the molecular geometry is bent due to the lone pair repulsions.

Step 3: Lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsion.

Step 4: The bond angle is reduced from the ideal tetrahedral angle of $109.5^0$ to approximately $104.5^0$ .

Why other options are wrong:
- Option A: $109^{\circ} 28^{\prime}$ is the tetrahedral bond angle, which is not the case for water due to lone pair repulsion.
- Option B: $107^0 10^{\prime}$ is close to the bond angle in ammonia, not water.
- Option D: $92^{\circ}$ is too small for the bond angle in water.

Key Concepts:
- VSEPR theory
- Lone pair repulsion
- Molecular geometry

Memory Aids:
- Water is bent, not linear!

Quick Tips:
- Remember that lone pairs affect bond angles.

207 Easy

207 of 220Question permalink

Analyzing +4 | -1

Which of the following sets of species does not follow octet rule?

View answer and explanation

Correct option: C

\mathrm{AlCl}_3, \mathrm{BF}_3, \mathrm{PCl}_5, \mathrm{SF}_6

Option C (

\mathrm{AlCl}_3, \mathrm{BF}_3, \mathrm{PCl}_5, \mathrm{SF}_6

) is correct because in these molecules, Al and B have less than 8 electrons, while P and S have more than 8 electrons around the central atom.

Solution approach

Correct Answer + Core Reason: $\mathrm{AlCl}_3, \mathrm{BF}_3, \mathrm{PCl}_5, \mathrm{SF}_6$ . These molecules have central atoms with incomplete or expanded octets.

Step-by-Step Solution:

Step 1: Draw the Lewis structure for each molecule/ion in each set.

Step 2: Count the number of electrons around the central atom in each molecule/ion.
- Set A: CO (8), PCl5 (10), PCl3 (8), AlCl3 (6)
- Set B: CO (8), CCl4 (8), NH3 (8), H2O (8)
- Set C: AlCl3 (6), BF3 (6), PCl5 (10), SF6 (12)
- Set D: H2O (8), CCl4 (8), CO2 (8), AlCl3 (6)

Step 3: Identify the set where all species do not follow the octet rule (i.e., have less than or more than 8 electrons around the central atom).

Why other options are wrong:
- **Option A ( $\mathrm{CO}, \mathrm{PCl}_5, \mathrm{PCl}_3, \mathrm{AlCl}_3$ ): CO and PCl3 follow the octet rule.
- Option B ( $\mathrm{CO}, \mathrm{CCl}_4, \mathrm{NH}_3, \mathrm{H}_2 \mathrm{O}$ ): All molecules in this set follow the octet rule.
- Option D ( $\mathrm{H}_2 \mathrm{O}, \mathrm{CCl}_4, \mathrm{CO}_2, \mathrm{AlCl}_3$ ): H2O, CCl4, and CO2 follow the octet rule.

Key Concepts:
- Octet rule
- Lewis structures

Memory Aids:
- Octet rule: atoms tend to gain, lose, or share electrons to achieve a full outer shell of 8 electrons.

Quick Tips:**
- Draw Lewis structures to determine the number of electrons around the central atom.
- Be aware of exceptions to the octet rule, such as electron-deficient and expanded octets.

208 Easy

208 of 220Question permalink

Remembering +4 | -1

Which of the following silver salts is insoluble in water?

View answer and explanation

Correct option: B

\mathrm{A}{\mathrm{g}_2}\mathrm{S}{\mathrm{O}_4}

Option B (

\mathrm{A}{\mathrm{g}_2}\mathrm{S}{\mathrm{O}_4}

) is correct because silver sulfate is known to be sparingly soluble in water, while the other options are generally soluble.

Solution approach

Correct Answer + Core Reason: $\mathrm{A}{\mathrm{g}_2}\mathrm{S}{\mathrm{O}_4}$ . Silver sulfate is sparingly soluble in water, while other silver salts like perchlorate, fluoride, and nitrate are generally soluble.

Step-by-Step Solution:

Step 1: Recall the solubility rules for ionic compounds:
- All nitrates are soluble.
- All perchlorates are soluble.
- All fluorides are soluble, except for those of group 2 metals (Ca, Sr, Ba) and $Al^{3+}$ .
- All sulfates are soluble, except for those of $Ca^{2+}$ , $Sr^{2+}$ , $Ba^{2+}$ , $Pb^{2+}$ , and $Ag^+$ .

Step 2: Apply the solubility rules to the given options:
- $\mathrm{AgCl}{\mathrm{O}_4}$ : Silver perchlorate - soluble.
- $\mathrm{A}{\mathrm{g}_2}\mathrm{S}{\mathrm{O}_4}$ : Silver sulfate - insoluble (sparingly soluble).
- $\mathrm{AgF}$ : Silver fluoride - soluble.
- $\mathrm{AgN}{\mathrm{O}_3}$ : Silver nitrate - soluble.

Step 3: Conclusion: $\mathrm{A}{\mathrm{g}_2}\mathrm{S}{\mathrm{O}_4}$ is insoluble in water.

Why other options are wrong:
- Option A: $\mathrm{AgCl}{\mathrm{O}_4}$ : Soluble.
- Option C: $\mathrm{AgF}$ : Soluble.
- Option D: $\mathrm{AgN}{\mathrm{O}_3}$ : Soluble.

Key Concepts:
- Solubility rules
- Ionic compounds

Memory Aids:
- Remember the common solubility rules.

Quick Tips:
- Memorize the solubility rules to quickly determine the solubility of ionic compounds.

209 Easy

209 of 220Question permalink

Understanding +4 | -1

Higher is the bond order, greater is:

View answer and explanation

Correct option: A

Bond dissociation energy

Option A (Bond dissociation energy) is correct because a higher bond order indicates a stronger bond, which requires more energy to break.

Solution approach

Correct Answer + Core Reason: Bond dissociation energy. Higher bond order implies a stronger bond, hence greater bond dissociation energy.

Step-by-Step Solution:

Step 1: Define bond order.

Step 2: Bond order is the number of chemical bonds between a pair of atoms.

Step 3: Relate bond order to bond strength.

Step 4: A higher bond order indicates a stronger bond.

Step 5: Relate bond strength to bond dissociation energy.

Step 6: Bond dissociation energy is the energy required to break a bond.

Step 7: Therefore, a higher bond order implies a greater bond dissociation energy.

Why other options are wrong:
- Option B (Covalent character): While bond order can influence covalent character, it's not the most direct relationship.
- Option C (Bond length): Higher bond order generally corresponds to shorter bond length, not greater.
- Option D (Paramagnetism): Bond order doesn't directly determine paramagnetism. Paramagnetism depends on the presence of unpaired electrons.

Key Concepts:
- Bond Order
- Bond Dissociation Energy
- Bond Strength

Memory Aids:
- Higher Bond Order = Stronger Bond = Higher Dissociation Energy

Quick Tips:
- Remember the relationship between bond order, bond strength, and bond dissociation energy.

210 Easy

210 of 220Question permalink

Understanding +4 | -1

S - P

overlapping is present in

View answer and explanation

Correct option: D

HF

Option D (

HF

) is correct because hydrogen uses its s orbital and fluorine uses its p orbital to form the bond.

Solution approach

Correct Answer + Core Reason: $HF$ . Hydrogen uses an s orbital, and fluorine uses a p orbital for bonding.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each atom in the given molecules.

Step 2: $B{r_2}$ : Both bromine atoms use p orbitals for bonding (p-p overlap).

Step 3: ${H_2}$ : Both hydrogen atoms use s orbitals for bonding (s-s overlap).

Step 4: ${O_2}$ : Oxygen atoms use p orbitals for bonding (p-p overlap).

Step 5: $HF$ : Hydrogen uses its 1s orbital, and fluorine uses a 2p orbital for bonding (s-p overlap).

Why other options are wrong:
- Option A: $B{r_2}$ is incorrect because it involves p-p overlapping.
- Option B: ${H_2}$ is incorrect because it involves s-s overlapping.
- Option C: ${O_2}$ is incorrect because it involves p-p overlapping.

Key Concepts:
- Overlapping of orbitals
- Electronic configuration

Memory Aids:
- Hydrogen always uses its s orbital for bonding.

Quick Tips:
- Consider the valence electrons and orbitals involved in bonding.

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