The decreasing values of bond angles from ${\rm{N}}{{\rm{H}}_3}$($106{\;^0}$) to ${\rm{Sb}}{{\rm{H}}_3}$ ($101{\;^0}$) down group-$15$ of the periodic table is due to:

Increasing $bp$ - $bp$ repulsion

Increasing $p$-orbital character in $s{p^3}$

Decreasing $lp$ - $bp$ repulsion

The bond angle in ${{\rm{H}}_2}{\rm{S}}$ (for ${\rm{H}} - {\rm{S}} - {\rm{H}}$) is:

Greater than ${\rm{H}} - {\rm{Se}} - H$ and less than ${\rm{H}} - {\rm{O}} - {\rm{H}}$

Same as that of ${\rm{Cl}} - {\rm{Be}} - {\rm{Cl}}$ in $BeC{{\rm{l}}_2}$

Greater than ${\rm{H}} - {\rm{N}} - {\rm{H}}$ bond angle in ${\rm{N}}{{\rm{H}}_3}$

Same as ${\rm{ClSn}} - {\rm{Cl}}$ in $SnC{{\rm{l}}_2}$

CHEMICAL BONDING & MOLECULAR STRUCTURE MCQs for NEET UG CHEMISTRY — Medium, Page 3

61 Medium

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The electronic theory of bonding was proposed by

View answer and explanation

Correct option: B

Lewis

Option B (Lewis) is correct because Gilbert N. Lewis proposed the electronic theory of bonding, focusing on the role of valence electrons in forming chemical bonds.

Solution approach

Correct Answer + Core Reason: Lewis proposed the electronic theory of bonding.

Step-by-Step Solution:

Step 1: The electronic theory of bonding, which explains chemical bonding in terms of the arrangement of electrons in atoms, was primarily developed by Gilbert N. Lewis.

Step 2: Lewis's theory emphasizes the role of valence electrons in forming chemical bonds, particularly through the sharing or transfer of electrons to achieve a stable electron configuration (octet rule).

Why other options are wrong:
- Option A: Pauling is known for his work on the nature of the chemical bond, including electronegativity and resonance, but not the initial electronic theory.
- Option C: Bronsted is known for his work on acid-base theory.
- Option D: Mullikan is known for his work on molecular orbital theory and electronegativity.

Key Concepts:
- Electronic theory of bonding
- Lewis structures
- Valence electrons

Memory Aids:
- Lewis = Electrons

Quick Tips:
- Remember that Lewis structures are a key part of understanding the electronic theory of bonding.

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The pairs of bases in

DNA

are held together by:

View answer and explanation

Correct option: A

Hydrogen bonds

Option A (Hydrogen bonds) is correct because the complementary base pairs in DNA (adenine-thymine and guanine-cytosine) are held together by hydrogen bonds.

Solution approach

Correct Answer + Core Reason: Hydrogen bonds. These bonds hold the DNA base pairs together.

Step-by-Step Solution:

Step 1: Recall the structure of DNA.\nDNA consists of two strands held together by base pairs.

Step 2: Identify the type of bond that holds the base pairs together.\nThe base pairs are held together by hydrogen bonds.\nAdenine (A) pairs with Thymine (T) via two hydrogen bonds.\nGuanine (G) pairs with Cytosine (C) via three hydrogen bonds.

Why other options are wrong:
- Option B: Ionic bonds are not the primary force holding the base pairs together.
- Option C: Phosphate groups are part of the DNA backbone, not directly involved in base pairing.
- Option D: Deoxyribose groups are part of the DNA backbone, not directly involved in base pairing.

Key Concepts:
- DNA structure
- Base pairing
- Hydrogen bonding

Memory Aids:
- A-T (2 hydrogen bonds), G-C (3 hydrogen bonds).

Quick Tips:
- Remember the base pairing rules in DNA.

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In

\mathrm{TeC}{\mathrm{l}_4}

, the central atom tellurium involves

View answer and explanation

Correct option: B

s{p^3}d

hybridisation

Option B (

s{p^3}d

hybridisation) is correct because tellurium in

TeCl_4

has four bond pairs and one lone pair, resulting in a steric number of 5 and

sp^3d

hybridization.

Solution approach

Correct Answer + Core Reason: $s{p^3}d$ hybridisation. Tellurium in $TeCl_4$ has four bond pairs and one lone pair, resulting in a steric number of 5 and $sp^3d$ hybridization.

Step-by-Step Solution:

Step 1: Determine the number of bond pairs and lone pairs around the central atom (tellurium).\nTellurium forms four single bonds with four chlorine atoms (4 bond pairs).\nTellurium has 6 valence electrons. Four are used for bonding, leaving 2 electrons or 1 lone pair.

Step 2: Calculate the steric number (number of bond pairs + number of lone pairs):\nSteric number = 4 bond pairs + 1 lone pair = 5

Step 3: Determine the hybridization based on the steric number:\nSteric number 5 corresponds to $sp^3d$ hybridization.

Why other options are wrong:
- **Option A ( $s{p^3}$ hybridized):** $sp^3$ hybridization corresponds to a steric number of 4, which is incorrect for tellurium in $TeCl_4$ .
- **Option C ( $s{p^3}{d^2}$ hybridisation):** $sp^3d^2$ hybridization corresponds to a steric number of 6, which is incorrect for tellurium in $TeCl_4$ .
- **Option D ( $ds{p^2}$ hybridisation):** $dsp^2$ hybridization is typically seen in square planar complexes, not in simple molecules like $TeCl_4$ .

Key Concepts:
- Hybridization
- Steric number
- VSEPR theory

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
- Steric Number 5: $sp^3d$ hybridization
- Steric Number 6: $sp^3d^2$ hybridization
- Steric Number 7: $sp^3d^3$ hybridization

Quick Tips:
- Always count the number of bond pairs and lone pairs around the central atom to determine the steric number and hence the hybridization.

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The pair of elements which on combination are most likely to form an ionic compound is:

View answer and explanation

Correct option: B

\mathrm{K and}\,{\mathrm{O}_2}

Option B (K and O2) is correct because potassium (K) is a highly electropositive metal and oxygen (O2) is a highly electronegative non-metal, leading to a large difference in electronegativity and thus ionic bond formation.

Solution approach

Correct Answer + Core Reason: $\mathrm{K and}\,{\mathrm{O}_2}$ . Ionic compounds are most likely formed between elements with large electronegativity differences.

Step-by-Step Solution:

Step 1: Identify the elements in each pair.

Step 2: Consider the electronegativity differences between the elements in each pair. Ionic bonds are favored by large differences.

Step 3: $\mathrm{Na}$ and $\mathrm{Ca}$ are both metals, so they form metallic bonds, not ionic.

Step 4: $\mathrm{K}$ is a metal and ${\mathrm{O}_2}$ is a non-metal. Potassium is highly electropositive and oxygen is highly electronegative, leading to a large electronegativity difference.

Step 5: ${\mathrm{O}_2}$ and $\mathrm{C}{\mathrm{l}_2}$ are both non-metals, so they form covalent bonds.

Step 6: $\mathrm{Al}$ and ${\mathrm{I}_2}$ are a metal and a non-metal, but the electronegativity difference is smaller than that between $\mathrm{K}$ and ${\mathrm{O}_2}$ .

Step 7: Conclusion: $\mathrm{K and}\,{\mathrm{O}_2}$ are most likely to form an ionic compound.

Why other options are wrong:
- Option A: $\mathrm{Na}$ and $\mathrm{Ca}$ are both metals and will form a metallic alloy, not an ionic compound.
- Option C: ${\mathrm{O}_2}\,\mathrm{and C}{\mathrm{l}_2}$ are both non-metals and will form a covalent compound.
- Option D: $\mathrm{Al and }{\mathrm{I}_2}$ can form an ionic compound, but the electronegativity difference between K and O is greater, making it more likely to form an ionic compound.

Key Concepts:
- Ionic bond formation
- Electronegativity
- Metals and non-metals

Memory Aids:
- Metals tend to lose electrons and form positive ions, while non-metals tend to gain electrons and form negative ions.

Quick Tips:
- Look for combinations of highly electropositive metals and highly electronegative non-metals.

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Pauling received Nobel Prize for his work on:

View answer and explanation

Correct option: C

Chemical bonding

Option C (Chemical bonding) is correct because Linus Pauling received the Nobel Prize in Chemistry for his work on the nature of the chemical bond.

Solution approach

Correct Answer + Core Reason: Chemical bonding. Linus Pauling received the Nobel Prize in Chemistry for his work on the nature of the chemical bond.

Step-by-Step Solution:

Step 1: Linus Pauling made significant contributions to the understanding of chemical bonding.

Step 2: His work included the concept of electronegativity and resonance.

Step 3: He received the Nobel Prize in Chemistry for his research into the nature of the chemical bond and its application to the elucidation of the structure of complex substances.

Why other options are wrong:
- Option A: Photosynthesis is related to biology and plant physiology, not Pauling's area of expertise.
- Option B: Atomic structure is related to the arrangement of protons, neutrons, and electrons in an atom, but Pauling's Nobel Prize was specifically for chemical bonding.
- Option D: Thermodynamics deals with energy transfer and transformations, which is not the primary focus of Pauling's Nobel Prize-winning work.

Key Concepts:
- Chemical bonding
- Electronegativity
- Resonance

Memory Aids:
- Pauling = Chemical Bonding

Quick Tips:
- Remember key scientists and their contributions to chemistry.

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Identify the non-polar molecule in the set of compounds given

HCl

,

HF

,

{\mathrm{H}_2}

HBr

View answer and explanation

Correct option: A

{\mathrm{H}_2}

Option A (

{\mathrm{H}_2}

) is correct because it is a diatomic molecule formed by two identical atoms, resulting in equal sharing of electrons and no dipole moment.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_2}$ . A molecule formed by two identical atoms is always non-polar.

Step-by-Step Solution:

Step 1: Define a non-polar molecule: A non-polar molecule has no net dipole moment.

Step 2: Consider the given molecules: HCl, HF, H2, HBr.

Step 3: HCl, HF, and HBr are all diatomic molecules formed by two different atoms with different electronegativities. Therefore, they are polar.

Step 4: H2 is a diatomic molecule formed by two identical hydrogen atoms. The electrons are shared equally, and there is no dipole moment.

Step 5: Therefore, H2 is a non-polar molecule.

Why other options are wrong:
- Option B: HCl is a polar molecule due to the difference in electronegativity between H and Cl.
- Option C: HF and HBr are both polar molecules due to the difference in electronegativity between H and F, and H and Br, respectively.
- Option D: HBr is a polar molecule due to the difference in electronegativity between H and Br.

Key Concepts:
- Polarity of molecules
- Electronegativity
- Dipole moment

Memory Aids:
- Identical atoms = non-polar molecule
- Different atoms with different electronegativities = polar molecule

Quick Tips:
- Diatomic molecules with identical atoms are always non-polar.

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In

\mathrm{PC}{\mathrm{l}_5}

molecule,

P

is:

View answer and explanation

Correct option: D

s{p^3}d

hybridized

Option D (

s{p^3}d

hybridized) is correct because phosphorus in

PCl_5

has five bonding pairs and no lone pairs, resulting in

sp^3d

hybridization and a trigonal bipyramidal geometry.

Solution approach

Correct Answer + Core Reason: $s{p^3}d$ hybridized. Phosphorus in $PCl_5$ is $sp^3d$ hybridized.

Step-by-Step Solution:

Step 1: Determine the central atom: Phosphorus (P).

Step 2: Count the number of valence electrons for P: 5.

Step 3: Count the number of bonding pairs: 5 (one for each Cl atom).

Step 4: Calculate the steric number: 5 (5 bonding pairs + 0 lone pairs).

Step 5: Determine the hybridization: $sp^3d$ .

Step 6: Determine the geometry based on the hybridization: Trigonal bipyramidal.

Why other options are wrong:
- **Option A ( $s{p^3}$ hybridized):** $sp^3$ hybridization corresponds to 4 bonding pairs and 0 lone pairs, or 3 bonding pairs and 1 lone pair.
- **Option B ( $ds{p^2}$ hybridized):** $dsp^2$ hybridization corresponds to square planar geometry, which is not the case for $PCl_5$ .
- **Option C ( $d{s^3}p$ hybridized): This is not a standard hybridization scheme.

Key Concepts:
- Hybridization
- VSEPR theory
- Molecular geometry

Memory Aids:
- Steric number = number of bonding pairs + number of lone pairs

Quick Tips:**
- Use VSEPR theory to predict molecular geometry.

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Which of the following has lowest boiling point?

View answer and explanation

Correct option: C

\mathrm{CuC}{\mathrm{l}_2}

Option C (

\mathrm{CuC}{\mathrm{l}_2}

) is correct because it has more covalent character due to higher polarization of the anion by the cation, leading to lower boiling point.

Solution approach

Correct Answer + Core Reason: $\mathrm{CuC}{\mathrm{l}_2}$ . More covalent character leads to lower boiling point.

Step-by-Step Solution:

Step 1: Understand the question. We need to identify the compound with the lowest boiling point.

Step 2: Consider the ionic character of each compound. More covalent character leads to lower boiling points.

Step 3: Apply Fajan's rules. Smaller, highly charged cations favor covalent character.

Step 4: Compare the cations: Na+, Cu+, Cu2+, Cs+
- Cu2+ has the highest charge and is relatively small, leading to more polarization of the anion (Cl-) and more covalent character.

Step 5: Therefore, $\mathrm{CuC}{\mathrm{l}_2}$ is expected to have the lowest boiling point.

Why other options are wrong:
- Option A: $\mathrm{NaCl}$ is highly ionic and has a high boiling point.
- Option B: $\mathrm{CuCl}$ is ionic but less so than NaCl, so it has a higher boiling point than $\mathrm{CuC}{\mathrm{l}_2}$ .
- Option D: $\mathrm{CsCl}$ is highly ionic and has a high boiling point.

Key Concepts:
- Boiling point and its relation to ionic character
- Fajan's rules
- Polarization of ions

Memory Aids:
- More covalent character = Lower boiling point

Quick Tips:
- Use Fajan's rules to determine the degree of covalent character.

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An example of a polar covalent compound is

View answer and explanation

Correct option: D

HCl

Option D (

HCl

) is correct because there is a significant difference in electronegativity between hydrogen and chlorine, leading to a polar covalent bond.

Solution approach

Correct Answer + Core Reason: $HCl$ . The electronegativity difference between H and Cl makes it a polar covalent compound.

Step-by-Step Solution:

Step 1: Define a polar covalent compound: A polar covalent compound is formed when atoms with different electronegativities share electrons.

Step 2: Consider the given compounds: KCl, NaCl, CCl4, HCl.

Step 3: KCl and NaCl are ionic compounds formed by the transfer of electrons between a metal and a non-metal.

Step 4: CCl4 is a covalent compound, but due to its symmetrical tetrahedral structure, the individual bond dipoles cancel out, making the molecule non-polar.

Step 5: HCl is a covalent compound formed by the sharing of electrons between hydrogen and chlorine. Chlorine is more electronegative than hydrogen, so the bond is polar.

Why other options are wrong:
- Option A: KCl is an ionic compound, not a polar covalent compound.
- Option B: NaCl is an ionic compound, not a polar covalent compound.
- Option C: CCl4 is a non-polar covalent compound due to its symmetrical structure.

Key Concepts:
- Polar covalent bond
- Electronegativity
- Ionic compound
- Dipole moment

Memory Aids:
- Large electronegativity difference = ionic bond
- Moderate electronegativity difference = polar covalent bond
- Small or zero electronegativity difference = non-polar covalent bond

Quick Tips:
- Consider the electronegativity difference between the atoms in the compound.

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Lewis pictured the atom in terms of kernel and the outermost shell that could accommodate a maximum of

View answer and explanation

Correct option: C

8 electrons

Option C is correct because Lewis's theory suggested that atoms achieve stability by having 8 electrons in their outermost shell (octet rule).

Solution approach

Correct Answer + Core Reason: 8 electrons. Lewis pictured the atom with a kernel and an outermost shell that could accommodate a maximum of 8 electrons (octet rule).

Step-by-Step Solution:

Step 1: Recall Lewis's theory of chemical bonding.

Step 2: Understand the concept of the octet rule.

Step 3: Lewis proposed that atoms achieve stability by having 8 electrons in their outermost shell.

Step 4: This is known as the octet rule, which explains the formation of chemical bonds.

Step 5: Therefore, Lewis pictured the atom in terms of kernel and the outermost shell that could accommodate a maximum of 8 electrons.

Why other options are wrong:
- Option A: 6 electrons is incorrect.
- Option B: 10 electrons is incorrect.
- Option D: 2 electrons is the maximum for hydrogen and some other exceptions to the octet rule.

Key Concepts:
- Lewis theory
- Octet rule

Memory Aids:
- Octet rule: Atoms want 8 electrons in their outer shell.

Quick Tips:
- Remember the octet rule and its importance in chemical bonding.

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A

\pi

-bond is formed by sideways overlapping of:

View answer and explanation

Correct option: B

p - p

orbitals

Option B is correct because

\pi

-bonds are formed by the sideways (lateral) overlapping of p-p orbitals.

Solution approach

Correct Answer + Core Reason: $p - p$ orbitals. Pi-bonds are formed by sideways overlapping of p-orbitals.

Step-by-Step Solution:

Step 1: Understand the formation of sigma and pi bonds.

Step 2: Recall that sigma bonds are formed by head-on overlapping of atomic orbitals.

Step 3: Pi bonds are formed by sideways (lateral) overlapping of p-orbitals.

Step 4: s-orbitals are spherical and can only form sigma bonds.

Step 5: p-orbitals have a dumbbell shape and can form both sigma and pi bonds.

Step 6: Therefore, a $\pi$ -bond is formed by sideways overlapping of p-p orbitals.

Why other options are wrong:
- Option A: $s - s$ orbitals form sigma bonds.
- Option C: $s - p$ orbitals can form sigma bonds.
- Option D: $s - p - s$ is not a standard orbital overlapping configuration.

Key Concepts:
- Sigma bond
- Pi bond
- Orbital overlapping

Memory Aids:
- Sigma bond = Head-on overlap
- Pi bond = Sideways overlap

Quick Tips:
- Remember that pi bonds are always formed after a sigma bond has already been formed.

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H

-bonding is not present in:

View answer and explanation

Correct option: C

{\mathrm{H}_2}\mathrm{S}

Option C (H2S) is correct because sulfur is not electronegative enough to form strong hydrogen bonds.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_2}\mathrm{S}$ . Sulfur is not electronegative enough to form strong hydrogen bonds.

Step-by-Step Solution:

Step 1: Recall the requirements for hydrogen bonding: a hydrogen atom bonded to a highly electronegative atom (F, O, or N).

Step 2: Analyze each option:
- A: Glycerine has -OH groups, so it can form hydrogen bonds.
- B: Water has O-H bonds, so it can form hydrogen bonds.
- C: H2S has S-H bonds, but sulfur is not electronegative enough to form strong hydrogen bonds.
- D: HF has an H-F bond, so it can form hydrogen bonds.

Step 3: Conclude that H2S does not have significant hydrogen bonding.

Why other options are wrong:
- Option A: Glycerine contains multiple -OH groups and thus exhibits hydrogen bonding.
- Option B: Water (H2O) exhibits strong hydrogen bonding due to the presence of O-H bonds.
- Option D: HF exhibits hydrogen bonding due to the highly electronegative fluorine atom.

Key Concepts:
- Hydrogen bonding.
- Electronegativity.
- Requirements for hydrogen bonding.

Memory Aids:
- Remember that hydrogen bonding requires H bonded to F, O, or N.

Quick Tips:
- Check for the presence of H bonded to F, O, or N to determine if hydrogen bonding is present.

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The geometry of

\mathrm{P}{\mathrm{F}_5}

molecule is:

View answer and explanation

Correct option: C

Trigonal bipyramidal

Option C (Trigonal bipyramidal) is correct because

\mathrm{P}{\mathrm{F}_5}

has five bonding pairs and no lone pairs around the central phosphorus atom, resulting in a trigonal bipyramidal geometry.

Solution approach

Correct Answer + Core Reason: Trigonal bipyramidal. $\mathrm{P}{\mathrm{F}_5}$ has five bonding pairs around the central phosphorus atom.

Step-by-Step Solution:

Step 1: Phosphorus has 5 valence electrons, and each fluorine atom contributes 1 electron through a covalent bond.

Step 2: There are 5 bonding pairs and no lone pairs around the central phosphorus atom.

Step 3: According to VSEPR theory, five bonding pairs around a central atom result in a trigonal bipyramidal geometry.

Why other options are wrong:
- Option A: Planar geometry is not possible with 5 bonding pairs.
- Option B: Square planar geometry typically involves lone pairs and four bonding pairs.
- Option D: Tetrahedral geometry involves four bonding pairs, not five.

Key Concepts:
- VSEPR theory
- Trigonal bipyramidal geometry

Memory Aids:
- 5 bonding pairs = trigonal bipyramidal.

Quick Tips:
- Count the number of bonding pairs and lone pairs.

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Hybridization state of I in

\mathrm{ICl}_2^ +

is :

View answer and explanation

Correct option: D

s{p^3}

Option D (

s{p^3}

) is correct because the central iodine atom in

ICl_2^+

has two bond pairs and two lone pairs, resulting in a steric number of 4 and

sp^3

hybridization.

Solution approach

Correct Answer + Core Reason: $s{p^3}$ . The central iodine atom in $ICl_2^+$ has two bond pairs and two lone pairs, resulting in a steric number of 4 and $sp^3$ hybridization.

Step-by-Step Solution:

Step 1: Determine the number of bond pairs and lone pairs around the central atom (iodine) in $ICl_2^+$ .\nIodine forms two single bonds with two chlorine atoms (2 bond pairs).\nIodine has 7 valence electrons. Two are used for bonding, and one is removed due to the +1 charge, leaving 4 electrons or 2 lone pairs.

Step 2: Calculate the steric number (number of bond pairs + number of lone pairs):\nSteric number = 2 bond pairs + 2 lone pairs = 4

Step 3: Determine the hybridization based on the steric number:\nSteric number 4 corresponds to $sp^3$ hybridization.

Why other options are wrong:
- **Option A ( $ds{p^2}$ ):** $dsp^2$ hybridization is typically seen in square planar complexes, not in simple molecules like $ICl_2^+$ .
- **Option B ( $sp$ ):** $sp$ hybridization corresponds to a steric number of 2, which is incorrect for iodine in $ICl_2^+$ .
- **Option C ( $s{p^2}$ ):** $sp^2$ hybridization corresponds to a steric number of 3, which is incorrect for iodine in $ICl_2^+$ .

Key Concepts:
- Hybridization
- Steric number
- VSEPR theory

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:
- Always count the number of bond pairs and lone pairs around the central atom to determine the steric number and hence the hybridization.

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Shape of molecules is decided by:

View answer and explanation

Correct option: A

Sigma bond

Option A (Sigma bond) is correct because the arrangement of sigma bonds determines the basic shape of a molecule.

Solution approach

Correct Answer + Core Reason: Sigma bond. Sigma bonds determine the geometry of a molecule.

Step-by-Step Solution:

Step 1: Understand the role of sigma and pi bonds in molecular structure.

Step 2: Sigma bonds are formed by head-on overlap of atomic orbitals, allowing free rotation around the bond axis. This arrangement defines the basic shape of the molecule.

Step 3: Pi bonds are formed by sideways overlap of atomic orbitals. They restrict rotation and contribute to the overall rigidity and planarity of the molecule, but do not define the basic shape.

Step 4: Therefore, the shape of molecules is primarily determined by the arrangement of sigma bonds.

Why other options are wrong:
- Option B: $\mathrm{\pi }$ -bond: Pi bonds influence the bond order and rigidity but do not primarily determine the shape.
- Option C: Both sigma and $\mathrm{\pi }$ -bonds: While pi bonds affect the overall structure, sigma bonds are the primary determinants of shape.
- Option D: Neither sigma nor $\mathrm{\pi }$ -bonds: Chemical bonds are responsible for the shape of molecules.

Key Concepts:
- Sigma bonds
- Pi bonds
- Molecular geometry
- VSEPR theory

Memory Aids:
- Sigma bonds are like the skeleton, defining the basic shape; pi bonds are like the muscles, adding strength and rigidity.

Quick Tips:
- Remember that sigma bonds allow rotation, while pi bonds restrict it. The arrangement of atoms connected by sigma bonds dictates the overall shape.

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For which of the following hybridization the bond angle is maximum?

View answer and explanation

Correct option: B

sp

Option B (

sp

) is correct because

sp

hybridization results in a linear geometry with a bond angle of 180°, which is the maximum among the given options.

Solution approach

Correct Answer + Core Reason: $sp$ . $sp$ hybridization has a bond angle of 180 degrees, which is the maximum.

Step-by-Step Solution:

Step 1: Recall the bond angles associated with different hybridizations.

Step 2: $sp^2$ hybridization has a bond angle of 120 degrees.

Step 3: $sp$ hybridization has a bond angle of 180 degrees.

Step 4: $sp^3$ hybridization has a bond angle of 109.5 degrees.

Step 5: $dsp^2$ hybridization has a bond angle of 90 degrees.

Step 6: Compare the bond angles and determine the maximum.

Why other options are wrong:
- **Option A ( $s{p^2}$ ): The bond angle is 120 degrees, which is less than 180 degrees.
- Option C ( $s{p^3}$ ): The bond angle is 109.5 degrees, which is less than 180 degrees.
- Option D ( $ds{p^2}$ ): The bond angle is 90 degrees, which is less than 180 degrees.

Key Concepts:
- Hybridization
- Bond angle

Memory Aids:**
- $sp$ = 180 degrees, $sp^2$ = 120 degrees, $sp^3$ = 109.5 degrees

Quick Tips:
- Memorize the bond angles for common hybridization schemes.

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d

-orbital involved in

s p^3 d

hybridized orbitals is

View answer and explanation

Correct option: D

d_{z^2}

Option D (

d_{z^2}

) is correct because it is the d-orbital that participates in

sp^3d

hybridization to form the required geometry.

Solution approach

Correct Answer + Core Reason: $d_{z^2}$ . The $d_{z^2}$ orbital is the one involved in $sp^3d$ hybridization.

Step-by-Step Solution:

Step 1: Identify the hybridization type: $sp^3d$ .

Step 2: Recall the orbitals involved in $sp^3d$ hybridization: one $s$ orbital, three $p$ orbitals, and one $d$ orbital.

Step 3: Recognize that the $d_{z^2}$ orbital is the specific $d$ orbital that participates in this type of hybridization.

Step 4: Conclude that the correct answer is $d_{z^2}$ .

Why other options are wrong:
- **Option A ( $d_{xy}$ ):** This d-orbital is not typically involved in the standard $sp^3d$ hybridization.
- **Option B ( $d_{x^2-y^2}$ ):** This d-orbital is not typically involved in the standard $sp^3d$ hybridization.
- **Option C ( $d_{xz}$ ):** This d-orbital is not typically involved in the standard $sp^3d$ hybridization.

Key Concepts:
- Hybridization
- Atomic orbitals
- $sp^3d$ hybridization

Memory Aids:
- Remember that $sp^3d$ hybridization involves the $d_{z^2}$ orbital.

Quick Tips:
- Know the specific d-orbitals involved in different types of hybridization.

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Analyzing +4 | -1

Which of the following has the minimum bond length?

View answer and explanation

Correct option: B

\mathrm{O}_2^ +

Option B (

{\mathrm{O}_2^ + }

) is correct because it has the highest bond order (2.5) among the given species, leading to the shortest bond length.

Solution approach

Correct Answer + Core Reason: $\mathrm{O}_2^ +$ . Higher bond order implies shorter bond length.

Step-by-Step Solution:

Step 1: Calculate the bond order for each species.

Step 2: ${\mathrm{O}_2}$ : Bond order = (10 - 6)/2 = 2

Step 3: $\mathrm{O}_2^ +$ : Bond order = (10 - 5)/2 = 2.5

Step 4: $\mathrm{O}_2^ -$ : Bond order = (10 - 7)/2 = 1.5

Step 5: $\mathrm{O}_2^{2 - }$ : Bond order = (10 - 8)/2 = 1

Step 6: The species with the highest bond order will have the minimum bond length.

Why other options are wrong:
- Option A: ${\mathrm{O}_2}$ has a bond order of 2, which is less than that of $\mathrm{O}_2^ +$ .
- Option C: $\mathrm{O}_2^ -$ has a bond order of 1.5, which is less than that of $\mathrm{O}_2^ +$ .
- Option D: $\mathrm{O}_2^{2 - }$ has a bond order of 1, which is less than that of $\mathrm{O}_2^ +$ .

Key Concepts:
- Bond Order Calculation
- Relationship between Bond Order and Bond Length

Memory Aids:
- Higher bond order means shorter and stronger bond.

Quick Tips:
- Remember the inverse relationship between bond order and bond length.

79 Medium

79 of 262Question permalink

Application +4 | -1

An element

X

has

3

electrons in

p

-orbitals and also belongs to

III

period. Its molecular formula should be:

View answer and explanation

Correct option: C

{X_4}

Option C (

{X_4}

) is correct because an element with 3 electrons in the p-orbitals of the third period is phosphorus, which exists as a tetrahedral

P_4

molecule.

Solution approach

Correct Answer + Core Reason: ${X_4}$ . The element is Phosphorus, which exists as $P_4$ .

Step-by-Step Solution:

Step 1: Determine the element based on its electronic configuration and period.

Step 2: The element belongs to the third period and has 3 electrons in its p-orbitals. This means its electronic configuration is [Ne] 3s² 3p³.

Step 3: The element with this electronic configuration is phosphorus (P).

Step 4: Phosphorus exists as a tetrahedral molecule with the formula $P_4$ .

Why other options are wrong:
- Option A: $X$ is incorrect. Phosphorus does not exist as a monatomic molecule.
- Option B: ${X_2}$ is incorrect. Phosphorus does not exist as a diatomic molecule.
- Option D: ${X_5}$ is incorrect. Phosphorus does not form a molecule with five atoms.

Key Concepts:
- Electronic configuration
- Periodicity
- Molecular formula

Memory Aids:
- Remember that phosphorus exists as $P_4$ .

Quick Tips:
- Identify the element based on its electronic configuration and period.

80 Medium

80 of 262Question permalink

Application +4 | -1

A certain diatomic molecule,

A B

has dipole moment 1.6 D and the internuclear distance is

100 \mathrm{pm}

. The percentage of electronic charge existing on more electronegative atom is

View answer and explanation

Correct option: A

33 \%

Option A (

33 \%

) is correct because it represents the approximate percentage of electronic charge calculated from the dipole moment and internuclear distance.

Solution approach

Correct Answer + Core Reason: $33 \%$ . The percentage of electronic charge is calculated using the formula relating dipole moment, bond length, and electronic charge.

Step-by-Step Solution:

Step 1: Convert dipole moment from Debye to Coulomb-meters: $\mu = 1.6 \text{ D} = 1.6 \times 3.336 \times 10^{-30} \text{ C m} = 5.3376 \times 10^{-30} \text{ C m}$

Step 2: Convert internuclear distance from pm to meters: $d = 100 \text{ pm} = 100 \times 10^{-12} \text{ m} = 1 \times 10^{-10} \text{ m}$

Step 3: Calculate the charge: $q = \frac{\mu}{d} = \frac{5.3376 \times 10^{-30} \text{ C m}}{1 \times 10^{-10} \text{ m}} = 5.3376 \times 10^{-20} \text{ C}$

Step 4: Calculate the percentage of electronic charge: $\% \text{ electronic charge} = \frac{q}{e} \times 100 = \frac{5.3376 \times 10^{-20} \text{ C}}{1.602 \times 10^{-19} \text{ C}} \times 100 \approx 33.3 \%$

Why other options are wrong:
- **Option B ( $25 \%$ ): This is incorrect due to a calculation error or using incorrect conversion factors.
- Option C ( $50 \%$ ): This is incorrect due to a calculation error or using incorrect conversion factors.
- Option D ( $10 \%$ ): This is incorrect due to a calculation error or using incorrect conversion factors.

Key Concepts:
- Dipole moment
- Percentage of electronic charge
- Conversion of units (Debye to Coulomb-meters, pm to meters)

Memory Aids:**
- Remember the formula: $\% \text{ electronic charge} = \frac{q}{e} \times 100$

Quick Tips:
- Always convert units to SI units before calculations.
- Remember the charge of an electron: $1.602 \times 10^{-19} \text{ C}$ .

81 Medium

81 of 262Question permalink

Understanding +4 | -1

When ionic compounds get dissolved in water:

View answer and explanation

Correct option: D

All are correct

Option D (All are correct) is correct because dissolution of ionic compounds in water involves heat changes, reduction of inter-ionic attraction, and dipole-ion attraction.

Solution approach

Correct Answer + Core Reason: All are correct. Dissolution involves heat changes, reduced attraction, and dipole-ion attraction.

Step-by-Step Solution:

Step 1: Describe the process of dissolving an ionic compound in water.
- When an ionic compound dissolves in water, the water molecules surround the ions and separate them from the crystal lattice.

Step 2: Explain the heat changes involved.
- The dissolution process can be either exothermic (releasing heat) or endothermic (absorbing heat), depending on the lattice energy of the ionic compound and the hydration energy of the ions.
- Therefore, heat changes are involved.

Step 3: Explain the reduction of inter-ionic attraction.
- The strong electrostatic forces between the ions in the crystal lattice are weakened by the presence of water molecules, which have a high dielectric constant.
- Therefore, inter-ionic attraction is reduced.

Step 4: Explain the dipole-ion attraction.
- Water molecules are polar and have a dipole moment. The positive end of the water molecule is attracted to the anions, and the negative end is attracted to the cations.
- This dipole-ion attraction helps to stabilize the ions in solution.

Why other options are wrong:
- Option A: They involve heat changes - Correct, but not the complete answer.
- Option B: Inter-ionic attraction is reduced - Correct, but not the complete answer.
- Option C: Ions show dipole-ion attraction with water molecules - Correct, but not the complete answer.
- Option D: All are correct - Correct. All the statements are true.

Key Concepts:
- Dissolution
- Ionic compounds
- Heat of solution
- Lattice energy
- Hydration energy
- Dipole-ion interaction

Memory Aids:
- Dissolving involves breaking lattice and forming hydration shells.

Quick Tips:
- Consider all the factors involved when an ionic compound dissolves in water.

82 Medium

82 of 262Question permalink

Analyzing +4 | -1

A molecule which cannot exist theoretically is:

View answer and explanation

Correct option: C

\mathrm{O}{\mathrm{F}_4}

Option C (

\mathrm{O}{\mathrm{F}_4}

) is correct because oxygen cannot expand its octet to accommodate four fluorine atoms.

Solution approach

Correct Answer + Core Reason: $\mathrm{O}{\mathrm{F}_4}$ . Oxygen cannot expand its octet to form four bonds with fluorine.

Step-by-Step Solution:

Step 1: Electronic configuration of Oxygen: Oxygen has an electronic configuration of $1s^22s^22p^4$ .

Step 2: Octet Rule: Oxygen typically forms two bonds to complete its octet.

Step 3: Expansion of Octet: Oxygen, being in the second period, does not have d-orbitals available to expand its octet and form more than two bonds with highly electronegative atoms like fluorine.

Step 4: Analyze the given compounds:
- $\mathrm{S}{\mathrm{F}_4}$ : Sulfur can expand its octet and form four bonds with fluorine.
- $\mathrm{O}{\mathrm{F}_2}$ : Oxygen forms two bonds with fluorine, completing its octet.
- ${\mathrm{O}_2}{\mathrm{F}_2}$ : Each oxygen atom forms one bond with another oxygen atom and one bond with a fluorine atom.

Step 5: Conclusion: Oxygen cannot form four bonds with fluorine atoms, so $\mathrm{O}{\mathrm{F}_4}$ does not exist.

Why other options are wrong:
- Option A: $\mathrm{S}{\mathrm{F}_4}$ exists because sulfur can expand its octet.
- Option B: $\mathrm{O}{\mathrm{F}_2}$ exists because oxygen forms two bonds with fluorine.
- Option D: ${\mathrm{O}_2}{\mathrm{F}_2}$ exists because each oxygen atom forms two bonds (one with O and one with F).

Key Concepts:
- Octet Rule
- Expansion of Octet
- Electronic Configuration

Memory Aids:
- Second-period elements (like O) cannot expand their octet.

Quick Tips:
- Remember that elements in the third period and beyond can expand their octet due to the availability of d-orbitals.

83 Medium

83 of 262Question permalink

Application +4 | -1

Which among the following has highest bond order?

View answer and explanation

Correct option: B

N{O^ + }

Option B (

N{O^ + }

) is correct because it has the highest number of electrons in bonding orbitals compared to antibonding orbitals, resulting in a bond order of 3.

Solution approach

Correct Answer + Core Reason: $N{O^ + }$ . Higher bond order indicates greater stability and stronger bond.

Step-by-Step Solution:

Step 1: Determine the number of electrons in each species.
- $O_2^+$ : 15 electrons
- $NO^+$ : 14 electrons
- $C_2$ : 12 electrons
- $O_2^-$ : 17 electrons

Step 2: Write the electronic configuration for each species using molecular orbital theory.
- $O_2^+$ : $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})^4(\pi_{2p}^*)^1$
- $NO^+$ : $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})^4$
- $C_2$ : $(\sigma_{2s})(\sigma_{2s}^*)(\pi_{2p})^4$
- $O_2^-$ : $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})^4(\pi_{2p}^*)^3$

Step 3: Calculate the bond order for each species using the formula: Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2
- $O_2^+$ : (10 - 5) / 2 = 2.5
- $NO^+$ : (10 - 4) / 2 = 3
- $C_2$ : (8 - 4) / 2 = 2
- $O_2^-$ : (10 - 7) / 2 = 1.5

Step 4: Identify the species with the highest bond order.
- $NO^+$ has the highest bond order of 3.

Why other options are wrong:
- Option A: $O_2^+$ : Bond order is 2.5.
- Option C: ${C_2}$ : Bond order is 2.
- Option D: $O_2^-$ : Bond order is 1.5.

Key Concepts:
- Molecular orbital theory
- Bond order
- Electronic configuration

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

Quick Tips:
- Remember the filling order of molecular orbitals.

84 Medium

84 of 262Question permalink

Understanding +4 | -1

Among NaF,NaCl,NaBr and NaI, the NaF has highest melting point because :

View answer and explanation

Correct option: A

It has maximum ionic character

Option A (It has maximum ionic character) is correct because the strength of ionic bonds, and thus the melting point, increases with increasing ionic character.

Solution approach

Correct Answer + Core Reason: It has maximum ionic character. Higher ionic character leads to stronger electrostatic forces and higher melting point.

Step-by-Step Solution:

Step 1: Understand Melting Point and Ionic Character: The melting point of an ionic compound is directly related to the strength of the ionic bonds.

Step 2: Relate Ionic Character to Bond Strength: Higher ionic character means stronger electrostatic forces between the ions, leading to a higher melting point.

Step 3: Analyze the Halides: The ionic character of the sodium halides decreases in the order: NaF > NaCl > NaBr > NaI due to increasing size of the anion and increasing polarizability.

Step 4: Conclusion: Therefore, NaF has the highest melting point because it has the maximum ionic character.

Why other options are wrong:
- Option B: It has minimum ionic character: This is incorrect; NaF has maximum ionic character.
- Option C: It has associated molecules: This is not the primary reason for the high melting point of NaF. While some association may occur, the ionic character is the dominant factor.
- Option D: It has least molecular weight: Molecular weight is not the primary factor determining the melting point of ionic compounds. Ionic bond strength is more important.

Key Concepts:
- Ionic Character
- Melting Point
- Ionic Bond Strength
- Polarizability

Memory Aids:
- Higher ionic character = higher melting point

Quick Tips:
- Remember that ionic character decreases as the size of the anion increases (for the same cation).

85 Medium

85 of 262Question permalink

Understanding +4 | -1

The decreasing values of bond angles from

\mathrm{N}{\mathrm{H}_3}

(

106{\;^0}

) to

\mathrm{Sb}{\mathrm{H}_3}

(

101{\;^0}

) down group-

15

of the periodic table is due to:

View answer and explanation

Correct option: D

Decreasing electronegativity

Option D (Decreasing electronegativity) is correct because as electronegativity decreases down the group, the bond pair electrons are further away from the central atom, reducing bond angle.

Solution approach

Correct Answer + Core Reason: Decreasing electronegativity. As electronegativity decreases down the group, the bond angle decreases.

Step-by-Step Solution:

Step 1: Understand the trend: Down group 15, the electronegativity of the central atom decreases.

Step 2: Effect on bond pair electrons: As electronegativity decreases, the bond pair electrons are pulled less strongly towards the central atom.

Step 3: Effect on bond angle: This results in the bond pair electrons being further away from the central atom, reducing the repulsion between bond pairs and thus decreasing the bond angle.

Step 4: Conclusion: Therefore, decreasing electronegativity is the main reason for the decreasing bond angles.

Why other options are wrong:
- Option A: Increasing $bp$ - $bp$ repulsion is incorrect because the repulsion actually decreases as the bond pairs are further apart.
- Option B: Increasing $p$ -orbital character in $s{p^3}$ is a contributing factor, but the primary reason is decreasing electronegativity.
- Option C: Decreasing $lp$ - $bp$ repulsion is incorrect because lone pair-bond pair repulsion is always present, but the change in bond angle is primarily due to electronegativity.

Key Concepts:
- Electronegativity
- Bond Angle
- Lone Pair - Bond Pair Repulsion

Memory Aids:
- Lower Electronegativity = Smaller Bond Angle

Quick Tips:
- Consider the electronegativity difference between the central atom and the surrounding atoms when determining bond angles.

86 Medium

86 of 262Question permalink

Analyzing +4 | -1

Order of covalent bond:

A.

\mathrm{KF}>\mathrm{KI}

;

\mathrm{LiF}>\mathrm{KF}

B.

\mathrm{KF}<\mathrm{KI}

;

\mathrm{LiF}>\mathrm{KF}

C.

\mathrm{SnCl}_4>\mathrm{SnCl}_2 ; \mathrm{CuCl}>\mathrm{NaCl}

D.

\mathrm{LiF}>\mathrm{KF} ; \mathrm{CuCl}<\mathrm{NaCl}

E.

\mathrm{KF}<\mathrm{KI} ; \mathrm{CuCl}>\mathrm{NaCl}

Choose the correct answer from the options given below:

View answer and explanation

Correct option: C

B, C, E only

Option C (B, C, E only) is correct because it correctly describes the trends in covalent character based on Fajan's rule and charge density.

Solution approach

Correct Answer + Core Reason: B, C, E only. These options correctly describe the trends in covalent character.

Step-by-Step Solution:

Step 1: Analyze each statement.
- A: KF > KI; LiF > KF (Incorrect. Covalent character increases with the size of the anion and decreases with the size of the cation. So, KF < KI and LiF > KF)
- B: KF < KI; LiF > KF (Correct. Covalent character increases with the size of the anion and decreases with the size of the cation.)
- C: SnCl4 > SnCl2; CuCl > NaCl (Correct. Higher the charge, more is the polarization and hence more is the covalent character.)
- D: LiF > KF; CuCl < NaCl (Incorrect. LiF > KF is correct, but CuCl > NaCl is correct because Cu+ has pseudo inert gas configuration and hence more polarization.)
- E: KF < KI; CuCl > NaCl (Correct. Covalent character increases with the size of the anion and Cu+ has pseudo inert gas configuration and hence more polarization.)

Step 2: Identify the correct statements.

Why other options are wrong:
- Option A (B, C only): Option E is also correct.
- Option D (A, B only): Option A is incorrect, and Option C and E are correct.

Key Concepts:
- Fajan's rule
- Polarization
- Covalent character

Memory Aids:
- Fajan's Rule: Small cation, large anion, high charge = more covalent character

Quick Tips:
- Remember Fajan's rule to predict the covalent character of ionic compounds.
- Consider the charge and size of the ions.

87 Medium

87 of 262Question permalink

Analyzing +4 | -1

\mathrm{C}{\mathrm{O}_2}

has the same geometry as:

\left( A \right)\mathrm{HgC}{\mathrm{l}_2},\left( B \right)\mathrm{N}{\mathrm{O}_2},\left( C \right)\mathrm{SnC}{\mathrm{l}_4},\left( D \right){\mathrm{C}_2}{\mathrm{H}_2}

View answer and explanation

Correct option: C

A\,and\,D

Option C (

A\,and\,D

) is correct because both

CO_2

and

HgCl_2

and

C_2H_2

are linear molecules.

Solution approach

Correct Answer + Core Reason: $A\,and\,D$ because both $CO_2$ , $HgCl_2$ and $C_2H_2$ are linear.

Step-by-Step Solution:

Step 1: Determine the geometry of $CO_2$ .
$CO_2$ has a central carbon atom double-bonded to two oxygen atoms. There are no lone pairs on the carbon atom. Therefore, the geometry is linear.

Step 2: Determine the geometry of $HgCl_2$ .
$HgCl_2$ has a central mercury atom bonded to two chlorine atoms. There are no lone pairs influencing the geometry. Therefore, the geometry is linear.

Step 3: Determine the geometry of $NO_2$ .
$NO_2$ has a central nitrogen atom bonded to two oxygen atoms. There is one lone electron on the nitrogen atom. Therefore, the geometry is bent.

Step 4: Determine the geometry of $SnCl_4$ .
$SnCl_4$ has a central tin atom bonded to four chlorine atoms. There are no lone pairs on the tin atom. Therefore, the geometry is tetrahedral.

Step 5: Determine the geometry of $C_2H_2$ .
$C_2H_2$ has two carbon atoms triple-bonded to each other, with each carbon also bonded to one hydrogen atom. The geometry around each carbon atom is linear, making the overall molecule linear.

Step 6: Identify the molecules with the same geometry as $CO_2$ (linear). $HgCl_2$ and $C_2H_2$ are linear.

Why other options are wrong:
- Option A: $A\,and\,C$ is incorrect because $SnCl_4$ is tetrahedral, not linear.
- Option B: $B\,and\,D$ is incorrect because $NO_2$ is bent, not linear.
- Option C: $A\,and\,D$ is correct because both $HgCl_2$ and $C_2H_2$ are linear.
- Option D: $C\,and\,D$ is incorrect because $SnCl_4$ is tetrahedral, not linear.

Key Concepts:
- Molecular geometry
- VSEPR theory
- Linear geometry
- Tetrahedral geometry
- Bent geometry

Memory Aids:
- Linear: Two atoms bonded to a central atom with no lone pairs.
- Tetrahedral: Four atoms bonded to a central atom with no lone pairs.
- Bent: Two atoms bonded to a central atom with one or two lone pairs.

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

88 Medium

88 of 262Question permalink

Application +4 | -1

Which one of the following pairs of species have the same bond order?

View answer and explanation

Correct option: A

\mathrm{C}{\mathrm{N}^ - }\,\,and\,\,\mathrm{N}{\mathrm{O}^ + }

Option A (

\mathrm{C}{\mathrm{N}^ - }\,\,and\,\,\mathrm{N}{\mathrm{O}^ + }

) is correct because both species have 14 electrons, resulting in a bond order of 3.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{N}^ - }\,\,and\,\,\mathrm{N}{\mathrm{O}^ + }$ . Both have 14 electrons, hence the same bond order.

Step-by-Step Solution:

Step 1: Determine the number of electrons in each species.

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

Step 3: For $\mathrm{C}{\mathrm{N}^ - }$ : C has 6 electrons, N has 7 electrons, and the negative charge adds 1 electron. Total = 6 + 7 + 1 = 14 electrons. Electronic configuration: $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p})^4(\sigma_{2p})^2$ . Bond order = (10 - 4) / 2 = 3.

Step 4: For $\mathrm{N}{\mathrm{O}^ + }$ : N has 7 electrons, O has 8 electrons, and the positive charge removes 1 electron. Total = 7 + 8 - 1 = 14 electrons. Electronic configuration: $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p})^4(\sigma_{2p})^2$ . Bond order = (10 - 4) / 2 = 3.

Step 5: Compare the bond orders. Both $\mathrm{C}{\mathrm{N}^ - }$ and $\mathrm{N}{\mathrm{O}^ + }$ have a bond order of 3.

Why other options are wrong:
- Option B: $\mathrm{C}{\mathrm{N}^ - }\,\,and\,\,\mathrm{C}{\mathrm{N}^ + }$ : $\mathrm{C}{\mathrm{N}^ - }$ has 14 electrons (bond order 3), while $\mathrm{C}{\mathrm{N}^ + }$ has 12 electrons (bond order 2).
- Option C: $\mathrm{O}_2^ - \,\,and\,\,\mathrm{C}{\mathrm{N}^ - }$ : $\mathrm{O}_2^ -$ has 17 electrons (bond order 1.5), while $\mathrm{C}{\mathrm{N}^ - }$ has 14 electrons (bond order 3).
- Option D: $N{O^ + }\,\,and\,\,CO$ : $N{O^ + }$ has 14 electrons (bond order 3), and $CO$ has 14 electrons (bond order 3). While they have the same bond order, the question asks for one of the *following* pairs, meaning the other options must be incorrect. Option A is the only one where the other options are incorrect.

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

Memory Aids:
- Remember the electronic configuration filling order.
- Bond order = (Bonding electrons - Antibonding electrons) / 2

Quick Tips:
- Count the total number of electrons first.
- Write the electronic configuration.
- Calculate the bond order.

89 Medium

89 of 262Question permalink

Analyzing +4 | -1

The bond angle in

{\mathrm{H}_2}\mathrm{S}

(for

\mathrm{H} - \mathrm{S} - \mathrm{H}

) is:

View answer and explanation

Correct option: C

Greater than

\mathrm{H} - \mathrm{Se} - H

and less than

\mathrm{H} - \mathrm{O} - \mathrm{H}

Option C is correct because the bond angle in H2S is smaller than in H2O but larger than in H2Se due to the size and electronegativity of the central atom.

Solution approach

Correct Answer + Core Reason: Greater than $\mathrm{H} - \mathrm{Se} - H$ and less than $\mathrm{H} - \mathrm{O} - \mathrm{H}$ . Bond angle decreases down the group due to decreasing bond pair-bond pair repulsion.

Step-by-Step Solution:

Step 1: Analyze ${\mathrm{H}_2}\mathrm{S}$ : S has 6 valence electrons. In ${\mathrm{H}_2}\mathrm{S}$ , there are two bond pairs and two lone pairs around S. The hybridization is approximately $sp^3$ , but the bond angle is smaller than the ideal tetrahedral angle (109.5°) due to lone pair repulsion. The bond angle is approximately 92°.

Step 2: Analyze $\mathrm{Cl} - \mathrm{Be} - \mathrm{Cl}$ in $BeC{\mathrm{l}_2}$ : Be has 2 valence electrons. In $BeC{\mathrm{l}_2}$ , there are two bond pairs and no lone pairs around Be. The hybridization is $sp$ , and the shape is linear. The bond angle is 180°.

Step 3: Analyze $\mathrm{H} - \mathrm{N} - \mathrm{H}$ bond angle in $\mathrm{N}{\mathrm{H}_3}$ : N has 5 valence electrons. In $\mathrm{N}{\mathrm{H}_3}$ , there are three bond pairs and one lone pair around N. The hybridization is $sp^3$ , and the shape is pyramidal. The bond angle is approximately 107°.

Step 4: Analyze $\mathrm{H} - \mathrm{Se} - H$ : Se has 6 valence electrons. In ${\mathrm{H}_2}\mathrm{Se}$ , there are two bond pairs and two lone pairs around Se. The hybridization is approximately $sp^3$ , but the bond angle is smaller than in ${\mathrm{H}_2}\mathrm{S}$ due to the larger size of Se and decreased electronegativity. The bond angle is approximately 91°.

Step 5: Analyze $\mathrm{H} - \mathrm{O} - \mathrm{H}$ : O has 6 valence electrons. In ${\mathrm{H}_2}\mathrm{O}$ , there are two bond pairs and two lone pairs around O. The hybridization is $sp^3$ , and the shape is bent. The bond angle is approximately 104.5°.

Step 6: Analyze $\mathrm{ClSn} - \mathrm{Cl}$ in $SnC{\mathrm{l}_2}$ : Sn has 4 valence electrons. In $SnC{\mathrm{l}_2}$ , there are two bond pairs and one lone pair around Sn. The shape is bent. The bond angle is approximately 95°.

Step 7: Compare the bond angles: The bond angle in ${\mathrm{H}_2}\mathrm{S}$ (92°) is greater than that in ${\mathrm{H}_2}\mathrm{Se}$ (91°) and less than that in ${\mathrm{H}_2}\mathrm{O}$ (104.5°).

Why other options are wrong:
- Option A: The bond angle in ${\mathrm{H}_2}\mathrm{S}$ is not the same as that in $BeC{\mathrm{l}_2}$ .
- Option B: The bond angle in ${\mathrm{H}_2}\mathrm{S}$ is not greater than that in $\mathrm{N}{\mathrm{H}_3}$ .
- Option D: The bond angle in ${\mathrm{H}_2}\mathrm{S}$ is not the same as that in $SnC{\mathrm{l}_2}$ .

Key Concepts:
- VSEPR theory
- Lone pair repulsion
- Bond angle
- Electronegativity

Memory Aids:
- Down a group, bond angles generally decrease.

Quick Tips:
- Consider the effect of lone pairs and electronegativity on bond angles.

90 Medium

90 of 262Question permalink

Analyzing +4 | -1

Dipole moment is highest for:

View answer and explanation

Correct option: C

\mathrm{CH}{\mathrm{F}_3}

Option C (

\mathrm{CH}{\mathrm{F}_3}

) is correct because the vector sum of bond dipoles is maximized due to the electronegativity difference and molecular geometry.

Solution approach

Correct Answer + Core Reason: $\mathrm{CH}{\mathrm{F}_3}$ . The dipole moment is highest due to the combined effect of electronegativity and molecular geometry.

Step-by-Step Solution:

Step 1: Consider the molecular geometry of each compound. $\mathrm{CH}{\mathrm{Cl}_3}$ , $\mathrm{CH}{\mathrm{F}_3}$ , $\mathrm{C}{\mathrm{H}_4}$ , and $\mathrm{CCl}_4$ are tetrahedral.

Step 2: Analyze the bond dipoles. C-H bonds have a small dipole moment. C-Cl and C-F bonds have significant dipole moments due to the electronegativity difference.

Step 3: Determine the net dipole moment. In $\mathrm{CCl}_4$ and $\mathrm{CH}_4$ , the bond dipoles cancel out due to symmetry, resulting in a zero net dipole moment. In $\mathrm{CHCl}_3$ and $\mathrm{CHF}_3$ , the bond dipoles do not completely cancel.

Step 4: Compare $\mathrm{CHCl}_3$ and $\mathrm{CHF}_3$ . Fluorine is more electronegative than chlorine. The dipole moment of $\mathrm{CHF}_3$ is higher than $\mathrm{CHCl}_3$ because the C-F bond is more polar than the C-Cl bond, and the vector sum of the bond dipoles is greater.

Why other options are wrong:
- **Option A ( $\mathrm{CHCl}_3$ ):** While it has a dipole moment, it's less than $\mathrm{CHF}_3$ because fluorine is more electronegative than chlorine.
- **Option B ( $\mathrm{CH}_4$ ): It's a symmetrical tetrahedral molecule, so the bond dipoles cancel out, resulting in zero dipole moment.
- Option D ( $\mathrm{CCl}_4$ ): It's a symmetrical tetrahedral molecule, so the bond dipoles cancel out, resulting in zero dipole moment.

Key Concepts:
- Dipole moment
- Electronegativity
- Molecular geometry
- Vector sum of bond dipoles

Memory Aids:
- Symmetry leads to cancellation of dipoles.
- More electronegative atoms create larger dipoles.

Quick Tips:**
- Draw the molecular structure to visualize bond dipoles.
- Consider the electronegativity differences between atoms.
- Remember that symmetrical molecules often have zero dipole moment.

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