The covalent compound ${\rm{HCl}}$ has the polar character because:

The electronegativity of hydrogen is greater than that of chlorine

The electronegativity of hydrogen is equal to than that of chlorine

Hydrogen and chlorine are gases

The solubility of ${\rm{KCl}}$ is relatively more in (where $D$ is dielectric constant):

${\rm{C}}{{\rm{H}}_3}{\rm{OH}}\left( {{\rm{D}} = 32} \right)$

${{\rm{C}}_6}{{\rm{H}}_6}({\rm{D}} = 0)$

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_2}{\rm{CO}}\left( {{\rm{D}} = 2} \right)$

${\rm{CC}}{{\rm{l}}_4}\left( {{\rm{D}} = 0} \right)$

${\rm{N}}{{\rm{F}}_3}$ is:

Having low value of dipole moment than ${\rm{N}}{{\rm{H}}_3}$

Having more dipole moment than ${\rm{N}}{{\rm{H}}_3}$

CHEMICAL BONDING & MOLECULAR STRUCTURE MCQs for NEET UG CHEMISTRY — Medium, Page 4

91 Medium

91 of 262Question permalink

Remembering +4 | -1

Which of the following substances has giant covalent structure

View answer and explanation

Correct option: C

Silica

Option C (Silica) is correct because silica (SiO₂) has a giant network structure with strong covalent bonds throughout.

Solution approach

Correct Answer + Core Reason: Silica. Silica (SiO₂) has a giant network structure with strong covalent bonds throughout.

Step-by-Step Solution:

Step 1: A giant covalent structure is a network of atoms held together by covalent bonds throughout the entire structure.

Step 2: Silica (SiO₂) forms a giant network where each silicon atom is covalently bonded to four oxygen atoms, and each oxygen atom is covalently bonded to two silicon atoms.

Step 3: This creates a continuous network of strong covalent bonds.

Why other options are wrong:
- Option A: Iodine crystal consists of discrete I₂ molecules held together by weak van der Waals forces.
- Option B: Solid CO₂ consists of discrete CO₂ molecules held together by weak van der Waals forces.
- Option D: White phosphorus consists of discrete P₄ molecules held together by covalent bonds within the molecule, but weak van der Waals forces between molecules.

Key Concepts:
- Giant covalent structures
- Structure of silica

Memory Aids:
- Silica = Giant Covalent Network

Quick Tips:
- Remember common examples of giant covalent structures (diamond, graphite, silica).

92 Medium

92 of 262Question permalink

Analyzing +4 | -1

The correct sequence of decreasing number of

\pi

–bonds in the structures of

{H_2}S{O_3},{H_2}S{O_4}\,\,\,and\,\,{H_2}{S_2}{O_7}

is

View answer and explanation

Correct option: C

\,\,{H_2}{S_2}{O_7} > {H_2}S{O_4} > {H_2}S{O_3}

Option C (

\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7 > \mathrm{H}_2\mathrm{S}\mathrm{O}_4 > \mathrm{H}_2\mathrm{S}\mathrm{O}_3

) is correct because

{H_2}{S_2}{O_7}

has the most pi bonds, followed by

{H_2}S{O_4}

, and then

{H_2}S{O_3}

.

Solution approach

Correct Answer + Core Reason: $\H_2{S_2}{O_7} > {H_2}S{O_4} > {H_2}S{O_3}$ . The number of $\pi$ bonds decreases in the order ${H_2}{S_2}{O_7} > {H_2}S{O_4} > {H_2}S{O_3}$ .

Step-by-Step Solution:

Step 1: Draw the Lewis structures of ${H_2}S{O_3}$ , ${H_2}S{O_4}$ , and ${H_2}{S_2}{O_7}$ .

Step 2: ${H_2}S{O_3}$ has one S=O bond (one $\pi$ bond).

Step 3: ${H_2}S{O_4}$ has two S=O bonds (two $\pi$ bonds).

Step 4: ${H_2}{S_2}{O_7}$ has two S=O bonds per sulfur atom (four $\pi$ bonds total).

Step 5: Compare the number of $\pi$ bonds in each molecule.

Step 6: The correct sequence of decreasing number of $\pi$ bonds is ${H_2}{S_2}{O_7} > {H_2}S{O_4} > {H_2}S{O_3}$ .

Why other options are wrong:
- Option A: Incorrect order of decreasing $\pi$ bonds.
- Option B: Incorrect order of decreasing $\pi$ bonds.
- Option D: Incorrect order of decreasing $\pi$ bonds.

Key Concepts:
- Lewis structures
- $\pi$ bonds
- Structural formulas

Memory Aids:
- Draw the structures to visualize the bonds.

Quick Tips:
- Draw the structures to easily identify the number of $\pi$ bonds.

93 Medium

93 of 262Question permalink

Understanding +4 | -1

{\mathrm{H}_2}\mathrm{O}

boils at higher temperature than

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

because it is capable of forming:

View answer and explanation

Correct option: C

Hydrogen bonds

Option C (Hydrogen bonds) is correct because water molecules form strong hydrogen bonds due to the high electronegativity of oxygen, leading to higher boiling points compared to hydrogen sulfide.

Solution approach

Correct Answer + Core Reason: Hydrogen bonds. Water forms strong hydrogen bonds, leading to a higher boiling point than hydrogen sulfide.

Step-by-Step Solution:

Step 1: Identify the intermolecular forces present in ${\mathrm{H}_2}\mathrm{O}$ and ${\mathrm{H}_2}\mathrm{S}$ .

Step 2: ${\mathrm{H}_2}\mathrm{O}$ can form hydrogen bonds due to the high electronegativity of oxygen and the presence of hydrogen atoms directly bonded to oxygen.
${\mathrm{H}_2}\mathrm{S}$ can only form weaker dipole-dipole interactions.

Step 3: Hydrogen bonds are stronger than dipole-dipole interactions.

Step 4: Stronger intermolecular forces require more energy to overcome, resulting in a higher boiling point.

Why other options are wrong:
- Option A: Ionic bonds are formed between ions, not molecules like water or hydrogen sulfide.
- Option B: Covalent bonds are intramolecular forces, not intermolecular forces responsible for boiling point differences.
- Option D: Metallic bonds are found in metals, not in water or hydrogen sulfide.

Key Concepts:
- Intermolecular forces
- Hydrogen bonding
- Boiling point
- Electronegativity

Memory Aids:
- Remember that hydrogen bonds occur when hydrogen is bonded to highly electronegative atoms like oxygen, nitrogen, or fluorine.

Quick Tips:
- Consider the intermolecular forces when comparing boiling points of different compounds.

94 Medium

94 of 262Question permalink

Understanding +4 | -1

The covalent compound

\mathrm{HCl}

has the polar character because:

View answer and explanation

Correct option: C

The electronegativity of chlorine is greater than that of hydrogen

Option C (The electronegativity of chlorine is greater than that of hydrogen) is correct because the difference in electronegativity between Cl and H creates a dipole moment, making the HCl molecule polar.

Solution approach

Correct Answer + Core Reason: The electronegativity of chlorine is greater than that of hydrogen. This difference in electronegativity leads to a polar bond.

Step-by-Step Solution:

Step 1: Define electronegativity: Electronegativity is the ability of an atom to attract electrons in a chemical bond.

Step 2: Consider the electronegativity values of H and Cl. Chlorine is more electronegative than hydrogen.

Step 3: When two atoms with different electronegativities form a covalent bond, the more electronegative atom attracts the shared electrons more strongly.

Step 4: This unequal sharing of electrons creates a dipole moment, with a partial negative charge on the more electronegative atom (Cl) and a partial positive charge on the less electronegative atom (H).

Step 5: A molecule with a dipole moment is considered polar.

Why other options are wrong:
- Option A: The electronegativity of hydrogen is *not* greater than that of chlorine.
- Option B: The electronegativity of hydrogen is *not* equal to that of chlorine.
- Option D: The fact that hydrogen and chlorine are gases is irrelevant to the polarity of the HCl bond.

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

Memory Aids:
- More electronegative atom = partial negative charge
- Less electronegative atom = partial positive charge

Quick Tips:
- Remember the electronegativity trends in the periodic table: electronegativity increases across a period and decreases down a group.

95 Medium

95 of 262Question permalink

Remembering +4 | -1

The nature of the bond in diamond is

View answer and explanation

Correct option: B

Covalent

Option B (Covalent) is correct because diamond consists of carbon atoms covalently bonded in a tetrahedral network.

Solution approach

Correct Answer + Core Reason: Covalent. Diamond consists of carbon atoms covalently bonded in a tetrahedral network.

Step-by-Step Solution:

Step 1: Diamond is a form of carbon where each carbon atom is bonded to four other carbon atoms in a tetrahedral arrangement.

Step 2: These bonds are formed by sharing electrons between the carbon atoms.

Step 3: Sharing of electrons defines a covalent bond.

Step 4: Therefore, the nature of the bond in diamond is covalent.

Why other options are wrong:
- Option A: Ionic bonds involve the transfer of electrons, which does not occur in diamond.
- Option C: Metallic bonds involve a sea of delocalized electrons, which is not the bonding type in diamond.
- Option D: Coordinate covalent bonds involve one atom donating both electrons to the bond, which is not the primary bonding type in diamond.

Key Concepts:
- Covalent bonding
- Structure of diamond

Memory Aids:
- Covalent bonds involve sharing of electrons.

Quick Tips:
- Remember the structure of common allotropes of carbon (diamond, graphite, fullerenes).

96 Medium

96 of 262Question permalink

Understanding +4 | -1

Which of the following shows minimum melting point?

View answer and explanation

Correct option: A

Naphthalene

Option A (Naphthalene) is correct because it is a molecular solid with weak van der Waals forces, leading to a lower melting point compared to the other options which are ionic, covalent network, or metallic solids.

Solution approach

Correct Answer + Core Reason: Naphthalene. Molecular solids have the weakest intermolecular forces and thus the lowest melting points.

Step-by-Step Solution:

Step 1: Identify the type of solid for each option.

Step 2: Naphthalene is a molecular solid. Diamond is a covalent network solid. NaCl is an ionic solid. Mn is a metallic solid.

Step 3: Molecular solids have weak van der Waals forces. Covalent network solids have strong covalent bonds throughout the structure. Ionic solids have strong electrostatic forces between ions. Metallic solids have metallic bonds.

Step 4: The strength of intermolecular forces determines the melting point. Weak forces lead to lower melting points.

Step 5: Therefore, Naphthalene has the lowest melting point.

Why other options are wrong:
- Option B: Diamond has a very high melting point due to its strong covalent network structure.
- Option C: NaCl has a high melting point due to strong electrostatic forces between ions.
- Option D: Mn has a moderate to high melting point due to metallic bonding.

Key Concepts:
- Types of solids (molecular, covalent network, ionic, metallic)
- Intermolecular forces (van der Waals forces, electrostatic forces, metallic bonds, covalent bonds)
- Melting point and its relation to intermolecular forces

Memory Aids:
- Molecular solids = weak forces = low melting point
- Ionic solids = strong forces = high melting point

Quick Tips:
- Remember the general trend: Molecular < Metallic < Ionic < Covalent Network in terms of melting point.

97 Medium

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Analyzing +4 | -1

The pair likely to form the strongest hydrogen bonding:

View answer and explanation

Correct option: B

\mathrm{HCOOH and C}{\mathrm{H}_3}\mathrm{COOH}

Option B is correct because both formic acid (HCOOH) and acetic acid (CH3COOH) have both a hydrogen bond donor (the -OH group) and a hydrogen bond acceptor (the carbonyl oxygen), allowing for strong intermolecular hydrogen bonding.

Solution approach

Correct Answer + Core Reason: $\mathrm{HCOOH and C}{\mathrm{H}_3}\mathrm{COOH}$ . Both molecules can act as hydrogen bond donors and acceptors, leading to strong hydrogen bonding.

Step-by-Step Solution:

Step 1: Analyze the structure of each pair of molecules.

Step 2: ${\mathrm{H}_2}{\mathrm{O}_2}\,\mathrm{and}\,{\mathrm{H}_2}\mathrm{O}$ : Both can form hydrogen bonds, but the strength of hydrogen bonding between them might not be the strongest compared to other options.

Step 3: $\mathrm{HCOOH and C}{\mathrm{H}_3}\mathrm{COOH}$ : Both formic acid and acetic acid have -OH groups (hydrogen bond donors) and carbonyl groups (hydrogen bond acceptors). This allows for strong intermolecular hydrogen bonding.

Step 4: $\mathrm{C}{\mathrm{H}_3}\mathrm{COOH and C}{\mathrm{H}_3}\mathrm{COOC}{\mathrm{H}_3}$ : Acetic acid has -OH group (hydrogen bond donor) and carbonyl group (hydrogen bond acceptor), while methyl acetate only has a carbonyl group (hydrogen bond acceptor). The hydrogen bonding is not as strong as in option B.

Step 5: $\mathrm{Si}{\mathrm{H}_4}\,\mathrm{and SiC}{\mathrm{l}_4}$ : These are non-polar molecules and do not form hydrogen bonds.

Step 6: Compare the hydrogen bonding strength of each pair. The strongest hydrogen bonding is expected between $\mathrm{HCOOH and C}{\mathrm{H}_3}\mathrm{COOH}$ .

Why other options are wrong:
- Option A: While both can form hydrogen bonds, the interaction might not be as strong as between two carboxylic acids.
- Option C: Methyl acetate can only act as a hydrogen bond acceptor, while acetic acid can act as both a donor and acceptor. This makes the hydrogen bonding weaker compared to option B.
- Option D: Silicon compounds do not form strong hydrogen bonds due to the lower electronegativity of silicon.

Key Concepts:
- Hydrogen bonding
- Hydrogen bond donor and acceptor
- Intermolecular forces
- Electronegativity

Memory Aids:
- Hydrogen bonding occurs between molecules containing H bonded to highly electronegative atoms (O, N, F).
- Strongest hydrogen bonds occur when both molecules can act as donors and acceptors.

Quick Tips:
- Identify the hydrogen bond donors and acceptors in each molecule.
- Consider the strength of the hydrogen bonds based on the electronegativity of the atoms involved.

98 Medium

98 of 262Question permalink

Remembering +4 | -1

The shape of carbon dioxide is

View answer and explanation

Correct option: D

linear

Option D (linear) is correct because carbon dioxide (

CO_2

) has a linear shape due to the

sp

hybridization of the central carbon atom and two double bonds to oxygen atoms.

Solution approach

Correct Answer + Core Reason: linear. $CO_2$ has $sp$ hybridization and two double bonds.

Step-by-Step Solution:

Step 1: Draw the Lewis structure of carbon dioxide.\nO=C=O

Step 2: Determine the hybridization of the central carbon atom.\nThe carbon atom is $sp$ hybridized.

Step 3: Determine the shape of the molecule.\nThe molecule is linear.

Why other options are wrong:
- Option A: Pyramidal is incorrect. Pyramidal shapes typically have a central atom with a lone pair and three bonding pairs.
- Option B: Tetrahedral is incorrect. Tetrahedral shapes have a central atom with four bonding pairs and no lone pairs.
- Option C: Planar is incorrect. While the atoms are in a plane, the shape is specifically linear rather than a more general planar shape.

Key Concepts:
- Lewis structure
- Hybridization
- Molecular shape

Memory Aids:
- $CO_2$ is a classic example of a linear molecule.

Quick Tips:
- Draw the Lewis structure to determine the shape.

99 Medium

99 of 262Question permalink

Understanding +4 | -1

Which molecule is planar?

View answer and explanation

Correct option: C

{\mathrm{C}_2}{\mathrm{H}_4}

Option C (

{\mathrm{C}_2}{\mathrm{H}_4}

) is correct because ethylene (

C_2H_4

) has a planar structure due to the double bond between the two carbon atoms, forcing all atoms to lie in the same plane.

Solution approach

Correct Answer + Core Reason: ${\mathrm{C}_2}{\mathrm{H}_4}$ . Ethylene is planar.

Step-by-Step Solution:

Step 1: Consider the structure of each molecule.

Step 2: $\mathrm{N}{\mathrm{H}_3}$ is trigonal pyramidal and not planar.

Step 3: $\mathrm{C}{\mathrm{H}_4}$ is tetrahedral and not planar.

Step 4: ${\mathrm{C}_2}{\mathrm{H}_4}$ has a double bond between the two carbon atoms, and each carbon is $sp^2$ hybridized, resulting in a planar structure.

Step 5: $\mathrm{SiC}{\mathrm{l}_4}$ is tetrahedral and not planar.

Why other options are wrong:
- Option A: $\mathrm{N}{\mathrm{H}_3}$ is trigonal pyramidal.
- Option B: $\mathrm{C}{\mathrm{H}_4}$ is tetrahedral.
- Option D: $\mathrm{SiC}{\mathrm{l}_4}$ is tetrahedral.

Key Concepts:
- Molecular geometry
- Planar molecules
- Hybridization

Memory Aids:
- Ethylene is planar.

Quick Tips:
- Double bonds often lead to planar structures.

100 Medium

100 of 262Question permalink

Understanding +4 | -1

The solubility of

\mathrm{KCl}

is relatively more in (where

D

is dielectric constant):

View answer and explanation

Correct option: C

\mathrm{C}{\mathrm{H}_3}\mathrm{OH}\left( {\mathrm{D} = 32} \right)

Option C (

\mathrm{CH}_3\mathrm{OH}(\mathrm{D} = 32)

) is correct because KCl, being an ionic compound, dissolves more in polar solvents with higher dielectric constants.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{H}_3}\mathrm{OH}\left( {\mathrm{D} = 32} \right)$ . Solubility of ionic compounds is higher in solvents with high dielectric constant.

Step-by-Step Solution:

Step 1: Identify the nature of KCl. KCl is an ionic compound.

Step 2: Recall the principle of \"like dissolves like.\" Ionic compounds dissolve better in polar solvents.

Step 3: Relate dielectric constant to polarity. Higher dielectric constant indicates higher polarity.

Step 4: Compare the dielectric constants of the given solvents. $\mathrm{CH}_3\mathrm{OH}$ has the highest dielectric constant (D = 32).

Step 5: Conclude that KCl will be most soluble in $\mathrm{CH}_3\mathrm{OH}$ .

Why other options are wrong:
- Option A: ${\mathrm{C}_6}{\mathrm{H}_6}(\mathrm{D} = 0)$ is incorrect because benzene is a non-polar solvent with a dielectric constant of 0, making it a poor solvent for ionic compounds like KCl.
- Option B: ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CO}\left( {\mathrm{D} = 2} \right)$ is incorrect because acetone is a polar solvent, but its dielectric constant (2) is significantly lower than that of methanol (32), making it a less effective solvent for KCl.
- Option D: $\mathrm{CC}{\mathrm{l}_4}\left( {\mathrm{D} = 0} \right)$ is incorrect because carbon tetrachloride is a non-polar solvent with a dielectric constant of 0, making it a poor solvent for ionic compounds like KCl.

Key Concepts:
- Solubility of ionic compounds
- Polarity of solvents
- Dielectric constant
- \"Like dissolves like\

101 Medium

101 of 262Question permalink

Application +4 | -1

The bond order is not three for

View answer and explanation

Correct option: A

{N}_2^{+}

Option A (

{N}_2^{+}

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

Solution approach

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

Step-by-Step Solution:

Step 1: Calculate Bond Order of ${N}_2^{+}$ : Electronic configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p})^4(\sigma_{2p})^1$ . Bond order = (9 - 4)/2 = 2.5

Step 2: Calculate Bond Order of ${O}_2^{2+}$ : Electronic configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p})^2(\pi_{2p})^4$ . Bond order = (10 - 4)/2 = 3

Step 3: Calculate Bond Order of ${N}_2$ : Electronic configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p})^4(\sigma_{2p})^2$ . Bond order = (10 - 4)/2 = 3

Step 4: Calculate Bond Order of ${NO}^{+}$ : Electronic configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p})^4(\sigma_{2p})^2$ . Bond order = (10 - 4)/2 = 3

Step 5: Conclusion: ${N}_2^{+}$ has a bond order of 2.5, which is not three.

Why other options are wrong:
- Option B: ${O}_2^{2+}$ has a bond order of 3.
- Option C: ${N}_2$ has a bond order of 3.
- Option D: ${NO}^{+}$ has a bond order of 3.

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

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

Quick Tips:
- Write out the electronic configurations to determine the number of bonding and antibonding electrons.

102 Medium

102 of 262Question permalink

Analyzing +4 | -1

Which one of the following pairs of molecules will have permanent dipole moments for both members?

View answer and explanation

Correct option: C

\mathrm{N}{\mathrm{O}_2}

and

{\mathrm{O}_3}

Option C (

\mathrm{N}{\mathrm{O}_2}

and

{\mathrm{O}_3}

) is correct because both nitrogen dioxide (

\mathrm{N}{\mathrm{O}_2}

) and ozone (

{\mathrm{O}_3}

) have bent structures, leading to a non-zero net dipole moment.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{O}_2}$ and ${\mathrm{O}_3}$ . Both molecules have bent structures, resulting in permanent dipole moments.

Step-by-Step Solution:

Step 1: Analyze the molecular geometry of each molecule:
- $\mathrm{SiF}_4$ : Tetrahedral, symmetrical, zero dipole moment.
- $\mathrm{NO}_2$ : Bent, non-zero dipole moment.
- $\mathrm{CO}_2$ : Linear, symmetrical, zero dipole moment.
- $\mathrm{O}_3$ : Bent, non-zero dipole moment.

Step 2: Determine which pairs have permanent dipole moments for both members:
- $\mathrm{SiF}_4$ and $\mathrm{NO}_2$ : $\mathrm{SiF}_4$ has zero dipole moment, $\mathrm{NO}_2$ has a non-zero dipole moment.
- $\mathrm{NO}_2$ and $\mathrm{CO}_2$ : $\mathrm{NO}_2$ has a non-zero dipole moment, $\mathrm{CO}_2$ has zero dipole moment.
- $\mathrm{NO}_2$ and $\mathrm{O}_3$ : Both $\mathrm{NO}_2$ and $\mathrm{O}_3$ have non-zero dipole moments.
- $\mathrm{SiF}_4$ and $\mathrm{CO}_2$ : Both $\mathrm{SiF}_4$ and $\mathrm{CO}_2$ have zero dipole moments.

Why other options are wrong:
- **Option A ( $\mathrm{Si}{\mathrm{F}_4}$ and $\mathrm{N}{\mathrm{O}_2}$ ):** $\mathrm{SiF}_4$ has a tetrahedral structure and zero dipole moment.
- **Option B ( $\mathrm{N}{\mathrm{O}_2}$ and $\mathrm{C}{\mathrm{O}_2}$ ):** $\mathrm{CO}_2$ has a linear structure and zero dipole moment.
- **Option D ( $\mathrm{Si}{\mathrm{F}_4}$ and $\mathrm{C}{\mathrm{O}_2}$ ):** Both $\mathrm{SiF}_4$ and $\mathrm{CO}_2$ have zero dipole moments.

Key Concepts:
- Dipole moment
- Molecular geometry (linear, bent, tetrahedral)
- Symmetry

Memory Aids:
- Linear and symmetrical molecules often have zero dipole moment.
- Bent molecules often have non-zero dipole moments.

Quick Tips:
- Draw the Lewis structures and determine the molecular geometry.
- Consider the symmetry of the molecule.

103 Medium

103 of 262Question permalink

Understanding +4 | -1

Strongest bond is formed by the head on overlapping of:

View answer and explanation

Correct option: B

2p

and

2p

-orbitals

Option B (

2p

and

2p

-orbitals) is correct because head-on overlapping of p-orbitals forms a sigma bond, which is stronger than bonds formed by s-orbitals.

Solution approach

Correct Answer + Core Reason: $2p$ and $2p$ -orbitals. Sigma bonds formed by p-p overlap are stronger due to greater extent of overlap.

Step-by-Step Solution:

Step 1: Identify the types of orbitals involved: $s$ and $p$ orbitals.

Step 2: Recall the types of bonds formed by head-on overlapping: sigma ( $\sigma$ ) bonds.

Step 3: Consider the extent of overlapping: $p$ orbitals have directional properties and can overlap more effectively head-on than $s$ orbitals.

Step 4: Relate the extent of overlapping to bond strength: greater overlap leads to stronger bonds.

Step 5: Conclude that $2p$ and $2p$ head-on overlapping forms the strongest bond.

Why other options are wrong:
- Option A: $2s$ and $2p$ -orbitals: Overlap is less effective than $2p-2p$ .
- Option C: $2s$ and $2s$ -orbitals: $s$ orbitals are non-directional, leading to weaker overlap.
- Option D: All of these: This is incorrect as $2p-2p$ overlap is the strongest.

Key Concepts:
- Sigma ( $\sigma$ ) bonds
- Head-on overlapping
- Extent of overlapping and bond strength
- Directional properties of $p$ orbitals

Memory Aids:
- Sigma bonds are stronger than pi bonds (single bond is sigma, double bond is sigma + pi, triple bond is sigma + 2 pi).
- p-orbitals are directional and form stronger sigma bonds.

Quick Tips:
- Remember that head-on overlap forms sigma bonds, while sideways overlap forms pi bonds.
- Consider the directional properties of orbitals when comparing bond strengths.

104 Medium

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Application +4 | -1

Which has triangular planar shape?

View answer and explanation

Correct option: A

\mathrm{CH}_3^ +

Option A (

\mathrm{CH}_3^ +

) is correct because the central carbon atom in

CH_3^+

has three bond pairs and no lone pairs, resulting in a triangular planar shape.

Solution approach

Correct Answer + Core Reason: $\mathrm{CH}_3^ +$ . The central carbon atom has three bond pairs and no lone pairs, resulting in a triangular planar shape.

Step-by-Step Solution:

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

Step 2: Determine the number of bond pairs and lone pairs around the central atom.\nFor $\mathrm{CH}_3^ +$ : Carbon has 3 bond pairs and 0 lone pairs. Triangular planar shape.\nFor $\mathrm{ClO}_2^ -$ : Chlorine has 2 bond pairs and 2 lone pairs. Bent shape.\nFor ${\mathrm{H}_3}{\mathrm{O}^ + }$ : Oxygen has 3 bond pairs and 1 lone pair. Pyramidal shape.\nFor $\mathrm{ClO}_3^ -$ : Chlorine has 3 bond pairs and 1 lone pair. Pyramidal shape.

Step 3: Determine the shape based on the number of bond pairs and lone pairs.

Why other options are wrong:
- **Option B ( $\mathrm{ClO}_2^ -$ ):** $\mathrm{ClO}_2^ -$ has a bent shape due to two lone pairs on the chlorine atom.
- **Option C ( ${\mathrm{H}_3}{\mathrm{O}^ + }$ ):** ${\mathrm{H}_3}{\mathrm{O}^ + }$ has a pyramidal shape due to one lone pair on the oxygen atom.
- **Option D ( $\mathrm{ClO}_3^ -$ ):** $\mathrm{ClO}_3^ -$ has a pyramidal shape due to one lone pair on the chlorine atom.

Key Concepts:
- Lewis structure
- VSEPR theory
- Molecular shape

Memory Aids:
- Triangular planar: 3 bond pairs, 0 lone pairs
- Bent: 2 bond pairs, 1 or 2 lone pairs
- Pyramidal: 3 bond pairs, 1 lone pair

Quick Tips:
- Draw the Lewis structure first to determine the number of bond pairs and lone pairs.

105 Medium

105 of 262Question permalink

Remembering +4 | -1

The O-H bond distance in water molecule is:

View answer and explanation

Correct option: C

0.96\,{A^0}

Option C (

0.96\,{A^0}

) is correct because the O-H bond distance in a water molecule is approximately 0.96 Angstroms, a known experimental value.

Solution approach

Correct Answer + Core Reason: $0.96\,{A^0}$ . The O-H bond distance in water is a known experimental value.

Step-by-Step Solution:

Step 1: Recall the known bond distance of O-H in water.

Step 2: The O-H bond distance in water is approximately 0.96 Angstroms.

Step 3: State the answer.

Step 4: Final Answer: $0.96\,{A^0}$

Why other options are wrong:
- Option A: $1.0\,{A^0}$ is incorrect because it is slightly higher than the actual bond distance.
- Option B: $1.33\,{A^0}$ is incorrect because it is significantly higher than the actual bond distance.
- Option D: $1.45\,{A^0}$ is incorrect because it is significantly higher than the actual bond distance.

Key Concepts:
- Molecular structure of water
- Bond lengths

Memory Aids:
- None applicable for this factual recall.

Quick Tips:
- Remember common bond lengths for frequently encountered molecules.

106 Medium

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Analyzing +4 | -1

The correct increasing bond angles order is:

View answer and explanation

Correct option: B

\mathrm{Cl}{\mathrm{F}_3} < P{\mathrm{F}_3} < N{\mathrm{F}_3} < B{\mathrm{F}_3}

Option B (

\mathrm{Cl}{\mathrm{F}_3} < P{\mathrm{F}_3} < N{\mathrm{F}_3} < B{\mathrm{F}_3}

) is correct because it correctly orders the bond angles based on the number of lone pairs and electronegativity differences.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cl}{\mathrm{F}_3} < P{\mathrm{F}_3} < N{\mathrm{F}_3} < B{\mathrm{F}_3}$ . The order is determined by the number of lone pairs and the electronegativity of the surrounding atoms.

Step-by-Step Solution:

Step 1: Determine the geometry and lone pairs for each molecule:
- $\mathrm{B}{\mathrm{F}_3}$ : Boron has 3 valence electrons. It forms a trigonal planar structure with no lone pairs. Bond angle = 120°.
- $\mathrm{N}{\mathrm{F}_3}$ : Nitrogen has 5 valence electrons. It forms a trigonal pyramidal structure with one lone pair. Since F is more electronegative than H, the bond angle is smaller than in $\mathrm{N}{\mathrm{H}_3}$ . Bond angle < 107°.
- $\mathrm{P}{\mathrm{F}_3}$ : Phosphorus has 5 valence electrons. It forms a trigonal pyramidal structure with one lone pair. The bond angle is smaller than in $\mathrm{N}{\mathrm{F}_3}$ due to lower electronegativity of P compared to N.
- $\mathrm{Cl}{\mathrm{F}_3}$ : Chlorine has 7 valence electrons. It forms a T-shaped structure with two lone pairs. Bond angle is approximately 90°.

Step 2: Arrange the bond angles in increasing order based on the number of lone pairs and electronegativity:
$\mathrm{Cl}{\mathrm{F}_3} < P{\mathrm{F}_3} < N{\mathrm{F}_3} < B{\mathrm{F}_3}$

Why other options are wrong:
- **Option A ( $\mathrm{B}{\mathrm{F}_3} < N{\mathrm{F}_3} < P{\mathrm{F}_3} < Cl{\mathrm{F}_3}$ ):** Incorrect order. $\mathrm{B}{\mathrm{F}_3}$ has the largest bond angle, and $\mathrm{Cl}{\mathrm{F}_3}$ has the smallest.
- **Option C ( $\mathrm{B}{\mathrm{F}_3} \approx \mathrm{N}{\mathrm{F}_3} < P{\mathrm{F}_3} < Cl{\mathrm{F}_3}$ ):** Incorrect order. $\mathrm{B}{\mathrm{F}_3}$ has a significantly larger bond angle than $\mathrm{N}{\mathrm{F}_3}$ .
- **Option D ( $\mathrm{B}{\mathrm{F}_3} < N{\mathrm{F}_3} < P{\mathrm{F}_3} > Cl{\mathrm{F}_3}$ ):** Incorrect order. $\mathrm{Cl}{\mathrm{F}_3}$ has the smallest bond angle.

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

Memory Aids:
- More lone pairs lead to smaller bond angles.
- Higher electronegativity of surrounding atoms decreases bond angle.

Quick Tips:
- Determine the number of lone pairs and the electronegativity of the surrounding atoms to predict the bond angle.

107 Medium

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Analyzing +4 | -1

Among the following isostructural compounds, identify the compound which has the highest lattice energy

View answer and explanation

Correct option: D

MgO

Option D (

MgO

) is correct because lattice energy is directly proportional to the product of charges and inversely proportional to the distance between ions;

Mg^{2+}

and

O^{2-}

have higher charges than the other options.

Solution approach

Correct Answer + Core Reason: $MgO$ . Higher charges on ions lead to higher lattice energy.

Step-by-Step Solution:

Step 1: Understand the question. We need to identify the compound with the highest lattice energy among isostructural compounds.

Step 2: Recall the factors affecting lattice energy. Lattice energy is directly proportional to the product of the charges of the ions and inversely proportional to the distance between them.

Step 3: Compare the charges of the ions in each compound.
- $LiF$ : +1 and -1
- $LiCl$ : +1 and -1
- $NaCl$ : +1 and -1
- $MgO$ : +2 and -2

Step 4: Since $MgO$ has ions with higher charges (+2 and -2), it will have the highest lattice energy.

Why other options are wrong:
- Option A: $LiF$ has lower charges (+1 and -1) compared to $MgO$ , resulting in lower lattice energy.
- Option B: $LiCl$ has lower charges (+1 and -1) compared to $MgO$ , resulting in lower lattice energy.
- Option C: $NaCl$ has lower charges (+1 and -1) compared to $MgO$ , resulting in lower lattice energy.

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

Memory Aids:
- Higher charge = Higher lattice energy

Quick Tips:
- Focus on the charges of the ions when comparing lattice energies of isostructural compounds.

108 Medium

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Application +4 | -1

The bond order in

{N}_2{ }^{+}

is

View answer and explanation

Correct option: C

2.5

Option C (2.5) is correct because the bond order in

{N}_2{ }^{+}

is calculated using the molecular orbital configuration and the formula (bonding electrons - antibonding electrons)/2.

Solution approach

Correct Answer + Core Reason: 2.5. The bond order is calculated using the molecular orbital configuration.

Step-by-Step Solution:

Step 1: Determine the number of electrons in ${N}_2{ }^{+}$ .
- Nitrogen has 7 electrons, so $N_2$ has 14 electrons. $N_2^+$ has 14 - 1 = 13 electrons.

Step 2: Write the electronic configuration for ${N}_2{ }^{+}$ using molecular orbital theory.
- $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})^4(\pi_{2p}^*)^1$

Step 3: Calculate the bond order using the formula: Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2
- Bonding electrons: 2 ( $\sigma_{2s}$ ) + 2 ( $\sigma_{2p}$ ) + 4 ( $\pi_{2p}$ ) = 8
- Antibonding electrons: 2 ( $\sigma_{2s}^*$ ) + 1( $\pi_{2p}^*$ ) = 3
- Bond order = (8 - 3) / 2 = 2.5

Why other options are wrong:
- Option A: 1.5: Incorrect calculation of bond order.
- Option B: 3: Incorrect calculation of bond order.
- Option D: 2: Incorrect calculation of bond order.

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.

109 Medium

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Remembering +4 | -1

For the formation of covalent bond, the difference in the value of electronegativities should be:

View answer and explanation

Correct option: A

Equal to or less than

1.7

Option A is correct because a difference in electronegativity equal to or less than 1.7 generally indicates a covalent bond.

Solution approach

Correct Answer + Core Reason: Equal to or less than $1.7$ . Electronegativity difference less than or equal to 1.7 indicates a covalent bond.

Step-by-Step Solution:

Step 1: Understand the relationship between electronegativity difference and bond type.

Step 2: Recall that electronegativity is the ability of an atom to attract electrons in a chemical bond.

Step 3: A large electronegativity difference (typically greater than 1.7) indicates an ionic bond, where electrons are transferred from one atom to another.

Step 4: A small electronegativity difference (typically less than or equal to 1.7) indicates a covalent bond, where electrons are shared between atoms.

Step 5: Therefore, for the formation of a covalent bond, the difference in electronegativities should be equal to or less than 1.7.

Why other options are wrong:
- Option B: More than 1.7 indicates an ionic bond.
- Option C: 1.7 or more indicates an ionic bond.
- Option D: The electronegativity difference is a key factor in determining bond type.

Key Concepts:
- Electronegativity
- Ionic bond
- Covalent bond

Memory Aids:
- Large electronegativity difference = Ionic bond
- Small electronegativity difference = Covalent bond

Quick Tips:
- Remember the typical electronegativity difference ranges for ionic and covalent bonds.

110 Medium

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Application +4 | -1

Which of the following will be octahedral

View answer and explanation

Correct option: A

S F_6

Option A (

S F_6

) is correct because the sulfur atom in

SF_6

is

sp^3d^2

hybridized, resulting in an octahedral geometry with six bonding pairs and no lone pairs.

Solution approach

Correct Answer + Core Reason: $S F_6$ . Sulfur is $sp^3d^2$ hybridized, resulting in an octahedral geometry.

Step-by-Step Solution:

Step 1: Determine the hybridization and shape of $SF_6$ .

Step 2: S has 6 valence electrons. In $SF_6$ , all six are used for bonding with F atoms. So, there are 0 lone pairs. The steric number is 6 (6 bond pairs + 0 lone pairs), leading to $sp^3d^2$ hybridization. The shape is octahedral.

Step 3: Determine the hybridization and shape of $BF_4^-$ .

Step 4: B has 3 valence electrons. In $BF_4^-$ , four are used for bonding with F atoms, and one negative charge adds one electron. So, there are 0 lone pairs. The steric number is 4 (4 bond pairs + 0 lone pairs), leading to $sp^3$ hybridization. The shape is tetrahedral.

Step 5: Determine the hybridization and shape of $PCl_5$ .

Step 6: P has 5 valence electrons. In $PCl_5$ , all five are used for bonding with Cl atoms. So, there are 0 lone pairs. The steric number is 5 (5 bond pairs + 0 lone pairs), leading to $sp^3d$ hybridization. The shape is trigonal bipyramidal.

Step 7: Determine the hybridization and shape of $BO_3^{3-}$

Step 8: B has 3 valence electrons. In $BO_3^{3-}$ , three are used for bonding with O atoms, and three negative charges add three electrons. So, there are 0 lone pairs. The steric number is 3 (3 bond pairs + 0 lone pairs), leading to $sp^2$ hybridization. The shape is trigonal planar.

Why other options are wrong:
- Option B: $BF_4^-$ is tetrahedral.
- Option C: $PCl_5$ is trigonal bipyramidal.
- Option D: $BO_3^{3-}$ is trigonal planar.

Key Concepts:
- VSEPR theory
- Hybridization
- Molecular shapes

Memory Aids:
- Steric number = Number of bond pairs + Number of lone pairs

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

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Understanding +4 | -1

Chlorine atom differs from chloride ion in the number of:

View answer and explanation

Correct option: C

Electrons

Option C (Electrons) is correct because a chloride ion is formed when a chlorine atom gains an electron.

Solution approach

Correct Answer + Core Reason: Electrons. A chloride ion has one more electron than a chlorine atom.

Step-by-Step Solution:

Step 1: Define a chlorine atom: A neutral atom with a specific number of protons, neutrons, and electrons.

Step 2: Define a chloride ion: A chlorine atom that has gained an electron, resulting in a negative charge.

Step 3: Compare the number of subatomic particles: The number of protons and neutrons remains the same, but the number of electrons differs.

Step 4: Conclude that chlorine atom differs from chloride ion in the number of electrons.

Why other options are wrong:
- Option A: Protons: The number of protons defines the element and does not change during ion formation.
- Option B: Neutrons: The number of neutrons can vary (isotopes), but it's not the difference between an atom and its ion.
- Option D: Protons and electrons: Only the number of electrons changes.

Key Concepts:
- Atomic structure
- Ions (anions and cations)
- Electron gain and loss

Memory Aids:
- Anions are negatively charged ions formed by gaining electrons.
- Cations are positively charged ions formed by losing electrons.

Quick Tips:
- Remember that gaining electrons results in a negative charge, and losing electrons results in a positive charge.

112 Medium

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Understanding +4 | -1

Which of the following statement is correct?

View answer and explanation

Correct option: A

Polarization of an anion is maximum by high charged cation

Option A (Polarization of an anion is maximum by high charged cation) is correct because according to Fajan's rule, a highly charged cation will cause greater polarization of the anion.

Solution approach

Correct Answer + Core Reason: Polarization of an anion is maximum by high charged cation. According to Fajan's rule.

Step-by-Step Solution:

Step 1: Define polarization.
- Polarization is the distortion of the electron cloud of an anion by a cation.

Step 2: State Fajan's rules.
- Fajan's rules describe the factors that affect the polarization of ions in ionic compounds.
- The rules state that polarization is favored by:
- Small cation
- Large anion
- High charge on either ion

Step 3: Apply Fajan's rules to the given statements.
- A high charged cation will cause greater polarization of the anion.
- A small sized cation maximizes the polarization.
- A large anion undergoes a high degree of polarization.

Why other options are wrong:
- Option A: Polarization of an anion is maximum by high charged cation - Correct.
- Option B: Small sized cation minimises the polarisation - Incorrect. Small sized cation maximizes the polarization.
- Option C: A small anion brings about a large degree of polarization - Incorrect. A large anion undergoes a large degree of polarization.
- Option D: A small anion undergoes a high degree of polarization - Incorrect. A large anion undergoes a high degree of polarization.

Key Concepts:
- Polarization
- Fajan's rules
- Cation
- Anion

Memory Aids:
- Fajan's rules: Small cation, large anion, high charge = more polarization

Quick Tips:
- Remember Fajan's rules to determine the extent of polarization.

113 Medium

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Analyzing +4 | -1

\mathrm{N}{\mathrm{F}_3}

is:

View answer and explanation

Correct option: C

Having low value of dipole moment than

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

Option C is correct because the dipole moments of N-F bonds partially cancel each other due to the direction of the bond dipoles and the lone pair on nitrogen.

Solution approach

Correct Answer + Core Reason: Having low value of dipole moment than $\mathrm{N}{\mathrm{H}_3}$ . The dipole moment of $\mathrm{NF}_3$ is lower than $\mathrm{NH}_3$ due to the opposing directions of bond dipoles and lone pair dipole.

Step-by-Step Solution:

Step 1: Analyze the structure of $\mathrm{NF}_3$ and $\mathrm{NH}_3$ .

Step 2: In $\mathrm{NH}_3$ , the dipole moment of each N-H bond points towards N, and the lone pair's dipole moment also points in the same direction. These dipole moments add up, resulting in a significant net dipole moment.

Step 3: In $\mathrm{NF}_3$ , the dipole moment of each N-F bond points towards F (more electronegative), which is opposite to the direction of the lone pair's dipole moment. These dipole moments partially cancel each other, resulting in a smaller net dipole moment.

Step 4: Therefore, $\mathrm{NF}_3$ has a lower dipole moment than $\mathrm{NH}_3$ .

Why other options are wrong:
- Option A: $\mathrm{NF}_3$ is a polar compound because of the electronegativity difference between N and F and its asymmetrical structure.
- Option B: $\mathrm{NF}_3$ is a covalent compound, not an electrovalent (ionic) compound.
- Option D: $\mathrm{NF}_3$ has a lower dipole moment than $\mathrm{NH}_3$ , not more.

Key Concepts:
- Dipole moment
- Electronegativity
- Molecular geometry
- Vector addition of dipole moments

Memory Aids:
- Remember that the direction of bond dipole is from less electronegative to more electronegative atom.
- Lone pair dipole moment contributes to the overall dipole moment.

Quick Tips:
- Draw the Lewis structure and consider the electronegativity difference to determine the direction of bond dipoles.
- Consider the geometry of the molecule to determine if the bond dipoles cancel each other or add up.

114 Medium

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Understanding +4 | -1

Which compound shows hydrogen bonding?

View answer and explanation

Correct option: A

R\mathrm{C}{\mathrm{H}_2}\mathrm{NHC}{\mathrm{H}_3}

Option A (RCH2NHCH3) is correct because it contains an N-H bond, which is necessary for hydrogen bonding.

Solution approach

Correct Answer + Core Reason: $R\mathrm{C}{\mathrm{H}_2}\mathrm{NHC}{\mathrm{H}_3}$ . It contains an N-H bond, which is necessary for hydrogen bonding.

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: $R\mathrm{C}{\mathrm{H}_2}\mathrm{NHC}{\mathrm{H}_3}$ (secondary amine) has an N-H bond, so it can form hydrogen bonds.
- B: $R\mathrm{C}{\mathrm{H}_2}\mathrm{CHO}$ (aldehyde) has a C=O bond but no O-H or N-H bonds, so it cannot form hydrogen bonds.
- C: ${\mathrm{C}_2}{\mathrm{H}_6}$ (ethane) has only C-H bonds, so it cannot form hydrogen bonds.
- D: $HCl$ (hydrogen chloride) can form dipole-dipole interactions but not strong hydrogen bonds in the same way as with F, O, or N.

Step 3: Conclude that the secondary amine can form hydrogen bonds.

Why other options are wrong:
- Option B: Aldehydes do not have N-H or O-H bonds and thus cannot form hydrogen bonds.
- Option C: Ethane is a nonpolar molecule and cannot form hydrogen bonds.
- Option D: While HCl is polar, it doesn't form strong hydrogen bonds like those with F, O, or N.

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.

115 Medium

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Remembering +4 | -1

Energy required to completely separate one mole of a solid ionic compound into gaseous constituent ions is called ...........

View answer and explanation

Correct option: D

Lattice enthalpy

Option D (Lattice enthalpy) is correct because lattice enthalpy is the energy required to completely separate one mole of a solid ionic compound into its gaseous ions.

Solution approach

Correct Answer + Core Reason: Lattice enthalpy is the energy to separate one mole of solid ionic compound into gaseous ions.

Step-by-Step Solution:

Step 1: Lattice enthalpy is defined as the energy required to completely separate one mole of a solid ionic compound into its gaseous constituent ions.

Step 2: This energy represents the strength of the ionic bonds within the crystal lattice.

Why other options are wrong:
- Option A: Ionization enthalpy is the energy required to remove an electron from a gaseous atom or ion.
- Option B: Electron gain enthalpy is the energy change when an electron is added to a gaseous atom or ion.
- Option C: Bond dissociation enthalpy is the energy required to break one mole of a specific bond in the gaseous phase.

Key Concepts:
- Lattice enthalpy
- Ionization enthalpy
- Electron gain enthalpy
- Bond dissociation enthalpy

Memory Aids:
- Lattice = Crystal = Separation into ions

Quick Tips:
- Remember that lattice enthalpy is specifically related to ionic compounds in the solid state.

116 Medium

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Understanding +4 | -1

Which statement is wrong?

View answer and explanation

Correct option: D

ds{p^3}

-hybrid orbitals are directed towards the corners of a regular tetrahedron

Option D (

ds{p^3}

-hybrid orbitals are directed towards the corners of a regular tetrahedron) is wrong because

sp^3

hybrid orbitals are directed towards the corners of a regular tetrahedron, not

dsp^3

which would have a different geometry.

Solution approach

Correct Answer + Core Reason: $ds{p^3}$ -hybrid orbitals are directed towards the corners of a regular tetrahedron. $sp^3$ orbitals form a tetrahedron, not $dsp^3$ .

Step-by-Step Solution:

Step 1: Analyze each statement.

Step 2: Statement A: Hybridization is the mixing of atomic orbitals prior to their combining into molecular orbitals. This is correct.

Step 3: Statement B: $s{p^2}$ -hybrid orbitals are formed from two p-atomic orbitals and one s-atomic orbitals. This is correct.

Step 4: Statement C: $ds{p^2}$ - hybrid orbitals are all at 90° to one another. This is correct, they form a square planar geometry.

Step 5: Statement D: $ds{p^3}$ -hybrid orbitals are directed towards the corners of a regular tetrahedron. This is incorrect. $sp^3$ hybrid orbitals are directed towards the corners of a regular tetrahedron. $dsp^3$ hybrid orbitals are directed towards the corners of a trigonal bipyramid.

Why other options are wrong:
- Option A: This statement is correct.
- Option B: This statement is correct.
- Option C: This statement is correct.

Key Concepts:
- Hybridization
- Molecular geometry

Memory Aids:
- Remember the geometries associated with different hybridizations.

Quick Tips:
- Know the geometries associated with different hybridization schemes.

117 Medium

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Understanding +4 | -1

Which bond angle results in the minimum dipole moment for the triatomic molecule

X{Y_2}

shown below?

View answer and explanation

Correct option: D

180^\circ

Option D (

180^\circ

) is correct because the dipole moments of the two Y atoms cancel each other out, resulting in a minimum (zero) net dipole moment.

Solution approach

Correct Answer + Core Reason: $180^\circ$ . The dipole moments cancel out at 180 degrees.

Step-by-Step Solution:

Step 1: Consider the geometry of the $XY_2$ molecule. The dipole moment is a vector quantity, and the net dipole moment is the vector sum of the individual bond dipole moments.

Step 2: If the bond angle is $180^\circ$ , the molecule is linear. The two bond dipole moments are equal in magnitude and opposite in direction, so they cancel each other out.

Step 3: As the bond angle decreases from $180^\circ$ , the vector sum of the bond dipole moments increases, resulting in a larger net dipole moment.

Step 4: Therefore, the minimum dipole moment occurs when the bond angle is $180^\circ$ .

Why other options are wrong:
- **Option A ( $90^\circ$ ):** A bond angle of $90^\circ$ results in a significant net dipole moment because the bond dipoles are not directly opposing each other.
- **Option B ( $120^\circ$ ):** A bond angle of $120^\circ$ also results in a non-zero net dipole moment, although smaller than at $90^\circ$ .
- **Option C ( $150^\circ$ ):** A bond angle of $150^\circ$ results in a smaller net dipole moment compared to $90^\circ$ and $120^\circ$ , but it is still greater than zero.

Key Concepts:
- Dipole moment as a vector quantity
- Vector addition of dipole moments
- Molecular geometry and its effect on dipole moment

Memory Aids:
- Linear molecules with identical bonds have zero dipole moment.

Quick Tips:
- Visualize the vector addition of bond dipoles to determine the net dipole moment.

118 Medium

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Understanding +4 | -1

As a result of resonance:

View answer and explanation

Correct option: D

All are correct

Option D (All are correct) is correct because resonance leads to decreased bond length, decreased energy, and increased stability of the molecule.

Solution approach

Correct Answer + Core Reason: All are correct. Resonance leads to a more stable structure with decreased energy and bond length.

Step-by-Step Solution:

Step 1: Define Resonance: Resonance is a phenomenon where multiple Lewis structures can be drawn for a single molecule.

Step 2: Impact on Bond Length: Resonance results in bond lengths that are intermediate between single and multiple bonds.

Step 3: Impact on Energy: Resonance structures contribute to a resonance hybrid, which is more stable and has lower energy than any individual contributing structure.

Step 4: Impact on Stability: The delocalization of electrons in resonance increases the stability of the molecule.

Step 5: Conclusion: Therefore, bond length decreases, energy decreases, and stability increases as a result of resonance.

Why other options are wrong:
- Option A: Bond length decreases is only partially correct; the other effects of resonance are not considered.
- Option B: Energy of the molecules decreases is only partially correct; the other effects of resonance are not considered.
- Option C: Stability of the molecule increases is only partially correct; the other effects of resonance are not considered.
- Option D: All are correct is the only option that encompasses all the effects of resonance.

Key Concepts:
- Resonance
- Bond Length
- Molecular Stability
- Energy Levels

Memory Aids:
- Resonance = Increased Stability

Quick Tips:
- Remember that resonance always leads to a more stable molecule.

119 Medium

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Remembering +4 | -1

The bond length between

C\mathrm{ }--\mathrm{ }C

bond in

s{p^2}

hybridised molecule is

View answer and explanation

Correct option: B

1.39{A^0}

Option C (

1.33{A^0}

) is the closest to the actual C=C bond length in a

sp^2

hybridized molecule, which is approximately 1.34 Å. Option B (

1.39{A^0}

) is the C-C bond length in benzene.

Solution approach

Correct Answer + Core Reason: $1.33{A^0}$ . The C=C bond length in $sp^2$ hybridized molecules is around 1.34 Å.

Step-by-Step Solution:

Step 1: Understand the question. We need to determine the bond length between C--C in an $sp^2$ hybridized molecule.

Step 2: Recall the typical bond lengths for different types of carbon-carbon bonds.
- C-C (single bond): 1.54 Å
- C=C (double bond): 1.34 Å
- C≡C (triple bond): 1.20 Å

Step 3: The question refers to a double bond in an $sp^2$ hybridized molecule. The closest value to 1.34 Å is $1.33{A^0}$ .

Why other options are wrong:
- Option A: $1.2{A^0}$ is closer to the length of a C≡C triple bond.
- Option B: $1.39{A^0}$ is the C-C bond length in benzene, which has resonance and partial double bond character.
- Option D: $1.54{A^0}$ is the length of a C-C single bond.

Key Concepts:
- Bond length
- Hybridization
- Carbon-carbon bond lengths

Memory Aids:
- Single bond > Double bond > Triple bond (bond length order)

Quick Tips:
- Memorize the typical bond lengths for common bonds.

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Remembering +4 | -1

Among the following the weakest one is

View answer and explanation

Correct option: C

Van der Waal's force

Option C (Van der Waal's force) is correct because Van der Waals forces are weak intermolecular forces arising from temporary dipoles.

Solution approach

Correct Answer + Core Reason: Van der Waal's force. Van der Waals forces are weak intermolecular forces arising from temporary dipoles.

Step-by-Step Solution:

Step 1: Metallic bonds involve the attraction between metal ions and delocalized electrons.

Step 2: Ionic bonds involve the electrostatic attraction between oppositely charged ions.

Step 3: Covalent bonds involve the sharing of electrons between atoms.

Step 4: Van der Waals forces are weak intermolecular forces arising from temporary dipoles.

Step 5: Comparing the strengths, Van der Waals forces are the weakest.

Why other options are wrong:
- Option A: Metallic bonds are generally stronger than Van der Waals forces.
- Option B: Ionic bonds are generally stronger than Van der Waals forces.
- Option D: Covalent bonds are generally stronger than Van der Waals forces.

Key Concepts:
- Metallic bonding
- Ionic bonding
- Covalent bonding
- Van der Waals forces

Memory Aids:
- Van der Waals = Weakest

Quick Tips:
- Remember the relative strengths of different types of chemical bonds and intermolecular forces.

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