According to molecular orbital theory for ${\rm{O}}_2^ + $:

Bond order is more than ${{\rm{O}}_2}$ and ${\rm{O}}_2^ + $ is paramagnetic

Bond order is less than ${{\rm{O}}_2}$ and ${\rm{O}}_2^ + $ is paramagnetic

Bond order is less than ${{\rm{O}}_2}$ and ${\rm{O}}_2^ + $ is diamagnetic

Bond order is more than ${{\rm{O}}_2}$ and ${\rm{O}}_2^ + $ is diamagnetic

Magnesium reacts with an element $\left( {\rm{X}} \right)$ to form an ionic compound. If the ground state electronic configuration of $\left( {\rm{X}} \right)$ is ${\rm{1}}{{\rm{s}}^{\rm{2}}}\,{\rm{2}}{{\rm{s}}^{\rm{2}}}\,{\rm{2}}{{\rm{p}}^{\rm{3}}}$ , the simplest formula for this compound is:

${\rm{M}}{{\rm{g}}_{\rm{3}}}\,{{\rm{X}}_{\rm{2}}}$

${\rm{M}}{{\rm{g}}_2}\,{{\rm{X}}_3}$

${\rm{M}}{{\rm{g}}_2}\,{\rm{X}}$

CHEMICAL BONDING & MOLECULAR STRUCTURE MCQs for NEET UG CHEMISTRY — Easy, Page 5

121 Easy

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Understanding +4 | -1

Which of the following molecules has zero bond order

View answer and explanation

Correct option: D

B{e_2}

Option D (

B{e_2}

) is correct because its bond order is zero, calculated as (4-4)/2 = 0.

Solution approach

Correct Answer + Core Reason: $B{e_2}$ . The bond order of Be2 is zero.

Step-by-Step Solution:

Step 1: Calculate the bond order for each molecule using molecular orbital theory.

Step 2: ${F_2}$ : Bond order = (10 - 8) / 2 = 1

Step 3: $L{i_2}$ : Bond order = (2 - 0) / 2 = 1

Step 4: ${N_2}$ : Bond order = (10 - 4) / 2 = 3

Step 5: $B{e_2}$ : Bond order = (4 - 4) / 2 = 0

Why other options are wrong:
- Option A: ${F_2}$ has a bond order of 1.
- Option B: $L{i_2}$ has a bond order of 1.
- Option C: ${N_2}$ has a bond order of 3.

Key Concepts:
- Molecular Orbital Theory
- Bond Order

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

Quick Tips:
- Calculate the bond order to determine the stability of a molecule.

122 Easy

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Two ice cubes are pressed over each other and unite to form one cube. Which force is responsible for holding them together?

View answer and explanation

Correct option: C

Hydrogen bond formation

Option C (Hydrogen bond formation) is correct because the ice cubes unite due to the formation of hydrogen bonds between water molecules on the surfaces.

Solution approach

Correct Answer + Core Reason: Hydrogen bond formation. Hydrogen bonds between water molecules cause the ice cubes to stick together.

Step-by-Step Solution:

Step 1: Identify the forces present between water molecules.

Step 2: Water molecules are held together by hydrogen bonds.

Step 3: When two ice cubes are pressed together, the water molecules on the surfaces can form hydrogen bonds with each other.

Step 4: These hydrogen bonds hold the ice cubes together, causing them to unite.

Why other options are wrong:
- Option A (van der Waals forces): Van der Waals forces are weaker than hydrogen bonds and are not the primary force responsible for holding the ice cubes together.
- Option B (Covalent attraction): Covalent bonds are strong bonds within the water molecule itself (between H and O), not between different water molecules.
- Option D (Dipole-dipole attraction): Dipole-dipole attractions are present, but hydrogen bonds are a stronger and more significant factor in this case.

Key Concepts:
- Hydrogen Bonding
- Intermolecular Forces
- van der Waals Forces
- Dipole-Dipole Interactions

Memory Aids:
- Hydrogen bonds are strong intermolecular forces in water.

Quick Tips:
- Remember that hydrogen bonds are responsible for many of water's unique properties.

123 Easy

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Which of the following statement is correct

View answer and explanation

Correct option: A

Maximum polarization is brought about by a cation of high charge

Option A (Maximum polarization is brought about by a cation of high charge) is correct because a higher charge on the cation increases its polarizing power, leading to greater distortion of the anion's electron cloud.

Solution approach

Correct Answer + Core Reason: Maximum polarization is brought about by a cation of high charge. Higher charge increases polarizing power.

Step-by-Step Solution:

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

Step 2: Identify Factors Affecting Polarization: The polarizing power of a cation is directly proportional to its charge. A higher charge means greater polarizing power.

Step 3: Relate Charge to Polarization: Therefore, a cation with a high charge will cause maximum polarization of an anion.

Why other options are wrong:
- Option B: Minimum Polarization is brought about by a cation of low charge: This is correct, but the question asks for the *correct* statement among the options, and option A is a more direct and complete statement about polarization.
- Option C: A large cation is likely to bring about a large degree of polarization: A *small* cation is more likely to bring about a large degree of polarization.
- Option D: A small anion is likely to undergo a large degree of polarization: A *large* anion is more likely to undergo a large degree of polarization.

Key Concepts:
- Polarization
- Cation Charge
- Anion Size

Memory Aids:
- High charge = High polarization

Quick Tips:
- Remember that higher charge on the cation leads to increased polarization of the anion.

124 Easy

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The nature of bonding in graphite is

View answer and explanation

Correct option: A

Covalent

Option A (Covalent) is correct because graphite consists of carbon atoms covalently bonded in layers.

Solution approach

Correct Answer + Core Reason: Covalent. Graphite consists of carbon atoms covalently bonded in layers.

Step-by-Step Solution:

Step 1: Graphite is a form of carbon where each carbon atom is bonded to three other carbon atoms in a hexagonal arrangement, forming layers.

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 bonding in graphite is covalent.

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

Key Concepts:
- Covalent bonding
- Structure of graphite

Memory Aids:
- Covalent bonds involve sharing of electrons.

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

125 Easy

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In an ice crystal, each water molecule is hydrogen
bonded to_____ neighbouring molecules.

View answer and explanation

Correct option: A

4

Option A (4) is correct because each water molecule in ice forms hydrogen bonds with four other water molecules in a tetrahedral arrangement.

Solution approach

Correct Answer + Core Reason: 4. Each water molecule forms 4 hydrogen bonds in ice.

Step-by-Step Solution:

Step 1: Describe the structure of ice.\nIce has a tetrahedral structure.

Step 2: Explain the hydrogen bonding in ice.\nEach water molecule is hydrogen bonded to four other water molecules.\nTwo hydrogen atoms of a water molecule form hydrogen bonds with two other water molecules, and the two lone pairs of electrons on the oxygen atom form hydrogen bonds with two other water molecules.

Why other options are wrong:
- Option B: Incorrect. Water molecules are hydrogen bonded to 4 neighboring molecules.
- Option C: Incorrect. Water molecules are hydrogen bonded to 4 neighboring molecules.
- Option D: Incorrect. Water molecules are hydrogen bonded to 4 neighboring molecules.

Key Concepts:
- Hydrogen bonding
- Structure of ice
- Tetrahedral arrangement

Memory Aids:
- Ice has an open structure due to hydrogen bonding.

Quick Tips:
- Visualize the tetrahedral arrangement of water molecules in ice.

126 Easy

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One debye (D) equal to

View answer and explanation

Correct option: B

1 \times {10^{ - 18}}\,esu.\,cm

Option B (

1 \times {10^{ - 18}}\,esu.\,cm

) is correct because 1 debye is defined as

1 \times {10^{ - 18}}

electrostatic units times centimeter.

Solution approach

Correct Answer + Core Reason: $1 \times {10^{ - 18}}\,esu.\,cm$ . This is the definition of one Debye unit.

Step-by-Step Solution:

Step 1: The debye (D) is a unit of electric dipole moment.

Step 2: By definition, 1 debye (D) = $1 \times {10^{ - 18}}$ esu.cm, where esu is the electrostatic unit.

Why other options are wrong:
- Option A: $1 \times {10^{ - 4}}\,esu.\,cm$ is incorrect. This is not the correct conversion factor for Debye.
- Option C: $1 \times {10^{ - 10}}\,esu.\,cm$ is incorrect. This is not the correct conversion factor for Debye.
- Option D: $1 \times {10^{ - 16}}\,esu.\,cm$ is incorrect. This is not the correct conversion factor for Debye.

Key Concepts:
- Electric dipole moment
- Debye unit

Memory Aids:
- Remember the definition of Debye: $1 D = 1 \times 10^{-18} esu.cm$

Quick Tips:
- Memorize the definition of the Debye unit.

127 Easy

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Which of the following pairs will from the most stable ionic bond?

View answer and explanation

Correct option: B

Mg\mathrm{ }\, and \mathrm{ }\,F

Option B (

Mg\mathrm{ }\, and \mathrm{ }\,F

) is correct because magnesium (Mg) readily loses two electrons to form Mg²⁺, and fluorine (F) readily gains one electron to form F⁻. The higher charges lead to a stronger electrostatic attraction and a more stable ionic bond compared to the other options.

Solution approach

Correct Answer + Core Reason: $Mg\mathrm{ }\, and \mathrm{ }\,F$ . Higher charges lead to a more stable ionic bond.

Step-by-Step Solution:

Step 1: Consider the charges of the ions formed by each pair of elements:
- Na and Cl: Na+ and Cl- (charges of +1 and -1)
- Mg and F: Mg2+ and F- (charges of +2 and -1)
- Li and F: Li+ and F- (charges of +1 and -1)
- Na and F: Na+ and F- (charges of +1 and -1)

Step 2: The strength of an ionic bond is directly proportional to the product of the charges of the ions.

Step 3: Mg and F form ions with charges of +2 and -1, respectively. The product of the charges is 2, which is higher than the product of the charges for the other options (which is 1).

Step 4: Therefore, the ionic bond formed between Mg and F is the most stable.

Why other options are wrong:
- Option A: Na and Cl form ions with charges of +1 and -1, resulting in a less stable ionic bond compared to Mg and F.
- Option C: Li and F form ions with charges of +1 and -1, resulting in a less stable ionic bond compared to Mg and F.
- Option D: Na and F form ions with charges of +1 and -1, resulting in a less stable ionic bond compared to Mg and F.

Key Concepts:
- Ionic bond
- Ionic charge
- Electrostatic attraction

Memory Aids:
- Higher charges = stronger ionic bond.

Quick Tips:
- Consider the charges of the ions when comparing the stability of ionic bonds.

128 Easy

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In

\mathrm{Br}{\mathrm{F}_3}

molecule, the lone pair occupy equatorial position to minimize :

View answer and explanation

Correct option: C

Lone pair-lone pair repulsion and lone pair-bond pair repulsion

Option C (Lone pair-lone pair repulsion and lone pair-bond pair repulsion) is correct because lone pairs occupy equatorial positions to minimize both types of repulsions.

Solution approach

Correct Answer + Core Reason: Lone pair-lone pair repulsion and lone pair-bond pair repulsion. Lone pairs occupy equatorial positions to minimize repulsions.

Step-by-Step Solution:

Step 1: Draw the structure of BrF3. Br is the central atom, and there are three F atoms bonded to it.

Step 2: Determine the number of valence electrons for Br and F. Br has 7 valence electrons, and each F has 7.

Step 3: Calculate the total number of valence electrons: 7 + 3(7) = 28.

Step 4: Determine the number of electron pairs: 28/2 = 14.

Step 5: Determine the number of bond pairs: 3 (since there are 3 F atoms bonded to Br).

Step 6: Determine the number of lone pairs: (28 - 6)/2 = 2 (two lone pairs on Br).

Step 7: Apply VSEPR theory. With 3 bond pairs and 2 lone pairs, the electron geometry is trigonal bipyramidal. Lone pairs prefer equatorial positions to minimize lone pair-lone pair and lone pair-bond pair repulsions.

Why other options are wrong:
- Option A: While lone pair-bond pair repulsion is a factor, it's not the only one. Lone pair-lone pair repulsion is also significant.
- Option B: Bond pair-bond pair repulsion is always present but is less significant than lone pair repulsions.
- Option D: Lone pair-lone pair repulsion is important, but lone pair-bond pair repulsion also contributes to the overall minimization of repulsion.

Key Concepts:
- VSEPR theory.
- Lone pair repulsion.
- Bond pair repulsion.
- Trigonal bipyramidal geometry.

Memory Aids:
- Lone pairs take up more space than bond pairs.

Quick Tips:
- Lone pairs always try to occupy positions that minimize overall repulsion in a molecule.

129 Easy

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Covalency of chlorine atom after the excitation of two electrons will be

View answer and explanation

Correct option: B

5

Option B (5) is correct because after the excitation of two electrons, chlorine can form five covalent bonds by using one s, three p, and one d orbitals.

Solution approach

Correct Answer + Core Reason: 5. Chlorine can form 5 bonds after excitation of two electrons.

Step-by-Step Solution:

Step 1: Write the electronic configuration of chlorine in the ground state: [Ne] 3s² 3p⁵.

Step 2: In the ground state, chlorine has one unpaired electron, so its covalency is 1.

Step 3: After the excitation of one electron, the electronic configuration becomes [Ne] 3s² 3p⁴ 3d¹. Now, chlorine has three unpaired electrons, so its covalency is 3.

Step 4: After the excitation of two electrons, the electronic configuration becomes [Ne] 3s¹ 3p³ 3d². Now, chlorine has five unpaired electrons, so its covalency is 5.

Why other options are wrong:
- Option A: 2 is incorrect. Chlorine cannot have a covalency of 2 after excitation.
- Option C: 3 is incorrect. Chlorine has a covalency of 3 after the excitation of *one* electron, not two.
- Option D: 7 is incorrect. Chlorine cannot achieve a covalency of 7 with just two electron excitations.

Key Concepts:
- Covalency
- Electronic configuration
- Excitation of electrons

Memory Aids:
- Covalency = number of unpaired electrons

Quick Tips:
- Determine the number of unpaired electrons after each excitation.

130 Easy

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Analyzing +4 | -1

In the electronic structure of acetic acid there are

View answer and explanation

Correct option: A

16 shared and 8 unshared valence electrons

Option A (16 shared and 8 unshared valence electrons) is correct because acetic acid has 8 bonding pairs (16 shared electrons) and 4 lone pairs (8 unshared electrons).

Solution approach

Correct Answer + Core Reason: 16 shared and 8 unshared valence electrons. Acetic acid has 8 bonding pairs (16 shared electrons) and 4 lone pairs (8 unshared electrons).

Step-by-Step Solution:

Step 1: Draw the Lewis structure of acetic acid (CH3COOH).

Step 2: Count the number of shared (bonding) valence electrons. Each single bond consists of 2 shared electrons. Acetic acid has 8 single bonds. Therefore, there are 8 * 2 = 16 shared electrons.

Step 3: Count the number of unshared (lone pair) valence electrons. Each oxygen atom has 2 lone pairs, and each lone pair consists of 2 electrons. Acetic acid has 2 oxygen atoms, so there are 2 * 2 * 2 = 8 unshared electrons.

Why other options are wrong:
- Option B (8 shared and 16 unshared valence electrons): This is incorrect because the number of shared electrons is higher than the number of unshared electrons in acetic acid.
- Option C (12 shared and 12 unshared valence electrons): This is incorrect because the number of shared and unshared electrons is not equal in acetic acid.
- Option D (18 shared and 6 unshared valence electrons): This is incorrect because the number of shared electrons is lower and the number of unshared electrons is higher than the actual number in acetic acid.

Key Concepts:
- Lewis structures
- Shared electrons (bonding pairs)
- Unshared electrons (lone pairs)

Memory Aids:
- Each single bond has 2 shared electrons.
- Each lone pair has 2 unshared electrons.

Quick Tips:
- Draw the Lewis structure carefully to avoid mistakes in counting electrons.
- Remember that each bond consists of 2 shared electrons and each lone pair consists of 2 unshared electrons.

131 Easy

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Remembering +4 | -1

Who introduced the term covalent bond ?

View answer and explanation

Correct option: B

Langmuir

Option B is correct. Langmuir introduced the term \"covalent bond.\"

Solution approach

Correct Answer + Core Reason: Langmuir. Langmuir introduced the term covalent bond.

Step-by-Step Solution:

Step 1: Recall the history of chemical bonding theories.

Step 2: Identify the contributions of each scientist:
- Lewis: Developed the Lewis dot structures and the concept of electron sharing.
- Langmuir: Introduced the term \"covalent bond.\"
- Nyholm and Gillespie: Developed the VSEPR theory.
- Heitler and London: Developed the valence bond theory.

Step 3: Conclude that Langmuir introduced the term covalent bond.

Why other options are wrong:
- Option A: Lewis contributed to the understanding of electron sharing but did not coin the term covalent bond.
- Option C: Nyholm and Gillespie are known for VSEPR theory, not the term covalent bond.
- Option D: Heitler and London developed valence bond theory, not the term covalent bond.

Key Concepts:
- History of chemical bonding theories
- Covalent bond

Memory Aids:
- Langmuir = Covalent bond (LC)

Quick Tips:
- Remember the key contributions of each scientist to the development of chemical bonding theories.

132 Easy

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What is the dominant intermolecular force or bond that must be overcome in converting liquid

\mathrm{C}{\mathrm{H}_3}\mathrm{OH}

to a gas?

View answer and explanation

Correct option: B

Hydrogen bonding

Option B (Hydrogen bonding) is correct because methanol (CH3OH) has a strong hydrogen bonding network that must be overcome to convert it to a gas.

Solution approach

Correct Answer + Core Reason: Hydrogen bonding. Methanol has strong hydrogen bonds that need to be overcome to change it to gas.

Step-by-Step Solution:

Step 1: Identify the intermolecular forces present in liquid $\mathrm{C}{\mathrm{H}_3}\mathrm{OH}$ (methanol).

Step 2: Methanol has the following intermolecular forces:
- London dispersion forces (present in all molecules).
- Dipole-dipole interactions (due to the polar C-O and O-H bonds).
- Hydrogen bonding (due to the O-H bond).

Step 3: Compare the strengths of these forces. Hydrogen bonding is significantly stronger than London dispersion forces and dipole-dipole interactions in methanol.

Step 4: Covalent bonds are intramolecular forces and are not overcome during phase changes.

Step 5: Conclude that hydrogen bonding is the dominant intermolecular force that must be overcome.

Why other options are wrong:
- Option A: London dispersion forces are present but are much weaker than hydrogen bonding in methanol.
- Option C: Dipole-dipole interactions are present but are weaker than hydrogen bonding in methanol.
- Option D: Covalent bonds are intramolecular forces and are not broken during phase changes.

Key Concepts:
- Intermolecular forces: London dispersion forces, dipole-dipole interactions, hydrogen bonding.
- Relative strengths of intermolecular forces.
- Phase changes.

Memory Aids:
- Hydrogen bonding is stronger than dipole-dipole and London dispersion forces.

Quick Tips:
- When determining the dominant intermolecular force, consider hydrogen bonding first if H is bonded to F, O, or N.

133 Easy

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Analyzing +4 | -1

Which of the following molecules is covalent and shows expanded octet in its formation?

View answer and explanation

Correct option: D

\mathrm{Cl}{\mathrm{F}_3}

Option D (

\mathrm{Cl}{\mathrm{F}_3}

) is correct because Cl has 7 valence electrons, and in ClF3, it forms 3 bonds and has 2 lone pairs, resulting in 10 electrons around Cl.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cl}{\mathrm{F}_3}$ . Chlorine has 7 valence electrons and forms 3 bonds and has 2 lone pairs, resulting in 10 electrons around Cl.

Step-by-Step Solution:

Step 1: Determine the number of valence electrons for the central atom in each molecule.
- HF: Hydrogen has 1 valence electron, Fluorine has 7.
- NF3: Nitrogen has 5 valence electrons, Fluorine has 7.
- BF3: Boron has 3 valence electrons, Fluorine has 7.
- ClF3: Chlorine has 7 valence electrons, Fluorine has 7.

Step 2: Draw the Lewis structure for each molecule and count the number of electrons around the central atom.
- HF: H-F (2 electrons around H, 2 electrons around F)
- NF3: N has 3 bonds and 1 lone pair (8 electrons around N)
- BF3: B has 3 bonds and no lone pairs (6 electrons around B)
- ClF3: Cl has 3 bonds and 2 lone pairs (10 electrons around Cl)

Step 3: Identify the molecule with an expanded octet (more than 8 electrons around the central atom).

Why other options are wrong:
- **Option A ( $\mathrm{HF}$ ): HF is a simple diatomic molecule and does not have an expanded octet. Both H and F follow the duet and octet rule, respectively.
- Option B ( $\mathrm{N}{\mathrm{F}_3}$ ): NF3 follows the octet rule. Nitrogen has 5 valence electrons and forms 3 bonds and has 1 lone pair, resulting in 8 electrons around N.
- Option C ( $\mathrm{B}{\mathrm{F}_3}$ ): BF3 is electron deficient. Boron has 3 valence electrons and forms 3 bonds and no lone pairs, resulting in 6 electrons around B. It does not follow the octet rule but does not have an expanded octet either.

Key Concepts:
- Octet rule
- Expanded octet
- Lewis structures

Memory Aids:
- Expanded octet: central atom has more than 8 electrons

Quick Tips:**
- Draw the Lewis structure to determine the number of electrons around the central atom.
- Elements in the third period and beyond can have expanded octets.

134 Easy

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Understanding +4 | -1

Among HX, the maximum dipole moment is of:

View answer and explanation

Correct option: A

HF

Option A is correct because HF has the highest electronegativity difference between H and F, leading to the largest dipole moment.

Solution approach

Correct Answer + Core Reason: HF. Fluorine is the most electronegative element, leading to the largest dipole moment in HF.

Step-by-Step Solution:

Step 1: Recall the definition of dipole moment. Dipole moment is proportional to the electronegativity difference between the atoms in a bond.

Step 2: Consider the electronegativity of the halogens. Electronegativity decreases down the group: F > Cl > Br > I.

Step 3: Determine the electronegativity difference between H and each halogen.

Step 4: HF has the largest electronegativity difference.

Step 5: Conclusion: HF has the maximum dipole moment.

Why other options are wrong:
- Option B: HCl: This is incorrect because the electronegativity difference between H and Cl is less than that between H and F.
- Option C: HBr: This is incorrect because the electronegativity difference between H and Br is less than that between H and F.
- Option D: HI: This is incorrect because the electronegativity difference between H and I is less than that between H and F.

Key Concepts:
- Dipole Moment
- Electronegativity

Memory Aids:
- Remember that dipole moment increases with increasing electronegativity difference.

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

135 Easy

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Remembering +4 | -1

The hybridisation state of carbon in fullerene is

View answer and explanation

Correct option: B

s p^2

Option B (

s p^2

) is correct because each carbon atom in fullerene is bonded to three other carbon atoms, similar to graphite, resulting in

sp^2

hybridization.

Solution approach

Correct Answer + Core Reason: $s p^2$ . Each carbon atom in fullerene is bonded to three other carbon atoms.

Step-by-Step Solution:

Step 1: Recall the structure of fullerene (e.g., $C_{60}$ ).

Step 2: Each carbon atom in fullerene is bonded to three other carbon atoms.

Step 3: This arrangement corresponds to $sp^2$ hybridization.

Step 4: Final Answer: $s p^2$

Why other options are wrong:
- Option A: $sp$ hybridization corresponds to linear geometry.
- Option C: $sp^3$ hybridization corresponds to tetrahedral geometry.
- Option D: $sp^3 d$ hybridization is not typical for carbon.

Key Concepts:
- Hybridization
- Fullerene structure
- $sp^2$ hybridization

Memory Aids:
- Fullerene: $sp^2$ (3 bonds)

Quick Tips:
- Count the number of sigma bonds around the carbon atom.
- Relate the number of sigma bonds to the hybridization.

136 Easy

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Remembering +4 | -1

According to Pauling, the atomic orbitals combine to form new set of equivalent atomic orbitals known as

View answer and explanation

Correct option: B

Hybrid orbitals

Option B (Hybrid orbitals) is correct because, according to Pauling's theory, atomic orbitals combine to form a new set of equivalent orbitals called hybrid orbitals.

Solution approach

Correct Answer + Core Reason: Hybrid orbitals. Pauling defined hybrid orbitals as the result of atomic orbital combination.

Step-by-Step Solution:

Step 1: Recall Pauling's theory of hybridization.

Step 2: According to Pauling, atomic orbitals combine to form a new set of equivalent atomic orbitals called hybrid orbitals.

Step 3: Final Answer: Hybrid orbitals

Why other options are wrong:
- Option A: Molecular orbitals are formed by the combination of atomic orbitals from different atoms.
- Option C: Pure orbitals are the original atomic orbitals before hybridization.
- Option D: This is incorrect because hybrid orbitals are formed.

Key Concepts:
- Hybridization
- Hybrid orbitals
- Pauling's theory

Memory Aids:
- Pauling = Hybrid orbitals

Quick Tips:
- Remember the definition of hybrid orbitals.

137 Easy

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Understanding +4 | -1

Specify the geometry and hybridisation of N and B atoms in a

1:1

complex of

B{F_3}\,\& \,N{H_3}

View answer and explanation

Correct option: A

N : tetrahedral ,

s{p^3}

B : tetrahedral ,

s{p^3}

Option A (N : tetrahedral ,

sp^3

B : tetrahedral ,

sp^3

) is correct because after forming the complex, both Nitrogen and Boron atoms become tetrahedrally coordinated and

sp^3

hybridized.

Solution approach

Correct Answer + Core Reason: N : tetrahedral , $sp^3$ B : tetrahedral , $sp^3$ . Both N and B become $sp^3$ hybridized and tetrahedral.

Step-by-Step Solution:

Step 1: Analyze the structure of $BF_3$ and $NH_3$ individually.

Step 2: $BF_3$ is $sp^2$ hybridized and trigonal planar. $NH_3$ is $sp^3$ hybridized and pyramidal.

Step 3: When $BF_3$ and $NH_3$ form a complex, the lone pair on N is donated to B, forming a coordinate covalent bond.

Step 4: This changes the hybridization of B from $sp^2$ to $sp^3$ , making it tetrahedral.

Step 5: The hybridization of N remains $sp^3$ , and its geometry becomes tetrahedral as well.

Why other options are wrong:
- **Option B (N : pyramidal , $sp^3$ B : pyramidal , $sp^3$ ): Boron changes from pyramidal to tetrahedral.
- Option C (N : tetrahedral , $sp^3$ B : tetrahedral , $sp^3$ ): This is the correct option.
- Option D (N : pyramidal , $sp^3$ B : tetrahedral , $sp^3$ ): Nitrogen becomes tetrahedral.

Key Concepts:
- Hybridization
- Coordinate covalent bond
- VSEPR theory

Memory Aids:
- Lone pair donation changes hybridization.

Quick Tips:**
- Consider the change in geometry upon complex formation.

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Remembering +4 | -1

Shape of methane molecule is

View answer and explanation

Correct option: A

Tetrahedral

Option A (Tetrahedral) is correct because methane (

CH_4

) has a tetrahedral shape due to

sp^3

hybridization of the carbon atom.

Solution approach

Correct Answer + Core Reason: Tetrahedral. Methane ( $CH_4$ ) has a tetrahedral shape due to $sp^3$ hybridization of the carbon atom.

Step-by-Step Solution:

Step 1: Draw the Lewis structure of methane ( $CH_4$ ).

Step 2: Determine the number of bond pairs and lone pairs around the central carbon atom.

Step 3: In methane, the carbon atom has 4 bond pairs and 0 lone pairs.

Step 4: Apply VSEPR theory to predict the shape of the molecule.

Step 5: According to VSEPR theory, 4 bond pairs and 0 lone pairs result in a tetrahedral shape.

Step 6: Therefore, the shape of the methane molecule is tetrahedral.

Why other options are wrong:
- Option B: Pyramidal shape is typically associated with molecules having 3 bond pairs and 1 lone pair around the central atom, like ammonia ( $NH_3$ ).
- Option C: Octahedral shape is typically associated with molecules having 6 bond pairs and 0 lone pairs around the central atom, like sulfur hexafluoride ( $SF_6$ ).
- Option D: Square planar shape is typically associated with molecules having 4 bond pairs and 2 lone pairs around the central atom, like xenon tetrafluoride ( $XeF_4$ ).

Key Concepts:
- Molecular shape
- VSEPR theory
- Tetrahedral geometry
- Hybridization

Memory Aids:
- Tetrahedral: 4 bond pairs, 0 lone pairs.
- Pyramidal: 3 bond pairs, 1 lone pair.
- Octahedral: 6 bond pairs, 0 lone pairs.
- Square planar: 4 bond pairs, 2 lone pairs.

Quick Tips:
- Draw the Lewis structure and apply VSEPR theory to predict the shape of the molecule.

139 Easy

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Analyzing +4 | -1

Which of the following has least bond energy?

View answer and explanation

Correct option: A

Xe{F_2}

Option A (

Xe{F_2}

) is correct because noble gas compounds generally have lower bond energies compared to covalent bonds in other molecules.

Solution approach

Correct Answer + Core Reason: $Xe{F_2}$ . Noble gas compounds generally have lower bond energies.

Step-by-Step Solution:

Step 1: Recall the general trend of bond energies.

Step 2: Noble gas compounds (like $Xe{F_2}$ ) are formed with relatively weak bonds compared to typical covalent bonds.

Step 3: Compare the bond types in the given options:
- $Xe{F_2}$ : Xenon-Fluorine bond (noble gas compound)
- $C{H_3}Cl$ : Carbon-Chlorine bond (covalent)
- $S{O_2}$ : Sulfur-Oxygen bond (covalent)
- $N{H_3}$ : Nitrogen-Hydrogen bond (covalent)

Step 4: Since $Xe{F_2}$ is a noble gas compound, it will have the least bond energy.

Why other options are wrong:
- Option B: $C{H_3}Cl$ : Covalent bond, higher bond energy than $Xe{F_2}$ .
- Option C: $S{O_2}$ : Covalent bond, higher bond energy than $Xe{F_2}$ .
- Option D: $N{H_3}$ : Covalent bond, higher bond energy than $Xe{F_2}$ .

Key Concepts:
- Bond energy
- Noble gas compounds

Memory Aids:
- Noble gas compounds have relatively weak bonds.

Quick Tips:
- Consider the type of bond when comparing bond energies.

140 Easy

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Understanding +4 | -1

When metals react with non-metals, the metal atoms tend to

View answer and explanation

Correct option: B

Lose electrons

Option B (Lose electrons) is correct because metals tend to lose electrons to achieve a stable electron configuration, forming positive ions in the process.

Solution approach

Correct Answer + Core Reason: Lose electrons. Metals lose electrons to form positive ions when reacting with non-metals.

Step-by-Step Solution:

Step 1: Understand Metal-Nonmetal Reactions: Metals react with non-metals through the transfer of electrons.

Step 2: Describe Electron Transfer: Metals, having low ionization energies, tend to lose electrons to achieve a stable electron configuration.

Step 3: Explain Ion Formation: By losing electrons, metals form positive ions (cations).

Step 4: Conclusion: Therefore, when metals react with non-metals, the metal atoms tend to lose electrons.

Why other options are wrong:
- Option A: Share electrons: Sharing electrons is characteristic of covalent bonding, which typically occurs between non-metals.
- Option C: Gain electrons: Non-metals tend to gain electrons, not metals.
- Option D: None of the above: This is incorrect because metals do tend to lose electrons.

Key Concepts:
- Metal-Nonmetal Reactions
- Electron Transfer
- Ionization Energy
- Cations

Memory Aids:
- Metals lose electrons (think of \"metals are positive\")

Quick Tips:
- Remember that metals typically form positive ions by losing electrons.

141 Easy

141 of 220Question permalink

Application +4 | -1

Two hybrid orbitals have a bond angle of

120^\circ

. The percentage ofs character in the hybrid orbital is nearly

View answer and explanation

Correct option: B

33\%

Option B is correct because for a bond angle of 120°, the hybridization is sp2, which has 33% s character.

Solution approach

Correct Answer + Core Reason: $33\%$ . A bond angle of $120^\circ$ corresponds to $sp^2$ hybridization, which has one s orbital and two p orbitals.

Step-by-Step Solution:

Step 1: Relate bond angle to hybridization: A bond angle of $120^\circ$ indicates $sp^2$ hybridization.

Step 2: Determine the composition of $sp^2$ hybrid orbitals: $sp^2$ hybrid orbitals are formed by mixing one s orbital and two p orbitals.

Step 3: Calculate the percentage of s character: The percentage of s character is the ratio of the number of s orbitals to the total number of orbitals multiplied by 100.
$\text{{Percentage of s character}} = \frac{{\text{{Number of s orbitals}}}}{{\text{{Total number of orbitals}}}} \times 100\%$

Step 4: Substitute the values: In $sp^2$ hybridization, there is 1 s orbital and a total of 3 orbitals (1 s + 2 p).
$\text{{Percentage of s character}} = \frac{{1}}{{3}} \times 100\% \approx 33\%$

Why other options are wrong:
- Option A: $25\%$ corresponds to $sp^3$ hybridization (tetrahedral geometry).
- Option C: $50\%$ corresponds to $sp$ hybridization (linear geometry).
- Option D: $66\%$ is incorrect. It does not correspond to any common hybridization.

Key Concepts:
- Hybridization
- Bond angle
- Percentage s character

Memory Aids:
- $sp$ hybridization: 50% s character, 180

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142 of 220Question permalink

Understanding +4 | -1

N{H_4}CN

contains

View answer and explanation

Correct option: D

All

Option D (All) is correct because

N{H_4}CN

contains an ionic bond between

N{H_4}^+

and

CN^-

, covalent bonds within the

CN^-

ion, and a dative bond within the

N{H_4}^+

ion.

Solution approach

Correct Answer + Core Reason: All. $N{H_4}CN$ contains ionic, covalent, and dative bonds.

Step-by-Step Solution:

Step 1: Analyze the structure of $N{H_4}CN$ .

Step 2: Identify the types of bonds present:
- Ionic bond: Between the $N{H_4}^+$ cation and the $CN^-$ anion.
- Covalent bond: Within the $CN^-$ ion (C≡N).
- Dative bond: In the formation of $N{H_4}^+$ , a lone pair from nitrogen is donated to a proton ( $H^+$ ).

Why other options are wrong:
- Option A (Ionic bond): While an ionic bond is present, it's not the only type of bond.
- Option B (covalent bond): While covalent bonds are present, it's not the only type of bond.
- Option C (dative bond): While a dative bond is present, it's not the only type of bond.

Key Concepts:
- Ionic Bond
- Covalent Bond
- Dative Bond (Coordinate Covalent Bond)

Memory Aids:
- $N{H_4}CN$ is a complex compound containing multiple types of bonds.

Quick Tips:
- Break down the compound into its constituent ions to identify the types of bonds present.

143 Easy

143 of 220Question permalink

Understanding +4 | -1

Compounds formed by

s{p^3}{d^2}

-hybridization will have configuration:

View answer and explanation

Correct option: B

Octahedral

Option B (Octahedral) is correct because

s{p^3}{d^2}

hybridization results in six hybrid orbitals, which are arranged in an octahedral geometry.

Solution approach

Correct Answer + Core Reason: Octahedral. $s{p^3}{d^2}$ hybridization leads to octahedral geometry.

Step-by-Step Solution:

Step 1: Understand the relationship between hybridization and geometry.

Step 2: $s{p^3}{d^2}$ hybridization involves one s orbital, three p orbitals, and two d orbitals, resulting in six hybrid orbitals.

Step 3: Six hybrid orbitals arrange themselves in an octahedral geometry to minimize repulsion.

Why other options are wrong:
- Option A: Square planar is associated with $dsp^2$ hybridization.
- Option C: Trigonal bipyramidal is associated with $s{p^3}d$ hybridization.
- Option D: Pentagonal bipyramidal is associated with $s{p^3}{d^3}$ hybridization.

Key Concepts:
- Hybridization
- Octahedral geometry

Memory Aids:
- $s{p^3}{d^2}$ = Octahedral.

Quick Tips:
- Memorize the hybridization-geometry relationships.

144 Easy

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Analyzing +4 | -1

Which one of the following arrangements of molecules is correct on the basis of their dipole moments

View answer and explanation

Correct option: D

N{H_3} > N{F_3} > B{F_3}

Option D (

N{H_3} > N{F_3} > B{F_3}

) is correct because the dipole moment of

NH_3

is higher than

NF_3

due to the direction of bond dipoles and lone pair dipole, and

BF_3

has zero dipole moment due to its symmetrical trigonal planar structure.

Solution approach

Correct Answer + Core Reason: $N{H_3} > N{F_3} > B{F_3}$ . This order is based on the vector sum of bond dipoles and molecular geometry.

Step-by-Step Solution:

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

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

Step 3: Compare the magnitudes of dipole moments.
$N{H_3}$ has the largest dipole moment because the bond dipoles and lone pair dipole reinforce each other.
$N{F_3}$ has a smaller dipole moment because the bond dipoles and lone pair dipole oppose each other.
$B{F_3}$ has zero dipole moment due to its symmetrical structure.

Step 4: Arrange the molecules in decreasing order of dipole moments.
$N{H_3} > N{F_3} > B{F_3}$

Why other options are wrong:
- Option A: $B{F_3} > N{H_3} > N{F_3}$ is incorrect because $BF_3$ has zero dipole moment and $NH_3$ has a larger dipole moment than $NF_3$ .
- Option B: $B{F_3} > N{F_3} > N{H_3}$ is incorrect because $BF_3$ has zero dipole moment and $NH_3$ has a larger dipole moment than $NF_3$ .
- Option C: $N{H_3} > B{F_3} > N{F_3}$ is incorrect because $NF_3$ has a non-zero dipole moment, while $BF_3$ has zero. Therefore, $NF_3$ has a larger dipole moment than $BF_3$ .

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

Memory Aids:
- Remember that symmetrical molecules like $B{F_3}$ have zero dipole moment.
- Consider the direction of bond dipoles and lone pair dipoles.

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

145 Easy

145 of 220Question permalink

Application +4 | -1

According to molecular orbital theory for

\mathrm{O}_2^ +

:

View answer and explanation

Correct option: B

Bond order is more than

{\mathrm{O}_2}

and

\mathrm{O}_2^ +

is paramagnetic

Option B (Bond order is more than

{\mathrm{O}_2}

and

\mathrm{O}_2^ +

is paramagnetic) is correct because removing an electron from

O_2

to form

O_2^+

increases the bond order and leaves an unpaired electron, making it paramagnetic.

Solution approach

Correct Answer + Core Reason: Bond order is more than ${\mathrm{O}_2}$ and $\mathrm{O}_2^ +$ is paramagnetic. Removing an electron increases bond order and leaves an unpaired electron.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of $\mathrm{O}_2$ and $\mathrm{O}_2^ +$ using molecular orbital theory.

Step 2: The electronic configuration of $O_2$ is $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^1(\pi_{2p_y}^*)^1$ .\nThe electronic configuration of $O_2^+$ is $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^1$ .

Step 3: Calculate the bond order for both species.\nBond order of $O_2$ = (10 - 6) / 2 = 2\nBond order of $O_2^+$ = (10 - 5) / 2 = 2.5

Step 4: Determine the magnetic properties.
$O_2$ has two unpaired electrons, so it is paramagnetic.
$O_2^+$ has one unpaired electron, so it is paramagnetic.

Step 5: Compare the bond order and magnetic properties.

Why other options are wrong:
- Option A: The bond order is more than $O_2$ .
- Option C: The bond order is more than $O_2$ and it is paramagnetic.
- Option D: It is paramagnetic.

Key Concepts:
- Molecular orbital theory
- Electronic configuration
- Bond order
- Paramagnetism and diamagnetism

Memory Aids:
- Remember the filling order of molecular orbitals.

Quick Tips:
- Calculate the bond order and check for unpaired electrons to determine magnetic properties.

146 Easy

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Analyzing +4 | -1

The hybrid orbital having only 20%

d

– character

View answer and explanation

Correct option: C

S{F_4}

Option C (

S{F_4}

) is correct because it has

sp^3d

hybridization, which means one d orbital is involved in the hybridization, resulting in approximately 20% d-character.

Solution approach

Correct Answer + Core Reason: $S{F_4}$ . $SF_4$ has $sp^3d$ hybridization, implying 20% d-character.

Step-by-Step Solution:

Step 1: Determine the hybridization of each molecule.

Step 2: For $CC{l_4}$ , the central atom C has 4+4(7) = 32 valence electrons. The Lewis structure has four C-Cl bonds and no lone pair on C. Steric number = 4, so it is $sp^3$ hybridized (0% d-character).

Step 3: For $S{F_6}$ , the central atom S has 6+6(7) = 48 valence electrons. The Lewis structure has six S-F bonds and no lone pair on S. Steric number = 6, so it is $sp^3d^2$ hybridized (33.3% d-character).

Step 4: For $S{F_4}$ , the central atom S has 6+4(7) = 34 valence electrons. The Lewis structure has four S-F bonds and one lone pair on S. Steric number = 5, so it is $sp^3d$ hybridized (20% d-character).

Step 5: For $C{l_2}O$ , the central atom O has 6+2(7) = 20 valence electrons. The Lewis structure has two O-Cl bonds and two lone pairs on O. Steric number = 4, so it is $sp^3$ hybridized (0% d-character).

Step 6: Identify the molecule with approximately 20% d-character.

Why other options are wrong:
- Option A: $CC{l_4}$ has $sp^3$ hybridization, so it has 0% d-character.
- Option B: $S{F_6}$ has $sp^3d^2$ hybridization, so it has 33.3% d-character.
- Option D: $C{l_2}O$ has $sp^3$ hybridization, so it has 0% d-character.

Key Concepts:
- Hybridization
- Steric number
- Percentage of d-character

Memory Aids:
- $sp^3$ = 0% d-character
- $sp^3d$ = 20% d-character
- $sp^3d^2$ = 33.3% d-character

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

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Understanding +4 | -1

Linear combination of two hybridized orbitals, belonging to two atoms and each having one electron leads to:

View answer and explanation

Correct option: A

Sigma-bond

Option A (Sigma-bond) is correct because the linear combination of two hybridized orbitals, each containing one electron, forms a sigma bond.

Solution approach

Correct Answer + Core Reason: Sigma-bond. Head-on overlap forms a sigma bond.

Step-by-Step Solution:

Step 1: Consider the formation of chemical bonds.

Step 2: Sigma bonds are formed by head-on overlap of atomic orbitals or hybridized orbitals.

Step 3: Pi bonds are formed by sideways overlap of p orbitals.

Step 4: Coordinate covalent bonds involve one atom donating both electrons to the bond.

Step 5: The linear combination of two hybridized orbitals, each having one electron, leads to a head-on overlap, forming a sigma bond.

Why other options are wrong:
- Option B: Double bonds consist of one sigma bond and one pi bond.
- Option C: Coordinate covalent bonds involve donation of electrons from one atom.
- Option D: Pi bonds are formed by sideways overlap.

Key Concepts:
- Sigma bonds
- Pi bonds
- Hybridization

Memory Aids:
- Sigma bonds are stronger than pi bonds.

Quick Tips:
- Head-on overlap = sigma bond.

148 Easy

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Application +4 | -1

Magnesium reacts with an element

\left( \mathrm{X} \right)

to form an ionic compound. If the ground state electronic configuration of

\left( \mathrm{X} \right)

is

\mathrm{1}{\mathrm{s}^\mathrm{2}}\,\mathrm{2}{\mathrm{s}^\mathrm{2}}\,\mathrm{2}{\mathrm{p}^\mathrm{3}}

, the simplest formula for this compound is:

View answer and explanation

Correct option: A

\mathrm{M}{\mathrm{g}_\mathrm{3}}\,{\mathrm{X}_\mathrm{2}}

Option A (

\mathrm{M}{\mathrm{g}_\mathrm{3}}\,{\mathrm{X}_\mathrm{2}}

) is correct because Magnesium forms Mg²⁺ ions and X forms X³⁻ ions, leading to the formula Mg₃X₂.

Solution approach

Correct Answer + Core Reason: $\mathrm{M}{\mathrm{g}_\mathrm{3}}\,{\mathrm{X}_\mathrm{2}}$ . Magnesium forms Mg²⁺ ions and X forms X³⁻ ions.

Step-by-Step Solution:

Step 1: Determine the charge of the magnesium ion. Magnesium (Mg) is in group 2, so it forms Mg²⁺ ions.

Step 2: Determine the charge of the element X. The electronic configuration is $\mathrm{1}{\mathrm{s}^\mathrm{2}}\,\mathrm{2}{\mathrm{s}^\mathrm{2}}\,\mathrm{2}{\mathrm{p}^\mathrm{3}}$ . This means X needs 3 more electrons to complete its octet, so it forms X³⁻ ions.

Step 3: Determine the simplest formula for the ionic compound formed between Mg²⁺ and X³⁻. To balance the charges, we need 3 Mg²⁺ ions and 2 X³⁻ ions. Therefore, the formula is Mg₃X₂.

Why other options are wrong:
- Option B: $\mathrm{M}{\mathrm{g}_2}\,{\mathrm{X}_3}$ : Incorrect ratio of Mg and X.
- Option C: $\mathrm{M}{\mathrm{g}_2}\,\mathrm{X}$ : Incorrect ratio of Mg and X.
- Option D: $\,\mathrm{Mg}\,{\mathrm{X}_2}$ : Incorrect ratio of Mg and X.

Key Concepts:
- Formation of ionic compounds
- Charges of ions based on group number and electronic configuration

Memory Aids:
- Group 2 elements form +2 ions
- Elements needing 3 electrons to complete octet form -3 ions

Quick Tips:
- Balance the charges of the ions to determine the correct formula of the ionic compound.

149 Easy

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Understanding +4 | -1

All bond angles are exactly equal to

109^\circ 28`

in

View answer and explanation

Correct option: D

Carbon tetrachloride

Option D (Carbon tetrachloride) is correct because carbon tetrachloride (

CCl_4

) has four identical bonds and a tetrahedral geometry, resulting in all bond angles being exactly

109^\circ 28`

.

Solution approach

Correct Answer + Core Reason: Carbon tetrachloride. $CCl_4$ has perfect tetrahedral geometry.

Step-by-Step Solution:

Step 1: Consider the structure of each molecule.

Step 2: Methyl chloride ( $CH_3Cl$ ) has slightly distorted tetrahedral geometry due to the different electronegativities of H and Cl.

Step 3: Iodoform ( $CHI_3$ ) has slightly distorted tetrahedral geometry due to the different electronegativities of H and I.

Step 4: Chloroform ( $CHCl_3$ ) has slightly distorted tetrahedral geometry due to the different electronegativities of H and Cl.

Step 5: Carbon tetrachloride ( $CCl_4$ ) has perfect tetrahedral geometry because all four substituents are identical.

Why other options are wrong:
- Option A: Methyl chloride has a slightly distorted tetrahedral geometry.
- Option B: Iodoform has a slightly distorted tetrahedral geometry.
- Option C: Chloroform has a slightly distorted tetrahedral geometry.

Key Concepts:
- Tetrahedral geometry
- Molecular symmetry

Memory Aids:
- Identical substituents lead to perfect symmetry.

Quick Tips:
- Look for molecules with identical substituents.

150 Easy

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Application +4 | -1

The following element forms a molecule with eight its own weight atoms

View answer and explanation

Correct option: B

S

Option B (

S

) is correct because sulfur exists as

S_8

molecules, meaning each molecule contains eight sulfur atoms.

Solution approach

Correct Answer + Core Reason: $S$ . Sulfur exists as $S_8$ molecules.

Step-by-Step Solution:

Step 1: Recall the common allotropes and molecular formulas of the given elements.

Step 2: Silicon (Si) forms network covalent structures.

Step 3: Sulfur (S) exists as $S_8$ molecules.

Step 4: Chlorine (Cl) exists as $Cl_2$ molecules.

Step 5: Phosphorus (P) exists as $P_4$ molecules.

Step 6: Therefore, sulfur forms a molecule with eight of its own atoms.

Why other options are wrong:
- Option A: Silicon forms giant covalent structures, not discrete molecules with eight atoms.
- Option C: Chlorine forms diatomic molecules ( $Cl_2$ ).
- Option D: Phosphorus forms tetra-atomic molecules ( $P_4$ ).

Key Concepts:
- Allotropes
- Molecular formulas

Memory Aids:
- Sulfur exists as $S_8$ .
- Phosphorus exists as $P_4$ .

Quick Tips:
- Remember the common molecular formulas of elements.

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