Here below some basic mCQ’s about “Valence Bond Theory” with answer which is explained in details. Let’s check one by one.
- Which of the following bonds is best explained by Valence Bond Theory?
A. Ionic bond
B. Covalent bond
C. Metallic bond
D. Hydrogen bond
Answer: B. Covalent bond
Explanation: VBT is used to explain the formation of covalent bonds through the overlap of atomic orbitals. It doesn’t apply to ionic, metallic, or hydrogen bonds.
- According to VBT, a sigma (σ) bond forms by:
A. Side-to-side overlap of p-orbitals
B. End-to-end overlap of orbitals
C. Overlap of d-orbitals
D. Overlap of f-orbitals
Answer: B. End-to-end overlap of orbitals
Explanation: A sigma bond forms when atomic orbitals overlap directly along the internuclear axis, which is end-to-end overlap, providing maximum overlap.
- In a double bond between two atoms, the bond consists of:
A. One sigma bond and two pi bonds
B. Two sigma bonds
C. One sigma bond and one pi bond
D. Two pi bonds
Answer: C. One sigma bond and one pi bond
Explanation: In a double bond, one bond is a sigma bond (from head-on overlap) and the other is a pi bond (from side-to-side overlap of p-orbitals).
- Which of the following molecules is sp³ hybridized according to VBT?
A. CH₄
B. CO₂
C. BF₃
D. BeCl₂
Answer: A. CH₄
Explanation: In methane (CH₄), carbon undergoes sp³ hybridization, where one s and three p orbitals mix to form four equivalent sp³ hybrid orbitals, creating a tetrahedral shape.
- Valence Bond Theory explains that in an NH₃ molecule, the nitrogen atom forms:
A. Three sigma bonds and one lone pair
B. Four sigma bonds
C. Two sigma bonds and two lone pairs
D. Three pi bonds and one lone pair
Answer: A. Three sigma bonds and one lone pair
Explanation: In ammonia (NH₃), nitrogen forms three sigma bonds with hydrogen atoms and retains one lone pair of electrons.
- Which type of orbital overlap leads to the formation of a pi bond?
A. Overlap along the internuclear axis
B. Lateral overlap of p-orbitals
C. Head-on overlap of s-orbitals
D. Overlap of d-orbitals
Answer: B. Lateral overlap of p-orbitals
Explanation: A pi bond forms when two p-orbitals overlap sideways, leading to a bond above and below the plane of the atoms involved.
- According to VBT, the bond angle in water (H₂O) is approximately:
A. 109.5°
B. 104.5°
C. 120°
D. 180°
Answer: B. 104.5°
Explanation: Water (H₂O) has a bent shape due to sp³ hybridization of oxygen. Two lone pairs reduce the bond angle to approximately 104.5° instead of the ideal tetrahedral angle of 109.5°.
- According to VBT, in the formation of H₂ molecule:
A. The two hydrogen atoms do not overlap
B. There is no overlap of orbitals
C. The 1s orbitals of two hydrogen atoms overlap
D. The 2p orbitals of two hydrogen atoms overlap
Answer: C. The 1s orbitals of two hydrogen atoms overlap
Explanation: The H₂ molecule forms when the 1s orbitals of two hydrogen atoms overlap, resulting in the formation of a sigma bond.
- In Valence Bond Theory, hybridization helps to:
A. Minimize the bond angles
B. Explain the magnetic properties of molecules
C. Explain the shape and geometry of molecules
D. Create molecular orbitals
Answer: C. Explain the shape and geometry of molecules
Explanation: Hybridization explains the shape and geometry of molecules by mixing atomic orbitals to form hybrid orbitals, which determine the bond angles.
- In VBT, why is the bond angle in NH₃ smaller than the bond angle in CH₄?
A. Because nitrogen has lone pairs
B. Because hydrogen atoms are smaller in NH₃
C. Because nitrogen is more electronegative
D. Because ammonia has more pi bonds
Answer: A. Because nitrogen has lone pairs
Explanation: The lone pair-bond pair repulsion in NH₃ forces the bond angles to decrease from the ideal tetrahedral angle of 109.5° (in CH₄) to around 107°.
- In a molecule of ethene (C₂H₄), the carbon atoms are:
A. sp hybridized
B. sp² hybridized
C. sp³ hybridized
D. Not hybridized
Answer: B. sp² hybridized
Explanation: In ethene (C₂H₄), each carbon atom is sp² hybridized, which results in a trigonal planar arrangement. One p-orbital on each carbon atom remains unhybridized, forming a pi bond between them.
- What type of bond is formed in ethyne (C₂H₂)?
A. Two sigma bonds and two pi bonds
B. Three sigma bonds and one pi bond
C. One sigma bond and two pi bonds
D. One sigma bond and one pi bond
Answer: A. Two sigma bonds and two pi bonds
Explanation: In ethyne (C₂H₂), each carbon forms one sigma bond with hydrogen and one sigma bond with the other carbon. The remaining p-orbitals overlap to form two pi bonds, making a triple bond between the carbon atoms.
- According to VBT, the shape of a BF₃ molecule is:
A. Linear
B. Trigonal planar
C. Tetrahedral
D. Bent
Answer: B. Trigonal planar
Explanation: Boron in BF₃ is sp² hybridized, leading to a trigonal planar shape with bond angles of 120°.
- The bond angle in methane (CH₄) is:
A. 90°
B. 104.5°
C. 107°
D. 109.5°
Answer: D. 109.5°
Explanation: Methane (CH₄) has a tetrahedral geometry due to sp³ hybridization of carbon, which results in bond angles of 109.5°.
- Which of the following molecules does not involve hybridization of the central atom?
A. H₂O
B. CH₄
C. CO₂
D. O₂
Answer: D. O₂
Explanation: The oxygen molecule (O₂) involves a double bond with no hybridization since there is no central atom. In contrast, H₂O, CH₄, and CO₂ involve hybridization of the central atom.
- In a PCl₅ molecule, phosphorus exhibits which type of hybridization?
A. sp³
B. sp³d
C. sp²
D. sp³d²
Answer: B. sp³d
Explanation: Phosphorus in PCl₅ undergoes sp³d hybridization, which leads to a trigonal bipyramidal geometry with 90° and 120° bond angles.
- In Valence Bond Theory, overlap of atomic orbitals is essential because:
A. It helps distribute electrons evenly across the molecule
B. It leads to the formation of molecular orbitals
C. It results in the pairing of electrons between atoms
D. It is responsible for ion formation
Answer: C. It results in the pairing of electrons between atoms
Explanation: VBT describes bonding as the result of the overlap of atomic orbitals, leading to the pairing of electrons between atoms to form covalent bonds.
- In Valence Bond Theory, why does H₂O have a bent structure?
A. Due to the sp hybridization of oxygen
B. Because of lone pair-lone pair repulsion
C. Due to bond pair-bond pair repulsion
D. Because oxygen does not hybridize
Answer: B. Because of lone pair-lone pair repulsion
Explanation: The bent shape of water is a result of two lone pairs on oxygen repelling the bonded pairs, leading to a reduction in bond angles from 109.5° (ideal tetrahedral) to about 104.5°.
- What is the hybridization of the central atom in CO₂?
A. sp
B. sp²
C. sp³
D. sp³d
Answer: A. sp
Explanation: In CO₂, the carbon atom is sp hybridized, resulting in a linear structure with bond angles of 180°.
- Which of the following molecules does not exhibit sp³ hybridization?
A. CCl₄
B. NH₃
C. H₂O
D. BeCl₂
Answer: D. BeCl₂
Explanation: BeCl₂ is sp hybridized and has a linear geometry, whereas CCl₄, NH₃, and H₂O exhibit sp³ hybridization.
- The shape of SF₆ can be explained using which type of hybridization?
A. sp²
B. sp³d²
C. sp³d
D. sp³
Answer: B. sp³d²
Explanation: In sulfur hexafluoride (SF₆), sulfur is sp³d² hybridized, leading to an octahedral geometry with 90° bond angles.
- Which of the following molecules has no lone pair on its central atom?
A. NH₃
B. H₂O
C. CH₄
D. SO₂
Answer: C. CH₄
Explanation: In methane (CH₄), the carbon atom forms four sigma bonds and has no lone pairs. In contrast, NH₃ and H₂O have lone pairs on the central atom.
- In the Valence Bond Theory, a pi bond forms due to:
A. Head-to-head overlap of p-orbitals
B. Side-to-side overlap of p-orbitals
C. Overlap of s-orbitals
D. Overlap of d-orbitals
Answer: B. Side-to-side overlap of p-orbitals
Explanation: A pi bond is formed by the lateral (side-to-side) overlap of unhybridized p-orbitals.
- Which of the following statements is true about hybrid orbitals in VBT?
A. Hybrid orbitals have the same energy as the original atomic orbitals.
B. Hybrid orbitals are less directional than atomic orbitals.
C. Hybrid orbitals result in maximum overlap and stronger bonds.
D. Hybrid orbitals always form pi bonds.
Answer: C. Hybrid orbitals result in maximum overlap and stronger bonds.
Explanation: Hybridization allows for maximum overlap of orbitals, leading to stronger and more stable covalent bonds.
- The bond angle in BeCl₂ is:
A. 109.5°
B. 120°
C. 180°
D. 90°
Answer: C. 180°
Explanation: In BeCl₂, the central beryllium atom is sp hybridized, giving the molecule a linear shape with a bond angle of 180°.
- In sp² hybridization, the number of hybrid orbitals formed is:
A. 2
B. 3
C. 4
D. 5
Answer: B. 3
Explanation: In sp² hybridization, one s orbital and two p orbitals mix to form three sp² hybrid orbitals, which are arranged in a trigonal planar geometry.
- The bond order of a molecule can be predicted by VBT when:
A. Bonding and antibonding orbitals are equally occupied
B. There is maximum overlap of orbitals
C. Orbitals do not overlap
D. Only hybrid orbitals are involved
Answer: B. There is maximum overlap of orbitals
Explanation: In VBT, a higher bond order is associated with greater overlap of atomic orbitals, leading to stronger bonds.
- In PCl₃, the central phosphorus atom undergoes:
A. sp³ hybridization
B. sp hybridization
C. sp² hybridization
D. sp³d hybridization
Answer: A. sp³ hybridization
Explanation: In phosphorus trichloride (PCl₃), phosphorus undergoes sp³ hybridization, resulting in a trigonal pyramidal shape with one lone pair.
- In SF₄, the shape of the molecule can be explained using:
A. sp² hybridization
B. sp³ hybridization
C. sp³d hybridization
D. sp³d² hybridization
Answer: C. sp³d hybridization
Explanation: In sulfur tetrafluoride (SF₄), sulfur undergoes sp³d hybridization, which leads to a seesaw-shaped molecular geometry.
- How many pi bonds are present in a molecule of C₂H₂ (ethyne)?
A. 0
B. 1
C. 2
D. 3
Answer: C. 2
Explanation: In ethyne (C₂H₂), there is one sigma bond and two pi bonds between the two carbon atoms, forming a triple bond.
- Which of the following has sp hybridization in the central atom?
A. CH₄
B. CO₂
C. NH₃
D. H₂O
Answer: B. CO₂
Explanation: The carbon atom in CO₂ is sp hybridized, resulting in a linear structure with bond angles of 180°.
- In VBT, the term hybridization refers to:
A. The mixing of atomic orbitals to form new, equivalent orbitals
B. The sharing of electrons in molecular orbitals
C. The transfer of electrons between atoms
D. The formation of ionic bonds
Answer: A. The mixing of atomic orbitals to form new, equivalent orbitals
Explanation: Hybridization is the process where atomic orbitals mix to form new hybrid orbitals that are degenerate (equal in energy) and participate in bonding.
- Which of the following molecules involves sp³ hybridization?
A. BeCl₂
B. BF₃
C. H₂O
D. C₂H₂
Answer: C. H₂O
Explanation: In water (H₂O), oxygen undergoes sp³ hybridization, resulting in a bent molecular geometry due to the presence of two lone pairs.
- The shape of a molecule of NH₄⁺ is:
A. Trigonal planar
B. Linear
C. Tetrahedral
D. Square planar
Answer: C. Tetrahedral
Explanation: In NH₄⁺ (ammonium ion), nitrogen is sp³ hybridized, forming a tetrahedral structure with bond angles of 109.5°.
- According to VBT, the bond angle in ammonia (NH₃) is approximately:
A. 90°
B. 107°
C. 120°
D. 180°
Answer: B. 107°
Explanation: In ammonia, nitrogen undergoes sp³ hybridization. The presence of a lone pair compresses the bond angle from the ideal tetrahedral angle of 109.5° to about 107°.
- Which of the following molecules has no hybridization involved in its bonding?
A. N₂
B. H₂O
C. CH₄
D. NH₃
Answer: A. N₂
Explanation: The nitrogen molecule (N₂) involves a triple bond between two nitrogen atoms, with no need for hybridization since only p-orbital overlap is required for the bonding.
- In the molecule of BCl₃, boron is:
A. sp hybridized
B. sp² hybridized
C. sp³ hybridized
D. sp³d hybridized
Answer: B. sp² hybridized
Explanation: In boron trichloride (BCl₃), boron is sp² hybridized, leading to a trigonal planar structure with bond angles of 120°.
- According to VBT, multiple bonds (double or triple bonds) involve:
A. Only sigma bonds
B. Only pi bonds
C. One sigma bond and additional pi bonds
D. Two sigma bonds
Answer: C. One sigma bond and additional pi bonds
Explanation: Multiple bonds consist of one sigma bond (due to head-on overlap) and one or more pi bonds (due to side-on overlap of p-orbitals).
- Which of the following molecules involves sp³d² hybridization?
A. SF₆
B. PCl₅
C. BF₃
D. CO₂
Answer: A. SF₆
Explanation: In sulfur hexafluoride (SF₆), sulfur undergoes sp³d² hybridization, resulting in an octahedral molecular geometry with 90° bond angles.
- The shape of a molecule of XeF₂ can be explained by which hybridization?
A. sp
B. sp²
C. sp³d
D. sp³d²
Answer: C. sp³d
Explanation: In xenon difluoride (XeF₂), xenon undergoes sp³d hybridization, resulting in a linear structure due to the presence of three lone pairs on xenon.
- The bond order of a molecule with a double bond is:
A. 1
B. 2
C. 3
D. 4
Answer: B. 2
Explanation: A double bond consists of one sigma bond and one pi bond, which gives it a bond order of 2.
- In a triple bond, there are:
A. 3 sigma bonds
B. 1 sigma bond and 2 pi bonds
C. 2 sigma bonds and 1 pi bond
D. 3 pi bonds
Answer: B. 1 sigma bond and 2 pi bonds
Explanation: A triple bond consists of one sigma bond formed by head-on overlap and two pi bonds formed by side-on overlap of p-orbitals.
- Which of the following molecules has sp³d hybridization?
A. BeCl₂
B. SF₄
C. H₂O
D. CO₂
Answer: B. SF₄
Explanation: In sulfur tetrafluoride (SF₄), sulfur undergoes sp³d hybridization, which leads to a seesaw-shaped molecular geometry.
- Which of the following is a linear molecule?
A. H₂O
B. NH₃
C. CO₂
D. CH₄
Answer: C. CO₂
Explanation: In carbon dioxide (CO₂), the central carbon atom is sp hybridized, leading to a linear structure with a bond angle of 180°.
- The bond angles in BF₃ are:
A. 90°
B. 109.5°
C. 120°
D. 180°
Answer: C. 120°
Explanation: In boron trifluoride (BF₃), boron is sp² hybridized, leading to a trigonal planar shape with bond angles of 120°.
- In Valence Bond Theory, a sigma bond is formed due to:
A. Sideways overlap of p-orbitals
B. End-to-end overlap of atomic orbitals
C. Overlap of hybrid orbitals only
D. Overlap of d-orbitals
Answer: B. End-to-end overlap of atomic orbitals
Explanation: A sigma bond is formed by the head-on (end-to-end) overlap of atomic orbitals, which can be s-s, s-p, or p-p overlaps.
- In XeF₄, xenon is:
A. sp³ hybridized
B. sp² hybridized
C. sp³d² hybridized
D. sp³d hybridized
Answer: C. sp³d² hybridized
Explanation: In xenon tetrafluoride (XeF₄), xenon undergoes sp³d² hybridization, resulting in a square planar structure.
- The shape of BeCl₂ is:
A. Linear
B. Trigonal planar
C. Tetrahedral
D. Bent
Answer: A. Linear
Explanation: In beryllium chloride (BeCl₂), beryllium is sp hybridized, leading to a linear structure with bond angles of 180°.
- In the ammonium ion (NH₄⁺), the nitrogen atom is:
A. sp² hybridized
B. sp³ hybridized
C. sp hybridized
D. Not hybridized
Answer: B. sp³ hybridized
Explanation: In the ammonium ion (NH₄⁺), nitrogen is sp³ hybridized, giving the ion a tetrahedral geometry.
- Which of the following molecules has a trigonal planar geometry?
A. NH₃
B. H₂O
C. BF₃
D. CH₄
Answer: C. BF₃
Explanation: In boron trifluoride (BF₃), boron is sp² hybridized, which gives the molecule a trigonal planar geometry with bond angles of 120°.
- The bond angle in a molecule of XeF₂ is:
A. 90°
B. 109.5°
C. 180°
D. 120°
Answer: C. 180°
Explanation: In xenon difluoride (XeF₂), xenon is sp³d hybridized, and the molecule has a linear shape with a bond angle of 180°.
- According to VBT, lone pairs of electrons on the central atom:
A. Do not affect molecular geometry
B. Increase bond angles
C. Repel bonding pairs and reduce bond angles
D. Are involved in sigma bonds
Answer: C. Repel bonding pairs and reduce bond angles
Explanation: Lone pairs on the central atom repel bonding pairs, which leads to a reduction in bond angles. This explains why the bond angle in molecules like H₂O is less than the ideal tetrahedral angle of 109.5°.
- In a trigonal bipyramidal molecule, the bond angles are:
A. 90° and 180°
B. 90° and 120°
C. 109.5°
D. 120° and 180°
Answer: B. 90° and 120°
Explanation: In a trigonal bipyramidal structure, like PCl₅, the bond angles are 90° between axial and equatorial positions and 120° between the equatorial positions.
- In VBT, hybrid orbitals are:
A. Lower in energy than the original orbitals
B. Higher in energy than the original orbitals
C. Equivalent in energy to the original orbitals
D. Averaged energy between the original orbitals
Answer: D. Averaged energy between the original orbitals
Explanation: Hybrid orbitals have an energy level that is an average of the original atomic orbitals that combined to form them.
- The shape of NH₃ can be explained using which type of hybridization?
A. sp
B. sp²
C. sp³
D. sp³d
Answer: C. sp³
Explanation: In ammonia (NH₃), nitrogen undergoes sp³ hybridization, which results in a trigonal pyramidal shape due to the lone pair on nitrogen.
- Which of the following molecules exhibits sp hybridization?
A. C₂H₄
B. CH₄
C. CO₂
D. NH₃
Answer: C. CO₂
Explanation: In carbon dioxide (CO₂), the central carbon atom undergoes sp hybridization, resulting in a linear geometry with bond angles of 180°.
- In Valence Bond Theory, bond strength increases with:
A. Less overlap of atomic orbitals
B. Greater overlap of atomic orbitals
C. More lone pairs on the central atom
D. The number of pi bonds
Answer: B. Greater overlap of atomic orbitals
Explanation: In VBT, stronger bonds result from greater overlap of atomic orbitals, as this leads to a higher degree of electron sharing between atoms.
- Which of the following statements is true about a pi bond?
A. It forms by head-on overlap of atomic orbitals
B. It is weaker than a sigma bond
C. It is always the first bond to form between two atoms
D. It involves hybrid orbitals
Answer: B. It is weaker than a sigma bond
Explanation: A pi bond is generally weaker than a sigma bond because it results from the side-to-side overlap of p-orbitals, which is less effective than the head-on overlap in a sigma bond.
- In VSEPR theory, the presence of lone pairs on the central atom:
A. Does not affect the shape
B. Affects only the bond length
C. Changes the bond angles
D. Only affects molecules with pi bonds
Answer: C. Changes the bond angles
Explanation: According to VSEPR theory, lone pairs on the central atom exert more repulsion than bonding pairs, which leads to a change in bond angles, often reducing them.
- Which of the following molecules has a square planar shape?
A. NH₃
B. XeF₄
C. BF₃
D. H₂O
Answer: B. XeF₄
Explanation: Xenon in xenon tetrafluoride (XeF₄) is sp³d² hybridized, resulting in a square planar shape due to the presence of two lone pairs.
- The angle between sp³ hybrid orbitals is:
A. 90°
B. 109.5°
C. 120°
D. 180°
Answer: B. 109.5°
Explanation: The sp³ hybrid orbitals are arranged in a tetrahedral geometry, and the angle between the orbitals is 109.5°.
- Which molecule has both sigma and pi bonds?
A. CH₄
B. CO₂
C. NH₃
D. H₂O
Answer: B. CO₂
Explanation: In carbon dioxide (CO₂), each carbon-oxygen bond consists of one sigma bond and one pi bond, making it a double bond.
- Hybridization in SF₆ can be described as:
A. sp³
B. sp²
C. sp³d²
D. sp³d
Answer: C. sp³d²
Explanation: In sulfur hexafluoride (SF₆), sulfur undergoes sp³d² hybridization, leading to an octahedral structure with bond angles of 90°.
- Which molecule is an example of sp³ hybridization?
A. CO₂
B. CH₄
C. C₂H₂
D. BeCl₂
Answer: B. CH₄
Explanation: In methane (CH₄), carbon undergoes sp³ hybridization, resulting in a tetrahedral structure with bond angles of 109.5°.
- Which of the following is linear due to sp hybridization?
A. CH₄
B. BeCl₂
C. BF₃
D. H₂O
Answer: B. BeCl₂
Explanation: In beryllium chloride (BeCl₂), beryllium is sp hybridized, leading to a linear structure with bond angles of 180°.
- Which of the following molecules involves sp² hybridization?
A. C₂H₂
B. CH₄
C. BF₃
D. NH₃
Answer: C. BF₃
Explanation: In boron trifluoride (BF₃), boron is sp² hybridized, leading to a trigonal planar structure with bond angles of 120°.
- The bond angle in CH₄ is:
A. 90°
B. 109.5°
C. 120°
D. 180°
Answer: B. 109.5°
Explanation: In methane (CH₄), carbon is sp³ hybridized, resulting in a tetrahedral structure with bond angles of 109.5°.
- In VBT, a pi bond is formed by:
A. End-to-end overlap of orbitals
B. Sideways overlap of p-orbitals
C. Overlap of s-orbitals
D. Hybrid orbital overlap
Answer: B. Sideways overlap of p-orbitals
Explanation: Pi bonds result from the sideways overlap of p-orbitals, whereas sigma bonds are formed by end-to-end overlap.
- Which of the following has no pi bonds?
A. CO₂
B. C₂H₄
C. CH₄
D. C₂H₂
Answer: C. CH₄
Explanation: Methane (CH₄) contains only sigma bonds, as all carbon-hydrogen bonds are formed by the overlap of sp³ hybrid orbitals.
- In VBT, the term resonance refers to:
A. The mixing of orbitals to form hybrid orbitals
B. The existence of multiple bond structures for a molecule
C. The formation of pi bonds
D. The delocalization of electrons in molecular orbitals
Answer: B. The existence of multiple bond structures for a molecule
Explanation: Resonance occurs when a molecule can be represented by two or more valid Lewis structures, with the actual structure being a hybrid of these forms.
- Which of the following molecules exhibits sp hybridization?
A. BeF₂
B. NH₃
C. BF₃
D. H₂O
Answer: A. BeF₂
Explanation: In beryllium fluoride (BeF₂), beryllium is sp hybridized, resulting in a linear structure with bond angles of 180°.
- In SF₆, the sulfur atom undergoes:
A. sp hybridization
B. sp³ hybridization
C. sp³d hybridization
D. sp³d² hybridization
Answer: D. sp³d² hybridization
Explanation: In sulfur hexafluoride (SF₆), sulfur undergoes sp³d² hybridization, leading to an octahedral shape with bond angles of 90°.
- The hybridization of the central atom in PCl₅ is:
A. sp³
B. sp²
C. sp³d
D. sp³d²
Answer: C. sp³d
Explanation: In phosphorus pentachloride (PCl₅), phosphorus is sp³d hybridized, which results in a trigonal bipyramidal shape.
- The bond order in O₂ is:
A. 1
B. 2
C. 2.5
D. 3
Answer: B. 2
Explanation: In the oxygen molecule (O₂), the bond order is 2, indicating a double bond consisting of one sigma bond and one pi bond.
- Which of the following molecules has a trigonal pyramidal shape?
A. BF₃
B. NH₃
C. H₂O
D. CO₂
Answer: B. NH₃
Explanation: In ammonia (NH₃), nitrogen is sp³ hybridized, resulting in a trigonal pyramidal shape with a bond angle of about 107°.
- In VBT, the term bond dissociation energy refers to:
A. The energy required to break a bond
B. The energy released when a bond forms
C. The energy associated with pi bonds only
D. The energy required to form a hybrid orbital
Answer: A. The energy required to break a bond
Explanation: Bond dissociation energy is the energy required to break a chemical bond in a molecule, resulting in the formation of separate atoms.
- In a molecule of ethene (C₂H₄), the hybridization of each carbon atom is:
A. sp
B. sp²
C. sp³
D. sp³d
Answer: B. sp²
Explanation: In ethene (C₂H₄), each carbon atom is sp² hybridized, leading to a planar structure with one sigma bond and one pi bond between the carbons.
- The bond order in a triple bond is:
A. 1
B. 2
C. 3
D. 4
Answer: C. 3
Explanation: A triple bond has a bond order of 3, consisting of one sigma bond and two pi bonds.
- In VBT, the geometry of a molecule is primarily determined by:
A. The number of sigma and pi bonds
B. The hybridization of the central atom
C. The total number of valence electrons
D. The mass of the atoms
Answer: B. The hybridization of the central atom
Explanation: The geometry of a molecule is largely determined by the hybridization state of the central atom, which defines how the orbitals are arranged in space.
- Which of the following molecules involves sp³d² hybridization?
A. PCl₅
B. SF₆
C. BF₃
D. CH₄
Answer: B. SF₆
Explanation: In sulfur hexafluoride (SF₆), sulfur undergoes sp³d² hybridization, resulting in an octahedral structure with bond angles of 90°.
- The bond angle in water (H₂O) is approximately:
A. 109.5°
B. 107°
C. 104.5°
D. 90°
Answer: C. 104.5°
Explanation: In water (H₂O), the oxygen atom undergoes sp³ hybridization, and the bond angle is reduced to about 104.5° due to the lone pairs of electrons on oxygen.
- In VBT, delocalization of electrons is explained by:
A. Overlap of atomic orbitals
B. Resonance structures
C. Hybridization
D. Bond dissociation energy
Answer: B. Resonance structures
Explanation: Delocalization of electrons in VBT is often explained using resonance, where electrons are shared among multiple atoms across several valid Lewis structures.
- In benzene (C₆H₆), the hybridization of each carbon atom is:
A. sp
B. sp²
C. sp³
D. sp³d
Answer: B. sp²
Explanation: In benzene, each carbon atom is sp² hybridized, and the delocalized pi electrons form a conjugated ring system, giving the molecule its planar structure.
- The bond angle in a molecule of CO₂ is:
A. 90°
B. 104.5°
C. 120°
D. 180°
Answer: D. 180°
Explanation: In carbon dioxide (CO₂), the central carbon atom is sp hybridized, resulting in a linear structure with a bond angle of 180°.
- In VBT, the strength of a pi bond is:
A. Equal to that of a sigma bond
B. Greater than that of a sigma bond
C. Less than that of a sigma bond
D. Dependent on hybridization
Answer: C. Less than that of a sigma bond
Explanation: A pi bond is generally weaker than a sigma bond because it arises from the sideways overlap of p-orbitals, which is less effective than the head-on overlap that forms a sigma bond.
- The bond order of nitrogen (N₂) is:
A. 1
B. 2
C. 3
D. 4
Answer: C. 3
Explanation: In nitrogen (N₂), the bond order is 3, meaning the two nitrogen atoms are connected by a triple bond (one sigma and two pi bonds).
- Which of the following molecules has a tetrahedral structure?
A. NH₃
B. H₂O
C. CH₄
D. CO₂
Answer: C. CH₄
Explanation: Methane (CH₄) has a tetrahedral structure, as the carbon atom is sp³ hybridized, and the bond angles are 109.5°.
- In VBT, the bond angle in a molecule depends primarily on:
A. The number of pi bonds
B. The number of lone pairs
C. The hybridization of the central atom
D. The size of the atoms involved
Answer: C. The hybridization of the central atom
Explanation: The bond angle in a molecule is largely determined by the hybridization of the central atom, which dictates the spatial arrangement of the bonding orbitals.
- The hybridization of carbon in C₂H₂ is:
A. sp
B. sp²
C. sp³
D. sp³d
Answer: A. sp
Explanation: In acetylene (C₂H₂), each carbon atom is sp hybridized, resulting in a linear structure with bond angles of 180°.
- In VBT, the formation of a sigma bond is always due to:
A. Sideways overlap of orbitals
B. End-to-end overlap of orbitals
C. Overlap of d-orbitals
D. Overlap of pi orbitals
Answer: B. End-to-end overlap of orbitals
Explanation: A sigma bond is formed by the head-on (end-to-end) overlap of atomic orbitals, which results in stronger bonding than a pi bond.
- The bonding in methane (CH₄) can be explained by:
A. sp hybridization
B. sp² hybridization
C. sp³ hybridization
D. sp³d hybridization
Answer: C. sp³ hybridization
Explanation: In methane (CH₄), carbon undergoes sp³ hybridization, resulting in a tetrahedral structure with bond angles of 109.5°.
- The geometry of a molecule with sp² hybridization is:
A. Linear
B. Trigonal planar
C. Tetrahedral
D. Trigonal bipyramidal
Answer: B. Trigonal planar
Explanation: sp² hybridization leads to a trigonal planar structure with bond angles of 120°. This occurs in molecules like BF₃.
- The overlap of two p-orbitals along the internuclear axis forms a:
A. Sigma bond
B. Pi bond
C. Ionic bond
D. Hydrogen bond
Answer: A. Sigma bond
Explanation: A sigma bond is formed when two atomic orbitals overlap along the internuclear axis, resulting in head-on overlap. This is typically stronger than a pi bond, which is a sideways overlap.
- Which of the following molecules has sp³d hybridization?
A. SF₆
B. NH₃
C. PCl₅
D. BF₃
Answer: C. PCl₅
Explanation: In phosphorus pentachloride (PCl₅), phosphorus undergoes sp³d hybridization, resulting in a trigonal bipyramidal shape with bond angles of 90° and 120°.
- Which molecule involves delocalized pi electrons?
A. CH₄
B. NH₃
C. H₂O
D. Benzene (C₆H₆)
Answer: D. Benzene (C₆H₆)
Explanation: In benzene, the six carbon atoms are sp² hybridized, and the pi electrons are delocalized over the entire ring, leading to resonance and extra stability.
- Which of the following molecules has a linear shape?
A. CO₂
B. H₂O
C. NH₃
D. CH₄
Answer: A. CO₂
Explanation: In carbon dioxide (CO₂), the carbon atom is sp hybridized, and the molecule has a linear structure with bond angles of 180°.
- In VBT, resonance is used to explain:
A. Hybridization of orbitals
B. The delocalization of electrons in certain molecules
C. The mixing of atomic orbitals
D. The formation of sigma bonds
Answer: B. The delocalization of electrons in certain molecules
Explanation: Resonance in VBT refers to the delocalization of electrons across several atoms in a molecule, as seen in molecules like ozone (O₃) and benzene (C₆H₆).
- The bond angles in NH₃ are slightly less than:
A. 180°
B. 120°
C. 109.5°
D. 90°
Answer: C. 109.5°
Explanation: In ammonia (NH₃), the nitrogen atom is sp³ hybridized, and the bond angles are approximately 107° due to the lone pair-bond pair repulsion.
- The hybridization of carbon in ethyne (C₂H₂) is:
A. sp
B. sp²
C. sp³
D. sp³d
Answer: A. sp
Explanation: In ethyne (C₂H₂), each carbon atom undergoes sp hybridization, resulting in a linear structure with bond angles of 180°.
- Bond dissociation energy is highest for:
A. Pi bonds
B. Sigma bonds
C. Hydrogen bonds
D. Ionic bonds
Answer: B. Sigma bonds
Explanation: Sigma bonds are generally stronger than pi bonds because they involve end-to-end overlap of atomic orbitals, which results in greater bond strength
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