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101. A covalent bond is called polar when:
ⓐ. Electrons are shared equally between atoms
ⓑ. Electrons are transferred completely from one atom to another
ⓒ. Electrons are shared unequally due to difference in electronegativity
ⓓ. The atoms involved are identical
Correct Answer: Electrons are shared unequally due to difference in electronegativity
Explanation: In a polar covalent bond, one atom attracts the shared electron pair more strongly, creating partial charges (δ⁺ and δ⁻). Example: H–Cl, where Cl is more electronegative. If atoms are identical, the bond is non-polar, and if electrons are transferred completely, the bond is ionic.
102. Which of the following molecules contains a non-polar covalent bond?
ⓐ. HCl
ⓑ. HF
ⓒ. O₂
ⓓ. H₂O
Correct Answer: O₂
Explanation: In O₂, both atoms have the same electronegativity, so electrons are shared equally. This makes the bond purely non-polar. HCl and HF are polar because of electronegativity difference. H₂O is polar overall due to bent geometry.
103. The bond between carbon and hydrogen (C–H) is considered:
ⓐ. Purely ionic
ⓑ. Polar covalent with significant polarity
ⓒ. Weak polar covalent, nearly non-polar
ⓓ. Purely metallic
Correct Answer: Weak polar covalent, nearly non-polar
Explanation: The electronegativity difference between C (2.5) and H (2.1) is only 0.4, so the bond is slightly polar but often treated as non-polar in organic chemistry. This makes hydrocarbons generally non-polar compounds.
104. Which molecule has polar covalent bonds but is overall non-polar?
ⓐ. CO₂
ⓑ. H₂O
ⓒ. NH₃
ⓓ. HF
Correct Answer: CO₂
Explanation: Each C=O bond in CO₂ is polar, but the linear geometry (180° bond angle) causes dipole moments to cancel out, making the molecule non-polar overall. In contrast, H₂O and NH₃ are polar because their bond dipoles do not cancel. HF is polar due to large electronegativity difference.
105. Which of the following molecules is polar?
ⓐ. CCl₄
ⓑ. CH₄
ⓒ. H₂O
ⓓ. CO₂
Correct Answer: H₂O
Explanation: H₂O is polar because the O–H bonds are polar and the molecule is bent, so dipoles add up to give a net dipole. CCl₄ and CH₄ are symmetrical and thus non-polar. CO₂ is linear, so dipoles cancel.
106. The polarity of a covalent bond depends mainly on:
ⓐ. Bond length only
ⓑ. Electronegativity difference between atoms
ⓒ. Atomic mass of bonded atoms
ⓓ. Number of protons in nucleus
Correct Answer: Electronegativity difference between atoms
Explanation: Polarity arises when one atom attracts the shared pair more strongly than the other. Greater electronegativity difference increases polarity. Bond length and atomic mass do not determine polarity directly.
107. Which of the following bonds is the most polar?
ⓐ. H–Cl
ⓑ. H–F
ⓒ. H–Br
ⓓ. H–I
Correct Answer: H–F
Explanation: Fluorine is the most electronegative element (3.98), so the electronegativity difference between H (2.1) and F is the highest, making H–F the most polar bond. Polarity decreases down the halogen group.
108. A molecule with polar bonds may still be non-polar if:
ⓐ. The electronegativity difference is very large
ⓑ. The molecule is symmetrical and dipoles cancel
ⓒ. It contains hydrogen atoms only
ⓓ. It has more than one lone pair
Correct Answer: The molecule is symmetrical and dipoles cancel
Explanation: Example: CCl₄ and CO₂ have polar bonds, but symmetry cancels out dipole moments, resulting in non-polar molecules. This is why shape (VSEPR theory) must be considered alongside bond polarity.
109. Which best describes the difference between polar covalent and non-polar covalent bonds?
ⓐ. Polar covalent bonds have high bond enthalpy, non-polar have low
ⓑ. Polar covalent bonds involve unequal sharing, non-polar involve equal sharing
ⓒ. Polar covalent bonds form only between metals, non-polar between non-metals
ⓓ. Polar covalent bonds involve transfer of electrons, non-polar involve sharing
Correct Answer: Polar covalent bonds involve unequal sharing, non-polar involve equal sharing
Explanation: Polarity is about how equally electrons are shared. In polar covalent bonds (like H–Cl), one atom pulls harder, while in non-polar bonds (like Cl₂), electrons are shared equally. Options A, C, and D confuse ionic vs covalent characteristics.
110. Which property is strongly influenced by molecular polarity?
ⓐ. Atomic number
ⓑ. Melting point of metals
ⓒ. Solubility in polar or non-polar solvents
ⓓ. Nuclear stability
Correct Answer: Solubility in polar or non-polar solvents
Explanation: Polarity determines intermolecular forces. Polar molecules dissolve better in polar solvents (like H₂O), and non-polar molecules dissolve better in non-polar solvents (like benzene). This principle is called “like dissolves like.” Atomic number and nuclear stability are intrinsic atomic properties, not influenced by polarity.
111. Bond length in a covalent bond is defined as:
ⓐ. Distance between the nuclei of two bonded atoms at maximum repulsion
ⓑ. Distance between the outermost electrons of bonded atoms
ⓒ. Equilibrium distance between the nuclei of two bonded atoms where attractive and repulsive forces balance
ⓓ. Distance between the nuclei of two non-bonded atoms
Correct Answer: Equilibrium distance between the nuclei of two bonded atoms where attractive and repulsive forces balance
Explanation: Bond length is the average distance between the nuclei of two bonded atoms at the lowest potential energy state. If atoms come too close, nuclear repulsion dominates; if too far, attraction weakens. Thus, equilibrium is achieved at bond length, which is characteristic of each bond type.
112. Which of the following bonds has the shortest bond length?
ⓐ. C–C single bond
ⓑ. C=C double bond
ⓒ. C≡C triple bond
ⓓ. C–H bond
Correct Answer: C≡C triple bond
Explanation: Bond length decreases as the number of shared electron pairs increases because nuclei are pulled closer. Hence, C≡C (120 pm) < C=C (134 pm) < C–C (154 pm). The C–H bond is shorter than a single C–C bond but among C–C multiple bonds, the triple bond is the shortest.
113. The bond angle in methane (CH₄) is approximately:
ⓐ. 90°
ⓑ. 104.5°
ⓒ. 120°
ⓓ. 109.5°
Correct Answer: 109.5°
Explanation: In CH₄, carbon undergoes sp³ hybridization, forming a tetrahedral geometry. This results in equal bond angles of 109.5°. Options B and C correspond to bent (H₂O) and trigonal planar (BF₃) molecules, respectively.
114. Which of the following molecules has a bond angle of about 120°?
ⓐ. NH₃
ⓑ. H₂O
ⓒ. BF₃
ⓓ. CH₄
Correct Answer: BF₃
Explanation: In BF₃, boron undergoes sp² hybridization, forming a trigonal planar geometry with bond angles of \~120°. NH₃ has \~107° (sp³ with one lone pair), H₂O has \~104.5° (sp³ with two lone pairs), and CH₄ has 109.5° (sp³ tetrahedral).
115. Why is the bond angle in H₂O smaller than that in NH₃?
ⓐ. Oxygen is less electronegative than nitrogen
ⓑ. H₂O has two lone pairs, while NH₃ has only one lone pair
ⓒ. H₂O is linear, while NH₃ is bent
ⓓ. NH₃ has sp² hybridization, while H₂O has sp³ hybridization
Correct Answer: H₂O has two lone pairs, while NH₃ has only one lone pair
Explanation: Lone pairs repel more strongly than bonding pairs. In H₂O, two lone pairs push hydrogen atoms closer, reducing bond angle to 104.5°. In NH₃, only one lone pair is present, so bond angle is larger (\~107°). CH₄, with no lone pairs, maintains 109.5°.
116. Which of the following factors decreases bond length?
ⓐ. Increase in bond order
ⓑ. Decrease in bond order
ⓒ. Presence of bulky substituents
ⓓ. Lone pair–lone pair repulsion
Correct Answer: Increase in bond order
Explanation: Bond order measures the number of bonds between atoms. Higher bond order (triple > double > single) means greater electron density between nuclei, stronger attraction, and shorter bond length. Lone pairs and bulky groups increase repulsion, often lengthening bonds.
117. Which molecule shows the largest bond angle?
ⓐ. CO₂
ⓑ. CH₄
ⓒ. NH₃
ⓓ. H₂O
Correct Answer: CO₂
Explanation: CO₂ is linear due to sp hybridization of carbon, giving a bond angle of 180°. CH₄ (sp³) has 109.5°, NH₃ has 107°, and H₂O has 104.5°. Thus, CO₂ has the maximum bond angle.
118. The bond length of H–F (92 pm) is shorter than that of H–Cl (127 pm). This is mainly because:
ⓐ. Fluorine is more electronegative than chlorine
ⓑ. Fluorine has smaller atomic radius than chlorine
ⓒ. Hydrogen forms a stronger bond with chlorine
ⓓ. H–Cl bond order is higher than H–F
Correct Answer: Fluorine has smaller atomic radius than chlorine
Explanation: Bond length is directly related to atomic size. Since fluorine is much smaller than chlorine, the nuclei can come closer, making H–F bond shorter. Electronegativity affects bond polarity, but bond length is primarily determined by atomic radius.
119. Which of the following is the correct order of bond angles?
ⓐ. H₂O < NH₃ < CH₄ < CO₂
ⓑ. CO₂ < H₂O < NH₃ < CH₄
ⓒ. NH₃ < H₂O < CH₄ < CO₂
ⓓ. CH₄ < NH₃ < H₂O < CO₂
Correct Answer: H₂O < NH₃ < CH₄ < CO₂
Explanation: Bond angles: H₂O \~104.5°, NH₃ \~107°, CH₄ \~109.5°, CO₂ = 180°. Hence the correct order is H₂O < NH₃ < CH₄ < CO₂. Lone pair repulsion explains the smaller angles in H₂O and NH₃ compared to CH₄.
120. Bond length and bond angle together determine:
ⓐ. Hybridization only
ⓑ. Shape and geometry of the molecule
ⓒ. Type of orbitals present in the atom
ⓓ. Nuclear binding energy
Correct Answer: Shape and geometry of the molecule
Explanation: Bond length fixes the distances between atoms, while bond angles fix their relative orientation. Together, they determine molecular geometry (linear, bent, tetrahedral, etc.). Hybridization helps explain these values, but the actual shape is defined by bond length and bond angle.
121. Bond enthalpy is defined as:
ⓐ. Energy released when a bond is formed in gaseous state
ⓑ. Energy required to break one mole of a given bond in gaseous state
ⓒ. Energy required to ionize an atom
ⓓ. Energy released when an atom gains an electron
Correct Answer: Energy required to break one mole of a given bond in gaseous state
Explanation: Bond enthalpy (also called bond dissociation energy) is the average amount of energy needed to break one mole of bonds of the same type in the gaseous phase. For example, the bond enthalpy of H–H is 436 kJ mol⁻¹. Option A is bond formation enthalpy, option C is ionization energy, and option D is electron affinity.
122. Which of the following bonds has the highest bond enthalpy?
ⓐ. H–H
ⓑ. O=O
ⓒ. N≡N
ⓓ. C–C
Correct Answer: N≡N
Explanation: The triple bond in N₂ has very high bond enthalpy (\~946 kJ mol⁻¹), making nitrogen gas extremely stable and inert. O=O has lower bond enthalpy (\~498 kJ mol⁻¹), H–H has \~436 kJ mol⁻¹, and C–C single bond has \~348 kJ mol⁻¹.
123. Which factor increases bond enthalpy?
ⓐ. Larger bond length
ⓑ. Smaller bond order
ⓒ. Greater bond order and shorter bond length
ⓓ. Greater number of lone pairs
Correct Answer: Greater bond order and shorter bond length
Explanation: Bond enthalpy increases with bond order because multiple bonds (double, triple) have more electron density between atoms, making the bond shorter and stronger. Lone pairs often repel and weaken bonds, lowering bond enthalpy.
124. The bond enthalpy of O=O is less than that of N≡N because:
ⓐ. Oxygen is smaller than nitrogen
ⓑ. N₂ has a triple bond while O₂ has a double bond
ⓒ. O₂ is linear while N₂ is bent
ⓓ. O₂ has higher atomic number
Correct Answer: N₂ has a triple bond while O₂ has a double bond
Explanation: Bond enthalpy correlates with bond order. N₂ has bond order 3, making it very strong. O₂ has bond order 2, hence weaker. Options A, C, and D are unrelated.
125. Which is the correct formula to calculate bond order in Molecular Orbital Theory?
ⓐ. $\dfrac{N_b – N_a}{2}$
ⓑ. $\dfrac{N_b + N_a}{2}$
ⓒ. $\dfrac{N_a – N_b}{2}$
ⓓ. $N_b \times N_a$
Correct Answer: $\dfrac{N_b – N_a}{2}$
Explanation: Bond order = (Number of electrons in bonding orbitals − Number in antibonding orbitals)/2. This formula predicts stability and magnetic properties of molecules. Higher bond order means greater stability.
126. What is the bond order of O₂ molecule?
ⓐ. 1
ⓑ. 2
ⓒ. 3
ⓓ. 2.5
Correct Answer: 2
Explanation: O₂ has 8+8 = 16 electrons. MO filling gives 10 in bonding and 6 in antibonding orbitals. Bond order = (10 − 6)/2 = 2, indicating a double bond. This also matches Lewis structure.
127. The bond order of O₂⁺ ion is:
ⓐ. 1.5
ⓑ. 2.0
ⓒ. 2.5
ⓓ. 3.0
Correct Answer: 2.5
Explanation: O₂⁺ has one less electron than O₂, removed from an antibonding orbital. Bond order = (10 − 5)/2 = 2.5. Hence, O₂⁺ is stronger and has shorter bond length than O₂.
128. Which has the lowest bond enthalpy among the following?
ⓐ. N₂
ⓑ. O₂
ⓒ. F₂
ⓓ. H₂
Correct Answer: F₂
Explanation: F₂ has very low bond enthalpy (\~159 kJ mol⁻¹) despite being a covalent bond. This is due to strong lone pair–lone pair repulsions between small, highly electronegative fluorine atoms. N₂, O₂, and H₂ have much higher bond enthalpies.
129. Which of the following relationships is correct between bond order, bond length, and bond enthalpy?
ⓐ. Higher bond order → shorter bond length → higher bond enthalpy
ⓑ. Higher bond order → longer bond length → lower bond enthalpy
ⓒ. Lower bond order → shorter bond length → higher bond enthalpy
ⓓ. Bond order has no relation to bond length or enthalpy
Correct Answer: Higher bond order → shorter bond length → higher bond enthalpy
Explanation: Bond order directly reflects bond strength. Triple bonds are shortest and strongest, with highest enthalpy; single bonds are longest and weakest. Hence, bond order, bond length, and bond enthalpy are interdependent.
130. Which of the following species has the highest bond order?
ⓐ. O₂
ⓑ. O₂⁻
ⓒ. O₂²⁻
ⓓ. O₂⁺
Correct Answer: O₂⁺
Explanation: O₂⁺ has bond order 2.5, O₂ has 2, O₂⁻ has 1.5, and O₂²⁻ has 1. Hence, O₂⁺ has the strongest bond, shortest bond length, and highest bond enthalpy among the options.
131. Which of the following statements correctly relates bond order with bond length?
ⓐ. Higher bond order corresponds to longer bond length
ⓑ. Higher bond order corresponds to shorter bond length
ⓒ. Bond order and bond length are unrelated
ⓓ. Lower bond order corresponds to shorter bond length
Correct Answer: Higher bond order corresponds to shorter bond length
Explanation: As bond order increases (single → double → triple), electron density between nuclei increases, pulling atoms closer. Hence, triple bonds are shortest, double bonds are intermediate, and single bonds are longest.
132. What is the bond order of N₂ molecule and how does it relate to its bond length?
ⓐ. 1; longest bond length
ⓑ. 2; moderate bond length
ⓒ. 3; shortest bond length
ⓓ. 2.5; intermediate bond length
Correct Answer: 3; shortest bond length
Explanation: In MO theory, N₂ has bond order (10 − 4)/2 = 3, i.e., a triple bond. High bond order results in very short bond length (\~110 pm) and very high bond enthalpy, making N₂ exceptionally stable.
133. Which oxygen species has the weakest O–O bond?
ⓐ. O₂
ⓑ. O₂⁻
ⓒ. O₂²⁻
ⓓ. O₂⁺
Correct Answer: O₂²⁻
Explanation: Bond order decreases in the series: O₂⁺ (2.5) > O₂ (2) > O₂⁻ (1.5) > O₂²⁻ (1). A lower bond order corresponds to weaker bonding and longer bond length. Thus, O₂²⁻ has the weakest and longest O–O bond.
134. Which factor explains why CO has a shorter bond length than CO₂?
ⓐ. CO contains a triple bond, CO₂ contains double bonds
ⓑ. CO contains a single bond, CO₂ contains triple bonds
ⓒ. Both contain the same bond order
ⓓ. CO₂ is ionic while CO is covalent
Correct Answer: CO contains a triple bond, CO₂ contains double bonds
Explanation: In CO, bond order is approximately 3, giving a very short bond length (\~112 pm). In CO₂, each C=O is a double bond with bond order 2, so the bond length is longer (\~120 pm).
135. Which species has the highest bond order and therefore the shortest bond length?
ⓐ. O₂
ⓑ. O₂⁻
ⓒ. O₂²⁻
ⓓ. O₂⁺
Correct Answer: O₂⁺
Explanation: Bond order values: O₂⁺ = 2.5, O₂ = 2, O₂⁻ = 1.5, O₂²⁻ = 1. A higher bond order corresponds to greater bond strength and shorter bond length. Thus, O₂⁺ is strongest and shortest.
136. The relationship between bond order and bond stability is:
ⓐ. Higher bond order → greater stability
ⓑ. Lower bond order → greater stability
ⓒ. Bond order has no effect on stability
ⓓ. Maximum stability occurs at bond order 1
Correct Answer: Higher bond order → greater stability
Explanation: Greater bond order means more electron pairs are shared, increasing attraction between nuclei and strengthening the bond. Hence, triple bonds are most stable, followed by double, then single bonds.
137. Which of the following has the longest bond length?
ⓐ. N₂
ⓑ. NO
ⓒ. NO⁻
ⓓ. NO⁺
Correct Answer: NO⁻
Explanation: Bond orders: N₂ = 3, NO = 2.5, NO⁻ = 2, NO⁺ = 3. Thus, NO⁻ has the lowest bond order among these species, giving it the longest bond length.
138. If bond order = 0, what does it signify about a molecule?
ⓐ. The molecule is stable and has longest bond
ⓑ. The molecule does not exist as it has no bonding interaction
ⓒ. The molecule exists but has only ionic bonds
ⓓ. The molecule exists but has weak hydrogen bonding
Correct Answer: The molecule does not exist as it has no bonding interaction
Explanation: A bond order of 0 means the number of bonding and antibonding electrons are equal, leading to no net bond formation. This implies the species is unstable and cannot exist under normal conditions.
139. The stability order of the species B₂, B₂⁺, and B₂⁻ based on bond order is:
ⓐ. B₂⁺ > B₂ > B₂⁻
ⓑ. B₂⁻ > B₂ > B₂⁺
ⓒ. B₂ > B₂⁺ > B₂⁻
ⓓ. B₂ > B₂⁻ > B₂⁺
Correct Answer: B₂⁻ > B₂ > B₂⁺
Explanation: Bond orders: B₂ (1), B₂⁺ (0.5), B₂⁻ (1.5). The higher the bond order, the greater the stability. Thus, B₂⁻ is most stable, B₂ is moderately stable, and B₂⁺ is least stable.
140. Which trend correctly matches bond order, bond length, and bond enthalpy?
ⓐ. Higher bond order → longer bond length → weaker bond
ⓑ. Higher bond order → shorter bond length → stronger bond
ⓒ. Lower bond order → shorter bond length → stronger bond
ⓓ. Bond order has no effect on bond enthalpy
Correct Answer: Higher bond order → shorter bond length → stronger bond
Explanation: Multiple bonds increase electron density between nuclei, reducing bond length and increasing bond enthalpy. Thus, triple bonds are shortest and strongest, while single bonds are longest and weakest.
141. The central idea of VSEPR theory is that:
ⓐ. Bonds are formed by overlap of atomic orbitals
ⓑ. Electron pairs around a central atom repel each other and arrange themselves to minimize repulsion
ⓒ. Molecular shape depends only on bond length
ⓓ. Bond polarity determines geometry
Correct Answer: Electron pairs around a central atom repel each other and arrange themselves to minimize repulsion
Explanation: VSEPR (Valence Shell Electron Pair Repulsion) theory states that both bonding pairs and lone pairs of electrons arrange themselves to minimize repulsion, thereby determining the geometry of the molecule. Orbital overlap is part of Valence Bond Theory, not VSEPR.
142. According to VSEPR theory, which electron pair repulsions are the strongest?
ⓐ. Bond pair–bond pair
ⓑ. Lone pair–bond pair
ⓒ. Lone pair–lone pair
ⓓ. Nucleus–nucleus
Correct Answer: Lone pair–lone pair
Explanation: Repulsion order is: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. Lone pairs are localized closer to the nucleus and spread out more, causing stronger repulsion. This explains why molecules with lone pairs (like H₂O) have smaller bond angles than tetrahedral molecules without lone pairs (like CH₄).
143. Which of the following is a correct postulate of VSEPR theory?
ⓐ. Molecular geometry is determined only by the type of bonds (single, double, triple)
ⓑ. The shape of a molecule depends on the total number of valence electron pairs around the central atom
ⓒ. Bond enthalpy determines the shape of a molecule
ⓓ. Molecular mass is the deciding factor for geometry
Correct Answer: The shape of a molecule depends on the total number of valence electron pairs around the central atom
Explanation: According to VSEPR, both bonding and lone pairs contribute to molecular shape. The arrangement minimizes repulsion among these pairs. Bond enthalpy or mass are not part of this model.
144. In VSEPR theory, double and triple bonds are treated as:
ⓐ. Equivalent to one single electron pair region
ⓑ. Two or three separate electron pairs
ⓒ. Non-bonding regions
ⓓ. Not considered at all
Correct Answer: Equivalent to one single electron pair region
Explanation: A multiple bond is considered as one region of electron density because all shared pairs are localized between the same two atoms. However, multiple bonds exert slightly more repulsion than single bonds.
145. According to VSEPR theory, which arrangement of electron pairs gives rise to linear geometry?
ⓐ. 2 electron pairs around central atom
ⓑ. 3 electron pairs around central atom
ⓒ. 4 electron pairs around central atom
ⓓ. 5 electron pairs around central atom
Correct Answer: 2 electron pairs around central atom
Explanation: Two regions of electron density arrange themselves 180° apart to minimize repulsion, giving a linear geometry (e.g., BeCl₂, CO₂).
146. The order of repulsion strength according to VSEPR theory is:
ⓐ. Bond pair–bond pair > lone pair–bond pair > lone pair–lone pair
ⓑ. Lone pair–bond pair > lone pair–lone pair > bond pair–bond pair
ⓒ. Lone pair–lone pair > lone pair–bond pair > bond pair–bond pair
ⓓ. Bond pair–bond pair > lone pair–lone pair > lone pair–bond pair
Correct Answer: Lone pair–lone pair > lone pair–bond pair > bond pair–bond pair
Explanation: Lone pairs are localized closer to the nucleus and occupy more space, exerting the strongest repulsion. Bond pair–bond pair interactions are the weakest since the electron cloud is shared.
147. Which of the following molecules has a bent geometry according to VSEPR theory?
ⓐ. CO₂
ⓑ. CH₄
ⓒ. H₂O
ⓓ. BF₃
Correct Answer: H₂O
Explanation: Oxygen in H₂O has 2 bonding pairs and 2 lone pairs. According to VSEPR, the electron pairs arrange tetrahedrally, but lone pair repulsion reduces the H–O–H bond angle to 104.5°, giving a bent geometry.
148. According to VSEPR theory, a molecule with 4 bond pairs and no lone pairs has which geometry?
ⓐ. Trigonal planar
ⓑ. Tetrahedral
ⓒ. Square planar
ⓓ. Trigonal bipyramidal
Correct Answer: Tetrahedral
Explanation: Four electron pairs arrange symmetrically around the central atom at 109.5° bond angles, forming a tetrahedral structure (e.g., CH₄).
149. Why does NH₃ have a bond angle of about 107° instead of 109.5°?
ⓐ. Because nitrogen is less electronegative than carbon
ⓑ. Because NH₃ has one lone pair that repels bonding pairs more strongly
ⓒ. Because N–H bonds are longer than C–H bonds
ⓓ. Because nitrogen has only two valence electrons
Correct Answer: Because NH₃ has one lone pair that repels bonding pairs more strongly
Explanation: In NH₃, nitrogen has 3 bonding pairs and 1 lone pair. The lone pair exerts stronger repulsion, compressing the bond angle from 109.5° (tetrahedral) to \~107° (trigonal pyramidal).
150. Which of the following molecules is predicted by VSEPR theory to have a trigonal planar geometry?
ⓐ. BF₃
ⓑ. CH₄
ⓒ. NH₃
ⓓ. H₂O
Correct Answer: BF₃
Explanation: Boron in BF₃ has 3 valence electrons, forming 3 bonds with fluorine. With no lone pairs, 3 regions of electron density spread out at 120°, giving a trigonal planar structure.
151. According to VSEPR theory, the shape of CO₂ is:
ⓐ. Bent
ⓑ. Trigonal planar
ⓒ. Linear
ⓓ. Tetrahedral
Correct Answer: Linear
Explanation: The central atom (C) in CO₂ is bonded to two oxygen atoms via double bonds. These are counted as two regions of electron density. To minimize repulsion, they arrange themselves 180° apart, giving CO₂ a linear shape with bond angle 180°.
152. What is the shape of BF₃ molecule as predicted by VSEPR theory?
ⓐ. Tetrahedral
ⓑ. Trigonal planar
ⓒ. Trigonal pyramidal
ⓓ. Bent
Correct Answer: Trigonal planar
Explanation: In BF₃, boron has 3 bonding pairs and no lone pairs. The three regions of electron density spread out at 120° angles in a plane, giving a trigonal planar geometry. This is consistent with sp² hybridization.
153. The shape of NH₃ molecule according to VSEPR theory is:
ⓐ. Trigonal planar
ⓑ. Trigonal pyramidal
ⓒ. Bent
ⓓ. Linear
Correct Answer: Trigonal pyramidal
Explanation: Nitrogen in NH₃ has 3 bonding pairs and 1 lone pair. Lone pairs exert stronger repulsion than bond pairs, pushing the N–H bonds downward to give a trigonal pyramidal shape with bond angle \~107°.
154. The shape of H₂O molecule is best described as:
ⓐ. Linear
ⓑ. Tetrahedral
ⓒ. Bent (V-shaped)
ⓓ. Trigonal planar
Correct Answer: Bent (V-shaped)
Explanation: Oxygen has 2 bonding pairs and 2 lone pairs. According to VSEPR, this arrangement is tetrahedral in terms of electron pairs, but due to the lone pairs, the molecular shape becomes bent with \~104.5° bond angle.
155. The molecular shape of PCl₅ is:
ⓐ. Tetrahedral
ⓑ. Square pyramidal
ⓒ. Trigonal bipyramidal
ⓓ. Octahedral
Correct Answer: Trigonal bipyramidal
Explanation: Phosphorus in PCl₅ has 5 bonding pairs and no lone pairs. According to VSEPR, 5 electron pairs arrange in a trigonal bipyramidal geometry with bond angles of 120° (equatorial) and 90° (axial).
156. What is the shape of SF₆ molecule?
ⓐ. Tetrahedral
ⓑ. Octahedral
ⓒ. Trigonal bipyramidal
ⓓ. Pentagonal bipyramidal
Correct Answer: Octahedral
Explanation: Sulfur in SF₆ has 6 bonding pairs and no lone pairs. According to VSEPR, these pairs arrange themselves symmetrically at 90° to each other, giving an octahedral shape.
157. According to VSEPR theory, the shape of XeF₂ is:
ⓐ. Trigonal planar
ⓑ. Tetrahedral
ⓒ. Linear
ⓓ. Trigonal bipyramidal
Correct Answer: Linear
Explanation: Xenon in XeF₂ has 5 electron pairs (2 bonding + 3 lone). To minimize repulsion, the 3 lone pairs occupy equatorial positions, and the 2 bond pairs remain at axial positions, giving a linear shape with 180° bond angle.
158. Which molecule has a square planar shape according to VSEPR theory?
ⓐ. PCl₅
ⓑ. SF₆
ⓒ. XeF₄
ⓓ. CH₄
Correct Answer: XeF₄
Explanation: Xenon in XeF₄ has 6 electron pairs (4 bonding + 2 lone). The lone pairs occupy opposite positions in an octahedral arrangement, resulting in a square planar molecular geometry.
159. The shape of BeCl₂ molecule is:
ⓐ. Linear
ⓑ. Bent
ⓒ. Tetrahedral
ⓓ. Trigonal planar
Correct Answer: Linear
Explanation: In BeCl₂, beryllium has 2 bonding pairs and no lone pairs. According to VSEPR, the two bonds arrange 180° apart, giving the molecule a linear geometry.
160. Why does CH₄ have a tetrahedral shape with 109.5° bond angles?
ⓐ. Carbon has one lone pair and three bonding pairs
ⓑ. Carbon is sp² hybridized
ⓒ. Carbon forms 4 equivalent sp³ hybrid orbitals, minimizing electron repulsion
ⓓ. CH₄ is planar due to delocalized bonding
Correct Answer: Carbon forms 4 equivalent sp³ hybrid orbitals, minimizing electron repulsion
Explanation: In CH₄, carbon undergoes sp³ hybridization. The four bonding electron pairs arrange themselves at equal distances (109.5°) in 3D space to minimize repulsion, producing a tetrahedral shape.
161. According to VSEPR theory, the molecular shape of BeCl₂ is:
ⓐ. Trigonal planar
ⓑ. Linear
ⓒ. Tetrahedral
ⓓ. Bent
Correct Answer: Linear
Explanation: Beryllium in BeCl₂ has 2 bonding pairs and no lone pairs. The two electron domains arrange themselves at 180° to minimize repulsion, giving a linear geometry with bond angle 180°.
162. What is the molecular shape of BF₃ according to VSEPR theory?
ⓐ. Tetrahedral
ⓑ. Trigonal planar
ⓒ. Trigonal pyramidal
ⓓ. Bent
Correct Answer: Trigonal planar
Explanation: Boron in BF₃ has 3 bonding pairs and no lone pairs. The electron pairs arrange themselves in a plane at 120° angles, resulting in a trigonal planar geometry.
163. In methane (CH₄), the H–C–H bond angle is:
ⓐ. 180°
ⓑ. 120°
ⓒ. 107°
ⓓ. 109.5°
Correct Answer: 109.5°
Explanation: Carbon in CH₄ undergoes sp³ hybridization, producing four equivalent orbitals. To minimize electron pair repulsion, they orient themselves in a tetrahedral arrangement with 109.5° bond angles.
164. The molecular geometry of NH₃ is:
ⓐ. Trigonal planar
ⓑ. Trigonal pyramidal
ⓒ. Tetrahedral
ⓓ. Bent
Correct Answer: Trigonal pyramidal
Explanation: Nitrogen in NH₃ has 3 bonding pairs and 1 lone pair. The lone pair repels the bond pairs more strongly, slightly reducing the bond angle to \~107°, giving a trigonal pyramidal shape.
165. Which of the following best describes the shape of H₂O according to VSEPR theory?
ⓐ. Linear
ⓑ. Trigonal planar
ⓒ. Bent (V-shaped)
ⓓ. Trigonal bipyramidal
Correct Answer: Bent (V-shaped)
Explanation: Oxygen in H₂O has 2 bonding pairs and 2 lone pairs. The tetrahedral arrangement of electron pairs distorts into a bent molecular shape, with bond angle reduced to \~104.5° due to strong lone pair–bond pair repulsion.
166. The geometry of PCl₅ as predicted by VSEPR theory is:
ⓐ. Trigonal planar
ⓑ. Tetrahedral
ⓒ. Trigonal bipyramidal
ⓓ. Octahedral
Correct Answer: Trigonal bipyramidal
Explanation: Phosphorus in PCl₅ has 5 bonding pairs and no lone pairs. The electron pairs arrange themselves as a trigonal bipyramid with 120° equatorial and 90° axial bond angles.
167. Which molecule has an octahedral shape according to VSEPR theory?
ⓐ. SF₆
ⓑ. CH₄
ⓒ. NH₃
ⓓ. XeF₂
Correct Answer: SF₆
Explanation: Sulfur in SF₆ has 6 bonding pairs and no lone pairs. The six electron domains are oriented at 90° to each other, resulting in an octahedral geometry.
168. According to VSEPR theory, the molecular shape of XeF₂ is:
ⓐ. Bent
ⓑ. Linear
ⓒ. Trigonal planar
ⓓ. Tetrahedral
Correct Answer: Linear
Explanation: Xenon in XeF₂ has 5 electron pairs (2 bonding + 3 lone). The three lone pairs occupy equatorial positions in a trigonal bipyramidal arrangement, while the two F atoms occupy axial positions, giving a linear geometry (180°).
169. Which of the following correctly matches the molecule with its shape?
ⓐ. BeCl₂ – Trigonal planar
ⓑ. BF₃ – Linear
ⓒ. CH₄ – Tetrahedral
ⓓ. NH₃ – Linear
Correct Answer: CH₄ – Tetrahedral
Explanation: BeCl₂ is linear, BF₃ is trigonal planar, CH₄ is tetrahedral, and NH₃ is trigonal pyramidal. Hence only CH₄ is matched correctly in this list.
170. In PCl₅, the bond pairs occupy positions such that:
ⓐ. All five are equivalent and at 90° to each other
ⓑ. Three occupy equatorial positions at 120° and two occupy axial positions at 90°
ⓒ. Four are planar and one above the plane
ⓓ. They alternate between lone and bond pairs
Correct Answer: Three occupy equatorial positions at 120° and two occupy axial positions at 90°
Explanation: In trigonal bipyramidal geometry, 3 P–Cl bonds are placed equatorially at 120° to each other, and 2 are placed axially at 90° to the equatorial plane. This arrangement minimizes repulsion and is characteristic of 5 bond pairs.
171. Which of the following molecules is linear according to VSEPR theory?
ⓐ. H₂O
ⓑ. XeF₂
ⓒ. BF₃
ⓓ. CH₄
Correct Answer: XeF₂
Explanation: XeF₂ has 5 electron pairs (2 bonding + 3 lone). To minimize repulsion, the three lone pairs occupy equatorial positions in a trigonal bipyramid, and the two fluorine atoms stay in axial positions. This results in a linear structure with 180° bond angle.
172. The bond angle in BF₃ is approximately:
ⓐ. 90°
ⓑ. 107°
ⓒ. 120°
ⓓ. 109.5°
Correct Answer: 120°
Explanation: BF₃ has 3 regions of electron density and no lone pairs on the central atom. According to VSEPR theory, this gives a trigonal planar shape with bond angles of 120°.
173. In NH₃, the bond angle is slightly less than tetrahedral because:
ⓐ. Nitrogen has a lower electronegativity
ⓑ. There is one lone pair on nitrogen that compresses bond angles
ⓒ. Hydrogen atoms repel each other strongly
ⓓ. N–H bonds are purely ionic
Correct Answer: There is one lone pair on nitrogen that compresses bond angles
Explanation: Lone pairs repel more strongly than bond pairs. In NH₃, this pushes the N–H bonds closer together, reducing the bond angle from 109.5° to \~107°.
174. What is the electron-pair geometry of H₂O according to VSEPR theory?
ⓐ. Trigonal planar
ⓑ. Tetrahedral
ⓒ. Octahedral
ⓓ. Linear
Correct Answer: Tetrahedral
Explanation: H₂O has 2 bonding pairs and 2 lone pairs around oxygen, for a total of 4 regions of electron density. These adopt a tetrahedral electron-pair geometry, but the molecular shape is bent due to the lone pairs.
175. Which molecule has a bond angle of exactly 180°?
ⓐ. BeCl₂
ⓑ. NH₃
ⓒ. H₂O
ⓓ. BF₃
Correct Answer: BeCl₂
Explanation: Beryllium in BeCl₂ has two bonding pairs and no lone pairs. The two regions of electron density arrange linearly at 180°, making BeCl₂ linear.
176. The geometry of SF₆ is described as:
ⓐ. Tetrahedral
ⓑ. Trigonal bipyramidal
ⓒ. Octahedral
ⓓ. Square planar
Correct Answer: Octahedral
Explanation: Sulfur in SF₆ has 6 bonding pairs and no lone pairs. The electron pairs adopt 90° orientations in a symmetrical arrangement, producing an octahedral geometry.
177. Which molecule is trigonal bipyramidal according to VSEPR theory?
ⓐ. PCl₅
ⓑ. SF₆
ⓒ. CH₄
ⓓ. XeF₂
Correct Answer: PCl₅
Explanation: PCl₅ has 5 bonding pairs and no lone pairs. According to VSEPR, these arrange in a trigonal bipyramidal geometry: three equatorial bonds at 120° and two axial bonds at 90°.
178. Why is CH₄ perfectly tetrahedral while NH₃ is trigonal pyramidal?
ⓐ. Because CH₄ has only single bonds and NH₃ has double bonds
ⓑ. Because carbon has no lone pairs while nitrogen has one lone pair
ⓒ. Because hydrogen is larger than nitrogen
ⓓ. Because CH₄ is ionic and NH₃ is covalent
Correct Answer: Because carbon has no lone pairs while nitrogen has one lone pair
Explanation: In CH₄, four bonding pairs arrange symmetrically in a tetrahedral geometry. In NH₃, one lone pair distorts the tetrahedron, giving a trigonal pyramidal shape with bond angle \~107°.
179. Which of the following correctly matches the molecule with its predicted geometry?
ⓐ. H₂O – Tetrahedral
ⓑ. XeF₂ – Linear
ⓒ. SF₆ – Trigonal bipyramidal
ⓓ. BF₃ – Tetrahedral
Correct Answer: XeF₂ – Linear
Explanation: H₂O has bent shape, XeF₂ is linear, SF₆ is octahedral, and BF₃ is trigonal planar. Only XeF₂ matches correctly with its shape.
180. The reduction in bond angle from 109.5° in CH₄ to 104.5° in H₂O is due to:
ⓐ. Presence of lone pairs on oxygen
ⓑ. Weaker O–H bonds compared to C–H bonds
ⓒ. Oxygen being heavier than carbon
ⓓ. Covalent character of O–H bonds
Correct Answer: Presence of lone pairs on oxygen
Explanation: Lone pairs occupy more space than bonding pairs. Oxygen in H₂O has 2 lone pairs, which repel the bonding pairs strongly, compressing the H–O–H angle to 104.5°. In CH₄, there are no lone pairs, so the bond angle remains 109.5°.
181. According to Valence Bond Theory, a covalent bond is formed when:
ⓐ. Two nuclei fuse together completely
ⓑ. Orbitals of two atoms overlap and electrons are shared between them
ⓒ. Electrons are transferred completely from one atom to another
ⓓ. Atomic masses of two atoms become equal
Correct Answer: Orbitals of two atoms overlap and electrons are shared between them
Explanation: VBT explains covalent bond formation by orbital overlap. When two half-filled atomic orbitals overlap, they share a pair of electrons of opposite spins. This results in bond formation with maximum stability when the overlap is maximum.
182. Which condition is essential for overlap of orbitals according to VBT?
ⓐ. Orbitals must be completely filled
ⓑ. Orbitals must have the same energy level and opposite spin electrons
ⓒ. Orbitals must be empty
ⓓ. Orbitals must belong to the same atom
Correct Answer: Orbitals must have the same energy level and opposite spin electrons
Explanation: For effective overlap, the orbitals must be half-filled with electrons of opposite spins. The energy levels of the overlapping orbitals should be comparable to allow efficient sharing. Completely filled or empty orbitals cannot form covalent bonds.
183. Which type of overlap forms the strongest bond according to VBT?
ⓐ. s–p overlap
ⓑ. p–p sideways overlap
ⓒ. s–s overlap
ⓓ. Head-on overlap (end-to-end)
Correct Answer: Head-on overlap (end-to-end)
Explanation: Head-on (axial) overlap of orbitals forms sigma (σ) bonds, which are stronger than pi (π) bonds formed by sideways overlap. This is because sigma bonds have greater orbital overlap along the internuclear axis.
184. In the formation of an H₂ molecule, the type of overlap is:
ⓐ. s–s overlap
ⓑ. s–p overlap
ⓒ. p–p overlap
ⓓ. d–p overlap
Correct Answer: s–s overlap
Explanation: Each hydrogen atom has a half-filled 1s orbital. When they approach each other, these 1s orbitals overlap head-on, forming a sigma (σ) bond. This is the simplest covalent bond described by VBT.
185. Which of the following best explains why covalent bond formation is exothermic?
ⓐ. Because nuclei repel each other strongly
ⓑ. Because overlap of orbitals lowers potential energy of the system
ⓒ. Because atomic radii decrease drastically
ⓓ. Because electron transfer occurs
Correct Answer: Because overlap of orbitals lowers potential energy of the system
Explanation: Orbital overlap increases electron density between nuclei, stabilizing the system. As potential energy decreases, energy is released, making bond formation exothermic. Repulsion occurs at very close distances, but stable overlap balances attraction and repulsion.
186. Which orbitals overlap in the bond formation of Cl₂ molecule?
ⓐ. s–s overlap
ⓑ. p–p sideways overlap
ⓒ. p–p head-on overlap
ⓓ. s–p overlap
Correct Answer: p–p head-on overlap
Explanation: Each chlorine atom has 7 valence electrons (3p⁵). The unpaired electron in the 3p orbital overlaps head-on with that of another chlorine atom, forming a sigma (σ) bond. Sideways overlap would result in a pi bond, not observed in Cl₂.
187. Why is the extent of overlap important in VBT?
ⓐ. Larger overlap decreases bond strength
ⓑ. Smaller overlap increases bond strength
ⓒ. Greater overlap results in stronger and more stable bonds
ⓓ. Overlap has no effect on bond strength
Correct Answer: Greater overlap results in stronger and more stable bonds
Explanation: The stability of a covalent bond depends on the extent of orbital overlap. The larger the region of electron density shared between atoms, the stronger the attraction between nuclei, leading to higher bond enthalpy and shorter bond length.
188. Which of the following bonds is formed due to sideways overlap of orbitals?
ⓐ. σ (sigma) bond
ⓑ. π (pi) bond
ⓒ. Ionic bond
ⓓ. Metallic bond
Correct Answer: π (pi) bond
Explanation: Pi bonds are formed by sideways overlap of p-orbitals above and below the internuclear axis. They are weaker than sigma bonds since overlap is less effective. Sigma bonds are formed by head-on overlap. Ionic and metallic bonds are not explained by orbital overlap.
189. In a C=C double bond, how many sigma and pi bonds are present?
ⓐ. 2 sigma, 0 pi
ⓑ. 1 sigma, 1 pi
ⓒ. 0 sigma, 2 pi
ⓓ. 2 sigma, 2 pi
Correct Answer: 1 sigma, 1 pi
Explanation: A carbon–carbon double bond consists of one sigma bond formed by head-on overlap of sp² orbitals and one pi bond formed by sideways overlap of unhybridized p orbitals. Thus, C=C contains 1 σ and 1 π bond.
190. Which of the following is a limitation of VBT regarding orbital overlap?
ⓐ. It explains covalent bond formation by orbital overlap
ⓑ. It explains shapes of molecules correctly in all cases
ⓒ. It explains bond strength qualitatively
ⓓ. It cannot explain paramagnetic nature of O₂
Correct Answer: It cannot explain paramagnetic nature of O₂
Explanation: While VBT successfully explains bond formation through orbital overlap, it fails to account for magnetic properties and delocalization. For example, it predicts O₂ as diamagnetic, but experimentally O₂ is paramagnetic. This limitation was addressed by Molecular Orbital Theory.
191. In the formation of H₂ molecule, the type of orbital overlap is:
ⓐ. s–s overlap
ⓑ. s–p overlap
ⓒ. p–p overlap
ⓓ. d–p overlap
Correct Answer: s–s overlap
Explanation: Each hydrogen atom has a half-filled 1s orbital. When these 1s orbitals approach each other and overlap head-on, a sigma bond is formed. This is the simplest type of overlap explained by VBT.
192. Which molecule involves s–p overlap in bond formation?
ⓐ. H₂
ⓑ. HCl
ⓒ. Cl₂
ⓓ. O₂
Correct Answer: HCl
Explanation: Hydrogen has a half-filled 1s orbital, while chlorine has a half-filled 3p orbital. The overlap between these two orbitals forms a sigma bond, which is classified as an s–p overlap.
193. In F₂ molecule, the type of overlap responsible for bond formation is:
ⓐ. s–s overlap
ⓑ. p–p head-on overlap
ⓒ. p–p sideways overlap
ⓓ. s–p overlap
Correct Answer: p–p head-on overlap
Explanation: Each fluorine atom has one unpaired electron in a 2p orbital. The head-on overlap of these 2p orbitals results in a sigma bond. Sideways overlap of p-orbitals forms pi bonds, which are not present in F₂.
194. Which overlap is present in the N–H bond of NH₃?
ⓐ. s–s overlap
ⓑ. s–p overlap
ⓒ. p–p overlap
ⓓ. d–p overlap
Correct Answer: s–p overlap
Explanation: In NH₃, hydrogen contributes its 1s orbital, while nitrogen contributes its sp³ hybrid orbital. The overlap between the 1s orbital of hydrogen and sp³ orbital of nitrogen is classified as an s–p overlap (σ bond).
195. The bond between two carbon atoms in C₂H₄ (ethene) contains:
ⓐ. One sigma (σ) bond from s–s overlap and one pi (π) bond from s–p overlap
ⓑ. One sigma (σ) bond from sp²–sp² overlap and one pi (π) bond from p–p sideways overlap
ⓒ. Two sigma bonds from p–p overlap
ⓓ. One sigma bond from s–s overlap and one pi bond from d–p overlap
Correct Answer: One sigma (σ) bond from sp²–sp² overlap and one pi (π) bond from p–p sideways overlap
Explanation: In ethene, each carbon atom undergoes sp² hybridization. The sp² orbitals overlap head-on to form a sigma bond, and the unhybridized p orbitals overlap sideways to form a pi bond, making a C=C double bond.
196. The O–H bond in water (H₂O) results from:
ⓐ. s–s overlap
ⓑ. s–p overlap
ⓒ. p–p overlap
ⓓ. d–p overlap
Correct Answer: s–p overlap
Explanation: Hydrogen contributes its 1s orbital, while oxygen contributes its sp³ hybrid orbitals. The head-on overlap forms sigma bonds, categorized as s–p overlaps.
197. In a C–H bond of CH₄, the overlap is:
ⓐ. s–s overlap
ⓑ. s–p overlap (σ bond)
ⓒ. p–p overlap
ⓓ. d–p overlap
Correct Answer: s–p overlap (σ bond)
Explanation: In CH₄, carbon undergoes sp³ hybridization. Each sp³ orbital overlaps with the 1s orbital of hydrogen to form σ bonds, which are examples of s–p overlaps.
198. Which type of overlap occurs in the σ bond of Cl₂ molecule?
ⓐ. s–s overlap
ⓑ. p–p head-on overlap
ⓒ. p–p sideways overlap
ⓓ. s–p overlap
Correct Answer: p–p head-on overlap
Explanation: Each chlorine atom has a half-filled 3p orbital. These orbitals overlap head-on to form a σ bond. Sideways p–p overlap would instead form a π bond, which Cl₂ does not have.
199. In acetylene (C₂H₂), the carbon–carbon triple bond consists of:
ⓐ. One σ bond from sp–sp overlap and two π bonds from p–p sideways overlap
ⓑ. Two σ bonds from s–s and one π bond from s–p overlap
ⓒ. One σ bond from p–p overlap and two π bonds from sp–sp overlap
ⓓ. Three σ bonds only
Correct Answer: One σ bond from sp–sp overlap and two π bonds from p–p sideways overlap
Explanation: In C₂H₂, each carbon is sp hybridized. The sp orbitals overlap head-on to form a σ bond, while the two unhybridized p orbitals overlap sideways to form two π bonds.
200. Which type of overlap is stronger?
ⓐ. s–s
ⓑ. s–p
ⓒ. p–p sideways
ⓓ. p–p head-on
Correct Answer: p–p head-on
Explanation: Head-on overlap (σ bond) is stronger than sideways overlap (π bond) because electron density is concentrated along the internuclear axis, maximizing attraction. Among the overlaps listed, p–p head-on overlap is the strongest.
You are now on Class 11 Chemistry MCQs – Chapter 4: Chemical Bonding and Molecular Structure (Part 2). This chapter is fundamental in Chemistry, as it explains how atoms combine and molecules are structured. It covers Ionic bond formation, Lattice enthalpy, Covalent bond properties, Polarity of bonds, and dipole moment. Students also study the Valence Bond Theory with concepts of Hybridization (sp, sp2, sp3, dsp2, sp3d, etc.) and how orbital overlaps explain bond strength and geometry. The VSEPR theory is applied to predict molecular geometries like linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral. Out of the full 395 MCQs available in this chapter, this part gives you the next 100 questions with solutions to enhance accuracy and speed. These are excellent for Class 11 board exam preparation as well as competitive exams like JEE, NEET, and other tests.
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