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201. Which of the following statements about sigma (σ) bonds is correct?
ⓐ. Sigma bonds are formed by sideways overlap of orbitals
ⓑ. Sigma bonds are stronger than pi bonds
ⓒ. Sigma bonds restrict rotation around the bond axis
ⓓ. Sigma bonds exist only in multiple bonds
Correct Answer: Sigma bonds are stronger than pi bonds
Explanation: Sigma bonds are formed by head-on (axial) overlap of orbitals, which allows greater overlap and higher bond strength. Pi bonds, formed by sideways overlap, are weaker. Sigma bonds do not generally restrict rotation (pi bonds do), and they are present in both single and multiple bonds.
202. In a double bond (C=C), the number of sigma and pi bonds are:
ⓐ. 2 sigma, 0 pi
ⓑ. 1 sigma, 1 pi
ⓒ. 0 sigma, 2 pi
ⓓ. 2 sigma, 2 pi
Correct Answer: 1 sigma, 1 pi
Explanation: Every covalent bond has one sigma bond as the first bond. In a double bond, one sigma bond is formed by head-on overlap, and one pi bond is formed by sideways overlap of unhybridized p orbitals.
203. In a triple bond (C≡C), the bonding consists of:
ⓐ. 1 sigma and 2 pi bonds
ⓑ. 2 sigma and 1 pi bonds
ⓒ. 3 sigma bonds only
ⓓ. 3 pi bonds only
Correct Answer: 1 sigma and 2 pi bonds
Explanation: In acetylene, carbon atoms undergo sp hybridization. One sigma bond is formed from sp–sp overlap, and two pi bonds are formed from sideways overlap of unhybridized p orbitals.
204. Which bond is always present in any covalent bond, whether single, double, or triple?
ⓐ. Pi bond
ⓑ. Sigma bond
ⓒ. Ionic bond
ⓓ. Metallic bond
Correct Answer: Sigma bond
Explanation: Every covalent bond starts with the formation of one sigma bond. Additional bonds (double, triple) involve pi bonds. Hence, sigma bonds are universal in covalent bonding.
205. Which of the following molecules contains only sigma bonds?
ⓐ. C₂H₂
ⓑ. C₂H₄
ⓒ. CH₄
ⓓ. O₂
Correct Answer: CH₄
Explanation: In CH₄, carbon forms four single bonds with hydrogen through sp³–1s overlaps, all of which are sigma bonds. In C₂H₂ and C₂H₄, pi bonds are present in addition to sigma bonds, and O₂ contains a double bond with one pi bond.
206. Which property is correctly associated with pi bonds?
ⓐ. Formed by head-on overlap
ⓑ. Stronger than sigma bonds
ⓒ. Electron density lies above and below the internuclear axis
ⓓ. Present in all single bonds
Correct Answer: Electron density lies above and below the internuclear axis
Explanation: Pi bonds arise from sideways overlap of unhybridized p orbitals. This produces regions of electron density above and below the bond axis. They are weaker than sigma bonds and only present in multiple bonds.
207. Why are sigma bonds stronger than pi bonds?
ⓐ. They are formed by overlap of filled d-orbitals
ⓑ. They are formed by greater extent of orbital overlap along the internuclear axis
ⓒ. They involve no electron sharing
ⓓ. They exist only in ionic compounds
Correct Answer: They are formed by greater extent of orbital overlap along the internuclear axis
Explanation: In sigma bonds, orbital overlap is maximum because it occurs directly between nuclei. Pi bonds involve lateral overlap, which is less effective, making them weaker.
208. In ethene (C₂H₄), the C=C bond consists of:
ⓐ. 1 sigma and 2 pi bonds
ⓑ. 1 sigma and 1 pi bond
ⓒ. 2 sigma bonds only
ⓓ. 2 pi bonds only
Correct Answer: 1 sigma and 1 pi bond
Explanation: Each carbon is sp² hybridized. The sp² orbitals form a sigma bond between carbons, and the unhybridized p orbitals overlap sideways to form one pi bond. Thus, the C=C bond contains one σ and one π bond.
209. Rotation around a C=C double bond is restricted because:
ⓐ. Sigma bond prevents free rotation
ⓑ. Pi bond electron cloud resists rotation
ⓒ. Lone pairs on carbon repel each other
ⓓ. Carbon atoms are too small to rotate
Correct Answer: Pi bond electron cloud resists rotation
Explanation: Pi bonds are formed by sideways overlap of parallel p orbitals. Rotation around the bond axis would break the parallel orientation, destroying the pi bond. Thus, double bonds restrict free rotation, unlike single bonds which allow it.
210. Which of the following molecules has both sigma and pi bonds?
ⓐ. HCl
ⓑ. CH₄
ⓒ. O₂
ⓓ. NaCl
Correct Answer: O₂
Explanation: Oxygen molecule (O₂) contains a double bond. One sigma bond is formed by head-on overlap, and one pi bond by sideways overlap of p orbitals. HCl and CH₄ have only sigma bonds, and NaCl is ionic.
211. In BeCl₂, the central atom beryllium is:
ⓐ. sp hybridized and linear
ⓑ. sp² hybridized and trigonal planar
ⓒ. sp³ hybridized and tetrahedral
ⓓ. sp³d hybridized and trigonal bipyramidal
Correct Answer: sp hybridized and linear
Explanation: Beryllium in BeCl₂ forms two sigma bonds with chlorine. To achieve this, its 2s and one 2p orbital mix to form two sp hybrid orbitals. These arrange 180° apart, giving BeCl₂ a linear geometry.
212. The bond angle in BeCl₂ according to hybridization is approximately:
ⓐ. 90°
ⓑ. 109.5°
ⓒ. 120°
ⓓ. 180°
Correct Answer: 180°
Explanation: sp hybridization leads to a linear arrangement of orbitals. Therefore, in BeCl₂, the Cl–Be–Cl bond angle is 180°.
213. Which of the following statements about sp hybridization is true?
ⓐ. It involves mixing of one s orbital and three p orbitals
ⓑ. It produces two equivalent hybrid orbitals oriented 180° apart
ⓒ. It always leads to tetrahedral geometry
ⓓ. It occurs only in transition metals
Correct Answer: It produces two equivalent hybrid orbitals oriented 180° apart
Explanation: sp hybridization arises from mixing one s and one p orbital, producing two hybrid orbitals. These are oriented linearly at 180° to minimize repulsion, as seen in BeCl₂ and CO₂.
214. In BF₃, the central atom boron is:
ⓐ. sp hybridized and linear
ⓑ. sp² hybridized and trigonal planar
ⓒ. sp³ hybridized and tetrahedral
ⓓ. sp³d hybridized and trigonal bipyramidal
Correct Answer: sp² hybridized and trigonal planar
Explanation: Boron in BF₃ has three bonding domains and no lone pairs. Its 2s and two 2p orbitals mix to form three sp² orbitals, arranged in a trigonal planar geometry with bond angles of 120°.
215. Which is the bond angle in BF₃?
ⓐ. 90°
ⓑ. 107°
ⓒ. 109.5°
ⓓ. 120°
Correct Answer: 120°
Explanation: In sp² hybridization, three orbitals arrange themselves at equal distances in a plane, resulting in bond angles of 120°. BF₃ is the classic example of a trigonal planar molecule.
216. Why does BF₃ violate the octet rule?
ⓐ. Boron has 3 valence electrons and can form only 3 bonds, resulting in 6 electrons around it
ⓑ. Boron uses d orbitals to expand its octet
ⓒ. Fluorine atoms donate electrons through coordinate bonds
ⓓ. Boron forms double bonds with fluorine
Correct Answer: Boron has 3 valence electrons and can form only 3 bonds, resulting in 6 electrons around it
Explanation: In BF₃, boron forms three covalent bonds with fluorine, leaving it electron-deficient (only 6 electrons in its valence shell). This makes BF₃ an incomplete-octet molecule and an exception to the octet rule.
217. Which of the following molecules shows sp hybridization?
ⓐ. BF₃
ⓑ. BeCl₂
ⓒ. CH₄
ⓓ. NH₃
Correct Answer: BeCl₂
Explanation: BeCl₂ is linear with 2 regions of electron density around beryllium, leading to sp hybridization. BF₃ is sp², CH₄ is sp³, and NH₃ is also sp³.
218. Which hybrid orbitals are formed when one s orbital mixes with two p orbitals?
ⓐ. sp
ⓑ. sp²
ⓒ. sp³
ⓓ. sp³d
Correct Answer: sp²
Explanation: Mixing one s and two p orbitals produces three sp² hybrid orbitals, oriented 120° apart in a trigonal planar arrangement. This is the case in BF₃ and C₂H₄.
219. The geometry of sp² hybridized orbitals is:
ⓐ. Linear
ⓑ. Tetrahedral
ⓒ. Trigonal planar
ⓓ. Trigonal pyramidal
Correct Answer: Trigonal planar
Explanation: sp² hybridization produces three equivalent orbitals lying in one plane at 120° to each other. Examples include BF₃, ethene (C₂H₄), and SO₃.
220. In BeCl₂, why does Be undergo sp hybridization?
ⓐ. To achieve an octet by forming coordinate bonds
ⓑ. To form two equivalent Be–Cl bonds with maximum overlap
ⓒ. Because chlorine forces hybridization
ⓓ. To minimize polarity in the bonds
Correct Answer: To form two equivalent Be–Cl bonds with maximum overlap
Explanation: In ground state, Be has 2s² configuration. One electron is promoted to 2p, and mixing of 2s and 2p orbitals gives sp hybridization. This ensures the two Be–Cl bonds are equivalent and arranged linearly to minimize electron pair repulsion.
221. In CH₄, the central carbon atom undergoes which type of hybridization?
ⓐ. sp
ⓑ. sp²
ⓒ. sp³
ⓓ. sp³d
Correct Answer: sp³
Explanation: Carbon in CH₄ has 4 valence electrons. It mixes its 2s and three 2p orbitals to form four equivalent sp³ hybrid orbitals. These orbitals arrange themselves tetrahedrally at 109.5° bond angles, giving methane its stable structure.
222. The bond angle in CH₄ according to sp³ hybridization is:
ⓐ. 90°
ⓑ. 107°
ⓒ. 109.5°
ⓓ. 120°
Correct Answer: 109.5°
Explanation: Four sp³ hybrid orbitals spread out in 3D space to minimize electron repulsion, resulting in a perfect tetrahedral structure with bond angles of 109.5°.
223. Which geometry is predicted by sp³ hybridization when there are no lone pairs?
ⓐ. Linear
ⓑ. Trigonal planar
ⓒ. Tetrahedral
ⓓ. Trigonal pyramidal
Correct Answer: Tetrahedral
Explanation: Four regions of electron density (bonding pairs) arrange symmetrically in 3D space. This minimizes repulsion, giving a tetrahedral shape, as in CH₄.
224. In NH₃, the shape is not perfectly tetrahedral but:
ⓐ. Linear
ⓑ. Trigonal pyramidal
ⓒ. Trigonal planar
ⓓ. Bent
Correct Answer: Trigonal pyramidal
Explanation: NH₃ has sp³ hybridization. Nitrogen uses three sp³ orbitals to form bonds with hydrogens, and the fourth orbital contains a lone pair. Lone pair–bond pair repulsion compresses the bond angle to \~107°, producing a trigonal pyramidal shape.
225. The bond angle in NH₃ is smaller than in CH₄ because:
ⓐ. Lone pair–bond pair repulsion reduces the bond angle
ⓑ. N–H bonds are weaker than C–H bonds
ⓒ. Nitrogen is more electronegative than carbon
ⓓ. NH₃ is polar while CH₄ is non-polar
Correct Answer: Lone pair–bond pair repulsion reduces the bond angle
Explanation: In NH₃, the lone pair occupies more space than bonding pairs, pushing the N–H bonds closer together. This decreases the bond angle to \~107°, compared to 109.5° in CH₄.
226. The molecular geometry of H₂O according to sp³ hybridization is:
ⓐ. Tetrahedral
ⓑ. Linear
ⓒ. Bent (V-shaped)
ⓓ. Trigonal pyramidal
Correct Answer: Bent (V-shaped)
Explanation: Oxygen in H₂O has sp³ hybridization with 2 bonding pairs and 2 lone pairs. Lone pair repulsions distort the tetrahedral arrangement, compressing the bond angle to \~104.5° and producing a bent shape.
227. Which of the following is the correct bond angle in H₂O?
ⓐ. 90°
ⓑ. 104.5°
ⓒ. 109.5°
ⓓ. 120°
Correct Answer: 104.5°
Explanation: H₂O has sp³ hybridization. Two lone pairs on oxygen exert stronger repulsion, reducing the bond angle from 109.5° (ideal tetrahedral) to \~104.5°.
228. Which best describes the difference between sp³ hybridization in CH₄, NH₃, and H₂O?
ⓐ. CH₄ has 4 bond pairs, NH₃ has 3 bond pairs + 1 lone pair, H₂O has 2 bond pairs + 2 lone pairs
ⓑ. All three have identical geometry
ⓒ. CH₄ and NH₃ are planar while H₂O is tetrahedral
ⓓ. CH₄ is sp² hybridized while NH₃ and H₂O are sp³
Correct Answer: CH₄ has 4 bond pairs, NH₃ has 3 bond pairs + 1 lone pair, H₂O has 2 bond pairs + 2 lone pairs
Explanation: All three molecules involve sp³ hybridization. However, the presence of lone pairs changes the molecular geometry: CH₄ → tetrahedral, NH₃ → trigonal pyramidal, H₂O → bent.
229. Which of the following correctly represents the decreasing order of bond angle?
ⓐ. H₂O > NH₃ > CH₄
ⓑ. NH₃ > H₂O > CH₄
ⓒ. CH₄ > NH₃ > H₂O
ⓓ. H₂O > CH₄ > NH₃
Correct Answer: CH₄ > NH₃ > H₂O
Explanation: CH₄ has 109.5° (no lone pairs), NH₃ has 107° (one lone pair), and H₂O has 104.5° (two lone pairs). More lone pairs → more repulsion → smaller bond angle.
230. In sp³ hybridization, the number of equivalent hybrid orbitals formed is:
ⓐ. 2
ⓑ. 3
ⓒ. 4
ⓓ. 6
Correct Answer: 4
Explanation: One s and three p orbitals combine to form four equivalent sp³ hybrid orbitals. These arrange tetrahedrally to minimize repulsion, forming the basis for structures like CH₄, NH₃, and H₂O.
231. In PCl₅, the central phosphorus atom undergoes which type of hybridization?
ⓐ. sp³
ⓑ. sp³d
ⓒ. sp³d²
ⓓ. sp³d³
Correct Answer: sp³d
Explanation: Phosphorus in PCl₅ has 5 valence electrons. It mixes one s, three p, and one d orbital to form 5 sp³d hybrid orbitals. These orbitals arrange in a trigonal bipyramidal geometry, with 3 bonds in the equatorial plane and 2 bonds in the axial positions.
232. The geometry of PCl₅ according to sp³d hybridization is:
ⓐ. Tetrahedral
ⓑ. Trigonal bipyramidal
ⓒ. Octahedral
ⓓ. Square pyramidal
Correct Answer: Trigonal bipyramidal
Explanation: The five sp³d orbitals arrange themselves at 120° (equatorial) and 90° (axial) to minimize repulsion, giving PCl₅ a trigonal bipyramidal shape.
233. The bond angles in PCl₅ are:
ⓐ. 90° and 120°
ⓑ. 109.5°
ⓒ. 104.5°
ⓓ. 180° only
Correct Answer: 90° and 120°
Explanation: In PCl₅, the three equatorial bonds are 120° apart, while the two axial bonds are 90° to the equatorial plane. This arrangement minimizes repulsion in a trigonal bipyramidal geometry.
234. In SF₆, the sulfur atom undergoes:
ⓐ. sp³ hybridization
ⓑ. sp³d
ⓒ. sp³d²
ⓓ. sp³d³
Correct Answer: sp³d²
Explanation: Sulfur in SF₆ mixes one s, three p, and two d orbitals to form six sp³d² hybrid orbitals. These arrange octahedrally, with six S–F bonds oriented at 90° to each other.
235. The molecular geometry of SF₆ according to VSEPR and hybridization is:
ⓐ. Trigonal bipyramidal
ⓑ. Octahedral
ⓒ. Square pyramidal
ⓓ. Pentagonal bipyramidal
Correct Answer: Octahedral
Explanation: Six bonding pairs of electrons around sulfur in SF₆ arrange symmetrically at 90° to each other. This gives the molecule an octahedral shape.
236. The bond angle in SF₆ is:
ⓐ. 90°
ⓑ. 109.5°
ⓒ. 120°
ⓓ. 180°
Correct Answer: 90°
Explanation: In an octahedral geometry, all bond angles between adjacent bonds are 90°. Opposite bonds are 180°. Thus SF₆ has a perfectly symmetrical octahedral geometry.
237. In IF₇, the iodine atom undergoes which type of hybridization?
ⓐ. sp³d
ⓑ. sp³d²
ⓒ. sp³d³
ⓓ. sp²
Correct Answer: sp³d³
Explanation: Iodine in IF₇ mixes one s, three p, and three d orbitals to form seven sp³d³ hybrid orbitals. These arrange in a pentagonal bipyramidal geometry with 5 bonds in a plane and 2 bonds axial.
238. The geometry of IF₇ is described as:
ⓐ. Octahedral
ⓑ. Pentagonal bipyramidal
ⓒ. Trigonal bipyramidal
ⓓ. Square pyramidal
Correct Answer: Pentagonal bipyramidal
Explanation: Seven sp³d³ orbitals arrange with five F atoms in a pentagonal plane (72° apart) and two F atoms above and below the plane (axial), giving a pentagonal bipyramidal shape.
239. Which of the following correctly matches the molecule with its geometry?
ⓐ. PCl₅ – Octahedral
ⓑ. SF₆ – Trigonal bipyramidal
ⓒ. IF₇ – Pentagonal bipyramidal
ⓓ. BeCl₂ – Trigonal planar
Correct Answer: IF₇ – Pentagonal bipyramidal
Explanation: PCl₅ is trigonal bipyramidal, SF₆ is octahedral, IF₇ is pentagonal bipyramidal, and BeCl₂ is linear. Thus only IF₇ is correctly matched here.
240. The difference between the geometry of SF₆ and IF₇ is:
ⓐ. SF₆ is trigonal planar, IF₇ is square planar
ⓑ. SF₆ is octahedral (sp³d²), IF₇ is pentagonal bipyramidal (sp³d³)
ⓒ. SF₆ is tetrahedral, IF₇ is octahedral
ⓓ. SF₆ has 120° bond angles, IF₇ has 109.5°
Correct Answer: SF₆ is octahedral (sp³d²), IF₇ is pentagonal bipyramidal (sp³d³)
Explanation: Sulfur in SF₆ undergoes sp³d² hybridization, giving an octahedral geometry. Iodine in IF₇ undergoes sp³d³ hybridization, resulting in a pentagonal bipyramidal structure. The difference arises from the number of bonding domains around the central atom.
241. Which principle forms the basis of Molecular Orbital Theory (MOT)?
ⓐ. Orbitals of only the same atom can overlap
ⓑ. Atomic orbitals combine to form molecular orbitals that extend over the whole molecule
ⓒ. Bonding depends only on electronegativity difference
ⓓ. Electrons remain localized between two atoms
Correct Answer: Atomic orbitals combine to form molecular orbitals that extend over the whole molecule
Explanation: MOT states that atomic orbitals of comparable energy and symmetry combine to form molecular orbitals. These orbitals belong to the entire molecule rather than to individual atoms, delocalizing the electrons across the molecule.
242. In MOT, how many molecular orbitals are formed when two atomic orbitals combine?
ⓐ. 1
ⓑ. 2
ⓒ. 3
ⓓ. Infinite
Correct Answer: 2
Explanation: Combination of two atomic orbitals produces two molecular orbitals: one lower-energy bonding molecular orbital and one higher-energy antibonding molecular orbital.
243. Which molecular orbital is lower in energy when two atomic orbitals overlap constructively?
ⓐ. Bonding molecular orbital
ⓑ. Antibonding molecular orbital
ⓒ. Non-bonding orbital
ⓓ. Hybrid orbital
Correct Answer: Bonding molecular orbital
Explanation: Constructive overlap increases electron density between nuclei, leading to strong attraction and stabilization. This results in the formation of a lower-energy bonding molecular orbital.
244. What symbol is used to represent an antibonding molecular orbital?
ⓐ. σ
ⓑ. π
ⓒ. σ\* or π\*
ⓓ. δ
Correct Answer: σ\* or π\*
Explanation: Antibonding orbitals are denoted with an asterisk (*). For example, σ*1s is the antibonding counterpart of σ1s, and π\*2p is the antibonding counterpart of π2p.
245. According to MOT, bond order is given by the formula:
ⓐ. $\dfrac{\text{Number of bonding electrons}}{\text{Number of antibonding electrons}}$
ⓑ. $\dfrac{N_b + N_a}{2}$
ⓒ. $\dfrac{N_b – N_a}{2}$
ⓓ. $\dfrac{N_a – N_b}{2}$
Correct Answer: $\dfrac{N_b – N_a}{2}$
Explanation: Bond order = (Number of bonding electrons − Number of antibonding electrons)/2. It gives information about bond strength, bond length, and stability.
246. A molecule with bond order 0 according to MOT will be:
ⓐ. Very stable
ⓑ. Highly reactive but stable
ⓒ. Unstable and will not exist
ⓓ. Paramagnetic
Correct Answer: Unstable and will not exist
Explanation: If bond order is 0, the number of bonding and antibonding electrons are equal, leading to no net bond formation. Such species are unstable and do not exist under normal conditions.
247. Which of the following molecules is paramagnetic according to MOT?
ⓐ. N₂
ⓑ. O₂
ⓒ. F₂
ⓓ. Ne₂
Correct Answer: O₂
Explanation: O₂ has 16 electrons. MO configuration shows two unpaired electrons in π*2px and π*2py orbitals, making O₂ paramagnetic. This was a major success of MOT, as VBT incorrectly predicted O₂ to be diamagnetic.
248. What is the bond order of N₂ molecule using MOT?
ⓐ. 1
ⓑ. 2
ⓒ. 3
ⓓ. 4
Correct Answer: 3
Explanation: N₂ has 14 electrons. MO filling: 10 electrons in bonding orbitals, 4 in antibonding orbitals. Bond order = (10 − 4)/2 = 3, consistent with the triple bond in N₂.
249. According to MOT, which of the following species has the highest bond order?
ⓐ. O₂
ⓑ. O₂⁺
ⓒ. O₂⁻
ⓓ. O₂²⁻
Correct Answer: O₂⁺
Explanation: Bond orders: O₂⁺ = 2.5, O₂ = 2, O₂⁻ = 1.5, O₂²⁻ = 1. Thus O₂⁺ has the highest bond order, shortest bond length, and strongest bond among these.
250. Which molecular orbital has higher energy in diatomic molecules up to nitrogen (Z ≤ 7)?
ⓐ. σ2px
ⓑ. π2px and π2py
ⓒ. σ1s
ⓓ. σ2s
Correct Answer: π2px and π2py
Explanation: For lighter diatomic molecules (B₂, C₂, N₂), due to orbital mixing, π2px and π2py orbitals lie lower in energy than σ2px. For heavier molecules (O₂, F₂, Ne₂), this order reverses.
251. The Linear Combination of Atomic Orbitals (LCAO) approach states that:
ⓐ. Molecular orbitals are formed by simple addition of atomic numbers
ⓑ. Molecular orbitals are formed by combining atomic orbitals mathematically (addition or subtraction of wave functions)
ⓒ. Atomic orbitals remain unchanged in molecules
ⓓ. Orbitals combine only if they belong to different energy levels
Correct Answer: Molecular orbitals are formed by combining atomic orbitals mathematically (addition or subtraction of wave functions)
Explanation: In LCAO, the wave functions of atomic orbitals are combined linearly. Constructive interference (addition) forms bonding molecular orbitals, while destructive interference (subtraction) forms antibonding molecular orbitals.
252. Which condition must be satisfied for two atomic orbitals to combine effectively under LCAO?
ⓐ. They must have large energy difference
ⓑ. They must have similar energies and proper symmetry
ⓒ. They must belong to different principal shells
ⓓ. They must be completely filled orbitals
Correct Answer: They must have similar energies and proper symmetry
Explanation: For effective combination, atomic orbitals should be of comparable energy (e.g., 1s with 1s, 2p with 2p) and proper orientation symmetry. Large energy differences prevent effective overlap.
253. When two atomic orbitals combine constructively in LCAO, the result is:
ⓐ. A non-bonding orbital
ⓑ. An antibonding orbital with higher energy
ⓒ. A bonding orbital with lower energy
ⓓ. A d-orbital hybrid
Correct Answer: A bonding orbital with lower energy
Explanation: Constructive overlap increases electron density between nuclei, lowering potential energy. This stabilizes the molecule and forms a bonding molecular orbital.
254. When two atomic orbitals combine destructively in LCAO, the result is:
ⓐ. Bonding molecular orbital
ⓑ. Antibonding molecular orbital with higher energy
ⓒ. A hybrid orbital
ⓓ. A delocalized π orbital only
Correct Answer: Antibonding molecular orbital with higher energy
Explanation: Destructive overlap reduces electron density between nuclei and creates a node (region of zero electron density). This destabilizes the system, producing an antibonding orbital.
255. In LCAO, the number of molecular orbitals formed is always:
ⓐ. Half the number of atomic orbitals
ⓑ. Equal to the number of atomic orbitals combined
ⓒ. Twice the number of atomic orbitals
ⓓ. Independent of number of atomic orbitals
Correct Answer: Equal to the number of atomic orbitals combined
Explanation: If two atomic orbitals combine, they form two molecular orbitals: one bonding and one antibonding. Thus, the total number of orbitals remains conserved.
256. The energy difference between bonding and antibonding molecular orbitals depends mainly on:
ⓐ. The size of the nucleus
ⓑ. The extent of overlap between atomic orbitals
ⓒ. The mass of the atom
ⓓ. The number of lone pairs present
Correct Answer: The extent of overlap between atomic orbitals
Explanation: Greater orbital overlap increases stabilization of bonding orbitals and destabilization of antibonding orbitals, thus widening the energy gap. Poor overlap results in weaker bonding and smaller energy separation.
257. In H₂, the bonding molecular orbital is formed from:
ⓐ. Addition of two 1s orbitals (constructive overlap)
ⓑ. Subtraction of two 1s orbitals (destructive overlap)
ⓒ. Overlap of one 1s and one 2p orbital
ⓓ. Mixing of one filled and one empty orbital
Correct Answer: Addition of two 1s orbitals (constructive overlap)
Explanation: In H₂, two hydrogen atoms each contribute a 1s orbital. Constructive overlap of these orbitals forms the σ1s bonding orbital, while destructive overlap forms the σ\*1s antibonding orbital.
258. The presence of a node between two nuclei is characteristic of:
ⓐ. Bonding orbital
ⓑ. Antibonding orbital
ⓒ. Hybrid orbital
ⓓ. Lone pair orbital
Correct Answer: Antibonding orbital
Explanation: Antibonding orbitals (σ\* or π\*) are formed by destructive overlap. A node appears between the nuclei where electron density is zero, destabilizing the molecule.
259. In the case of Li₂ molecule (6 electrons), using LCAO, the bond order is:
ⓐ. 0
ⓑ. 0.5
ⓒ. 1
ⓓ. 2
Correct Answer: 1
Explanation: MO filling for Li₂: σ1s², σ\*1s², σ2s². Bond order = (Nb − Na)/2 = (4 − 2)/2 = 1. Hence, Li₂ has a single bond and exists as a stable molecule.
260. Which of the following correctly pairs LCAO results?
ⓐ. Constructive combination → antibonding MO
ⓑ. Destructive combination → bonding MO
ⓒ. Constructive combination → bonding MO, Destructive combination → antibonding MO
ⓓ. Both give bonding MO
Correct Answer: Constructive combination → bonding MO, Destructive combination → antibonding MO
Explanation: Constructive overlap enhances electron density between nuclei, producing bonding MOs. Destructive overlap reduces electron density between nuclei, forming antibonding MOs. This duality explains stability and instability trends in molecules.
261. The key difference between bonding and antibonding molecular orbitals is:
ⓐ. Bonding orbitals are localized, antibonding orbitals are delocalized
ⓑ. Bonding orbitals increase electron density between nuclei, antibonding orbitals decrease it
ⓒ. Bonding orbitals are always higher in energy, antibonding are lower
ⓓ. Bonding orbitals exist only in ionic bonds
Correct Answer: Bonding orbitals increase electron density between nuclei, antibonding orbitals decrease it
Explanation: In bonding MOs, constructive overlap raises electron density between nuclei, strengthening attraction. In antibonding MOs, destructive overlap creates a node between nuclei, weakening bonding and destabilizing the molecule.
262. Which of the following symbols represents an antibonding orbital?
ⓐ. σ
ⓑ. π
ⓒ. σ\*
ⓓ. δ
Correct Answer: σ\*
Explanation: Antibonding orbitals are denoted with an asterisk (*). For example, σ*1s and π\*2px represent antibonding orbitals. Bonding orbitals are written without the star.
263. Which statement about the energy levels of bonding vs antibonding orbitals is correct?
ⓐ. Bonding orbitals have higher energy than the atomic orbitals
ⓑ. Antibonding orbitals have lower energy than the atomic orbitals
ⓒ. Bonding orbitals have lower energy than atomic orbitals, while antibonding orbitals have higher energy
ⓓ. Both bonding and antibonding orbitals have equal energy
Correct Answer: Bonding orbitals have lower energy than atomic orbitals, while antibonding orbitals have higher energy
Explanation: Constructive overlap stabilizes bonding orbitals, making them lower in energy. Destructive overlap destabilizes antibonding orbitals, raising their energy above the parent atomic orbitals.
264. Which of the following molecules has electrons in antibonding orbitals?
ⓐ. H₂
ⓑ. He₂
ⓒ. Li₂
ⓓ. N₂
Correct Answer: He₂
Explanation: In He₂, MO filling is σ1s² σ*1s². The antibonding orbital σ*1s is occupied, canceling the bonding effect. This results in bond order = 0, making He₂ unstable.
265. The presence of electrons in antibonding orbitals:
ⓐ. Increases bond strength and stability
ⓑ. Decreases bond strength and stability
ⓒ. Has no effect on stability
ⓓ. Always leads to triple bonds
Correct Answer: Decreases bond strength and stability
Explanation: Antibonding orbitals reduce net bonding by increasing repulsion. Each pair of electrons in antibonding orbitals lowers bond order by 1, making the molecule weaker and less stable.
266. Which of the following species has a bond order of zero due to equal bonding and antibonding electrons?
ⓐ. H₂⁺
ⓑ. He₂
ⓒ. Li₂
ⓓ. B₂
Correct Answer: He₂
Explanation: In He₂, there are 2 bonding electrons and 2 antibonding electrons. Bond order = (2 − 2)/2 = 0. Hence, He₂ does not exist under normal conditions.
267. In bonding orbitals, the wave functions of atomic orbitals combine:
ⓐ. Destructively, leading to nodes between nuclei
ⓑ. Constructively, increasing electron density between nuclei
ⓒ. Randomly, without affecting energy
ⓓ. Only through hybridization
Correct Answer: Constructively, increasing electron density between nuclei
Explanation: Constructive overlap reinforces the wave functions, concentrating electron density between nuclei. This lowers potential energy and stabilizes the molecule.
268. Which of the following is true about the nodal plane in antibonding orbitals?
ⓐ. There is maximum electron density between nuclei
ⓑ. There is a region of zero probability (node) between nuclei
ⓒ. Electrons are localized on one nucleus only
ⓓ. It indicates the presence of resonance
Correct Answer: There is a region of zero probability (node) between nuclei
Explanation: Antibonding orbitals result from destructive overlap, producing a node (zero electron density) between nuclei. This weakens nuclear attraction and destabilizes bonding.
269. Which molecular orbital arrangement is correct for molecules like B₂, C₂, and N₂?
ⓐ. σ1s < σ*1s < σ2s < σ*2s < σ2px < π2px, π2py
ⓑ. σ1s < σ*1s < σ2s < σ*2s < π2px, π2py < σ2pz
ⓒ. σ1s < σ*1s < π2px, π2py < σ2s < σ*2s
ⓓ. σ1s < σ*1s < π2px < π*2py < σ2pz
Correct Answer: σ1s < σ*1s < σ2s < σ*2s < π2px, π2py < σ2pz
Explanation: Due to orbital mixing in lighter diatomics (Z ≤ 7), π2px and π2py orbitals lie lower in energy than σ2pz. For heavier diatomics (O₂ onwards), the order reverses.
270. Which best describes the occupancy of bonding and antibonding orbitals in O₂?
ⓐ. Bonding: σ1s, σ2s, π2px, π2py; Antibonding: σ*1s, σ*2s
ⓑ. Bonding: σ1s, σ2s, σ2pz, π2px, π2py; Antibonding: σ*1s, σ*2s, π*2px, π*2py
ⓒ. Bonding: σ1s, σ2s only; Antibonding: none
ⓓ. Bonding: σ2pz only; Antibonding: σ*2pz only
Correct Answer: Bonding: σ1s, σ2s, σ2pz, π2px, π2py; Antibonding: σ*1s, σ*2s, π*2px, π*2py
Explanation: In O₂ (16 electrons), bonding orbitals are filled up to π2px and π2py, and antibonding orbitals include π*2px and π\*2py. This leads to 2 unpaired electrons, explaining O₂’s paramagnetism.
271. According to MOT, the correct molecular orbital configuration for H₂ is:
ⓐ. σ1s²
ⓑ. σ1s² σ*1s²
ⓒ. σ1s¹ σ*1s¹
ⓓ. σ2s²
Correct Answer: σ1s²
Explanation: H₂ has 2 electrons. Both fill the bonding molecular orbital (σ1s). This gives bond order = (2 − 0)/2 = 1, explaining the stability of H₂.
272. What is the bond order of He₂ according to the molecular orbital energy-level diagram?
ⓐ. 1
ⓑ. 0
ⓒ. 2
ⓓ. 0.5
Correct Answer: 0
Explanation: He₂ has 4 electrons. MO filling: σ1s² σ\*1s². Bond order = (2 − 2)/2 = 0, meaning no net bond exists. This explains why He₂ does not exist under normal conditions.
273. In homonuclear diatomic molecules, the relative energy order of σ2pz and π2px/π2py differs for:
ⓐ. H₂ and He₂
ⓑ. B₂, C₂, N₂ versus O₂, F₂, Ne₂
ⓒ. O₂ and F₂ only
ⓓ. All molecules equally
Correct Answer: B₂, C₂, N₂ versus O₂, F₂, Ne₂
Explanation: For lighter diatomics (Z ≤ 7), π2px and π2py are lower than σ2pz due to orbital mixing. For heavier diatomics (Z ≥ 8), σ2pz is lower in energy than π2px, π2py.
274. What is the molecular orbital configuration of Li₂?
ⓐ. σ1s² σ*1s² σ2s²
ⓑ. σ1s² σ2s²
ⓒ. σ1s² σ*1s²
ⓓ. σ1s² σ*1s² σ2s² σ*2s²
Correct Answer: σ1s² σ*1s² σ2s²
Explanation: Li₂ has 6 electrons. After filling σ1s² and σ*1s², the next 2 go into σ2s². Bond order = (4 − 2)/2 = 1, so Li₂ is stable.
275. Which of the following molecules is paramagnetic according to its MO diagram?
ⓐ. B₂
ⓑ. C₂
ⓒ. N₂
ⓓ. F₂
Correct Answer: B₂
Explanation: B₂ has 10 electrons. MO filling: σ1s² σ*1s² σ2s² σ*2s² π2px¹ π2py¹. The two unpaired electrons in π orbitals make B₂ paramagnetic.
276. The bond order of C₂ molecule is:
ⓐ. 1
ⓑ. 2
ⓒ. 3
ⓓ. 4
Correct Answer: 2
Explanation: C₂ has 12 electrons. MO filling: σ1s² σ*1s² σ2s² σ*2s² π2px² π2py². Bond order = (8 − 4)/2 = 2, corresponding to a double bond.
277. According to MO theory, N₂ has:
ⓐ. Bond order 2, diamagnetic
ⓑ. Bond order 3, diamagnetic
ⓒ. Bond order 2.5, paramagnetic
ⓓ. Bond order 3, paramagnetic
Correct Answer: Bond order 3, diamagnetic
Explanation: N₂ has 14 electrons. MO filling: σ1s² σ*1s² σ2s² σ*2s² π2px² π2py² σ2pz². Bond order = (10 − 4)/2 = 3. All electrons are paired, so N₂ is diamagnetic.
278. Which statement about O₂ is correct according to its MO diagram?
ⓐ. Bond order 3, diamagnetic
ⓑ. Bond order 2, paramagnetic
ⓒ. Bond order 2, diamagnetic
ⓓ. Bond order 3, paramagnetic
Correct Answer: Bond order 2, paramagnetic
Explanation: O₂ has 16 electrons. MO filling: σ1s² σ*1s² σ2s² σ*2s² σ2pz² π2px² π2py² π*2px¹ π*2py¹. Bond order = (10 − 6)/2 = 2. Two unpaired electrons make O₂ paramagnetic.
279. The bond order of F₂ molecule according to MO theory is:
ⓐ. 1
ⓑ. 2
ⓒ. 3
ⓓ. 0
Correct Answer: 1
Explanation: F₂ has 18 electrons. MO filling: σ1s² σ*1s² σ2s² σ*2s² σ2pz² π2px² π2py² π*2px² π*2py². Bond order = (10 − 8)/2 = 1, consistent with a single bond.
280. Which of the following molecules is predicted to be unstable by MO theory?
ⓐ. H₂
ⓑ. Li₂
ⓒ. Be₂
ⓓ. O₂
Correct Answer: Be₂
Explanation: Be₂ has 8 electrons. MO filling: σ1s² σ*1s² σ2s² σ*2s². Bond order = (4 − 4)/2 = 0. Since bond order is zero, Be₂ is unstable and does not exist as a stable molecule.
281. The formula for calculating bond order in Molecular Orbital Theory is:
ⓐ. $\dfrac{N_b}{N_a}$
ⓑ. $\dfrac{N_b – N_a}{2}$
ⓒ. $\dfrac{N_b + N_a}{2}$
ⓓ. $\dfrac{N_a – N_b}{2}$
Correct Answer: $\dfrac{N_b – N_a}{2}$
Explanation: Bond order is calculated as half the difference between the number of bonding ($N_b$) and antibonding ($N_a$) electrons. A higher bond order means stronger bonds and shorter bond length.
282. What is the bond order of H₂⁺ ion?
ⓐ. 0
ⓑ. 0.5
ⓒ. 1
ⓓ. 2
Correct Answer: 0.5
Explanation: H₂⁺ has 1 electron. MO filling: σ1s¹. Bond order = (1 − 0)/2 = 0.5. This explains why H₂⁺ is weakly bound and less stable than H₂.
283. The bond order of N₂ molecule is:
ⓐ. 2
ⓑ. 2.5
ⓒ. 3
ⓓ. 3.5
Correct Answer: 3
Explanation: N₂ has 14 electrons. MO filling: σ1s² σ*1s² σ2s² σ*2s² π2px² π2py² σ2pz². Bond order = (10 − 4)/2 = 3, corresponding to a triple bond.
284. What is the bond order of O₂⁺ ion?
ⓐ. 1.5
ⓑ. 2
ⓒ. 2.5
ⓓ. 3
Correct Answer: 2.5
Explanation: O₂⁺ has 15 electrons. MO filling: σ1s² σ*1s² σ2s² σ*2s² σ2pz² π2px² π2py² π\*2px¹. Bond order = (10 − 5)/2 = 2.5, so O₂⁺ has stronger bonding and shorter bond length than O₂.
285. The bond order of O₂²⁻ ion is:
ⓐ. 1
ⓑ. 1.5
ⓒ. 2
ⓓ. 2.5
Correct Answer: 1
Explanation: O₂²⁻ has 18 electrons. MO filling: σ1s² σ*1s² σ2s² σ*2s² σ2pz² π2px² π2py² π*2px² π*2py². Bond order = (10 − 8)/2 = 1. Thus, the O–O bond is weak, consistent with peroxide ion.
286. What is the bond order of He₂⁺ ion?
ⓐ. 0
ⓑ. 0.5
ⓒ. 1
ⓓ. 1.5
Correct Answer: 0.5
Explanation: He₂⁺ has 3 electrons. MO filling: σ1s² σ\*1s¹. Bond order = (2 − 1)/2 = 0.5. The species is weakly bound but more stable than He₂.
287. Which species has the highest bond order?
ⓐ. O₂
ⓑ. O₂⁺
ⓒ. O₂⁻
ⓓ. O₂²⁻
Correct Answer: O₂⁺
Explanation: Bond orders: O₂⁺ = 2.5, O₂ = 2, O₂⁻ = 1.5, O₂²⁻ = 1. Thus, O₂⁺ has the strongest bond, highest stability, and shortest bond length.
288. The bond order of F₂ molecule is:
ⓐ. 0
ⓑ. 1
ⓒ. 2
ⓓ. 3
Correct Answer: 1
Explanation: F₂ has 18 electrons. MO filling: σ1s² σ*1s² σ2s² σ*2s² σ2pz² π2px² π2py² π*2px² π*2py². Bond order = (10 − 8)/2 = 1, matching the single bond in F₂.
289. What is the bond order of B₂ molecule according to MO theory?
ⓐ. 0
ⓑ. 0.5
ⓒ. 1
ⓓ. 2
Correct Answer: 1
Explanation: B₂ has 10 electrons. MO filling: σ1s² σ*1s² σ2s² σ*2s² π2px¹ π2py¹. Bond order = (6 − 4)/2 = 1. The two unpaired electrons explain its paramagnetic behavior.
290. Which of the following species will be diamagnetic based on bond order?
ⓐ. B₂
ⓑ. O₂
ⓒ. C₂
ⓓ. O₂⁻
Correct Answer: C₂
Explanation: C₂ has 12 electrons. MO filling: σ1s² σ*1s² σ2s² σ*2s² π2px² π2py². Bond order = (8 − 4)/2 = 2. All electrons are paired, so C₂ is diamagnetic, unlike B₂ and O₂ which are paramagnetic.
291. Which property determines whether a molecule is paramagnetic or diamagnetic?
ⓐ. The number of protons in the nucleus
ⓑ. Presence of unpaired electrons in molecular orbitals
ⓒ. Type of chemical bonds (ionic vs covalent)
ⓓ. The atomic mass of the elements
Correct Answer: Presence of unpaired electrons in molecular orbitals
Explanation: According to MOT, a molecule is **paramagnetic** if it has one or more unpaired electrons in its molecular orbitals, because these unpaired spins align with an external magnetic field. If all electrons are paired, the molecule is **diamagnetic** and slightly repelled by a magnetic field. For example, O₂ is paramagnetic due to two unpaired electrons in π\* orbitals, while N₂ is diamagnetic as all its orbitals are filled with paired electrons.
292. Which of the following molecules is paramagnetic?
ⓐ. N₂
ⓑ. O₂
ⓒ. F₂
ⓓ. Ne₂
Correct Answer: O₂
Explanation: O₂ has 16 electrons. Its MO filling is σ1s² σ*1s² σ2s² σ*2s² σ2pz² π2px² π2py² π*2px¹ π*2py¹. This configuration leaves **two unpaired electrons** in antibonding π\* orbitals. These unpaired electrons cause O₂ to be paramagnetic, a fact confirmed experimentally by its attraction to strong magnets. N₂, F₂, and Ne₂ all have paired electrons, making them diamagnetic.
293. Why is N₂ diamagnetic according to MOT?
ⓐ. Because nitrogen has low electronegativity
ⓑ. Because all molecular orbitals are fully filled with paired electrons
ⓒ. Because it has 14 neutrons
ⓓ. Because it forms a triple bond
Correct Answer: Because all molecular orbitals are fully filled with paired electrons
Explanation: N₂ has 14 electrons. MO filling: σ1s² σ*1s² σ2s² σ*2s² π2px² π2py² σ2pz². All electrons are paired, leaving no unpaired spins. Therefore, N₂ is diamagnetic and slightly repelled by a magnetic field. Its strong triple bond (bond order 3) further explains its stability, but diamagnetism arises specifically from the absence of unpaired electrons.
294. Which ion has higher magnetic moment due to unpaired electrons?
ⓐ. O₂
ⓑ. O₂⁺
ⓒ. O₂⁻
ⓓ. O₂²⁻
Correct Answer: O₂⁻
Explanation: In O₂⁻, there are 17 electrons. MO filling: σ1s² σ*1s² σ2s² σ*2s² σ2pz² π2px² π2py² π*2px² π*2py¹. This leaves **one unpaired electron** in the antibonding orbital. Hence O₂⁻ is paramagnetic. O₂ itself has two unpaired electrons, but O₂²⁻ (peroxide ion) has 18 electrons with all pairing, making it diamagnetic. O₂⁺ has 15 electrons and only one unpaired electron, so O₂ is more magnetic than O₂⁺ but weaker than O₂⁻ in terms of net unpaired electrons.
295. Which of the following molecules is diamagnetic?
ⓐ. O₂
ⓑ. B₂
ⓒ. C₂
ⓓ. O₂⁻
Correct Answer: C₂
Explanation: C₂ has 12 electrons. MO filling: σ1s² σ*1s² σ2s² σ*2s² π2px² π2py². This leaves no unpaired electrons, making C₂ diamagnetic. In contrast, O₂ has two unpaired electrons, B₂ has two unpaired electrons in π orbitals, and O₂⁻ has one unpaired electron—all of which are paramagnetic.
296. Which observation provided strong experimental support for MOT over VBT?
ⓐ. O₂ is colorless
ⓑ. O₂ is paramagnetic
ⓒ. N₂ has a triple bond
ⓓ. He₂ does not exist
Correct Answer: O₂ is paramagnetic
Explanation: VBT predicts O₂ should be diamagnetic, since it shows a double bond with paired electrons. However, experiments show that O₂ is **paramagnetic** and attracted to a magnet. MOT successfully explains this by showing two unpaired electrons in π\*2p orbitals. This agreement with experimental data was a major success of MOT.
297. Which molecule is paramagnetic according to MO theory but wrongly predicted as diamagnetic by VBT?
ⓐ. CO
ⓑ. N₂
ⓒ. O₂
ⓓ. F₂
Correct Answer: O₂
Explanation: VBT considers O₂ to have all paired electrons in a double bond, predicting diamagnetism. However, MOT shows that two unpaired electrons remain in π\* orbitals, making O₂ paramagnetic. This correction demonstrates the superiority of MOT in predicting magnetic behavior.
298. Which of the following correctly matches species with its magnetic property?
ⓐ. N₂ – Paramagnetic
ⓑ. O₂ – Diamagnetic
ⓒ. F₂ – Diamagnetic
ⓓ. B₂ – Diamagnetic
Correct Answer: F₂ – Diamagnetic
Explanation: F₂ has 18 electrons. MO filling: σ1s² σ*1s² σ2s² σ*2s² σ2pz² π2px² π2py² π*2px² π*2py². All electrons are paired, so F₂ is diamagnetic. In contrast, N₂ is diamagnetic, O₂ is paramagnetic, and B₂ is also paramagnetic due to unpaired electrons in π orbitals.
299. Why is B₂ paramagnetic?
ⓐ. Because it has an odd number of electrons
ⓑ. Because it has two unpaired electrons in π2p orbitals
ⓒ. Because boron has low ionization energy
ⓓ. Because B₂ contains ionic bonding
Correct Answer: Because it has two unpaired electrons in π2p orbitals
Explanation: B₂ has 10 electrons. MO filling: σ1s² σ*1s² σ2s² σ*2s² π2px¹ π2py¹. The two unpaired electrons in π orbitals make B₂ paramagnetic. This matches experimental evidence and could not be explained properly by VBT.
300. Which of the following best describes diamagnetism?
ⓐ. Strong attraction to a magnetic field due to unpaired electrons
ⓑ. Slight repulsion from a magnetic field due to all electrons being paired
ⓒ. Complete independence from a magnetic field
ⓓ. Attraction only at high temperatures
Correct Answer: Slight repulsion from a magnetic field due to all electrons being paired
Explanation: Diamagnetic substances have no unpaired electrons. The paired electrons produce opposite spins that cancel out magnetic moments. As a result, diamagnetic molecules are weakly repelled by an external magnetic field. Examples: N₂, F₂, Ne₂.
This section contains Class 11 Chemistry MCQs – Chapter 4: Chemical Bonding and Molecular Structure (Part 3). At this stage, the chapter dives deeper into advanced bonding theories such as Molecular Orbital Theory (MOT), explaining bonding and antibonding orbitals, electronic configurations of diatomic molecules, bond order, and stability. The chapter also focuses on Paramagnetism, Diamagnetism, Magnetic properties of molecules, and the effect of electron filling on molecular stability. Another essential part is Resonance structures and delocalization of π-electrons, which is frequently asked in both board and competitive exams. Out of the total 395 MCQs in this chapter, this part provides the third set of 100 questions with detailed solutions. Practicing these will give you an edge in exams like JEE, NEET, AIIMS, and state-level entrance tests, along with strong preparation for Class 11 Chemistry board exams.
- 👉 Total MCQs in this chapter: 395.
- 👉 This page contains: Third set of 100 solved MCQs with answers.
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