CO: triple bond (bond order = 3) — isoelectronic with N₂. NO⁺: triple bond (bond order = 3) — isoelectronic with N₂ (10 electrons). Sum \(= 3+3 = \mathbf{6}\).
Q2
Aluminium chloride in acidified aqueous solution forms an ion [Al(H₂O)₆]³⁺. Its geometry is:
ASquare planar
BOctahedral
CTrigonal bipyramidal
DTetrahedral
✅ Correct: B
[Al(H₂O)₆]³⁺ has 6 water ligands around Al³⁺ → octahedral geometry (sp³d² hybridisation). Al³⁺ coordinates 6 water molecules via lone pairs on O.
Q3
The difference in energy between the actual structure of a molecule and that of the lowest energy resonance structure is called:
AElectromeric energy
BResonance energy
CIonization energy
DHyperconjugation energy
✅ Correct: B
Resonance energy (or delocalization energy) is the stabilization energy obtained because the actual molecule is more stable than any individual resonance structure. For benzene, resonance energy ≈ 150 kJ/mol.
Q4
In which pair do both central atoms show sp² hybridization?
ABF₃ and NO₂⁻
BNH₂⁻ and BF₃
CSO₂ and CO₂
DClF₃ and BF₃
✅ Correct: A
BF₃: 3 bond pairs, 0 lone pairs → sp² (trigonal planar). NO₂⁻: 2 bond pairs, 1 lone pair → sp² (bent, 120°). Both show sp² hybridization.
Q5
The hybridization of nitrogen in pyridine (C₅H₅N) is:
Asp
Bsp²
Csp³
Dsp³d
✅ Correct: B
In pyridine, N is part of the aromatic ring. N has one lone pair in a p-orbital participating in aromaticity, and forms 2 sigma bonds to adjacent C. Hybridization of N is sp², with the lone pair in the sp² plane (not aromatic conjugation).
Q6
Which of the following has the maximum bond angle?
AH₂O
BNH₃
CBF₃
DPCl₃
✅ Correct: C
BF₃: sp² hybridization, 3 bond pairs, 0 lone pairs → bond angle = 120°. NH₃: 107°, H₂O: 104.5°, PCl₃: ~100°. BF₃ has maximum bond angle.
Q7
The number of lone pairs on the central atom in XeF₄ is:
A1
B2
C3
D4
✅ Correct: B
Xe in XeF₄: 8 valence electrons. 4 used in bonds with F. Remaining = 4 electrons = 2 lone pairs. Hybridization: sp³d² (octahedral arrangement: 4F + 2 lone pairs → square planar shape).
Q8
Which of the following is the most polar molecule?
ACCl₄
BCHCl₃
CCH₂Cl₂
DCH₄
✅ Correct: B
CCl₄ and CH₄ have zero dipole moment due to symmetry. CH₂Cl₂ has dipole moment but CHCl₃ (chloroform) has higher dipole moment (~1.87 D) because 3 electronegative Cl atoms pull in one direction while only one H opposes.
Q9
The shape of SF₄ molecule according to VSEPR theory is:
ATetrahedral
BSeesaw (see-saw)
CSquare planar
DTrigonal pyramidal
✅ Correct: B
SF₄: S has 4 bonding pairs + 1 lone pair = 5 electron pairs. Arrangement: trigonal bipyramidal with one equatorial position occupied by lone pair → seesaw (see-saw) shape. Hybridization: sp³d.
Q10
Which of the following statements about hydrogen bonding is correct?
AHydrogen bonding is stronger than a covalent bond
BHF has stronger hydrogen bonding than H₂O
CHydrogen bond is formed between H and highly electronegative atoms (F, O, N)
DHydrogen bonding does not affect boiling point
✅ Correct: C
Hydrogen bonding forms between H (covalently bonded to F, O, or N) and another electronegative atom (F, O, or N). H-bond is much weaker than covalent bond. H₂O has more extensive H-bonding network than HF (4 H-bonds per molecule vs 2).
Q11
Bond order of O₂⁻ (superoxide ion) is:
A1
B1.5
C2
D2.5
✅ Correct: B
O₂: bond order = 2. O₂⁻ (superoxide): adds one electron to π* orbital. Bonding electrons = 8, antibonding = 5. Bond order \(= (8-5)/2 = \mathbf{1.5}\). O₂²⁻ (peroxide) has bond order 1.
Q12
Which of the following has zero dipole moment?
ANH₃
BH₂O
CBF₃
DNF₃
✅ Correct: C
BF₃ has zero dipole moment because it is a symmetrical trigonal planar molecule — the three B−F bond dipoles exactly cancel. NH₃, H₂O, and NF₃ all have net dipole moments due to asymmetric shapes.
Q13
The correct order of C−C bond lengths in diamond, graphite, and benzene is:
ADiamond > Graphite > Benzene
BGraphite > Benzene > Diamond
CDiamond > Benzene > Graphite
DBenzene > Graphite > Diamond
✅ Correct: C
Diamond: C−C single bonds → 154 pm. Benzene: C−C bond order 1.5 (delocalized) → 140 pm. Graphite: C−C within layer (aromatic, bond order ~1.5) → 142 pm. Order: Diamond > Benzene ≈ Graphite. So Diamond > Benzene > Graphite.
Q14
Which molecule has linear geometry?
AH₂O
BSO₂
CCO₂
DNO₂
✅ Correct: C
CO₂: O=C=O with sp hybridization, 2 double bonds, no lone pairs → linear geometry (180°). H₂O: bent (2 lone pairs). SO₂: bent (1 lone pair). NO₂: bent (1 odd electron). CO₂ is linear.
Q15
The hybridization of central atom and the shape of ClF₃ is:
Asp² and trigonal planar
Bsp³ and pyramidal
Csp³d and T-shaped
Dsp³d² and square planar
✅ Correct: C
ClF₃: Cl has 3 bond pairs + 2 lone pairs = 5 electron pairs. Hybridization: sp³d. Arrangement: trigonal bipyramidal with 2 lone pairs at equatorial positions → T-shaped geometry. Bond angle ≈ 87°.
Q16
According to VSEPR theory, the shape of PCl₅ is:
ATetrahedral
BSquare pyramidal
CTrigonal bipyramidal
DOctahedral
✅ Correct: C
PCl₅: P has 5 bond pairs, 0 lone pairs. 5 electron pairs → trigonal bipyramidal shape. 3 equatorial Cl at 120° and 2 axial Cl at 90°. Hybridization: sp³d.
Q17
The correct statement about the nature of bond in NaCl is:
AIt is purely ionic with no covalent character
BIt has some covalent character according to Fajan's rules
CIt is a coordinate covalent bond
DIt is a pure covalent bond
✅ Correct: B
According to Fajan's rules, NaCl has some covalent character. As the polarising power of cation (Na⁺) and polarisability of anion (Cl⁻) increase, the ionic bond gains covalent character. Pure ionic bonds don't exist in reality.
Q18
In which of the following species is the octet rule violated?
ACO₂
BPCl₅
CNH₃
DH₂O
✅ Correct: B
PCl₅ violates the octet rule. P has 5 bonds = 10 electrons around it (expanded octet). Other molecules: CO₂ (8e around C), NH₃ (8e around N), H₂O (8e around O) — all follow octet rule.
Q19
The strength of hydrogen bonding decreases in the order:
AF−H···F > O−H···O > N−H···N
BN−H···N > O−H···O > F−H···F
CO−H···O > F−H···F > N−H···N
DF−H···F > N−H···N > O−H···O
✅ Correct: A
Hydrogen bond strength depends on electronegativity: F > O > N. The strongest H-bond is F−H···F. Order: F−H···F > O−H···O > N−H···N. This explains the anomalously high boiling point of HF and H₂O.
Q20
Which of the following has the maximum number of unpaired electrons?