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Chemical Bonding — NEET Chemistry MCQs with Solutions
Free NEET Chemistry Chemical Bonding MCQs with step-by-step solutions. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — easy
Which among the following shows the limitation of Lewis octet rule?
A. CH4
B. NO ✓ Correct
C. CO2
D. NH4+
Solution: NO has odd electrons (7+8=15), violating the octet rule. CH4 and CO2 follow octet rule. NH4+ has complete octets.
Q2 — easy
Which of the following is an example of expanded octet?
A. All of these ✓ Correct
B. H2SO4
C. SF6
D. PF5
Solution: SF6 (12e- around S), PF5 (10e- around P), and H2SO4 (12e- around S) all have expanded octets.
Q3 — easy
A sigma bond is formed by the overlapping of:
A. s-s orbital alone
B. s–s, s–p or p–p orbitals along internuclear axis ✓ Correct
C. p–p orbital along the sides
D. s and p orbitals alone
Solution: Sigma bonds are formed by direct overlap of orbitals along the internuclear axis. Can involve s-s, s-p, or p-p orbitals.
Q4 — easy
Which overlapping is involved in HCl molecule:
A. s–p overlap ✓ Correct
B. s–d overlap
C. s–s overlap
D. p–p overlap
Solution: HCl forms from H (1s orbital) and Cl (3p orbital), resulting in s-p overlap.
Q5 — easy
π bond is formed:
A. By p - p collateral overlapping ✓ Correct
B. Head on overlapping of p -p orbitals
C. Overlapping of s - s orbitals
D. By overlapping of hybridised orbitals
Solution: π bonds form through lateral (side-by-side) overlapping of p orbitals, not head-on overlapping.
Q6 — easy
p–p overlapping will be observed in the molecules of:
A. Hydrogen bromide
B. Hydrogen
C. Chlorine ✓ Correct
D. Hydrogen chloride
Solution: Cl2 has p-p overlap since both Cl atoms have p electrons. H, HBr, and HCl involve s or s-p overlaps.
Q7 — easy
Which compound of xenon is not possible?
A. XeF5 ✓ Correct
B. XeF4
C. XeF2
D. XeF6
Solution: XeF5 is not stable. Xe can form XeF2, XeF4, and XeF6 through sp3d and sp3d2 hybridization.
Q8 — easy
Higher is the bond order, greater is -
A. Bond length
B. Bond dissociation energy ✓ Correct
C. Paramagnetism
D. Covalent character
Solution: Higher bond order means stronger bonds with greater dissociation energy. Bond length decreases with higher order.
Q9 — easy
In the protonation of H2O, change occurs in
A. Hybridisation state of oxygen
B. Shape of molecule ✓ Correct
C. Hybridisation and shape both
D. None
Solution: In H3O+, oxygen still maintains sp3 hybridization but shape changes from bent to pyramidal.
Q10 — easy
Which of the following has pyramidal shape?
A. NO3−
B. H3O+ ✓ Correct
C. BF3
D. CO32−
Solution: H3O+ (hydronium ion) has pyramidal shape with 3 bonding pairs and 1 lone pair.
Q11 — easy
Which statement is correct:
A. PH3 is polar molecule having non polar bonds ✓ Correct
B. All the compounds having polar bonds, have dipole moment
C. SO2 is non-polar
D. H2O molecule is non polar, having polar bonds
Solution: PH3 has polar molecule character due to lone pair despite individual P-H bonds being weakly polar.
Q12 — easy
Which of the following molecule have zero dipole moment:
A. CH2Cl2
B. NF3
C. BF3 ✓ Correct
D. SO2
Solution: BF3 is trigonal planar and symmetrical, resulting in zero net dipole moment.
Q13 — easy
The molecule does not have zero dipole moment:
A. HCl ✓ Correct
B. CCl4
C. BF3
D. CO2
Solution: HCl is linear and polar with non-zero dipole moment. CO2, CCl4, and BF3 are all non-polar.
Q14 — easy
Which of the following compound possess dipole moment:
A. Water ✓ Correct
B. Benzene
C. Carbon tetra chloride
D. Boron trifluoride
Solution: Water (H2O) is bent and polar with a permanent dipole moment.
Q15 — easy
In Co-ordinate bond, the acceptor atoms must essentially contain in its valence shell an orbital:
A. With paired electron
B. With three electron
C. With single electron
D. With no electron ✓ Correct
Solution: The acceptor atom needs an empty orbital to receive the electron pair from the donor atom.
Q16 — easy
Intermolecular hydrogen bonds are not present in:
A. C2H5NH2
B. CH3OCH3 ✓ Correct
C. CH3COOH
D. CH3CH2OH
Solution: CH3OCH3 (dimethyl ether) has no H bonded to highly electronegative atoms, so no H-bonding.
Q17 — easy
Maximum no. of hydrogen bonds formed by a water molecule in ice is
A. 3
B. 1
C. 2
D. 4 ✓ Correct
Solution: Water can form 4 hydrogen bonds in ice structure (2 as donor, 2 as acceptor).
Q18 — easy
Which one is the correct statement with reference to solubility of MgSO4 in water:
A. Ionic potential of Mg2+ is very low
B. Size of Mg2+ and SO42– are similar
C. SO42– ion mainly contributes towards hydration energy
D. Hydration energy of MgSO4 is higher in comparison to its lattice energy ✓ Correct
Solution: MgSO4 is soluble because its hydration energy exceeds lattice energy.
Q19 — easy
The force responsible for dissolution of ionic compound in water is:
A. Ion – dipole force ✓ Correct
B. Dipole – dipole forces
C. Ion – ion force
D. Hydrogen bond
Solution: Ion-dipole forces between ions and polar water molecules dissolve ionic compounds.
Q20 — easy
Which of the following does not show electrical conduction?
A. diamond ✓ Correct
B. graphite
C. potassium
D. sodium chloride (fused)
Solution: Diamond has no free electrons and does not conduct electricity. Others have mobile charge carriers.
Q21 — easy
In the protonation of H₂O, change occurs in
A. None
B. Hybridisation state of oxygen
C. Hybridisation and shape both
D. Shape of molecule ✓ Correct
Solution: In H₃O⁺, oxygen still maintains sp³ hybridization but shape changes from bent to pyramidal.
Q22 — easy
Which one is the correct statement with reference to solubility of MgSO₄ in water:
A. Size of Mg²⁺ and SO₄²⁻ are similar
B. SO₄²⁻ ion mainly contributes towards hydration energy
C. Hydration energy of MgSO₄ is higher in comparison to its lattice energy ✓ Correct
D. Ionic potential of Mg²⁺ is very low
Solution: MgSO₄ is soluble because its hydration energy exceeds lattice energy.
Q23 — medium
The strength of bonds by 2s - 2s, 2p - 2p and 2p -2s overlapping has the order:
A. p – p > s – p > s – s
B. p – p > s – s > p –s
C. s – s > p – p > s – p ✓ Correct
D. s – s > p – s > p –p
Solution: s-s overlap is strongest, followed by p-p, then s-p due to orbital alignment and overlap efficiency.
Q24 — medium
In which of the excitation state of chlorine ClF3 is formed:
A. In first excitation state
B. In ground state
C. In second excitation state ✓ Correct
D. In third excitation state
Solution: ClF3 requires sp3d hybridization, needing Cl to be in second excited state with 7 unpaired electrons.
Q25 — medium
Which is not characteristic of π-bond:
A. π - bond is formed when a sigma bond already formed
B. π - bond are formed from hybrid orbitals ✓ Correct
C. π-bond results from lateral overlap of atomic orbitals
D. π - bond may be formed by the overlapping of p-orbitals
Solution: π-bonds are formed by pure p-orbitals, not hybrid orbitals. Hybrid orbitals form σ-bonds.
Q26 — medium
Which condition is not favourable for the combination of atomic orbitals:
A. The combining atomic orbitals nearly have the same energy
B. The combining orbital must overlap to the minimum extent ✓ Correct
C. The combining atomic orbitals must have the same symmetry about the molecular axis
D. The combining orbitals must overlap to the maximum extent
Solution: For effective orbital combination, maximum overlap is required, not minimum.
Q27 — medium
In the compound CH2=CH—CH2—CH2—C≡CH, the C2—C3 bond is formed by the overlapping of:
A. sp2 – sp3 ✓ Correct
B. sp3 – sp3
C. sp – sp2
D. sp – sp3
Solution: C2 is sp2 hybridized (in C=C) and C3 is sp3 hybridized (saturated), forming sp2-sp3 overlap.
Q28 — medium
Which of the following elements can not exhibit sp3d hybridisation state: (a) C (b) P (c) Cl (d) B
A. a, c
B. a, d ✓ Correct
C. b, d
D. b, c
Solution: C and B cannot form sp3d because they are in period 2 with no d-orbitals available.
Q29 — medium
Which of the following species are expected to be planar: (a) NH3 (b) CO32– (c) CH3+ (d) PCl3
A. c and d
B. b and d
C. b and c ✓ Correct
D. a and d
Solution: CO32– (trigonal planar) and CH3+ (trigonal planar) are planar. NH3 and PCl3 are pyramidal.
Q30 — medium
In which following set of compound/ion has linear shape
A. CO32−, NO3−, BF3
B. NO2+, CO2, XeF2 ✓ Correct
C. CH4, NH4+, BH4−
D. BeCl2, BCl3, CH4
Solution: NO2+ (linear), CO2 (linear), XeF2 (linear with sp3d hybridization) are all linear molecules.