Free practice › Class 11 › Chemistry › Chapter 4
Chemical Bonding and Molecular Structure — NEET & JEE MCQs with answers
Unit 4 explains how and why atoms combine, using Lewis theory (electron-pair sharing/transfer), VSEPR theory for molecular geometry, and Valence Bond (VB) theory including hybridisation and orbital overlap. It also introduces Molecular Orbital (MO) theory and hydrogen bonding. Together these theories predict bond type, bond parameters, and the 3D shapes of molecules.
Topics in this chapter
- 4.1 Kössel–Lewis Approach to Chemical Bonding
- 4.2 Ionic or Electrovalent Bond
- 4.3 Bond Parameters
- 4.4 The Valence Shell Electron Pair Repulsion (VSEPR) Theory
- 4.5 Valence Bond Theory
- 4.6 Hybridisation
- 4.7 Molecular Orbital Theory
- 4.8 Bonding in Some Homonuclear Diatomic Molecules
- 4.9 Hydrogen Bonding
Free practice questions
Which of the following pairs of enthalpy terms together determine the ease of formation of ions for an ionic compound?
- A. Sublimation enthalpy and atomization enthalpy
- B. Ionization enthalpy and electron gain enthalpy
- C. Lattice enthalpy and bond enthalpy
- D. Hydration enthalpy and bond enthalpy
Show answer
Answer: B — The formation of a positive ion involves ionization (ionization enthalpy) and that of a negative ion involves addition of electrons (electron gain enthalpy). These determine the ease of ion formation.
NCERT Class 11 Chemistry · Ch 4 · 4.2 Ionic or Electrovalent Bond
How many equivalent hybrid orbitals are formed in sp2 hybridisation?
- A. Two
- B. Three
- C. Five
- D. Four
Show answer
Answer: B — NCERT: in sp2 hybridisation, one s and two p-orbitals form three equivalent sp2 hybridised orbitals.
NCERT Class 11 Chemistry · Ch 4 · 4.6 Hybridisation
Bond angle is experimentally determined by which method, according to NCERT?
- A. Titration
- B. Calorimetry
- C. Conductometry
- D. Spectroscopic methods
Show answer
Answer: D — NCERT states the bond angle is expressed in degrees and can be experimentally determined by spectroscopic methods. It helps in determining the shape of a molecule.
NCERT Class 11 Chemistry · Ch 4 · 4.3.2 Bond Angle
Which of the following has no unpaired electron and is therefore diamagnetic?
- A. $O_2^-$
- B. $O_2$
- C. $O_2^+$
- D. $C_2$
Show answer
Answer: D — $C_2$ : $KK(\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x^2 = \pi 2p_y^2)$ — the four pi electrons fill both degenerate $\pi 2p$ orbitals completely, leaving no unpaired electron. $O_2$, $O_2^-$ and $O_2^+$ each have singly occupied $\pi^* 2p$ orbitals.
NCERT Class 11 Chemistry Ch 4.8
Which of the following relations between electron-pair repulsions is correct?
- A. Lone pair-bond pair repulsion > bond pair-bond pair repulsion
- B. Bond pair-bond pair repulsion > lone pair-bond pair repulsion
- C. Lone pair-bond pair repulsion > lone pair-lone pair repulsion
- D. Bond pair-bond pair repulsion > lone pair-lone pair repulsion
Show answer
Answer: A — Section 4.4 orders the repulsions as lp-lp > lp-bp > bp-bp, so lp-bp repulsion is greater than bp-bp repulsion.
NCERT Class 11 Chemistry Ch 4.4
The bond length of the H2 molecule according to NCERT is:
- A. 104 pm
- B. 120 pm
- C. 144 pm
- D. 74 pm
Show answer
Answer: D — NCERT states that two hydrogen atoms bond together to form a stable molecule having a bond length of 74 pm.
NCERT Class 11 Chemistry · Ch 4 · 4.5 Valence Bond Theory
When hydrogen is bonded to a strongly electronegative element X, the shared electron pair moves away from hydrogen. As a result, the hydrogen atom acquires:
- A. No charge at all
- B. A fractional negative charge (δ-)
- C. A fractional positive charge (δ+)
- D. A full positive charge (+1)
Show answer
Answer: C — NCERT explains that because of the displacement of electrons towards X, the hydrogen acquires a fractional positive charge (δ+) while X attains a fractional negative charge (δ-).
NCERT Class 11 Chemistry · Ch 4 · 4.9 Hydrogen Bonding
Why does the He2 molecule not exist?
- A. Its bond order is zero (Nb = Na)
- B. Its bond order is 1
- C. It is strongly paramagnetic
- D. It has too many unpaired electrons
Show answer
Answer: A — Bond order of He2 is ½(2−2) = 0. A zero bond order means an unstable molecule, therefore He2 does not exist.
NCERT Class 11 Chemistry · Ch 4 · 4.7 Molecular Orbital Theory
According to the steps for writing Lewis structures, in the CO3^2- ion the central atom is:
- A. Oxygen
- B. One of the oxygen atoms shared with carbon
- C. Carbon
- D. There is no central atom
Show answer
Answer: C — The text states that in CO3^2-, carbon is the central atom (being the least electronegative) while oxygen atoms occupy the terminal positions.
NCERT Class 11 Chemistry · Ch 4 · 4.1 Kossel-Lewis Approach to Chemical Bonding
In SF6, the six sp3d2 hybrid orbitals overlap with which orbitals of fluorine to form the S-F bonds?
- A. lone-pair orbitals of fluorine
- B. empty d orbitals of fluorine
- C. singly occupied orbitals of fluorine
- D. completely filled s orbitals of fluorine
Show answer
Answer: C — NCERT: the six sp3d2 hybrid orbitals overlap with singly occupied orbitals of fluorine atoms to form six S-F sigma bonds.
NCERT Class 11 Chemistry · Ch 4 · 4.6 Hybridisation
Practise all 847 questions of Chemical Bonding and Molecular Structure in timed NEET & JEE tests
+4/−1 marking like the real exam · see which topics cost you marks · notes, XP and leaderboards
Start your free trial →