Chemical Bonding and Molecular Structure

What you'll learn

A concept-first guide to ionic, covalent and coordinate bonds, VSEPR shapes, hybridisation and molecular orbital theory for JEE Chemistry.

This chapter has

Chemical Bonding and Molecular Structure — solved practice questions

8 JEE Chemistry questions with step-by-step solutions. Attempt each, then reveal the answer.

1. Q1easy

   According to VSEPR theory, what is the shape of the SF4 molecule?

   • ATetrahedral
   • BSee-saw
   • CSquare planar
   • DTrigonal bipyramidal
   Show answer & solution

   Correct answer: (B) See-saw

   SF4 has 4 bond pairs and 1 lone pair around S (sp3d, 5 electron domains). The lone pair occupies an equatorial position, giving a see-saw shape.

2. Q2medium

   The hybridization of the central atom and shape of ClF3 are respectively:

   • Asp3, pyramidal
   • Bsp3d2, square planar
   • Csp3d, T-shaped
   • Dsp3d, trigonal planar
   Show answer & solution

   Correct answer: (C) sp3d, T-shaped

   Cl in ClF3 has 3 bond pairs and 2 lone pairs (5 domains, sp3d). The two lone pairs go equatorial, giving a T-shaped molecule.

3. Q3medium

   Which of the following molecules has a non-zero dipole moment?

   • ABF3
   • BCO2
   • CCCl4
   • DNH3
   Show answer & solution

   Correct answer: (D) NH3

   In BF3 (trigonal planar), CO2 (linear) and CCl4 (tetrahedral) the bond dipoles cancel by symmetry. NH3 is pyramidal with a lone pair, so the dipoles do not cancel and it has a net dipole moment.

4. Q4medium

   According to molecular orbital theory, the bond order of the O2 molecule is:

   • A2
   • B1
   • C2.5
   • D3
   Show answer & solution

   Correct answer: (A) 2

   O2 has 16 electrons. Bond order = (bonding − antibonding)/2 = (10 − 6)/2 = 2. Its two unpaired electrons in the pi* orbitals also make it paramagnetic.

5. Q5hard

   Which species has the maximum number of unpaired electrons according to molecular orbital theory?

   • AO2^+
   • BO2
   • CO2^2-
   • DN2
   Show answer & solution

   Correct answer: (B) O2

   O2 has 2 unpaired electrons (in pi* orbitals); O2^+ has 1; O2^2- has 0; N2 has 0. Thus O2 has the maximum number of unpaired electrons.

6. Q6hard

   The correct order of bond order for O2, O2^+, O2^-, and O2^2- is:

   • AO2 > O2^+ > O2^- > O2^2-
   • BO2^2- > O2^- > O2 > O2^+
   • CO2^+ > O2 > O2^- > O2^2-
   • DO2^+ > O2^- > O2 > O2^2-
   Show answer & solution

   Correct answer: (C) O2^+ > O2 > O2^- > O2^2-

   Bond orders are O2^+ = 2.5, O2 = 2.0, O2^- = 1.5, O2^2- = 1.0. Adding antibonding pi* electrons lowers the bond order, so O2^+ > O2 > O2^- > O2^2-.

7. Q7hard

   Among the following, which pair is isostructural (same shape and hybridization)?

   • ANH3 and BF3
   • BCO2 and SO2
   • CPCl5 and BrF5
   • DSO4^2- and ClO4^-
   Show answer & solution

   Correct answer: (D) SO4^2- and ClO4^-

   SO4^2- and ClO4^- are both tetrahedral with sp3 hybridized central atoms and no lone pairs. NH3 (pyramidal) vs BF3 (trigonal planar) and others differ in shape or lone-pair count.

8. Q8medium

   The number of sigma and pi bonds in a benzene (C6H6) molecule are respectively:

   • A12 sigma and 3 pi
   • B6 sigma and 3 pi
   • C12 sigma and 6 pi
   • D9 sigma and 3 pi
   Show answer & solution

   Correct answer: (A) 12 sigma and 3 pi

   Benzene has 6 C–C sigma, 6 C–H sigma (12 sigma total) and 3 pi bonds from the delocalized ring. So 12 sigma and 3 pi bonds.

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