Previous year questions: Chemical Bonding and Molecular Structure

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

Two statements are given. Statement I: the correct sequence of bond lengths in the following species is O2+ < O2 < O2- < O2^2-. Statement II: the correct sequence of the number of unpaired electrons in the following species is O2 > O2+ > O2- > O2^2-. Which is correct?

Answer: Statement I is true. The bond orders are 2.5, 2.0, 1.5 and 1.0 for O2+, O2, O2- and O2^2-, and a higher bond order means a shorter bond, so the bond length rises in the order given. Statement II is false. O2 has two unpaired electrons, O2^2- has none, but O2+ and O2- each have exactly one unpaired electron, so they cannot be placed one above the other. The correct order is O2 greater than O2+ equal to O2- greater than O2^2-. The answer is that statement I is true and statement II is false.

Q2 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

According to Lewis theory, what is the total number of sigma bond pairs and lone pairs of electrons around the central atom of the XeO6^4- ion?

Answer: Count the valence electrons: xenon gives 8, six oxygen atoms give 36, and the four negative charges add 4, making 48 electrons. The six Xe-O sigma bonds use 12 electrons. Each oxygen needs three lone pairs to complete its octet, so the six oxygens take 36 more electrons. All 48 electrons are used, and no lone pair is left on xenon. The central atom has six sigma bond pairs and zero lone pairs, so the total asked for is 6. The ion is octahedral.

Q3 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

How many of these species have an sp3d hybridised central atom: BrF5, XeF5-, BF4-, ICl4-, XeF4, SF4, NH4+, ClF3, XeF2, ICl2-?

Answer: Count the electron domains around each central atom; five domains correspond to sp3d hybridisation. BrF5 has five bond pairs and one lone pair, six domains, so sp3d2. XeF5- has five bond pairs and two lone pairs, seven domains, so it is not sp3d. BF4- has four domains, sp3. ICl4- has four bond pairs and two lone pairs, six domains, sp3d2. XeF4 has four bond pairs and two lone pairs, six domains, sp3d2. SF4 has four bond pairs and one lone pair, five domains, sp3d. NH4+ has four domains, sp3. ClF3 has three bond pairs and two lone pairs, five domains, sp3d. XeF2 has two bond pairs and three lone pairs, five domains, sp3d. ICl2- has two bond pairs and three lone pairs, five domains, sp3d. The sp3d species are SF4, ClF3, XeF2 and ICl2-, so the count is 4.

Q4 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

Two atoms A and B have covalent radii equal to rA and rB respectively. State the bond length of the AB molecule and the total length of the AB molecule.

Answer: The covalent bond length is the distance between the two bonded nuclei, and it equals the sum of the two covalent radii, rA + rB. The total length of the molecule is the distance from the outer edge of A to the outer edge of B, passing through both atoms; that is rA + rA + rB + rB, which equals 2(rA + rB). The correct pair is (rA + rB) and 2(rA + rB).

Q5 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

Bromine trifluoride autoionises to form BrF2+ and BrF4-. What are the shapes of the cation and the anion respectively?

Answer: In BrF2+ the bromine atom has two bond pairs and two lone pairs, four domains, with a bent arrangement of the two fluorine atoms. In BrF4- the bromine atom has four bond pairs and two lone pairs, six domains, and the two lone pairs sit opposite each other so the four fluorines form a square plane. The shapes are bent and square planar respectively. A quick check confirms both assignments: the cation has four domains like water but with two bond pairs, while the anion has six domains like xenon tetrafluoride. The correct option is bent, square planar.

Q6 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

Statement I: F2O < H2O < Cl2O is the correct trend in terms of bond angle. Statement II: SiF4, SnF4 and PbF4 are ionic in nature. Are the statements true?

Answer: Statement I is true. The bond angle is set by the repulsion between the electron domains on the central atom. Oxygen difluoride has a very small central atom, and its two lone pairs push hard, so its angle is the smallest. Water has a slightly larger angle, and in dichlorine monoxide the central oxygen is bonded to two large chlorine atoms, which reduces the repulsion between the oxygen lone pairs and opens the angle further. Statement II is false. Silicon tetrafluoride is a covalent molecular gas, not an ionic solid; the ionic character appears only with the larger, less polarising cations of tin and lead. The correct option is that statement I is true and statement II is false.

Q7 (NCERT Exemplar, 3 marks)

Isostructural species have the same shape and hybridisation. Which of the following pairs is isostructural: NF3 and BF3, BF4- and NH4+, BCl3 and BrCl3, or NH3 and NO3-?

Answer: Compare the domain count and the hybridisation. NF3 has three bond pairs and one lone pair, so nitrogen is sp3 with a trigonal pyramidal shape, while BF3 has three bond pairs and no lone pair, so boron is sp2 and the shape is trigonal planar, so they are not isostructural. BF4- has four bond pairs, tetrahedral, and NH4+ also has four bond pairs, tetrahedral; both central atoms are sp3, so this pair is isostructural. BCl3 is trigonal planar sp2 while BrCl3 has two lone pairs and a T shape, so they differ. NH3 is pyramidal while NO3- is trigonal planar. The isostructural pair is BF4- and NH4+.

Source: NCERT Exemplar Class 11 Chemistry, Unit 4, MCQ 1.

Q8 (NCERT Exemplar, 3 marks)

Which of these has the highest dipole moment: CO2, HI, H2O or SO2?

Answer: The dipole moment depends on both the polarity of the bonds and the shape. CO2 is linear, so its two C-O bond moments cancel and its net moment is zero. HI has a single polar bond and a small moment of about 0.44 D. SO2 is bent, but the molecule is symmetric enough that its moment is about 1.6 D. Water is strongly bent and the two highly polar O-H bonds add to a moment of about 1.85 D. Water therefore has the highest dipole moment of the four.

Source: NCERT Exemplar Class 11 Chemistry, Unit 4, MCQ 2.

Q9 (NCERT Exemplar, 3 marks)

What are the hybrid orbitals of nitrogen in NO2+, NO3- and NH4+ respectively?

Answer: Count the domains in each ion. NO2+ has two sigma bonds and no lone pair on nitrogen, so it uses two hybrid orbitals and nitrogen is sp. NO3- has three sigma bonds and no lone pair, so nitrogen uses three hybrids and is sp2. NH4+ has four sigma bonds and no lone pair, so nitrogen uses four hybrids and is sp3. The order is sp, sp2 and sp3.

Source: NCERT Exemplar Class 11 Chemistry, Unit 4, MCQ 3.

Q10 (NCERT Exemplar, 3 marks)

Arrange H2O, HF and NH3 in decreasing order of boiling point and justify the order.

Answer: The boiling point depends on the strength and the number of hydrogen bonds per molecule. Water forms up to four hydrogen bonds per molecule through its two hydrogen atoms and two lone pairs, so its boiling point is the highest at 100 degrees Celsius. Hydrogen fluoride has only one hydrogen atom and three lone pairs, so it can form fewer hydrogen bonds per molecule, giving a boiling point of about 19.5 degrees Celsius. Ammonia has three hydrogen atoms but only one lone pair on nitrogen, and the N-H bond is less polar than the O-H and H-F bonds, so its hydrogen bonding is the weakest and it boils at about -33 degrees Celsius. The decreasing order is H2O, HF, NH3.

Source: NCERT Exemplar Class 11 Chemistry, Unit 4, MCQ 4.