JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 27

Multiple Choice Questions (MCQs) for JEE Chemistry: Chemical Bonding and Molecular Structure

Review structured MCQ sets for JEE Chemistry Chemical Bonding and Molecular Structure. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Practice Chemical Bonding and Molecular Structure MCQs for JEE Chemistry

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Multiple Choice Questions

Question 1: The compounds having two lone pairs on the central atom among \( \ce{BrF5} \), \( \ce{ClF3} \), \( \ce{XeF4} \) and \( \ce{SF4} \) are
(a) \( \ce{BrF5} \) and \( \ce{SF4} \)
(b) \( \ce{ClF3} \) and \( \ce{XeF4} \)
(c) \( \ce{BrF5} \) and \( \ce{ClF3} \)
(d) \( \ce{XeF4} \) and \( \ce{SF4} \)
Show Answer & Explanation

Answer: (b) \( \ce{ClF3} \) and \( \ce{XeF4} \)

Explanation:
1. \( \ce{BrF5} \): 1 lp
2. \( \ce{ClF3} \): 2 lp
3. \( \ce{XeF4} \): 2 lp
4. \( \ce{SF4} \): 1 lp.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Question 2: The sum of the lone pairs on the central atoms of \( \ce{[TeBr6]^2-} \), \( \ce{[BrF2]+} \), \( \ce{SNF3} \) and \( \ce{[XeF3]-} \) is
(a) 5
(b) 7
(c) 8
(d) 6
Show Answer & Explanation

Answer: (d) 6

Explanation:
1. \( \ce{[TeBr6]^2-} \): 1; \( \ce{[BrF2]+} \): 2; \( \ce{SNF3} \): 0; \( \ce{[XeF3]-} \): 3
2. Sum = 6.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Question 3: In the formate ion \( \ce{HCOO-} \) the two C-O bonds are of equal length because
(a) the ion has two resonating structures
(b) the carbon orbitals are hybridised
(c) the C=O bond is weaker than the C-O bond
(d) the ion is made by removing a proton from the acid
Show Answer & Explanation

Answer: (a) the ion has two resonating structures

Explanation:
1. Charge is delocalised between the two oxygen atoms, giving each C-O bond a bond order of 1.5.
2. Only option (a) agrees with this working.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Question 4: Which of the following exists as a covalent crystal in the solid state?
(a) Iodine
(b) Sulphur
(c) Silicon
(d) Phosphorus
Show Answer & Explanation

Answer: (c) Silicon

Explanation:
1. Silicon is a giant network of Si atoms joined by covalent bonds
2. The others are molecular solids.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Question 5: The total number of lone pairs on the three iodine atoms of \( \ce{I3-} \) is
(a) 9
(b) 12
(c) 3
(d) 6
Show Answer & Explanation

Answer: (a) 9

Explanation:
1. Central I has 3 lone pairs, each terminal I has 3: total 9.
2. Only option (a) agrees with this working.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Question 6: Count the compounds that violate the octet rule among \( \ce{CO2} \), \( \ce{PCl5} \), \( \ce{SiF4} \), \( \ce{BrF5} \), \( \ce{IF7} \), \( \ce{PCl3} \), \( \ce{H2SO4} \), \( \ce{BF3} \).
(a) 3
(b) 4
(c) 6
(d) 5
Show Answer & Explanation

Answer: (d) 5

Explanation:
1. \( \ce{PCl5} \), \( \ce{BrF5} \), \( \ce{IF7} \), \( \ce{H2SO4} \) (S with expanded shell) and \( \ce{BF3} \) violate it
2. \( \ce{CO2} \), \( \ce{SiF4} \) and \( \ce{PCl3} \) obey it.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Question 7: For how many of the species \( \ce{XeF4} \), \( \ce{XeF5-} \), \( \ce{F2SeO2} \), \( \ce{XeF3+} \), \( \ce{XeOF4} \), \( \ce{ClOF3} \), \( \ce{ICl4-} \), \( \ce{SCl2} \) and \( \ce{OSF4} \) does the central atom have two lone pairs?
(a) 4
(b) 5
(c) 6
(d) 3
Show Answer & Explanation

Answer: (b) 5

Explanation:
1. Two lone pairs: \( \ce{XeF4} \), \( \ce{XeF5-} \), \( \ce{XeF3+} \), \( \ce{ICl4-} \) and \( \ce{SCl2} \)
2. \( \ce{F2SeO2} \), \( \ce{XeOF4} \), \( \ce{ClOF3} \) and \( \ce{OSF4} \) have one or none.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Question 8: How many resonating structures does the carbonate ion have?
(a) 2
(b) 3
(c) 4
(d) 6
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. The double bond can be placed on any one of three equivalent oxygen atoms.
2. Only option (b) agrees with this working.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Question 9: The bond angle and hybridisation of the central oxygen in dimethyl ether are
(a) greater than 109 degrees 28 minutes, sp3
(b) 104 degrees 31 minutes, sp3
(c) 106 degrees 51 minutes, sp3
(d) 109 degrees 28 minutes, sp2
Show Answer & Explanation

Answer: (a) greater than 109 degrees 28 minutes, sp3

Explanation:
1. Oxygen has two bond pairs and two lone pairs (sp3), but the two bulky methyl groups open the angle to about 111 degrees.
2. Only option (a) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 10: Which species has bonds that are NOT all equal?
(a) \( \ce{XeF4} \)
(b) \( \ce{BeF4^2-} \)
(c) \( \ce{C2H4} \)
(d) \( \ce{SiF4} \)
Show Answer & Explanation

Answer: (c) \( \ce{C2H4} \)

Explanation:
1. In \( \ce{C2H4} \) there are C=C and C-H bonds of different type
2. The other three have identical bonds.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 11: The correct order of decreasing X-O-X bond angle in \( \ce{H2O} \), \( \ce{F2O} \) and \( \ce{Cl2O} \) is
(a) \( \ce{H2O} \) > \( \ce{Cl2O} \) > \( \ce{F2O} \)
(b) \( \ce{F2O} \) > \( \ce{Cl2O} \) > \( \ce{H2O} \)
(c) \( \ce{Cl2O} \) > \( \ce{H2O} \) > \( \ce{F2O} \)
(d) \( \ce{F2O} \) > \( \ce{H2O} \) > \( \ce{Cl2O} \)
Show Answer & Explanation

Answer: (c) \( \ce{Cl2O} \) > \( \ce{H2O} \) > \( \ce{F2O} \)

Explanation:
1. \( \ce{Cl2O} \) is about 111 degrees (large Cl pushes the bonds apart), \( \ce{H2O} \) 104.5 and \( \ce{F2O} \) 103.
2. Only option (c) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 12: Which of the following is paramagnetic?
(a) \( \ce{O2-} \)
(b) \( \ce{CN-} \)
(c) CO
(d) \( \ce{NO+} \)
Show Answer & Explanation

Answer: (a) \( \ce{O2-} \)

Explanation:
1. \( \ce{O2-} \) has 17 valence electrons, one in an antibonding pi* orbital, so it has an unpaired electron
2. \( \ce{CN-} \), CO and \( \ce{NO+} \) have 10 valence electrons, all paired.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 13: The species PF2Cl3 is non-polar because
(a) the P-Cl bond is non-polar
(b) P and Cl have equal electronegativity
(c) the P-F bond is non-polar
(d) its bond dipoles cancel and the net dipole moment is zero
Show Answer & Explanation

Answer: (d) its bond dipoles cancel and the net dipole moment is zero

Explanation:
1. In the trigonal bipyramidal shape with the two F atoms axial, the P-F and P-Cl dipoles cancel.
2. Only option (d) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 14: Which set of species has the same hybridisation but different shapes?
(a) \( \ce{NO2+} \), \( \ce{NO2} \), \( \ce{NO2-} \)
(b) \( \ce{NH4+} \), \( \ce{H3O+} \), \( \ce{OF2} \)
(c) \( \ce{ClO4-} \), \( \ce{SF4} \), \( \ce{XeF4} \)
(d) \( \ce{SO4^2-} \), \( \ce{PO4^3-} \), \( \ce{ClO4-} \)
Show Answer & Explanation

Answer: (b) \( \ce{NH4+} \), \( \ce{H3O+} \), \( \ce{OF2} \)

Explanation:
1. All of \( \ce{NH4+} \), \( \ce{H3O+} \) and \( \ce{OF2} \) are sp3 with shapes tetrahedral, pyramidal and bent
2. In the second set the hybridisation is sp, sp2, sp2; the fourth set are all tetrahedral.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 15: The bonds present in \( \ce{N2O5} \) are
(a) only ionic
(b) only covalent
(c) covalent and ionic
(d) covalent and coordinate
Show Answer & Explanation

Answer: (d) covalent and coordinate

Explanation:
1. \( \ce{N2O5} \) is a molecule with N=O, N-O and a N->O coordinate link, so it has both covalent and coordinate bonds.
2. Only option (d) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 16: When \( \ce{NH3} \) accepts \( \ce{H+} \) to form \( \ce{NH4+} \), the
(a) bond angle increases
(b) hybridisation changes
(c) NH3 acts as a Lewis acid
(d) geometry changes from tetrahedral to pyramidal
Show Answer & Explanation

Answer: (a) bond angle increases

Explanation:
1. The lone-pair/bond-pair repulsion disappears, so the angle goes from 107 to 109.5 degrees; hybridisation remains sp3 and NH3 acts as a Lewis base.
2. Only option (a) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 17: NH3 and BF3 combine readily because of the formation of a
(a) hydrogen bond
(b) pi back bond
(c) coordinate bond
(d) ionic bond
Show Answer & Explanation

Answer: (c) coordinate bond

Explanation:
1. NH3 donates its lone pair to the empty orbital of B in BF3.
2. Only option (c) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 18: The hybridisation of Xe in \( \ce{XeF2} \) is
(a) sp3d
(b) sp3
(c) sp3d2
(d) sp3d3
Show Answer & Explanation

Answer: (a) sp3d

Explanation:
1. Xe has 2 bond pairs and 3 lone pairs, so 5 electron pairs: sp3d.
2. Only option (a) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 19: Which of the following has the highest boiling point?
(a) \( \ce{CH4} \)
(b) \( \ce{CF4} \)
(c) \( \ce{CD4} \)
(d) \( \ce{CCl4} \)
Show Answer & Explanation

Answer: (d) \( \ce{CCl4} \)

Explanation:
1. Heavier, more polarisable molecules have stronger London forces; \( \ce{CCl4} \) has the largest electron cloud.
2. Only option (d) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 20: Which of the following is non-polar?
(a) \( \ce{NF3} \)
(b) \( \ce{BF3} \)
(c) \( \ce{PF3} \)
(d) \( \ce{SF4} \)
Show Answer & Explanation

Answer: (b) \( \ce{BF3} \)

Explanation:
1. \( \ce{BF3} \) is trigonal planar and symmetric, so the dipoles cancel
2. The others have a lone pair and a pyramidal or see-saw shape.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 21: Which is the strongest hydrogen bond?
(a) O-H...S
(b) F-H...F
(c) S-H...O
(d) F-H...O
Show Answer & Explanation

Answer: (b) F-H...F

Explanation:
1. Strength depends on the electronegativity of the atoms; the F-H...F bond has the highest energy (about 40 kJ/mol).
2. Only option (b) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 22: The paramagnetic nature of oxygen is best explained by
(a) valence bond theory
(b) resonance
(c) VSEPR theory
(d) molecular orbital theory
Show Answer & Explanation

Answer: (d) molecular orbital theory

Explanation:
1. MO theory places two electrons singly in the degenerate pi* orbitals, giving two unpaired electrons.
2. Only option (d) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

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