JEE Chemistry Chemical Bonding And Molecular Structure MCQs Set 22

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Chapter-wise Objective Questions: Chemical Bonding and Molecular Structure

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Multiple Choice Questions

Question 1: Which of the following forms a crystal of NaCl?
(a) NaCl molecules
(b) Na+ and Cl− ions
(c) Na and Cl atoms
(d) None of the above
Show Answer & Explanation

Answer: (b) Na+ and Cl− ions

Explanation:
1. Sodium chloride is an ionic solid.
2. The crystal is a three-dimensional lattice of Na+ and Cl− ions, with no discrete NaCl molecules.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 2: Which one is least ionic among AgCl, KCl, BaCl2 and CaCl2?
(a) AgCl
(b) KCl
(c) BaCl2
(d) CaCl2
Show Answer & Explanation

Answer: (a) AgCl

Explanation:
1. Ag+ has a pseudo-noble-gas (18-electron) core, which shields the nuclear charge poorly, so its polarising power is high.
2. By Fajan's rules AgCl has the most covalent character.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 3: The electronic configurations of four elements L, P, Q and R are L: 1s² 2s² 2p⁴, Q: 1s² 2s² 2p⁶ 3s² 3p⁵, P: 1s² 2s² 2p⁶ 3s¹ and R: 1s² 2s² 2p⁶ 3s². Ionic compounds that can be formed between these elements are
(a) L2P, RL, PQ and R2Q
(b) LP, RL, PQ and RQ
(c) P2L, RL, PQ and RQ2
(d) LP, R2L, P2Q and RQ
Show Answer & Explanation

Answer: (c) P2L, RL, PQ and RQ2

Explanation:
1. L = oxygen (needs 2 electrons), Q = chlorine (needs 1), P = sodium (gives 1), R = magnesium (gives 2).
2. Possible compounds: P2L (Na2O), RL (MgO), PQ (NaCl) and RQ2 (MgCl2).

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 4: Which of the following properties is not shown by ionic compounds?
(a) Hard and brittle nature
(b) High melting and boiling points
(c) Directional properties
(d) Solubility in polar solvents
Show Answer & Explanation

Answer: (c) Directional properties

Explanation:
1. Electrostatic attraction in an ionic crystal acts equally in all directions, so ionic bonds are non-directional.
2. Hardness, brittleness, high melting points and solubility in water are all typical of ionic solids.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 5: The formula of the phosphate of a metal M whose oxide is MO is
(a) M2(PO4)2
(b) M(PO4)
(c) M2PO4
(d) M3(PO4)2
Show Answer & Explanation

Answer: (d) M3(PO4)2

Explanation:
1. In MO the metal is M2+.
2. Phosphate is PO43−, so balancing charges gives M3(PO4)2.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 6: The phosphate of a metal has the formula MHPO4. The formula of its chloride is
(a) MCl
(b) MCl2
(c) MCl3
(d) M2Cl3
Show Answer & Explanation

Answer: (b) MCl2

Explanation:
1. HPO4 carries a charge of 2−, so the metal is M2+.
2. Its chloride is MCl2.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 7: Which of the following has the highest melting point?
(a) NaCl
(b) NaBr
(c) NaI
(d) NaF
Show Answer & Explanation

Answer: (d) NaF

Explanation:
1. For the same cation, lattice energy is greatest for the smallest anion.
2. F− is the smallest, so NaF has the highest lattice energy and the highest melting point.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 8: Among LiCl, NaCl and BeCl2, the correct order of increasing covalent character is
(a) LiCl < NaCl < BeCl2
(b) BeCl2 < NaCl < LiCl
(c) NaCl < LiCl < BeCl2
(d) BeCl2 < LiCl < NaCl
Show Answer & Explanation

Answer: (c) NaCl < LiCl < BeCl2

Explanation:
1. Covalent character increases with the polarising power of the cation (small size, high charge).
2. Na+ < Li+ < Be2+, so NaCl < LiCl < BeCl2.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 9: Which of the following compounds has the maximum covalent character?
(a) NaI
(b) MgI2
(c) AlCl3
(d) AlI3
Show Answer & Explanation

Answer: (d) AlI3

Explanation:
1. Higher charge on the cation and larger, more polarisable anion increase covalent character.
2. Al3+ is the most strongly polarising cation and I− is the most polarisable anion, so AlI3 is the most covalent.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 10: The bond between two identical non-metal atoms has a pair of electrons
(a) Unequally shared between the two
(b) Transferred fully from one atom to another
(c) With identical spins
(d) Equally shared between them
Show Answer & Explanation

Answer: (d) Equally shared between them

Explanation:
1. Identical atoms have equal electronegativity.
2. The bonding pair is therefore shared equally (a non-polar covalent bond), and by Pauli's principle the two electrons have opposite spins.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 11: Which one of the following molecules is electron deficient?
(a) ICl
(b) NH3
(c) BCl3
(d) PCl3
Show Answer & Explanation

Answer: (c) BCl3

Explanation:
1. Boron in BCl3 has only six electrons in its valence shell.
2. It is electron deficient and acts as a Lewis acid.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 12: The number of sigma and pi bonds in 1-butene-3-yne (CH2=CH–C≡CH) are
(a) 5 sigma and 5 pi
(b) 7 sigma and 3 pi
(c) 8 sigma and 2 pi
(d) 6 sigma and 4 pi
Show Answer & Explanation

Answer: (b) 7 sigma and 3 pi

Explanation:
1. Sigma bonds: three C–C (one in each of the double, single and triple links) and four C–H bonds, a total of 7.
2. Pi bonds: one in the double bond and two in the triple bond, a total of 3.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 13: The number of sigma bonds in P4O10 is
(a) 6
(b) 7
(c) 17
(d) 16
Show Answer & Explanation

Answer: (d) 16

Explanation:
1. P4O10 has 4 P atoms, each bonded to 3 bridging oxygens and one terminal oxygen (P=O).
2. There are 6 bridging O atoms giving 12 P–O sigma bonds, and 4 terminal P=O bonds giving 4 sigma bonds.
3. Total sigma = 16.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 14: Which of the following has pπ–dπ bonding?
(a) NO3−
(b) CO32−
(c) BO33−
(d) SO32−
Show Answer & Explanation

Answer: (d) SO32−

Explanation:
1. pπ–dπ bonding needs a vacant d orbital on the central atom.
2. Sulphur (third period) has d orbitals; N, C and B (second period) do not.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 15: The hybridisation of carbon in CO2 is
(a) sp
(b) sp²
(c) sp³
(d) None of these
Show Answer & Explanation

Answer: (a) sp

Explanation:
1. CO2 is O=C=O with two sigma bonds and no lone pairs on carbon.
2. Two electron domains correspond to sp hybridisation.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 16: The shape of the XeF4 molecule is
(a) Linear
(b) Pyramidal
(c) Tetrahedral
(d) Square planar
Show Answer & Explanation

Answer: (d) Square planar

Explanation:
1. Xe has 4 bond pairs and 2 lone pairs (6 electron pairs, sp³d² hybridisation).
2. The lone pairs sit opposite each other, giving a square planar shape.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 17: Which one of the following molecules is linear?
(a) NO2
(b) ClO2
(c) CO2
(d) H2S
Show Answer & Explanation

Answer: (c) CO2

Explanation:
1. CO2 has sp hybridisation and no lone pair on carbon, so it is linear.
2. NO2, ClO2 and H2S are bent.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 18: The percentage of s-character in an sp³ hybrid orbital is
(a) 25
(b) 50
(c) 66
(d) 75
Show Answer & Explanation

Answer: (a) 25

Explanation:
1. An sp³ hybrid orbital is made from one s and three p orbitals.
2. s-character = 1/4 = 25%.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 19: The bond angle in water is nearly
(a) 120°
(b) 180°
(c) 109° 28′
(d) 104° 30′
Show Answer & Explanation

Answer: (d) 104° 30′

Explanation:
1. Oxygen in H2O is sp³ hybridised with two lone pairs.
2. Lone pair–lone pair and lone pair–bond pair repulsions squeeze the bond angle to about 104.5° from 109.5°.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 20: Which one has the smallest bond angle among the following?
(a) NH3
(b) BeF2
(c) H2O
(d) CH4
Show Answer & Explanation

Answer: (c) H2O

Explanation:
1. Bond angles: BeF2 180°, CH4 109.5°, NH3 107° and H2O 104.5°.
2. Water has the smallest.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 21: The correct order of bond angles (smallest first) in H2S, NH3, BF3 and SiH4 is
(a) H2S < NH3 < SiH4 < BF3
(b) NH3 < H2S < SiH4 < BF3
(c) H2S < SiH4 < NH3 < BF3
(d) H2S < NH3 < BF3 < SiH4
Show Answer & Explanation

Answer: (a) H2S < NH3 < SiH4 < BF3

Explanation:
1. H2S ≈ 92°, NH3 ≈ 107°, SiH4 = 109.5° and BF3 = 120°.
2. So H2S < NH3 < SiH4 < BF3.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 22: The structure of PCl5 in the gaseous state is
(a) Tetrahedral
(b) Square planar
(c) Trigonal bipyramidal
(d) Pentagonal bipyramidal
Show Answer & Explanation

Answer: (c) Trigonal bipyramidal

Explanation:
1. P in PCl5 has five bond pairs and no lone pair (sp³d hybridisation).
2. The shape is trigonal bipyramidal.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

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