JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 51

Chemistry Objective Questions and Answers: Chemical Bonding and Molecular Structure

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Multiple Choice Questions

Question 1: The dipole moments of AX3, BX3 and CY3 are 4.97 × 10-30 C m, 0.60 × 10-30 C m and 0.00 C m respectively. The shapes of the molecules may be:
(a) pyramidal, T-shape, trigonal planar
(b) pyramidal, trigonal planar, T-shape
(c) T-shape, pyramidal, trigonal planar
(d) pyramidal, T-shape, linear
Show Answer & Explanation

Answer: (a) pyramidal, T-shape, trigonal planar

Explanation:
1. Zero dipole moment means a symmetrical trigonal planar CY3.
2. The largest value (about 1.5 D) fits a pyramidal AX3 (like NH3), and the small value (0.18 D) fits a T-shaped BX3 (like ClF3).
3. Answer: (a)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 2: The bond strength in O2+, O2, O2- and O22- follows the order:
(a) O22- > O2- > O2 > O2+
(b) O2+ > O2 > O2- > O22-
(c) O2 > O2- > O22- > O2+
(d) O2- > O22- > O2+ > O2
Show Answer & Explanation

Answer: (b) O2+ > O2 > O2- > O22-

Explanation:
1. Bond orders: O2+ 2.5, O2 2, O2- 1.5, O22- 1.
2. Bond strength rises with bond order.
3. Answer: (b)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 3: A compound that leaves behind no residue on heating is:
(a) Cu(NO3)2
(b) KNO3
(c) NH4NO3
(d) None of these
Show Answer & Explanation

Answer: (c) NH4NO3

Explanation:
1. NH4NO3 → N2O + 2H2O (entirely gaseous products).
2. Cu(NO3)2 leaves CuO and KNO3 leaves KNO2.
3. Answer: (c)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 4: Which of the following molecules is polar and non-planar?
(a) XeF4
(b) XeF5-
(c) CH2F2
(d) ClF3
Show Answer & Explanation

Answer: (c) CH2F2

Explanation:
1. CH2F2 is tetrahedral with unequal bond dipoles, so it is polar and non-planar.
2. XeF4 and XeF5- are planar and non-polar, and ClF3 is T-shaped, i.e. planar.
3. Answer: (c)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 5: The dipole moment of NH3 is more than that of NF3 because:
(a) The N–F bond is more polar than the N–H bond
(b) NH3 is pyramidal while NF3 is planar
(c) In NH3 the orbital dipole of the lone pair is in the same direction as the resultant dipole of the N–H bonds, while in NF3 it is opposite to that of the N–F bonds
(d) None of these
Show Answer & Explanation

Answer: (c) In NH3 the orbital dipole of the lone pair is in the same direction as the resultant dipole of the N–H bonds, while in NF3 it is opposite to that of the N–F bonds

Explanation:
1. In NH3 the lone-pair dipole adds to the N–H bond dipoles.
2. In NF3 the N–F bond dipoles point toward F and oppose the lone-pair dipole, so the net value is small.
3. Answer: (c)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 6: Ammonia is soluble in water but phosphine is insoluble because:
(a) Phosphine has a higher molecular mass than ammonia
(b) Ammonia is polar while phosphine is non-polar
(c) Ammonia forms intermolecular hydrogen bonds with water but phosphine does not
(d) Ammonia is ionic while phosphine is covalent
Show Answer & Explanation

Answer: (c) Ammonia forms intermolecular hydrogen bonds with water but phosphine does not

Explanation:
1. N is small and electronegative, so NH3 forms hydrogen bonds with water.
2. P is large and less electronegative, so PH3 cannot.
3. Answer: (c)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 7: Which of the following resist hydrolysis at room temperature?
(a) PCl3, SF6
(b) CCl4, NO2
(c) PCl5, XeF6
(d) SF6, CCl4
Show Answer & Explanation

Answer: (d) SF6, CCl4

Explanation:
1. SF6 and CCl4 resist hydrolysis because S and C have no available empty orbitals or are sterically protected (SF6) or lack d-orbitals (CCl4).
2. PCl3, PCl5, NO2 and XeF6 react readily with water.
3. Answer: (d)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 8: Which of the following is polar?
(a) p-Dichlorobenzene
(b) trans-1-Chloropropene
(c) Boron trifluoride
(d) Xenon tetrafluoride
Show Answer & Explanation

Answer: (b) trans-1-Chloropropene

Explanation:
1. trans-1-Chloropropene (CH3–CH=CHCl) has a net dipole moment since the groups are different.
2. p-Dichlorobenzene, BF3 and XeF4 are symmetrical and non-polar.
3. Answer: (b)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 9: Which of the following has the shortest bond length?
(a) O2+
(b) O22-
(c) O2
(d) All have the same bond length
Show Answer & Explanation

Answer: (a) O2+

Explanation:
1. O2+ has the highest bond order (2.5), so its bond is the shortest.
2. Answer: (a)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 10: Which of the following attractions is the strongest?
(a) Cl- ··· H–Cl
(b) CHCl3 ··· CHCl3
(c) CCl4 ··· H2O
(d) Cl- ··· H2O
Show Answer & Explanation

Answer: (d) Cl- ··· H2O

Explanation:
1. Cl- ··· H2O is an ion–dipole attraction, much stronger than the dipole–dipole, dipole–induced dipole and weaker interactions in the other options.
2. Answer: (d)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 11: Phosphorus pentachloride in the solid state exists as:
(a) PCl5
(b) [PCl4]+[Cl]-
(c) [PCl4]+[PCl6]-
(d) PCl5·Cl2
Show Answer & Explanation

Answer: (c) [PCl4]+[PCl6]-

Explanation:
1. In the solid state PCl5 is ionic: tetrahedral [PCl4]+ and octahedral [PCl6]-.
2. Answer: (c)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 12: The least stable hydride is:
(a) Stannane
(b) Silane
(c) Plumbane
(d) Germane
Show Answer & Explanation

Answer: (c) Plumbane

Explanation:
1. Thermal stability of the group 14 hydrides falls down the group as the E–H bond weakens.
2. Plumbane (PbH4) is the least stable.
3. Answer: (c)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 13: Solid NaCl is a bad conductor of electricity because:
(a) In solid NaCl there are no ions
(b) Solid NaCl is covalent
(c) In solid NaCl there is no mobility of ions
(d) In solid NaCl there are no electrons
Show Answer & Explanation

Answer: (c) In solid NaCl there is no mobility of ions

Explanation:
1. The Na+ and Cl- ions are locked in fixed lattice positions, so no charge carriers can move.
2. Answer: (c)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 14: Which of the following halides is inert towards hydrolysis at room temperature?
(a) SiCl4
(b) PCl3
(c) NCl3
(d) NF3
Show Answer & Explanation

Answer: (d) NF3

Explanation:
1. NF3 is hydrolysed neither by water nor by acid at room temperature, because the lone pair on N is held tightly by the electronegative fluorine atoms.
2. The other halides are hydrolysed.
3. Answer: (d)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 15: The correct order of dipole moment is:
(a) CH4 < NF3 < NH3 < H2O
(b) NF3 < CH4 < NH3 < H2O
(c) NH3 < NF3 < CH4 < H2O
(d) H2O < NH3 < NF3 < CH4
Show Answer & Explanation

Answer: (a) CH4 < NF3 < NH3 < H2O

Explanation:
1. Dipole moments: CH4 0, NF3 0.24 D, NH3 1.47 D, H2O 1.85 D.
2. Answer: (a)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 16: Which of the following orders are correct? (I) Thermal stability: BeSO4 < MgSO4 < CaSO4 < BaSO4. (II) Basic nature: ZnO > BeO > MgO > CaO. (III) Solubility in water: LiOH > NaOH > KOH > RbOH > CsOH. (IV) Melting point: NaCl > KCl > RbCl > CsCl > LiCl.
(a) (I) and (IV)
(b) (I), (II) and (IV)
(c) (II) and (III)
(d) All are correct
Show Answer & Explanation

Answer: (a) (I) and (IV)

Explanation:
1. Sulphate stability rises down the group, and the melting points are 801, 770, 718, 645 and 605 °C for NaCl, KCl, RbCl, CsCl and LiCl.
2. Basic nature rises from BeO to CaO (ZnO is amphoteric), and the hydroxide solubility increases down the group.
3. Answer: (a)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 17: In which of the following diatomic molecules is the bond dissociation energy maximum?
(a) H2
(b) F2
(c) Cl2
(d) I2
Show Answer & Explanation

Answer: (a) H2

Explanation:
1. Bond dissociation energies (kJ/mol): H2 436, Cl2 242, F2 159, I2 151.
2. H–H is the strongest.
3. Answer: (a)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 18: What is true about CO2 and SO2?
(a) Both turn acidified K2Cr2O7 solution green
(b) Both turn lime water milky
(c) Both are oxidising agents and not reducing agents
(d) All of these
Show Answer & Explanation

Answer: (b) Both turn lime water milky

Explanation:
1. Both gases form insoluble CaCO3 or CaSO3, which turns lime water milky.
2. Only SO2 reduces dichromate (turning it green), and SO2 can act as a reducing agent.
3. Answer: (b)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 19: Consider the statements. I: PCl3 on hydrolysis in the presence of moisture gives fumes of HCl. II: PCl5 exists as [PCl4]+[PCl6]- in the solid state. III: All the five bonds in PCl5 are equivalent. Choose the correct statement(s).
(a) II and III
(b) I, II and III
(c) Only I
(d) I and II
Show Answer & Explanation

Answer: (d) I and II

Explanation:
1. PCl3 + 3H2O → H3PO3 + 3HCl, so I is true.
2. Solid PCl5 is [PCl4]+[PCl6]-, so II is true; the axial and equatorial bonds differ, so III is false.
3. Answer: (d)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 20: According to Fajans' rule, ionic character increases for:
(a) A small cation and small charge on the cation
(b) A large cation and a small anion
(c) A small cation and a large anion
(d) A large anion and small charge on the anion
Show Answer & Explanation

Answer: (b) A large cation and a small anion

Explanation:
1. Polarisation of the anion, and hence covalent character, rises for a small cation and a large anion.
2. So a large cation with a small anion favours ionic character.
3. Answer: (b)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 21: Paragraph: The electronic configurations of three elements A, B and C are A: 1s2 2s2 2p6; B: 1s2 2s2 2p6 3s2 3p3; C: 1s2 2s2 2p6 3s2 3p5. The stable form of A may be represented by the formula:
(a) A
(b) A2
(c) A3
(d) A4
Show Answer & Explanation

Answer: (a) A

Explanation:
1. A is neon, a noble gas with a complete octet, so it exists as single atoms.
2. Answer: (a)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 22: Paragraph: The electronic configurations of three elements A, B and C are A: 1s2 2s2 2p6; B: 1s2 2s2 2p6 3s2 3p3; C: 1s2 2s2 2p6 3s2 3p5. The stable form of C may be represented by the formula:
(a) C
(b) C2
(c) C3
(d) C4
Show Answer & Explanation

Answer: (b) C2

Explanation:
1. C is chlorine, which needs one electron to complete its octet and so exists as the diatomic molecule Cl2.
2. Answer: (b)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 23: Paragraph: The electronic configurations of three elements A, B and C are A: 1s2 2s2 2p6; B: 1s2 2s2 2p6 3s2 3p3; C: 1s2 2s2 2p6 3s2 3p5. The molecular formula of the compound formed from B and C will be:
(a) BC
(b) B2C
(c) BC2
(d) BC3
Show Answer & Explanation

Answer: (d) BC3

Explanation:
1. B is phosphorus (valency 3) and C is chlorine (valency 1), so they form PCl3, i.e. BC3.
2. Answer: (d)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 24: Paragraph: The electronic configurations of three elements A, B and C are A: 1s2 2s2 2p6; B: 1s2 2s2 2p6 3s2 3p3; C: 1s2 2s2 2p6 3s2 3p5. The bond between B and C will be:
(a) Ionic
(b) Covalent
(c) Hydrogen
(d) Coordinate
Show Answer & Explanation

Answer: (b) Covalent

Explanation:
1. Phosphorus and chlorine are both non-metals with a small electronegativity difference, so they share electrons in P–Cl covalent bonds.
2. Answer: (b)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 25: Which of the following orders of energies of the molecular orbitals of N2 is correct?
(a) π(2py) < σ(2pz) < π*(2px) ≈ π*(2py)
(b) π(2py) > σ(2pz) > π*(2px) ≈ π*(2py)
(c) π(2py) < σ(2pz) > π*(2px) ≈ π*(2py)
(d) π(2py) > σ(2pz) < π*(2px) ≈ π*(2py)
Show Answer & Explanation

Answer: (a) π(2py) < σ(2pz) < π*(2px) ≈ π*(2py)

Explanation:
1. For N2 (14 electrons or fewer) the π2p orbitals lie below σ2pz because of s–p mixing.
2. The antibonding π*2p orbitals lie above σ2pz.
3. Answer: (a)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

Question 26: Formation of PH4+ is difficult compared with NH4+ because:
(a) The lone pair of phosphorus is delocalised
(b) The lone pair of phosphorus resides in an almost pure p-orbital
(c) The lone pair of phosphorus resides in an sp3 orbital
(d) The lone pair of phosphorus resides in an almost pure s-orbital
Show Answer & Explanation

Answer: (d) The lone pair of phosphorus resides in an almost pure s-orbital

Explanation:
1. In PH3 the bond angle is about 93.5°, which shows that bonding uses nearly pure p-orbitals.
2. The lone pair therefore stays in an almost pure s-orbital, which is held close to the nucleus and is poor at donating.
3. Answer: (d)

Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.

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Chemical Bonding and Molecular Structure Objective Questions & Solutions for JEE Chemistry

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