JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 49

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

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Practice Chemical Bonding and Molecular Structure MCQs for JEE Chemistry

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Multiple Choice Questions

Question 1: Which of the following species is iso-structural with the N3- ion?
(a) I3+
(b) I3-
(c) NH2-
(d) HCO2-
Show Answer & Explanation

Answer: (b) I3-

Explanation:
1. N3- is linear (16 valence electrons, like CO2).
2. I3- is linear (sp3d with three lone pairs on the central atom), while I3+ and NH2- are bent and HCO2- is planar triangular.
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 2: Which of the following pairs of species is iso-electronic?
(a) CN- and NO+
(b) N2- and N2+
(c) H2+ and H2-
(d) CO and NO-
Show Answer & Explanation

Answer: (a) CN- and NO+

Explanation:
1. CN- and NO+ both have 14 electrons.
2. N2- has 15 and N2+ 13; H2+ has 1 and H2- 3; CO has 14 and NO- 16.
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 3: What are the C–O bond order and the formal charge on each oxygen atom of HCO2- (formate ion), respectively?
(a) 1.5, –0.5
(b) 2, –0.5
(c) 1.33, –1.5
(d) 1.5, –1.33
Show Answer & Explanation

Answer: (a) 1.5, –0.5

Explanation:
1. Resonance makes the two C–O bonds equal, each being half a double bond: bond order = (1 + 2)/2 = 1.5.
2. The negative charge is shared by the two oxygen atoms, so each carries –0.5.
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 4: Which of the following is an incorrect order of bond angle?
(a) NH3 < NF3 < NCl3
(b) OF2 < OH2 < OCl2
(c) SF2 < SCl2 < SBr2 < SI2
(d) ClO2 > ClO2- > OCl2
Show Answer & Explanation

Answer: (a) NH3 < NF3 < NCl3

Explanation:
1. Actual angles: NH3 107°, NF3 102°, NCl3 107.5°.
2. NF3 has the smallest angle, so the order NH3 < NF3 < NCl3 is incorrect; the other orders follow electronegativity and size trends.
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 5: Which of the following molecules is non-polar and planar?
(a) XeF4
(b) NH2-
(c) PF3Cl2
(d) PCl3F2
Show Answer & Explanation

Answer: (a) XeF4

Explanation:
1. XeF4 is square planar and symmetrical, so its dipole moment is zero.
2. NH2- is bent and polar, while PF3Cl2 and PCl3F2 are trigonal bipyramidal and not planar.
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 6: Which of the following is a covalent solid?
(a) Solid CO2
(b) SiO2
(c) Diamond
(d) SiO2 and diamond both
Show Answer & Explanation

Answer: (d) SiO2 and diamond both

Explanation:
1. SiO2 and diamond are giant three-dimensional networks held by covalent bonds.
2. Solid CO2 is a molecular solid held by weak van der Waals forces.
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 7: The M–O bond orders in some ions are: ClO4- = 1/4, PO43- = 3/4, NO2- = 2/3, CO32- = 5/3. Which of these matches is correct?
(a) Only CO32-
(b) ClO4- and PO43-
(c) NO2- and CO32-
(d) None of the matches is correct
Show Answer & Explanation

Answer: (d) None of the matches is correct

Explanation:
1. Correct bond orders: ClO4- = 7/4, PO43- = 5/4, NO2- = 3/2 and CO32- = 4/3.
2. None of the listed values agrees, so none of the matches is correct.
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 groups contains iso-structural species?
(a) XeF2, ICl2-, ClF3
(b) ClF3, PCl3, NCl3
(c) CO2, XeF2, I3-
(d) PCl5, XeOF2, ICl5
Show Answer & Explanation

Answer: (c) CO2, XeF2, I3-

Explanation:
1. CO2, XeF2 and I3- are all linear.
2. ClF3 is T-shaped, PCl3 and NCl3 are pyramidal, PCl5 is trigonal bipyramidal and ICl5 square pyramidal, so the other groups are mixed.
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 9: Select the correct order out of the given options.
(a) BeCO3 < BaCO3 : covalent character
(b) BeO > SrO : lattice energy
(c) Be2+ < Li+ : hydration energy
(d) Be2+(aq) > Li+(aq) : ionic mobility
Show Answer & Explanation

Answer: (b) BeO > SrO : lattice energy

Explanation:
1. BeO has a small cation, so its lattice energy is higher than that of SrO.
2. Be2+ is hydrated far more strongly than Li+ and therefore moves slower; BeCO3 is the most covalent carbonate. The other three orders are reversed.
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 10: Which combination of true (T) and false (F) is correct for the following statements about PCl5? (a) Hybridisation is sp3d and the structure is trigonal pyramidal. (b) The P–Cl bond angle is 90°, the same for all P–Cl bonds. (c) The axial P–Cl bond is longer than the equatorial one. (d) PCl5 has zero dipole moment.
(a) TFTT
(b) FTTT
(c) FFTT
(d) TFFT
Show Answer & Explanation

Answer: (c) FFTT

Explanation:
1. PCl5 is sp3d hybridised and trigonal bipyramidal, not pyramidal, so (a) is false.
2. It has both 90° and 120° angles, so (b) is false. Axial bonds are longer (repulsion from three equatorial pairs) and the molecule is symmetrical with zero dipole, so (c) and (d) are true.
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 11: If AB4n species are tetrahedral, which of the following is incorrectly matched (A, B, n)?
(a) Xe, O, zero
(b) Se, F, zero
(c) P, O, –3
(d) N, H, +1
Show Answer & Explanation

Answer: (b) Se, F, zero

Explanation:
1. XeO4, PO43- and NH4+ are all tetrahedral.
2. SeF4 has a lone pair on Se and is see-saw shaped, so the match (Se, F, zero) is incorrect.
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 12: The species having no pπ–pπ bond but with bond order equal to that of O2- (1.5) is:
(a) ClO3-
(b) PO43-
(c) SO42-
(d) XeO3
Show Answer & Explanation

Answer: (c) SO42-

Explanation:
1. In SO42- there are two dπ–pπ bonds shared among four S–O bonds, so each bond has order 6/4 = 1.5 and no pπ–pπ bond.
2. Bond orders of the others: ClO3- 5/3, PO43- 5/4, XeO3 2.
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: How many π-bonds does C2 have (molecular orbital description)?
(a) 1
(b) 2
(c) 0
(d) 3
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. C2 has 12 electrons, filling σ2s2 σ*2s2 π2p4.
2. Both bonds come from the π2p orbitals, so there are 2 π-bonds and no σ-bond from 2p.
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 14: Which of the following is not true about the H2O molecule?
(a) The molecule has μ = 0
(b) The molecule can act as a base
(c) It shows an abnormally high boiling point compared with the hydrides of other elements of the oxygen group
(d) The molecule has a bent shape
Show Answer & Explanation

Answer: (a) The molecule has μ = 0

Explanation:
1. H2O is bent, so the bond dipoles do not cancel and μ ≠ 0 (1.85 D).
2. The other three statements are true.
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 15: Match the changes in Column I with the effects in Column II: (a) N2 → N2+, (b) N2 → N2-, (c) O2 → O2+, (d) O2 → O2-; (p) bond order increases, (q) bond order decreases, (r) paramagnetism increases, (s) paramagnetism decreases, (t) no change in bond order.
(a) a-(q, r), b-(q, r), c-(p, s), d-(q, s)
(b) a-(q, s), b-(q, s), c-(p, s), d-(q, r)
(c) a-(p, q), b-(q, s), c-(p, r), d-(q, t)
(d) a-(p, s), b-(q, p), c-(q, t), d-(q, t)
Show Answer & Explanation

Answer: (a) a-(q, r), b-(q, r), c-(p, s), d-(q, s)

Explanation:
1. N2 (BO 3, diamagnetic) → N2+ and N2- (BO 2.5, one unpaired electron): bond order falls, paramagnetism rises.
2. O2 (BO 2, two unpaired) → O2+ (BO 2.5, one unpaired): p, s. O2 → O2- (BO 1.5, one unpaired): q, s.
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 16: Which of the following pairs have the same bond angle? (a) BF3, BCl3 (b) PO43-, SO42- (c) BF3, PF3 (d) NO2+, N2O (e) N3-, NO2
(a) a, b, d
(b) b, d
(c) b, c, d
(d) a, d, e
Show Answer & Explanation

Answer: (a) a, b, d

Explanation:
1. BF3 and BCl3 are both 120°, the two tetrahedral ions are both 109.5°, and NO2+ and N2O are both linear (180°).
2. BF3 (120°) and PF3 (about 98°) differ, and N3- (180°) and NO2 (about 134°) differ.
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: Which of the following is correct?
(a) PbS > ZnS (solubility)
(b) Li2CO3 > Na2CO3 (thermal stability)
(c) NaF > KF (lattice energy)
(d) BaSO4 > MgSO4 (solubility)
Show Answer & Explanation

Answer: (c) NaF > KF (lattice energy)

Explanation:
1. The smaller Na+ gives NaF a higher lattice energy than KF.
2. Li2CO3 is less stable thermally than Na2CO3, and BaSO4 is much less soluble than MgSO4; PbS is less soluble than ZnS.
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 18: Which among the following attractions is the strongest?
(a) HF ··· H2O (hydrogen bond)
(b) Na+ ··· H–Cl (ion–dipole)
(c) H2O ··· Cl2 (dipole–induced dipole)
(d) Cl–Cl ··· Cl–Cl (London forces)
Show Answer & Explanation

Answer: (b) Na+ ··· H–Cl (ion–dipole)

Explanation:
1. Ion–dipole attraction is stronger than hydrogen bonding, which is stronger than dipole–induced dipole and London forces.
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 19: Among the following, the pair in which the two species are not iso-structural is:
(a) IO3- and NH3
(b) BH4- and NH4+
(c) PF6- and SF6
(d) SiF4 and SF4
Show Answer & Explanation

Answer: (d) SiF4 and SF4

Explanation:
1. SiF4 is tetrahedral while SF4 is see-saw (one lone pair).
2. The other pairs are pyramidal, tetrahedral and octahedral respectively in both members.
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: Which of the following is a correct statement? (a) AlCl3 is conducting in the fused state. (b) The mobility of the Li+ ion in water is greater than that of the Cs+ ion. (c) MCl2 is more volatile than MCl4. (d) BeSO4 is more soluble in water than BaSO4.
(a) (a) and (b)
(b) (b), (c) and (d)
(c) (b) and (d)
(d) Only (d)
Show Answer & Explanation

Answer: (d) Only (d)

Explanation:
1. Only (d) is true: BeSO4 has a very high hydration energy and dissolves readily, while BaSO4 is nearly insoluble.
2. Fused AlCl3 is covalent and non-conducting, hydrated Li+ is larger and slower than Cs+, and MCl4 is more covalent and so more volatile than MCl2.
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 21: Assertion: p-hydroxybenzoic acid has a lower boiling point than o-hydroxybenzoic acid. Reason: o-hydroxybenzoic acid has intermolecular hydrogen bonding. Choose the correct option.
(a) Both are true and the reason explains the assertion
(b) Both are true but the reason does not explain the assertion
(c) The assertion is true but the reason is false
(d) Both the assertion and the reason are false
Show Answer & Explanation

Answer: (d) Both the assertion and the reason are false

Explanation:
1. The para isomer has intermolecular hydrogen bonding and so boils higher, so the assertion is false.
2. The ortho isomer has intramolecular hydrogen bonding, so the reason is also 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 22: The total number of sp-hybridised carbon atoms in the hydrocarbon H3C–C≡C–CH=CH2 is:
(a) 5
(b) 4
(c) 2
(d) 1
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. Only the two carbon atoms of the triple bond are sp hybridised.
2. The CH=CH2 carbons are sp2 and the CH3 carbon is sp3.
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 23: Match Column I with Column II. (a) C2H2 (b) SO2 (c) I3- (d) NH4+; (P) sp3d (Q) sp3 (R) sp2 (S) sp.
(a) (a) S (b) P (c) R (d) Q
(b) (a) P (b) S (c) R (d) Q
(c) (a) S (b) R (c) P (d) Q
(d) (a) R (b) S (c) P (d) Q
Show Answer & Explanation

Answer: (c) (a) S (b) R (c) P (d) Q

Explanation:
1. C2H2 is sp, SO2 is sp2, I3- is sp3d and NH4+ is sp3.
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 24: Match the solids in Column I with examples in Column II. (a) Covalent (b) Molecular (c) Ionic (d) Metallic; (P) SiO2 (Q) CaO (R) CCl4 (S) Bronze.
(a) (a) P (b) Q (c) R (d) S
(b) (a) R (b) P (c) Q (d) S
(c) (a) S (b) P (c) Q (d) R
(d) (a) P (b) R (c) Q (d) S
Show Answer & Explanation

Answer: (d) (a) P (b) R (c) Q (d) S

Explanation:
1. SiO2 is a covalent network solid, CCl4 a molecular solid, CaO an ionic solid and bronze (an alloy) is metallic.
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 25: Match each compound with its number of σ and π bonds. (a) H2S2O3 (b) H2SO5 (c) H2S2O8 (d) H2S2O6; (P) 6σ and 2π (Q) 11σ and 4π (R) 9σ and 4π (S) 7σ and 2π.
(a) (a) S (b) P (c) Q (d) R
(b) (a) P (b) S (c) Q (d) R
(c) (a) P (b) Q (c) R (d) S
(d) (a) Q (b) S (c) P (d) R
Show Answer & Explanation

Answer: (b) (a) P (b) S (c) Q (d) R

Explanation:
1. H2S2O3: 6σ, 2π; H2SO5: 7σ, 2π (it has an O–O link); H2S2O8: 11σ, 4π; H2S2O6: 9σ, 4π (S–S bond).
2. This gives (a) P, (b) S, (c) Q, (d) R.
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 26: Match the species with their shapes. (a) XeO2F2 (b) XeF5- (c) I3- (d) XeF4; (P) linear (Q) square planar (R) see-saw (S) pentagonal planar.
(a) (a) R (b) S (c) P (d) Q
(b) (a) R (b) S (c) Q (d) P
(c) (a) P (b) S (c) Q (d) R
(d) (a) S (b) Q (c) P (d) R
Show Answer & Explanation

Answer: (a) (a) R (b) S (c) P (d) Q

Explanation:
1. XeO2F2 has one lone pair on a trigonal bipyramid: see-saw.
2. XeF5- is pentagonal planar, I3- is linear and XeF4 is square planar.
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.

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

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