JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 10

Practice MCQs for JEE Chemistry Chemical Bonding and Molecular Structure

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Multiple Choice Questions

Question 1: Which of the following is a molecular solid?
(a) Solid \( \ce{SO2} \)
(b) SiC
(c) C (graphite)
(d) ZnS
Show Answer & Explanation

Answer: (a) Solid \( \ce{SO2} \)

Explanation:
1. Solid SO2 consists of discrete molecules.

Teacher's Note:
1. SiC and graphite are covalent network solids; ZnS is ionic.

Question 2: The geometry and number of lone pairs on \( \ce{ClF2^-} \) and \( \ce{ClF4^-} \) respectively are
(a) Square planar, 2 and linear, 3
(b) Square planar, 3 and linear, 2
(c) Linear, 3 and square planar, 2
(d) Tetrahedral, 2 and linear, 3
Show Answer & Explanation

Answer: (c) Linear, 3 and square planar, 2

Explanation:
1. ClF2- has 2 bond pairs and 3 lone pairs: linear. ClF4- has 4 bond pairs and 2 lone pairs: square planar.

Teacher's Note:
1. Add up Cl electrons and the charge.

Question 3: The bond order of which pair is different?
(a) \( \ce{F2^-} \), \( \ce{O2^2-} \)
(b) \( \ce{N2^-} \), \( \ce{O2+} \)
(c) \( \ce{C2} \), \( \ce{B2^2-} \)
(d) \( \ce{N2} \), \( \ce{O2^2+} \)
Show Answer & Explanation

Answer: (a) \( \ce{F2^-} \), \( \ce{O2^2-} \)

Explanation:
1. F2^- has BO 0.5; O2(2-) has BO 1, so they differ. In the other pairs the orders are equal (2.5, 2, 3).

Teacher's Note:
1. Pairs in the other options are isoelectronic.

Question 4: Which is the correct order of bond angle?
(a) \( \ce{SO2} \) > \( \ce{NH3} \)
(b) \( \ce{NH3} \) < \( \ce{H2O} \)
(c) \( \ce{SO2} \) < \( \ce{H2O} \)
(d) \( \ce{H2S} \) > \( \ce{NH3} \)
Show Answer & Explanation

Answer: (a) \( \ce{SO2} \) > \( \ce{NH3} \)

Explanation:
1. SO2 about 119, NH3 107, H2O 104.5, H2S 92.

Teacher's Note:
1. Bond angle falls with a more electronegative terminal/central atom effects.

Question 5: Which two species have the same shape but different hybridisation?
(a) \( \ce{XeF2} \) and \( \ce{ICl2^-} \)
(b) \( \ce{NO2+} \) and \( \ce{BeCl2} \)
(c) \( \ce{SO2} \) and \( \ce{NH2^-} \)
(d) \( \ce{NO2+} \) and \( \ce{SO2} \)
Show Answer & Explanation

Answer: (c) \( \ce{SO2} \) and \( \ce{NH2^-} \)

Explanation:
1. SO2 (sp2) and NH2- (sp3) are both bent, but different in hybridisation.

Teacher's Note:
1. Pairs (a) and (b) share both shape and hybridisation.

Question 6: Which of the following has the maximum percentage of s-character in the N-H bond?
(a) \( \ce{N2H2} \)
(b) \( \ce{NH4+} \)
(c) \( \ce{NH3} \)
(d) \( \ce{N2H4} \)
Show Answer & Explanation

Answer: (a) \( \ce{N2H2} \)

Explanation:
1. N in N2H2 is sp2 (33 percent s); others are sp3 (25 percent).

Teacher's Note:
1. Larger s-character gives a higher percentage.

Question 7: Which of the following are isoelectronic and isostructural?
(a) \( \ce{NO3^-} \), \( \ce{CO3^2-} \)
(b) \( \ce{SO3} \), \( \ce{NO3^-} \)
(c) \( \ce{ClO3^-} \), \( \ce{CO3^2-} \)
(d) \( \ce{CO3^2-} \), \( \ce{SO3} \)
Show Answer & Explanation

Answer: (a) \( \ce{NO3^-} \), \( \ce{CO3^2-} \)

Explanation:
1. NO3- and CO3(2-) both have 32 electrons and are planar. The others differ in electrons or shape.

Teacher's Note:
1. Both are sp2 with no lone pair on the central atom.

Question 8: According to molecular orbital theory, which statement regarding the magnetic character and bond order of \( \ce{O2+} \) is correct?
(a) Paramagnetic and bond order less than \( \ce{O2} \)
(b) Paramagnetic and bond order greater than \( \ce{O2} \)
(c) Diamagnetic and bond order less than \( \ce{O2} \)
(d) Diamagnetic and bond order greater than \( \ce{O2} \)
Show Answer & Explanation

Answer: (b) Paramagnetic and bond order greater than \( \ce{O2} \)

Explanation:
1. O2+ has one unpaired electron in pi* (paramagnetic), with bond order 2.5, greater than 2.

Teacher's Note:
1. Removing an antibonding electron raises the order.

Question 9: The bond order in NO is 2.5 while that in \( \ce{NO+} \) is 3. Which statement is true for these species?
(a) Bond length in \( \ce{NO+} \) is equal to that in NO
(b) Bond length in NO is greater than that in \( \ce{NO+} \)
(c) Bond length in \( \ce{NO+} \) is greater than that in NO
(d) Bond length is unpredictable
Show Answer & Explanation

Answer: (b) Bond length in NO is greater than that in \( \ce{NO+} \)

Explanation:
1. Higher bond order means a shorter bond, so NO (2.5) is longer than NO+ (3).

Teacher's Note:
1. Remove the pi* electron to get NO+.

Question 10: The boiling point of methanol is greater than that of methylthiol because
(a) There is intramolecular hydrogen bonding in methanol and intermolecular in methylthiol
(b) There is intermolecular hydrogen bonding in methanol and none in methylthiol
(c) There is no hydrogen bonding in methanol and intermolecular in methylthiol
(d) There is intramolecular hydrogen bonding in methanol and none in methylthiol
Show Answer & Explanation

Answer: (b) There is intermolecular hydrogen bonding in methanol and none in methylthiol

Explanation:
1. O-H...O bonds link methanol molecules; S is too large and less electronegative for H bonding in CH3SH.

Teacher's Note:
1. Hydrogen bonding needs F, O or N.

Question 11: In silicon dioxide
(a) Each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms
(b) Each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bonded to two silicon atoms
(c) Silicon atom is bonded to two oxygen atoms
(d) There are double bonds between silicon and oxygen atoms
Show Answer & Explanation

Answer: (a) Each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms

Explanation:
1. SiO2 is a 3D network of SiO4 tetrahedra with Si-O-Si bridges.

Teacher's Note:
1. Formula SiO2 reflects the 1:2 ratio.

Question 12: The number and type of bonds between two carbon atoms in calcium carbide are
(a) One sigma, one pi
(b) One sigma, two pi
(c) Two sigma, one pi
(d) Two sigma, two pi
Show Answer & Explanation

Answer: (b) One sigma, two pi

Explanation:
1. CaC2 contains the acetylide ion C2(2-) with a C-triple bond: one sigma and two pi.

Teacher's Note:
1. Isoelectronic with N2.

Question 13: The hybridisation of the oxygen atom in an alcohol molecule is
(a) sp3
(b) sp
(c) sp2
(d) sp3d
Show Answer & Explanation

Answer: (a) sp3

Explanation:
1. O has two sigma bonds and two lone pairs.

Teacher's Note:
1. Same as in water.

Question 14: Which of the following hydrogen bonds is the strongest?
(a) F-H...F
(b) O-H...O
(c) O-H...F
(d) O-H...N
Show Answer & Explanation

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

Explanation:
1. The F-H...F hydrogen bond is the strongest because of the highest electronegativity of F.

Teacher's Note:
1. Strength increases with the electronegativity of the atoms.

Question 15: Which of the following species exhibits diamagnetic behaviour?
(a) \( \ce{O2+} \)
(b) \( \ce{O2} \)
(c) NO
(d) \( \ce{O2^2-} \)
Show Answer & Explanation

Answer: (d) \( \ce{O2^2-} \)

Explanation:
1. O2(2-) (18 electrons) has all electrons paired. O2+, O2 and NO have unpaired electrons.

Teacher's Note:
1. Peroxide ion is diamagnetic.

Question 16: The charge/size ratio of a cation determines its polarising power. Which sequence represents the increasing order of the polarising power of \( \ce{K+} \), \( \ce{Ca^2+} \), \( \ce{Mg^2+} \), \( \ce{Be^2+} \)?
(a) \( \ce{Be^2+} \) < \( \ce{K+} \) < \( \ce{Ca^2+} \) < \( \ce{Mg^2+} \)
(b) \( \ce{K+} \) < \( \ce{Ca^2+} \) < \( \ce{Mg^2+} \) < \( \ce{Be^2+} \)
(c) \( \ce{Ca^2+} \) < \( \ce{Mg^2+} \) < \( \ce{Be^2+} \) < \( \ce{K+} \)
(d) \( \ce{Mg^2+} \) < \( \ce{Be^2+} \) < \( \ce{K+} \) < \( \ce{Ca^2+} \)
Show Answer & Explanation

Answer: (b) \( \ce{K+} \) < \( \ce{Ca^2+} \) < \( \ce{Mg^2+} \) < \( \ce{Be^2+} \)

Explanation:
1. Charge/size: K+ (small) < Ca2+ < Mg2+ < Be2+ (largest).

Teacher's Note:
1. Be2+ has the highest charge density.

Question 17: Which of the following constitutes a group of isoelectronic species?
(a) \( \ce{C2^2-} \), \( \ce{O2^2-} \), CO, NO
(b) \( \ce{NO+} \), \( \ce{C2^2-} \), \( \ce{CN^-} \), \( \ce{N2} \)
(c) \( \ce{CN^-} \), \( \ce{N2} \), \( \ce{O2^2-} \), \( \ce{C2^2-} \)
(d) \( \ce{N2} \), \( \ce{O2^-} \), \( \ce{NO+} \), CO
Show Answer & Explanation

Answer: (b) \( \ce{NO+} \), \( \ce{C2^2-} \), \( \ce{CN^-} \), \( \ce{N2} \)

Explanation:
1. NO+, C2(2-), CN- and N2 each have 14 electrons.

Teacher's Note:
1. Count the total electrons.

Question 18: Among the following, the molecule with the highest dipole moment is
(a) \( \ce{CH3Cl} \)
(b) \( \ce{CH2Cl2} \)
(c) \( \ce{CHCl3} \)
(d) \( \ce{CCl4} \)
Show Answer & Explanation

Answer: (a) \( \ce{CH3Cl} \)

Explanation:
1. Dipole moments: CH3Cl 1.9 D; CH2Cl2 1.6 D; CHCl3 1.0 D; CCl4 0.

Teacher's Note:
1. Bond dipoles partly cancel with more Cl atoms.

Question 19: For a diatomic molecule the dipole moment is 1.2 D and the bond length is 1.0 A. The fractional charge on each atom is (1 D = 3.336 x 10^-30 C m)
(a) 12 percent
(b) 18 percent
(c) 25 percent
(d) 29 percent
Show Answer & Explanation

Answer: (c) 25 percent

Explanation:
1. Charge = mu/d = (1.2 x 3.336e-30)/(1.0e-10) = 4.0e-20 C.
2. Fraction of e = 4.0e-20/1.602e-19 = 0.25 = 25 percent.

Teacher's Note:
1. Dipole = charge x distance.

Question 20: Which of the following molecules is expected to exhibit diamagnetic behaviour?
(a) \( \ce{C2} \)
(b) \( \ce{N2^-} \)
(c) \( \ce{O2} \)
(d) \( \ce{S2} \)
Show Answer & Explanation

Answer: (a) \( \ce{C2} \)

Explanation:
1. C2 has all 12 valence electrons paired in pi bonding MOs.

Teacher's Note:
1. N2- has an odd electron and O2, S2 have two unpaired electrons.

Question 21: The boiling point of ICl is nearly 40 degrees higher than that of \( \ce{Br2} \), although the two substances have nearly the same relative molecular mass. This is because
(a) ICl is an ionic compound
(b) The I-Cl bond is stronger than the Br-Br bond
(c) ICl is a polar covalent molecule while \( \ce{Br2} \) is non-polar
(d) The ionisation potential of iodine is less than that of bromine
Show Answer & Explanation

Answer: (c) ICl is a polar covalent molecule while \( \ce{Br2} \) is non-polar

Explanation:
1. Dipole-dipole forces in polar ICl add to the dispersion forces.

Teacher's Note:
1. Similar masses give similar dispersion forces.

Question 22: Nitrogen does not form \( \ce{NF5} \) because
(a) Nitrogen is a member of the V group
(b) It contains no empty d-orbital
(c) The bond energy of N#N is very high
(d) The inert pair effect exists
Show Answer & Explanation

Answer: (b) It contains no empty d-orbital

Explanation:
1. N is in period 2 and cannot expand its octet.

Teacher's Note:
1. Phosphorus forms PF5.

Question 23: \( \ce{CO2} \) is a gas while \( \ce{SiO2} \) is a solid, but both are
(a) Covalent containing pi-bonds
(b) Molecules having p-pi and d-pi bonding
(c) Acidic
(d) Discrete molecules
Show Answer & Explanation

Answer: (c) Acidic

Explanation:
1. Both are acidic oxides, forming H2CO3 and silicates with bases. CO2 is molecular and SiO2 a network solid.

Teacher's Note:
1. Non-metal oxides are acidic.

Question 24: Which trihalide ion is unknown due to the absence of vacant d-orbitals?
(a) \( \ce{F3^-} \)
(b) \( \ce{Br3^-} \)
(c) \( \ce{I3^-} \)
(d) \( \ce{Cl3^-} \)
Show Answer & Explanation

Answer: (a) \( \ce{F3^-} \)

Explanation:
1. F has no d-orbitals so it cannot accommodate the extra electron pair for F3-.

Teacher's Note:
1. I3-, Br3- and Cl3- use d-orbitals.

Question 25: Experiment shows that \( \ce{H2O} \) has a dipole moment whereas \( \ce{CO2} \) has not. The structures which best illustrate these facts are
(a) O-C-O linear and H-O-H linear
(b) CO2 bent and H2O linear
(c) CO2 linear (O=C=O) and H2O bent
(d) CO2 with C=O and H2O linear
Show Answer & Explanation

Answer: (c) CO2 linear (O=C=O) and H2O bent

Explanation:
1. CO2 is linear (zero net dipole) and H2O is bent (net dipole).

Teacher's Note:
1. Shape determines the vector sum.

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