JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 20

Answer: (b) 4

Explanation:
1. Each water molecule has two H atoms that can donate hydrogen bonds.
2. The oxygen has two lone pairs that can accept two hydrogen bonds.
3. So the maximum is four.

Teacher's Note:
1. Two donated plus two accepted.
2. This gives the tetrahedral network in ice.

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

Explanation:
1. \( \ce{BF3} \) has three bond pairs and no lone pair, so it is trigonal planar.
2. \( \ce{NF3} \), \( \ce{NCl3} \) and \( \ce{PH3} \) have a lone pair and are pyramidal.

Teacher's Note:
1. Lone pair on the central atom makes a pyramid.
2. No lone pair gives a plane.

Answer: (b) \( \ce{SO2} \)

Explanation:
1. \( \ce{SO2} \) has sulphur with two bond pairs and one lone pair, so it is \( sp^2 \).
2. \( \ce{CO2} \), \( \ce{N2O} \) and CO have central atoms that are sp.

Teacher's Note:
1. Steric number 3 gives \( sp^2 \).
2. Linear molecules are sp.

Answer: (d) \( \ce{CO} < \ce{CO2} < \ce{CO3^2-} \)

Explanation:
1. The C-O bond order is 3 in CO (113 pm), 2 in \( \ce{CO2} \) (116 pm) and 1.33 in \( \ce{CO3^2-} \) (129 pm).
2. The bond length increases as the bond order falls: CO < \( \ce{CO2} \) < \( \ce{CO3^2-} \).

Teacher's Note:
1. Higher bond order, shorter bond.
2. Carbonate has resonance giving partial double bonds.

Answer: (a) \( \ce{CH4} < \ce{NF3} < \ce{NH3} < \ce{H2O} \)

Explanation:
1. \( \ce{CH4} \) is symmetrical, so its dipole moment is zero.
2. \( \ce{NF3} \) is 0.23 D, \( \ce{NH3} \) is 1.47 D and \( \ce{H2O} \) is 1.85 D.
3. The order is \( \ce{CH4} < \ce{NF3} < \ce{NH3} < \ce{H2O} \).

Teacher's Note:
1. In NF3 the lone pair moment opposes the bond moments.
2. In NH3 they add.

Answer: (b) \( sp^3 \), \( dsp^2 \), \( sp^3d \), \( sp^2 \)

Explanation:
1. \( \ce{NH3} \) has four electron pairs and is \( sp^3 \).
2. \( \ce{[PtCl4]^2-} \) is square planar, \( dsp^2 \).
3. \( \ce{PCl5} \) has five pairs, \( sp^3d \), and \( \ce{BCl3} \) has three, \( sp^2 \).

Teacher's Note:
1. Count electron pairs.
2. Square planar complexes use \( dsp^2 \).

Answer: (d) 0

Explanation:
1. \( \ce{S3O9} \) is a cyclic trimer of \( \ce{SO3} \) with a ring of alternating S and O atoms.
2. Each sulphur is linked to two ring oxygens and two terminal oxygens.
3. There are no S-S bonds.

Teacher's Note:
1. The ring is made of S-O-S links.
2. Sulphur is in the +6 state, so no S-S bond is needed.

Answer: (a) N: tetrahedral, \( sp^3 \); B: tetrahedral, \( sp^3 \)

Explanation:
1. In \( \ce{H3N->BF3} \) the nitrogen has four bonds (three N-H and one N-B), so it is tetrahedral and \( sp^3 \).
2. Boron has four bonds (three B-F and one B-N), so it is also tetrahedral and \( sp^3 \).

Teacher's Note:
1. The lone pair of N is used in the bond, so there is no lone pair left.
2. Both atoms reach four bonds.

Answer: (a) the molecular plane

Explanation:
1. The pi bond comes from sideways overlap of unhybridised p orbitals perpendicular to the molecular plane.
2. The electron density is above and below the plane.
3. The plane of the molecule itself is the nodal plane.

Teacher's Note:
1. Pi electron density has zero value in the plane of the atoms.
2. The sigma bond lies in the nodal plane.

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

Explanation:
1. The dipole moments are \( \ce{CH3Cl} \) 1.87 D, \( \ce{CH2Cl2} \) 1.60 D, \( \ce{CHCl3} \) 1.04 D and \( \ce{CCl4} \) zero.
2. The partial cancellation of C-Cl bond moments increases as more chlorines are added.

Teacher's Note:
1. Vector addition of bond moments in a tetrahedron.
2. CCl4 cancels completely.

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

Explanation:
1. \( \ce{NO3-} \) and \( \ce{CO3^2-} \) both have 32 electrons and are trigonal planar.
2. \( \ce{SO3} \) (40 electrons) is planar but not isoelectronic, and \( \ce{ClO3-} \) (42 electrons) is pyramidal.

Teacher's Note:
1. Isoelectronic needs equal total electrons.
2. Isostructural needs the same shape.

Answer: (b) 1

Explanation:
1. Xe has 8 valence electrons.
2. It forms bonds with four F atoms and one O (as Xe=O), using 6 electrons in the usual counting.
3. One lone pair remains on xenon.

Teacher's Note:
1. Lone pairs on Xe = (8 - number of electrons used in bonds) / 2.
2. The shape is square pyramidal.

Answer: (c) \( \ce{H2S2O8} \)

Explanation:
1. \( \ce{H2S2O8} \) is peroxodisulphuric acid, with a peroxide bridge (-O-O-) between two sulphate groups.
2. The others contain S-S links or S chains but no O-O bond.

Teacher's Note:
1. Names with "peroxo" indicate O-O bonds.
2. Marshall's acid is peroxodisulphuric acid.

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

Explanation:
1. \( \ce{O2+} \) has 15 electrons with one unpaired electron in an antibonding pi orbital, so it is paramagnetic.
2. Removing an antibonding electron raises the bond order from 2 to 2.5.

Teacher's Note:
1. Remove an electron from an antibonding orbital and the bond order goes up.
2. O2+ still has an unpaired electron.

Answer: (c) pyramidal

Explanation:
1. Xe has three bond pairs and one lone pair, giving four electron pairs.
2. The shape is trigonal pyramidal, like \( \ce{NH3} \) or \( \ce{ClO3-} \).

Teacher's Note:
1. XeO3 is isostructural with \( \ce{IO3-} \) and \( \ce{ClO3-} \).
2. AX3E means pyramidal.

Answer: (b) sigma-p

Explanation:
1. Hyperconjugation is the delocalisation of electrons of a C-H sigma bond into an adjacent empty or partially filled p orbital or pi orbital.
2. So it is a sigma-p interaction.

Teacher's Note:
1. It stabilises alkenes and carbocations with more alkyl groups.
2. Also called no-bond resonance.

Answer: (d) \( \ce{OSF2} \)

Explanation:
1. \( \ce{OSF2} \) has three bond pairs and one lone pair on sulphur, so it is pyramidal.
2. \( \ce{SO3} \) and \( \ce{SiO3^2-} \) are trigonal planar, and \( \ce{BrF3} \) is T-shaped.

Teacher's Note:
1. AX3E is pyramidal.
2. Count the lone pairs.

Answer: (b) sp and \( sp^2 \)

Explanation:
1. Allene is \( \ce{CH2=C=CH2} \).
2. The central carbon has two double bonds, so it is sp.
3. The two terminal carbons each have one double bond, so they are \( sp^2 \).

Teacher's Note:
1. Cumulated double bonds make the central carbon sp.
2. The two CH2 planes are perpendicular.

Answer: (a) (ii) and (iv)

Explanation:
1. 1,4-Dichlorobenzene and trans-1,2-dichloroethene are symmetrical, so their bond dipoles cancel.
2. The cis isomer has both C-Cl dipoles on one side.
3. The pentene derivative has two different substituents on the double bond, so the dipoles do not cancel.

Teacher's Note:
1. Check symmetry.
2. Dissimilar groups on the double bond remove the cancellation.

Answer: (a) (ii), (iii) and (iv)

Explanation:
1. \( \ce{CS2} \), \( \ce{NO2+} \) and \( \ce{NCO-} \) are all 16 valence electron species with two bond pairs and no lone pair on the central atom, so they are linear.
2. \( \ce{SnCl2} \) has a lone pair on tin and is bent.

Teacher's Note:
1. 16 valence electron triatomics are linear.
2. Isoelectronic with \( \ce{CO2} \).

Answer: (a) (ii) and (iii)

Explanation:
1. Neopentane and 2,2,3,3-tetramethylbutane are highly symmetrical, so they have zero dipole moment.
2. Trans-2-pentene has two different alkyl groups, giving a small net dipole.
3. Cis-3-hexene has both ethyl groups on the same side and a net dipole.

Teacher's Note:
1. Alkyl groups are weakly electron donating.
2. Symmetrical structures have no dipole.

Answer: (a) (i), (iii) and (iv)

Explanation:
1. Resonance makes the two O-O bonds equal (128 pm).
2. Ozone is bent and diamagnetic.
3. Its decomposition to \( \ce{O2} \) is exothermic, because ozone is less stable than oxygen, so (ii) is false.

Teacher's Note:
1. Ozone is less stable than \( \ce{O2} \).
2. Its formation from oxygen is endothermic.

Answer: (a) (i) and (iii)

Explanation:
1. \( \ce{C2^2-} \) (10 valence electrons) is like \( \ce{N2} \) with no unpaired electrons, so it is diamagnetic.
2. \( \ce{N2+} \) and \( \ce{N2-} \) both have bond order 2.5.
3. \( \ce{O2^2+} \) has bond order 3, which gives a shorter bond than \( \ce{O2} \), and \( \ce{He2+} \) is more stable than separate atoms.

Teacher's Note:
1. Count electrons and fill MOs.
2. Isoelectronic reasoning helps.

Answer: (a) (ii) and (iii)

Explanation:
1. \( \ce{ClF3} \) has 2 lone pairs on Cl and \( \ce{XeF4} \) has 2 lone pairs on Xe.
2. \( \ce{BrF5} \) and \( \ce{SF4} \) each have just 1 lone pair.

Teacher's Note:
1. Lone pairs = (valence electrons - bonds) / 2.
2. ClF3 is T-shaped and XeF4 is square planar.