CBSE Class 11 Chemistry Chemical Bonding and Molecular Structure MCQs Set 09

Answer: (d) linear, sp

Explanation:
1. Be has two valence electrons and forms two bonds, so there are 2 bond pairs and no lone pair (type AX₂).
2. Two electron pairs stay 180° apart, so the molecule is linear.
3. One 2s and one 2p orbital of Be mix to give two sp hybrid orbitals.

Teacher's Note:
1. Count the electron pairs on the central atom first.
2. Two pairs always mean sp and a linear shape.

Answer: (b) PCl₃

Explanation:
1. P in PCl₃ has 5 valence electrons, 3 of which form bonds, so it has 3 bond pairs and 1 lone pair (AX₃E).
2. The four pairs are arranged tetrahedrally, so the shape is trigonal pyramidal.
3. BF₃, CO₃²⁻ and SO₃ have three bond pairs and no lone pair, so they are trigonal planar.

Teacher's Note:
1. Three bond pairs plus one lone pair give a pyramid.
2. Three bond pairs with no lone pair give a flat triangle.

Answer: (c) H₂O

Explanation:
1. The bond angles are CH₄ 109.5°, NH₃ 107°, H₂O 104.5° and BF₃ 120°.
2. H₂O has two lone pairs, which repel the bond pairs the most, so its angle is the smallest.

Teacher's Note:
1. More lone pairs on the central atom reduce the bond angle.
2. BF₃ has no lone pair and is planar at 120°.

Answer: (b) square planar

Explanation:
1. Xe has 8 valence electrons. Pairs = (8 + 4) ÷ 2 = 6, of which 4 are bond pairs and 2 are lone pairs (AX₄E₂).
2. The six pairs form an octahedron, and the two lone pairs sit opposite each other to minimise repulsion.
3. The four F atoms therefore lie in one plane, giving a square planar shape.

Teacher's Note:
1. Lone pairs go opposite each other in an octahedral arrangement.
2. XeF₄ has bond angles of 90°.

Answer: (a) T-shaped

Explanation:
1. Cl has 7 valence electrons. Pairs = (7 + 3) ÷ 2 = 5, with 3 bond pairs and 2 lone pairs (AX₃E₂).
2. The five pairs form a trigonal bipyramid, with both lone pairs in the equatorial plane.
3. The three F atoms then give a T-shaped molecule.

Teacher's Note:
1. Lone pairs prefer the equatorial positions of a trigonal bipyramid.
2. AX₃E₂ is T-shaped; AX₄E is see-saw.

Answer: (d) 1, 0, 2

Explanation:
1. SF₄: S has 6 electrons, 4 are used in bonds, leaving 2 electrons, which is 1 lone pair.
2. CF₄: all four carbon electrons are used in bonds, so 0 lone pairs.
3. XeF₄: Xe has 8 electrons, 4 are used in bonds, leaving 4 electrons, which is 2 lone pairs.

Teacher's Note:
1. Lone pairs = (valence electrons − electrons used in bonds) ÷ 2.
2. SF₄ is see-saw, CF₄ tetrahedral and XeF₄ square planar.

Answer: (a) sp³d

Explanation:
1. Xe has 8 valence electrons. Two are used for two Xe–F bonds, so the central atom has 2 bond pairs and 3 lone pairs, 5 pairs in all.
2. Five electron pairs need five hybrid orbitals, which is sp³d.
3. The three lone pairs occupy the equatorial positions, so the molecule is linear.

Teacher's Note:
1. Total electron pairs (bond pairs + lone pairs) fix the hybridisation: 2 sp, 3 sp², 4 sp³, 5 sp³d, 6 sp³d².
2. Lone pairs count too.

Answer: (c) 4

Explanation:
1. Each carbon in ethene is sp² hybridised and each C–H bond is formed by overlap of an sp² orbital of C with the 1s orbital of H.
2. There are four C–H bonds, so four sp²–s σ bonds; the C–C σ bond is sp²–sp².

Teacher's Note:
1. Name the orbitals used for each bond.
2. Count only the C–H bonds for sp²–s overlap.

Answer: (b) 6 and 2

Explanation:
1. C–H bonds: 1 + 3 = 4 σ bonds.
2. C≡C: one σ and two π bonds. The C–CH₃ bond: one σ bond.
3. Total σ = 4 + 1 + 1 = 6 and total π = 2.

Teacher's Note:
1. A triple bond has one σ and two π bonds.
2. All single bonds are σ bonds.

Answer: (d) XeF₂ and CO₂

Explanation:
1. XeF₂ is AX₂E₃ and CO₂ is AX₂, and both are linear.
2. NH₃ is pyramidal but BF₃ is planar. SO₂ is bent but CO₂ is linear. ClF₃ is T-shaped but BF₃ is trigonal planar.

Teacher's Note:
1. Write the AXE type of each molecule before comparing.
2. Three lone pairs on the central atom of AX₂E₃ spread out in the equatorial plane, so the three atoms form a straight line.

Answer: (a) sp² and trigonal planar to sp³ and tetrahedral

Explanation:
1. B in BF₃ has three bond pairs, so it is sp² hybridised and trigonal planar, with an empty 2p orbital.
2. F⁻ donates a lone pair into this empty orbital, so BF₄⁻ has four bond pairs and boron becomes sp³ hybridised.
3. The shape changes from trigonal planar to tetrahedral.

Teacher's Note:
1. Count the electron pairs on B before and after.
2. BF₄⁻ is isoelectronic with CH₄ in its outer-shell arrangement.

Answer: (c) sp² and trigonal planar to sp³ and tetrahedral

Explanation:
1. In AlCl₃ aluminium has 3 bond pairs (sp², trigonal planar).
2. Cl⁻ donates a lone pair into the empty 3p orbital, so AlCl₄⁻ has 4 bond pairs (sp³, tetrahedral).

Teacher's Note:
1. Add the pair donated by Cl⁻ to the pairs around Al.
2. The same change happens for B in BF₃ → BF₄⁻.

Answer: (a) O₂

Explanation:
1. O₂ has two unpaired electrons in the π*2px and π*2py orbitals, so it is paramagnetic.
2. N₂, F₂ and C₂ have all their electrons paired, so they are diamagnetic.

Teacher's Note:
1. Write the molecular orbital configuration and look for singly filled orbitals.
2. Lewis structures cannot explain the paramagnetism of O₂.

Answer: (b) O₂ < O₂⁻ < O₂²⁻

Explanation:
1. The bond orders are O₂ = 2, O₂⁻ = 1.5 and O₂²⁻ = 1.
2. Bond length decreases as bond order increases, so O₂ is the shortest and O₂²⁻ the longest.

Teacher's Note:
1. Calculate bond orders from the molecular orbital configurations.
2. Add electrons to the π* orbitals to lower the bond order.

Answer: (d) 2.5

Explanation:
1. The electron is removed from the bonding σ2pz orbital, which has the highest energy among the filled orbitals of N₂.
2. N₂⁺ has N_b = 9 and N_a = 4, so the bond order is (9 − 4)/2 = 2.5.
3. N₂⁺ has one unpaired electron and is paramagnetic.

Teacher's Note:
1. Removing a bonding electron lowers the bond order.
2. Removing an antibonding electron raises it, as in O₂ → O₂⁺.

Answer: (b) He₂

Explanation:
1. He₂ would have the configuration (σ1s)² (σ*1s)², so N_b = N_a = 2.
2. Its bond order is (2 − 2)/2 = 0, so it is not stable.
3. H₂⁺ has bond order 0.5, Li₂ has 1 and H₂ has 1.

Teacher's Note:
1. A bond order of zero means no bond.
2. He₂⁺ (bond order 0.5) can exist, but He₂ cannot.

Answer: (c) O₂

Explanation:
1. In B₂, C₂ and N₂ the 2s and 2p orbitals are close in energy and mix, so π2p orbitals lie below σ2pz.
2. In O₂ (and F₂) the gap between 2s and 2p is larger, so there is no mixing and σ2pz lies below π2p.

Teacher's Note:
1. Li₂ to N₂ show the π before σ order.
2. O₂, F₂ and Ne₂ show the σ before π order.

Answer: (d) It has a nodal plane between the two nuclei and higher energy than the parent orbitals

Explanation:
1. An antibonding orbital results from the out-of-phase (subtractive) combination ψA − ψB.
2. The wave function changes sign between the nuclei, giving a nodal plane with zero electron density there.
3. Its energy is higher than that of the parent atomic orbitals, and electrons in it weaken the bond.

Teacher's Note:
1. Bonding: in phase and lower energy. Antibonding: out of phase and higher energy.
2. Compare the bond orders when electrons go into each type.

Answer: (a) it has intramolecular hydrogen bonding, so the molecules are not associated

Explanation:
1. In o-nitrophenol the –OH and –NO₂ groups are close, so a hydrogen bond forms within the same molecule.
2. The molecules do not associate with each other, so less energy is needed to vaporise them.
3. In p-nitrophenol the groups are far apart, so the molecules are joined by intermolecular hydrogen bonds.

Teacher's Note:
1. Intramolecular means inside one molecule.
2. Intermolecular bonding raises the boiling point; intramolecular bonding does not.

Answer: (c) the hydrogen-bonded open cage structure of ice makes it less dense than water

Explanation:
1. In ice each O atom is hydrogen bonded to four other water molecules in an open tetrahedral cage.
2. The cage contains empty space, so ice has a larger volume for the same mass and is less dense than water.
3. A less dense solid floats on its liquid.

Teacher's Note:
1. Most solids are denser than their liquids, but ice is an exception.
2. Water has its maximum density at 277 K.