JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 24

Answer: (d) \(\ce{CaI2}\)

Explanation:
1. Covalent character increases with the polarisability of the anion (Fajans' rules).
2. The size of the anion increases F- < Cl- < Br- < I-.
3. So CaI2 has the highest covalent character.

Teacher's Note:
1. A large anion is easily polarised by the same cation.
2. Compare the anions when the cation is the same.

Answer: (c) \(\ce{Te^{2-}}\)

Explanation:
1. Polarisability increases with the size and the charge of the anion.
2. Te2- is the largest and has the highest charge among the options.

Teacher's Note:
1. Large, highly charged anions are the most polarisable.
2. Se2- is smaller than Te2-.

Answer: (d) different because BF3 is non-polar and NF3 is polar

Explanation:
1. BF3 is sp2 hybridised, trigonal planar and symmetrical, so the bond dipoles cancel and it is non-polar.
2. NF3 is sp3 hybridised, pyramidal with one lone pair, so it has a net dipole and is polar.

Teacher's Note:
1. Lone pairs change the shape and hence the polarity.
2. Symmetric planar molecules with identical bonds are non-polar.

Answer: (a) 9 and 9

Explanation:
1. Sigma bonds: one C=C (1), four C-C(N) (4) and four C-N (4) = 9.
2. Pi bonds: one in C=C and two in each of the four C≡N triple bonds = 1 + 8 = 9.

Teacher's Note:
1. A double bond is 1 sigma + 1 pi and a triple bond is 1 sigma + 2 pi.
2. Count every bond between different atoms.

Answer: (d) BCl3 has no lone pair of electrons but NCl3 has a lone pair on the central atom

Explanation:
1. In BCl3 the central boron has three bond pairs and no lone pair, giving a trigonal planar shape.
2. In NCl3 nitrogen has three bond pairs and one lone pair, which repels the bond pairs and gives a pyramidal shape.

Teacher's Note:
1. VSEPR shapes depend on bond pairs and lone pairs on the central atom.
2. A lone pair compresses the bond angles.

Answer: (d) sp2

Explanation:
1. Each carbon is attached to three atoms through sigma bonds and is part of a pi system.
2. Each carbon therefore has three sigma bonds and is sp2 hybridised.

Teacher's Note:
1. The number of sigma bonds plus lone pairs gives the hybridisation.
2. In butadiene the central C-C bond has partial double bond character.

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

Explanation:
1. BF3 is sp2 hybridised and trigonal planar, with all four atoms in one plane.
2. CH4 is tetrahedral, PF5 is trigonal bipyramidal and NH3 is pyramidal.

Teacher's Note:
1. Only AB3 molecules with no lone pair on A are planar.
2. sp2 hybridisation gives a planar geometry.

Answer: (d) 2 and 8

Explanation:
1. Oxygen forms two bonds and keeps two lone pairs.
2. Each fluorine keeps three lone pairs, i.e. six for two F atoms.
3. Total lone pairs = 2 + 6 = 8.

Teacher's Note:
1. Count lone pairs on every atom, not only on the central atom.
2. The total valence electron count is 20, which gives 2 bond pairs and 8 lone pairs.

Answer: (b) \(\ce{H2O}\)

Explanation:
1. CO2 is linear and non-polar.
2. HI has a small dipole because of the small electronegativity difference.
3. H2O is bent with highly polar O-H bonds (1.85 D), larger than SO2 (1.62 D).

Teacher's Note:
1. Both the bond polarity and the shape decide the dipole moment.
2. Known values: H2O 1.85 D, SO2 1.62 D, HI 0.38 D.

Answer: (b) \(\ce{BH4-}\)

Explanation:
1. BH4- has four bond pairs and no lone pair, so it is tetrahedral.
2. NH2- is bent, CO3 2- is trigonal planar and H3O+ is pyramidal.

Teacher's Note:
1. Four bond pairs and no lone pairs give a tetrahedron.
2. Compare the number of electron pairs for each.

Answer: (c) \(2s^2 2p^5\)

Explanation:
1. \(2s^2 2p^5\) is fluorine.
2. Fluorine has the highest electronegativity (4.0) of all elements.

Teacher's Note:
1. Electronegativity increases across a period and decreases down a group.
2. F is at the top right of the halogens.

Answer: (d) \(\ce{SbCl3}\)

Explanation:
1. All are pyramidal with one lone pair on the central atom.
2. Down the group the central atom becomes larger and less electronegative, so the bond pairs are farther away and repel less.
3. The bond angle falls in the order NCl3 > PCl3 > AsCl3 > SbCl3.

Teacher's Note:
1. As the central atom's electronegativity falls, the lone pair occupies more space.
2. The angle approaches 90 degrees for large central atoms.

Answer: (c) \(\ce{PF5}\) and \(\ce{BrF5}\)

Explanation:
1. CO3 2- and NO3- are both trigonal planar; PCl4+ and SiCl4 are both tetrahedral; AlF6 3- and SF6 are both octahedral.
2. PF5 is trigonal bipyramidal while BrF5 is square pyramidal, so this pair is not iso-structural.

Teacher's Note:
1. Compare the number of electron pairs including lone pairs.
2. BrF5 has one lone pair on bromine.

Answer: (c) \(\ce{AlCl3}\)

Explanation:
1. Al3+ has a high charge and a small size, so its polarising power is the greatest.
2. By Fajans' rules, AlCl3 has the greatest covalent character.

Teacher's Note:
1. Higher charge and smaller size of the cation increase covalent character.
2. AlCl3 exists as a dimer Al2Cl6 in vapour.

Answer: (b) pentagonal bipyramidal

Explanation:
1. Iodine has seven bond pairs and no lone pair, so it is sp3d3 hybridised.
2. Seven electron pairs give a pentagonal bipyramidal structure.

Teacher's Note:
1. Seven bond pairs means a pentagonal bipyramid.
2. Learn: IF7 pentagonal bipyramid, SF6 octahedral, PF5 trigonal bipyramid.

Answer: (a) \(\ce{XeF2}\)

Explanation:
1. Lone pairs on Xe: XeF2 has 3, XeF4 has 2, XeF6 has 1 and XeO3 has 1.
2. The maximum is in XeF2.

Teacher's Note:
1. Xe has 8 valence electrons.
2. Subtract the electrons used in bonding and halve the remainder.

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

Explanation:
1. Hydrogen bond strength increases with the electronegativity of the atoms involved.
2. F is the most electronegative, so F-H...F is the strongest.

Teacher's Note:
1. Strengths are about 160 kJ/mol for F-H...F and about 20 kJ/mol for O-H...O.
2. Remember the order F > O > N.

Answer: (a) \(\ce{SF4}\)

Explanation:
1. SF4 is sp3d with a lone pair in an equatorial position, so it has two axial and two equatorial S-F bonds with different lengths.
2. SiF4 and BF4- are tetrahedral and XeF4 is square planar, so all their bonds are equal.

Teacher's Note:
1. Trigonal bipyramidal geometry has two types of positions.
2. Axial bonds are longer than equatorial bonds.

Answer: (b) charge on the ion and size of the ion

Explanation:
1. Lattice energy is proportional to the product of the charges and inversely proportional to the distance between the ions, i.e. the sum of ionic radii.
2. So it depends on both the charge and the size.

Teacher's Note:
1. Higher charge and smaller ions give a higher lattice energy.
2. That is why MgO has a much higher lattice energy than NaCl.

Answer: (d) different, with 1, 0 and 2 lone pairs on the central atom respectively

Explanation:
1. SF4 has one lone pair (see-saw).
2. CF4 has none (tetrahedral).
3. XeF4 has two lone pairs (square planar). So the shapes are different, with 1, 0 and 2 lone pairs.

Teacher's Note:
1. Count lone pairs as (valence electrons - electrons used in bonds)/2.
2. Square planar XeF4 places two lone pairs opposite each other.

Answer: (c) sp3d2

Explanation:
1. dsp3 and sp3d are trigonal bipyramidal with 6 angles of 90 degrees.
2. dsp2 is square planar with 4.
3. sp3d2 is octahedral and has 12 angles of 90 degrees, the maximum.

Teacher's Note:
1. Octahedral geometry has 12 such angles.
2. Count each pair of adjacent bonds.