JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 17

Answer: (a) \( \ce{H2S} < \ce{NH3} < \ce{SiH4} < \ce{BF3} \)

Explanation:
1. The angles are about 92 degrees in \( \ce{H2S} \), 107 in \( \ce{NH3} \), 109.5 in \( \ce{SiH4} \) and 120 in \( \ce{BF3} \).
2. So the order, smallest first, is \( \ce{H2S} < \ce{NH3} < \ce{SiH4} < \ce{BF3} \).

Teacher's Note:
1. Planar sp2 is 120 and tetrahedral is 109.5.
2. Lone pairs shrink the angle below those values.

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

Explanation:
1. A higher bond order gives a shorter bond.
2. \( \ce{NO+} \) has a bond order of 3, so its bond is shorter.
3. So the bond length in NO is greater than in \( \ce{NO+} \).

Teacher's Note:
1. Bond length and bond order vary inversely.
2. Removing an antibonding electron strengthens the bond.

Answer: (b) \( sp^2 \) and \( sp^3 \)

Explanation:
1. Boron has three sigma bonds (to three OH groups), so it is sp2.
2. Each oxygen has two sigma bonds and two lone pairs, so it is sp3.

Teacher's Note:
1. Count sigma bonds plus lone pairs.
2. The molecule \( \ce{B(OH)3} \) is trigonal planar around B.

Answer: (a) \( \ce{BF4-} \)

Explanation:
1. \( \ce{BF4-} \) has four bond pairs and no lone pair, so it is regular tetrahedral.
2. \( \ce{SF4} \) is see-saw, \( \ce{XeF4} \) is square planar, and \( \ce{[Ni(CN)4]^2-} \) is square planar.

Teacher's Note:
1. AX4 with no lone pair is tetrahedral.
2. Ni(II) with strong field ligands is square planar.

Answer: (d) \( sp^3d^2 \) hybridisation

Explanation:
1. Square planar \( dsp^2 \) has 4 such angles.
2. Trigonal bipyramidal \( sp^3d \) has 6 (axial-equatorial).
3. Octahedral \( sp^3d^2 \) has 12 right angles between adjacent bonds, the maximum.

Teacher's Note:
1. Octahedron has 12 right angles between neighbouring bond pairs.
2. Count adjacent pairs.

Answer: (b) \( \ce{H2} \)

Explanation:
1. \( \ce{H2} \) has two electrons in the bonding orbital, all paired.
2. \( \ce{He2+} \) (3 electrons), \( \ce{H2+} \) (1 electron) and \( \ce{H2-} \) (3 electrons) each have an unpaired electron.

Teacher's Note:
1. An odd number of electrons always means paramagnetic.
2. Fill molecular orbitals in order.

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

Explanation:
1. \( \ce{SF4} \) has 1 lone pair (see-saw), \( \ce{CF4} \) has none (tetrahedral) and \( \ce{XeF4} \) has 2 (square planar).
2. The shapes are therefore different.

Teacher's Note:
1. Count the electron pairs on the central atom, then subtract the bond pairs.
2. Each shape is different.

Answer: (c) \( \ce{SO3^2-} \), \( \ce{CO3^2-} \), \( \ce{NO3-} \)

Explanation:
1. \( \ce{SO3^2-} \) has 16 + 24 + 2 = 42 electrons.
2. \( \ce{CO3^2-} \) and \( \ce{NO3-} \) each have 32 electrons.
3. So the set in (c) is not isoelectronic.
4. The other sets each have equal numbers of electrons.

Teacher's Note:
1. Count electrons including the charge.
2. Period 2 and period 3 oxyanions differ by 10 electrons.

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

Explanation:
1. \( \ce{O2^2-} \) has 18 electrons, with all antibonding orbitals filled, so all electrons are paired.
2. \( \ce{N2+} \) has 13 electrons, \( \ce{O2} \) has 2 unpaired electrons, and \( \ce{B2} \) has 2 unpaired electrons in pi orbitals.

Teacher's Note:
1. Peroxide ion is diamagnetic.
2. Use molecular orbital filling to find unpaired electrons.

Answer: (d) acetonitrile and acetone

Explanation:
1. Acetonitrile and acetone are both polar molecules without hydrogen bond donors, so dipole-dipole interaction dominates.
2. KCl and water show ion-dipole forces, benzene and carbon tetrachloride show dispersion forces, and benzene and ethanol are weakly interacting.

Teacher's Note:
1. Match the forces to the species: ion-dipole, dipole-dipole, dispersion.
2. Polar molecules without O-H or N-H show dipole-dipole forces.

Answer: (a) \( \ce{MCl2} \) is more ionic than \( \ce{MCl4} \)

Explanation:
1. The M2+ ion has lower charge and larger size than M4+, so it polarises Cl- less and \( \ce{MCl2} \) is more ionic.
2. The more covalent \( \ce{MCl4} \) is more volatile, more easily hydrolysed and more soluble in organic solvents.

Teacher's Note:
1. Higher oxidation state means more covalent character.
2. SnCl2 and SnCl4 are a good example.

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

Explanation:
1. \( \ce{SF4} \) is see-saw, with two axial and two equatorial S-F bonds of different lengths.
2. \( \ce{XeF4} \) (square planar), \( \ce{BF4-} \) and \( \ce{SiF4} \) (tetrahedral) have all bonds equal.

Teacher's Note:
1. Trigonal bipyramidal arrangements have unequal axial and equatorial bonds.
2. Lone pairs sit in equatorial positions.

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

Explanation:
1. Fluorine is the most electronegative and smallest of these atoms.
2. So the F-H...F bond has the highest energy (about 155 kJ per mole).

Teacher's Note:
1. Strength: F-H...F > O-H...O > N-H...N.
2. Bond strength follows electronegativity and size.

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

Explanation:
1. Polarising power rises with charge and falls with size.
2. \( \ce{K+} \) has a charge of 1 and the largest size, so it is the lowest.
3. For the +2 ions the size decreases in the order Ca > Mg > Be, so polarising power rises in the same order.

Teacher's Note:
1. Charge density is charge divided by size.
2. Smaller and more highly charged means stronger polarising power.

Answer: (d) significant p(pi)-p(pi) back bonding between B and F in \( \ce{BF3} \), whereas there is no possibility of such interaction between C and F in \( \ce{CF4} \)

Explanation:
1. In \( \ce{BF3} \) boron has an empty p orbital, which accepts a lone pair from fluorine (back bonding).
2. This adds partial double bond character to the B-F bond.
3. Carbon in \( \ce{CF4} \) has no empty p orbital, so no such interaction is possible.

Teacher's Note:
1. Back bonding needs an empty orbital on one atom and a lone pair on the other.
2. It is most efficient between small atoms of period 2.

Answer: (c) \( \ce{O2^2+} \)

Explanation:
1. Bond orders are \( \ce{O2-} \) 1.5, \( \ce{O2^2-} \) 1, \( \ce{O2^2+} \) 3 and \( \ce{O2+} \) 2.5.
2. The highest bond order gives the shortest bond length, so \( \ce{O2^2+} \) is the shortest.

Teacher's Note:
1. Bond order is inversely related to bond length.
2. \( \ce{O2^2+} \) is isoelectronic with \( \ce{N2} \).

Answer: (b) o-nitrophenol shows intramolecular H-bonding

Explanation:
1. In o-nitrophenol the OH group hydrogen bonds with the adjacent nitro group inside the same molecule.
2. It cannot hydrogen bond with water as effectively, so its solubility in water is lower.
3. The m- and p-isomers form hydrogen bonds with water.

Teacher's Note:
1. Intramolecular hydrogen bonding lowers solubility in water and raises volatility.
2. That is why o-nitrophenol is steam volatile.

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

Explanation:
1. All are pyramidal with a lone pair on the central atom.
2. Down the group the central atom is less electronegative and bigger, so the bond angle decreases.
3. So \( \ce{SbCl3} \) has the smallest angle.

Teacher's Note:
1. The angle falls from N to Sb.
2. Same trend as the hydrides.

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

Explanation:
1. \( \ce{AlF6^3-} \) and \( \ce{SF6} \) are octahedral, \( \ce{CO3^2-} \) and \( \ce{NO3-} \) are trigonal planar, and \( \ce{PCl4+} \) and \( \ce{SiCl4} \) are tetrahedral.
2. \( \ce{PF5} \) is trigonal bipyramidal while \( \ce{BrF5} \) is square pyramidal.

Teacher's Note:
1. A lone pair on the central atom changes the shape.
2. Match bond pairs and lone pairs for each pair.

Answer: (b) it contains \( \ce{Cs+} \) and \( \ce{I3-} \) ions

Explanation:
1. Caesium forms only the +1 ion, and the polyhalide ion \( \ce{I3-} \) is stable with a large cation.
2. So \( \ce{CsI3} \) consists of \( \ce{Cs+} \) and \( \ce{I3-} \) ions.

Teacher's Note:
1. Large cations stabilise large polyhalide anions.
2. Cs only has a +1 state.

Answer: (c) it is diamagnetic in the gaseous state

Explanation:
1. NO has 11 valence electrons, so it has one unpaired electron and is paramagnetic.
2. Its bond order is 2.5, it reacts with oxygen to give \( \ce{NO2} \), and it is a neutral oxide.

Teacher's Note:
1. Odd-electron molecules are paramagnetic.
2. NO dimerises only in the liquid and solid states.

Answer: (c) one sigma bond and two pi bonds

Explanation:
1. \( \ce{C2^2-} \) is isoelectronic with \( \ce{N2} \), so it has a triple bond between the carbons.
2. A triple bond has one sigma and two pi bonds.

Teacher's Note:
1. Isoelectronic with \( \ce{N2} \) and \( \ce{CO} \).
2. Bond order is 3.

Answer: (b) \( \ce{CH3F} < \ce{CH3Cl} < \ce{CH3Br} < \ce{CH3I} \)

Explanation:
1. The bond length increases with the size of the halogen atom.
2. F < Cl < Br < I, so the C-X bond lengths are 139, 178, 193 and 214 pm in that order.

Teacher's Note:
1. Larger atoms form longer bonds.
2. Bond strength falls in the same direction.

Answer: (b) \( \ce{(SiH3)3N} \)

Explanation:
1. In \( \ce{(SiH3)3N} \) the lone pair on nitrogen is delocalised into vacant d orbitals of silicon (p-d pi back bonding).
2. So nitrogen becomes sp2 and the molecule is planar.
3. The other three are pyramidal.

Teacher's Note:
1. Back bonding with silicon flattens the nitrogen.
2. Trimethylamine is pyramidal.