JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 45

Answer: (b) 3

Explanation:
1. O=C=C=C=O: every carbon is sp-hybridised and the whole molecule is linear, so all 3 carbons lie in a line.

Teacher's Note:
1. sp-hybridised carbon gives 180 degree angles.

Answer: (a) Trigonal planar

Explanation:
1. Only sigma bonds and lone pairs fix the geometry: 2 sigma + 1 lone pair = 3 electron domains, so the arrangement is trigonal planar (as in SO2).

Teacher's Note:
1. Pi bonds do not add electron domains.

Answer: (c) See-saw

Explanation:
1. AX4E has 5 electron pairs (trigonal bipyramid) with the lone pair in an equatorial position, giving a see-saw shape.

Teacher's Note:
1. Lone pairs prefer equatorial positions.

Answer: (b) \( \ce{CO2} \) and \( \ce{SiO2} \) are both linear

Explanation:
1. SiO2 is a three-dimensional network with tetrahedral Si and V-shaped Si-O-Si links, not linear.

Teacher's Note:
1. SiO2 is a covalent network solid.

Answer: (b) 66%

Explanation:
1. A 120 degree angle corresponds to sp2 hybrids, which have one third s-character and so two thirds (about 66%) p-character.

Teacher's Note:
1. sp = 50%, sp2 = 66.7%, sp3 = 75% p-character.

Answer: (c) \( \ce{Li2-} \) < \( \ce{Li2+} \) < \( \ce{Li2} \)

Explanation:
1. Bond orders: Li2 = 1, Li2+ = 0.5 and Li2- = 0.5, but Li2- has an electron in an antibonding orbital, which makes it the least stable.

Teacher's Note:
1. Same bond order: the ion with antibonding electrons is less stable.

Answer: (b) p orbitals always have only sideways overlapping

Explanation:
1. p orbitals can overlap head-on (sigma) as well as sideways (pi), so the statement is wrong.

Teacher's Note:
1. Head-on p-p overlap gives sigma bonds.

Answer: (c) \( \ce{OF6} \)

Explanation:
1. Oxygen is a second-period element and cannot expand its octet, so OF6 does not exist.

Teacher's Note:
1. Period 2 elements cannot be hypervalent.

Answer: (a) \( \ce{LiNO3} \)

Explanation:
1. LiNO3 gives NO2 and O2, which do not turn lime water milky. ZnCO3 and MgCO3 give CO2, and ZnSO3 gives SO2, both of which turn lime water milky.

Teacher's Note:
1. CO2 and SO2 give turbidity with lime water.

Answer: (c) IV (bond energy 485)

Explanation:
1. Shorter bonds are stronger. The element forming the strongest C-X bond (485 kJ per mol) is the smallest.

Teacher's Note:
1. Bond energy increases as atomic size decreases (within a group).

Answer: (c) Six molecules out of the given compounds are non-planar

Explanation:
1. Non-planar ones are NH3, PH3, PCl3, P4 and IF7: five, not six. sp2: SO2, SO3, BF3 (three). Hypervalent: SO2, SO3 and IF7. Both SO2 and SO3 have the two types of pi bonding.

Teacher's Note:
1. Triatomic molecules such as H2S are always planar.

Answer: (d) \( \ce{Ag2CO3} \)

Explanation:
1. Ag2CO3 decomposes to Ag2O + CO2 and then Ag2O gives Ag + O2, since silver oxide is unstable at high temperature.

Teacher's Note:
1. Oxides of noble metals decompose to the metal.

Answer: (a) all of these

Explanation:
1. London dispersion force acts between all molecules, polar or not.

Teacher's Note:
1. It is the universal intermolecular force.

Answer: (c) All of these

Explanation:
1. SF2 (bent, about 98 degrees), KrF4 and ICl4- (square planar, 90 degrees) each have two lone pairs on the central atom and angles below 109.5 degrees.

Teacher's Note:
1. Two lone pairs in AX2E2 or AX4E2 compress the angle.

Answer: (b) \( \ce{ClF3} \)

Explanation:
1. ClF3 is T-shaped (planar) with unsymmetrical dipoles. PCl3 is pyramidal and SF4 is see-saw (both non-planar).

Teacher's Note:
1. T-shape is planar and polar.

Answer: (b) sp\(^3\)d

Explanation:
1. sp3d orbitals form a trigonal bipyramid with two axial and three equatorial orbitals which differ in energy and directions.

Teacher's Note:
1. Trigonal bipyramid has two types of position.

Answer: (d) \( \ce{I3-} \)

Explanation:
1. I3- has 2 bond pairs and 3 lone pairs on the central iodine. H2O, SCl2 and each O of O2F2 have 2 bond pairs and 2 lone pairs.

Teacher's Note:
1. Count electron pairs on the central atom.

Answer: (a) \( \ce{PF5} \)

Explanation:
1. Solid PCl5 is [PCl4]+[PCl6]-, solid PBr5 is [PBr4]+Br-, but PF5 remains a molecular trigonal bipyramid because F is too small to accommodate the ionic form.

Teacher's Note:
1. Only the larger halides ionise in the solid.

Answer: (a) \( \ce{AB2L2} \) (bent)

Explanation:
1. AB2L2 is bent (like H2O) and polar. AB2L3 (linear XeF2), AB4L2 (square planar XeF4) and AB4 (tetrahedral) are symmetrical and non-polar when substituents are identical.

Teacher's Note:
1. Symmetrical geometries cancel dipoles.

Answer: (a) HF, H2O, H2O2

Explanation:
1. H-bond strength is HF > H2O > H2O2, but boiling points are H2O2 (150 C) > H2O (100 C) > HF (19.5 C): opposite orders.

Teacher's Note:
1. Number of H bonds per molecule matters for boiling point.

Answer: (c) Melting of SiO2

Explanation:
1. SiO2 is a covalent network solid, so melting it breaks covalent Si-O bonds. The others only overcome intermolecular or ionic forces.

Teacher's Note:
1. Covalent solids melt by breaking covalent bonds.

Answer: (d) (i), (ii) and (iii)

Explanation:
1. Pb(NO3)2 and LiNO3 give NO2 (and O2); NaNO3 gives O2. All are paramagnetic. NH4NO2 gives N2, which is diamagnetic.

Teacher's Note:
1. NO2 and O2 are paramagnetic; N2 is not.

Answer: (c) Linear

Explanation:
1. I2 + I- gives \( \ce{I3-} \): central I has 2 bond pairs and 3 lone pairs (sp3d), the lone pairs occupy the equatorial positions, so the shape is linear.

Teacher's Note:
1. Draw the Lewis structure, count bond pairs and lone pairs, then decide.

Answer: (c) \( \ce{XeO3} \)

Explanation:
1. XeO3 has 3 sigma bonds and 1 lone pair: sp3. I3-, SeCl4 and AsF4- each have 5 electron pairs: sp3d.

Teacher's Note:
1. Five electron pairs means sp3d.