JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 11

Answer: (b) a and c

Explanation:
1. NO to NO+ (2.5 to 3) and O2 to O2+ (2 to 2.5) increase. CN- and CO have a bonding electron removed, so the order decreases.

Teacher's Note:
1. Remove from the highest occupied MO; check whether it is bonding or antibonding.

Answer: (c) \( \ce{NO+} \) < NO < \( \ce{NO^-} \)

Explanation:
1. Bond orders: NO+ = 3, NO = 2.5, NO- = 2.
2. Higher order gives a shorter bond, so NO+ < NO < NO-.

Teacher's Note:
1. Bond length varies inversely with bond order.

Answer: (b) In \( \ce{O2^-} \) the O-O bond order increases

Explanation:
1. O2 (BO 2) to O2- (BO 1.5): the bond order decreases, so the statement that it increases is wrong.

Teacher's Note:
1. An extra electron goes into an antibonding orbital.

Answer: (d) Both (a) and (c)

Explanation:
1. B in R3B has an empty p orbital (acceptor); N in R3N and NH3 has a lone pair (donor).
2. Both pairs give adducts with N -> B dative bonds.

Teacher's Note:
1. Donor and acceptor needed.

Answer: (c) Coordinate

Explanation:
1. NH3 donates its lone pair to the vacant orbital of Al: a coordinate (dative) bond.

Teacher's Note:
1. Lewis acid-base adduct formation.

Answer: (c) More extent of hydrogen bonding

Explanation:
1. Glycerol has three OH groups and forms more hydrogen bonds than glycol (two OH groups).

Teacher's Note:
1. More H bonding means stronger intermolecular forces.

Answer: (c) o-nitrophenol > p-nitrophenol

Explanation:
1. o-Nitrophenol has intramolecular H bonding and is less viscous than p-nitrophenol.

Teacher's Note:
1. Intermolecular H bonding raises viscosity.

Answer: (d) \( \ce{C2H6} \)

Explanation:
1. Ethane has no electronegative atom, so it cannot accept or donate H bonds.

Teacher's Note:
1. Acetone and acetonitrile can accept H bonds.

Answer: (c) Be

Explanation:
1. Be(OH)2 is amphoteric and dissolves in NaOH; BeSO4 is soluble.

Teacher's Note:
1. Mg(OH)2 does not dissolve in alkali.

Answer: (b) MnO

Explanation:
1. MnO has Mn in the +2 state, so the metal has a low charge and its oxide is the most ionic.

Teacher's Note:
1. Higher oxidation states give more covalent oxides.

Answer: (d) Both \( \ce{B2} \) and \( \ce{C2} \)

Explanation:
1. B2 and C2 have pi bonding MOs only (the sigma bonding is cancelled by the sigma* orbitals). C2(2-) has a sigma bond as well.

Teacher's Note:
1. MO configuration shows the type of net bonding.

Answer: (b) The bond order of \( \ce{HeH+} \) is 0.5

Explanation:
1. HeH+ has two electrons in the sigma bonding orbital, so BO = 1.
2. The other statements are correct.

Teacher's Note:
1. Bond order = (2 - 0)/2.

Answer: (a) A molecular solid

Explanation:
1. CH4 molecules are held by weak London forces in the solid state.

Teacher's Note:
1. Methane is a non-polar molecular solid.

Answer: (b) Neon (solid)

Explanation:
1. Solid neon has only weak dispersion forces.

Teacher's Note:
1. Covalent network and ionic solids have strong bonds.

Answer: (c) a and d

Explanation:
1. (a) and (d) are correct. (b) is wrong because CaO is the most basic; (c) is wrong because solubility of group 1 hydroxides increases down the group.

Teacher's Note:
1. Check each order separately.

Answer: (d) \( \ce{ZnCl2} \) < \( \ce{CdCl2} \) < \( \ce{HgCl2} \) (ionic character)

Explanation:
1. HgCl2 is the most covalent of the three, so ionic character cannot increase to HgCl2.

Teacher's Note:
1. Hg2+ is strongly polarising.

Answer: (b) Dimerisation of \( \ce{NO2} \)

Explanation:
1. In dimerisation of NO2 to N2O4, N stays sp2.
2. PCl5 becomes \( \ce{PCl4+} \)\( \ce{PCl6^-} \) (sp3 and sp3d2); SO3 to the trimer (sp2 to sp3); BF3 (sp2 to sp3).

Teacher's Note:
1. Check hybridisation before and after.

Answer: (c) A is true, but R is false.

Explanation:
1. Ionic bonds arise from complete transfer of electrons from the metal to the non-metal, so A is true and R is false.

Teacher's Note:
1. Sharing of electrons is the covalent bond.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Li+ is small and polarises Cl-, giving covalent character (A true). R is true but is not the reason.

Teacher's Note:
1. Fajans' rules relate to size and charge, not mass.

Answer: (c) A is true, but R is false.

Explanation:
1. The low polarising power of the large K+ ion keeps the carbonate stable, so A is true. R is false: the value of phi for K+ is low.

Teacher's Note:
1. Large cations of low charge density do not polarise anions.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. CO2 is linear (dipoles cancel) and SO2 is bent. R is true but not the explanation of the polarity difference.

Teacher's Note:
1. Shape decides the net dipole.

Answer: (c) A is true, but R is false.

Explanation:
1. Methanol dissolves because of hydrogen bonding with water. It is a covalent compound, so R is false.

Teacher's Note:
1. Polar covalent molecules with OH dissolve in water.

Answer: (c) A is true, but R is false.

Explanation:
1. Mg transfers two electrons to oxygen, so its electrovalency is 2 (A true). R is false because the electrons are transferred, not shared.

Teacher's Note:
1. Electrovalency counts electrons lost or gained.

Answer: (c) A is true, but R is false.

Explanation:
1. Ions are already present in solution and react instantly (A true). R is not a general fact: ionic bonds are not necessarily weaker than covalent bonds.

Teacher's Note:
1. The speed depends on the availability of ions.

Answer: (c) A is true, but R is false.

Explanation:
1. LiI has a low lattice energy relative to hydration energy because I- is large, so it is soluble (A true).
2. R is false: LiF has more ionic character, while LiI is more covalent.

Teacher's Note:
1. Fajans' rules.