JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 15

Answer: (b) ClH...Cl < NH...N < OH...O < FH...F

Explanation:
1. Hydrogen bond strength rises with the electronegativity and smallness of the atom bonded to H.
2. The order is Cl < N < O < F.
3. The F-H...F bond is the strongest (about 155 kJ per mole).

Teacher's Note:
1. Cl is large and weakly electronegative.
2. Strongest is F, then O, then N.

Answer: (c) \( \ce{N2H4} \)

Explanation:
1. Hydrazine has N-H bonds, so its molecules are linked by N-H...N hydrogen bonds.
2. The other three have no H bonded to N, O or F.

Teacher's Note:
1. Look for H bonded to N, O or F.
2. Se and S are too large and not electronegative enough.

Answer: (d) its strong hydrogen bonding

Explanation:
1. HF molecules are strongly associated through F-H...F hydrogen bonds.
2. A lot of energy is needed to separate them, so HF is not very volatile.

Teacher's Note:
1. Strong hydrogen bonding means a high boiling point and low volatility.
2. HF boils at about 20 C, far above HCl.

Answer: (c) hydrogen bond formation

Explanation:
1. Surface water molecules form hydrogen bonds across the interface.
2. This joins the two pieces of ice into one.

Teacher's Note:
1. Ice is held by an extended network of hydrogen bonds.
2. Regelation is the same effect under pressure.

Answer: (a) salicylaldehyde

Explanation:
1. In salicylaldehyde the O-H group sits next to the C=O group and forms a six-membered chelate ring by hydrogen bonding.
2. Water and phenol form intermolecular hydrogen bonds only, and acetaldehyde has no O-H.

Teacher's Note:
1. Intramolecular H-bonds need donor and acceptor close in one molecule.
2. Ortho isomers often show this.

Answer: (b) covalent bonds

Explanation:
1. Each molecule is O=C=O, with covalent double bonds.
2. Between molecules only weak van der Waals forces act.
3. There are no ions and no hydrogen bonds.

Teacher's Note:
1. Distinguish forces inside molecules and between molecules.
2. Dry ice sublimes because the attraction is weak.

Answer: (b) intramolecular hydrogen bonding between the OH and CHO groups

Explanation:
1. The OH and CHO groups are on neighbouring carbon atoms, so the O-H hydrogen bonds to the carbonyl oxygen within the same molecule.
2. This reduces association, so the ortho isomer is steam volatile and boils lower than the para isomer.

Teacher's Note:
1. Ortho isomers with chelation are more volatile.
2. Para isomers hydrogen bond between molecules.

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

Explanation:
1. Van der Waals forces include dispersion (instantaneous dipole-induced dipole), dipole-induced dipole and dipole-dipole interactions.
2. Small molecular size is not itself a cause of these forces.

Teacher's Note:
1. Van der Waals forces are all attractions between neutral molecules.
2. London forces are present in all molecules.

Answer: (d) melting of \( \ce{SiO2} \)

Explanation:
1. \( \ce{SiO2} \) is a covalent network solid, so melting breaks Si-O covalent bonds.
2. Boiling water and \( \ce{CF4} \) only overcome intermolecular forces, and melting KCN overcomes ionic attractions.

Teacher's Note:
1. Network solids have very high melting points.
2. Molecular solids and liquids change state without breaking covalent bonds.

Answer: (c) \( \ce{CH3OH} \) (32)

Explanation:
1. A high dielectric constant reduces the electrostatic attraction between ions in solution.
2. Methanol has the highest value given, so it dissolves the ionic salt KI best.
3. Benzene and carbon tetrachloride are non-polar.

Teacher's Note:
1. Ionic solids dissolve in solvents of high dielectric constant.
2. Like dissolves like.

Answer: (d) 0, +1, -1

Explanation:
1. The Lewis structure is O=O(+)-O(-).
2. The central oxygen has three bonds and one lone pair, giving +1.
3. The singly bonded terminal oxygen has -1 and the doubly bonded one has 0.

Teacher's Note:
1. Formal charge = valence electrons - non-bonded electrons - bonds.
2. The charges sum to zero for a neutral molecule.

Answer: (d) electrovalent, covalent and coordinate covalent

Explanation:
1. \( \ce{Cu^2+} \) and \( \ce{SO4^2-} \) are linked by an electrovalent bond.
2. The S-O bonds are covalent.
3. Water molecules are coordinated to the copper ion by coordinate bonds (and one water by hydrogen bonds).

Teacher's Note:
1. Hydrates have coordinate bonds between ions and water.
2. Polyatomic ions bring covalent bonds.

Answer: (b) \( \ce{Na2SO4.10H2O} \)

Explanation:
1. Glauber's salt melts in its own water of crystallisation at 32.4 C and changes to the anhydrous salt.
2. The anhydrous salt has a lower solubility that decreases with temperature above that point.

Teacher's Note:
1. The transition temperature of Glauber's salt is 32.4 C.
2. Solubility curves show a break at that point.

Answer: (a) \( \ce{CH3Cl} > \ce{CH3F} > \ce{CH3Br} > \ce{CH3I} \)

Explanation:
1. Measured dipole moments (in D) are \( \ce{CH3Cl} \) 1.87, \( \ce{CH3F} \) 1.85, \( \ce{CH3Br} \) 1.81 and \( \ce{CH3I} \) 1.62.
2. The dipole moment depends on both charge and bond length, so fluorine, though most electronegative, does not give the largest value.

Teacher's Note:
1. Dipole moment = charge x distance.
2. The C-F bond is short, so its moment is slightly lower than C-Cl.

Answer: (b) \( \ce{CO2} \)

Explanation:
1. \( \ce{CO2} \) is linear and symmetrical, so the two bond dipoles cancel, giving zero.
2. The others are bent or pyramidal and have a net dipole.

Teacher's Note:
1. Symmetrical molecules have zero dipole moment.
2. Zero is less than any non-zero value.

Answer: (c) 17

Explanation:
1. Dipole moment for 100 percent ionic character = e x d = 4.8 x 10^-10 esu x 1.275 x 10^-8 cm = 6.12 x 10^-18 esu cm = 6.12 D.
2. Percentage ionic character = (1.03 / 6.12) x 100.
3. This is about 16.8, i.e. 17 percent.

Teacher's Note:
1. 1 D = 10^-18 esu cm.
2. Ionic character = observed moment divided by the moment for full charge separation.

Answer: (a) 0 D

Explanation:
1. Hexachlorobenzene is a regular hexagon with six identical C-Cl bonds.
2. The six bond moments cancel in pairs, so the resultant is zero.

Teacher's Note:
1. Symmetrical substitution gives zero dipole moment.
2. The same is true of p-dichlorobenzene.

Answer: (a) (i), (iii) and (iv)

Explanation:
1. The square planar complex, the linear \( \ce{CO2} \) and p-dichlorobenzene are all symmetrical, so their bond dipoles cancel.
2. \( \ce{CHCl3} \) is tetrahedral with different atoms and has a net dipole.

Teacher's Note:
1. A symmetrical arrangement of identical groups cancels dipoles.
2. Replace one atom by a different one and the dipole reappears.

Answer: (a) water

Explanation:
1. Water is bent, so its two O-H bond dipoles add to give a net dipole.
2. Benzene, carbon tetrachloride and boron trifluoride are symmetrical (planar or tetrahedral) and have zero net dipole.

Teacher's Note:
1. Check the molecular shape for symmetry.
2. Bent and pyramidal molecules are polar.

Answer: (a) (i), (ii) and (iii) only

Explanation:
1. Ice has covalent O-H bonds within molecules and hydrogen bonds between them, so (i) is correct.
2. The open cage structure collapses between 0 C and 4 C, so the density rises.
3. Above 4 C ordinary thermal expansion takes over.
4. The density rise is due to structural change, not entropy, so (iv) is not an explanation.

Teacher's Note:
1. Maximum density of water is at 4 C.
2. This is why lakes freeze from the top.

Answer: (a) Bond order is proportional to 1 / bond length, and proportional to bond energy

Explanation:
1. A higher bond order means a shorter bond.
2. It also means a stronger bond, i.e. higher bond energy.
3. So bond order varies inversely with length and directly with energy.

Teacher's Note:
1. Triple bonds are shorter and stronger than double bonds.
2. Use C-C, C=C and C#C as a check.

Answer: (b) P-O > S-O > Cl-O

Explanation:
1. Bond length follows the covalent radius of the atom bonded to oxygen.
2. Radii decrease across the period: P > S > Cl.
3. So P-O > S-O > Cl-O.

Teacher's Note:
1. Atomic radius decreases from left to right.
2. Compare the other atoms when oxygen is common.

Answer: (c) \( sp^2 \) hybrid

Explanation:
1. A symmetrical molecule with zero dipole moment and three bonds must be trigonal planar.
2. The central atom, such as B (\( \ce{BF3} \)), is then \( sp^2 \) hybridised.

Teacher's Note:
1. Zero dipole needs symmetry: planar triangular.
2. That means \( sp^2 \).

Answer: (a) nine sigma and nine pi

Explanation:
1. Sigma bonds: 1 (C=C) + 4 (C-C between the C=C carbon and cyanide carbon) + 4 (C-N) = 9.
2. Pi bonds: 1 (C=C) + 4 x 2 (in the four C#N triple bonds) = 9.

Teacher's Note:
1. Triple bond = one sigma + two pi.
2. Add the C=C pi bond as well.

Answer: (b) \( \ce{Cl2} \)

Explanation:
1. Bond energy decreases from \( \ce{Cl2} \) (243 kJ per mole) to \( \ce{I2} \) (151 kJ per mole) as atoms get bigger.
2. \( \ce{F2} \) (158 kJ per mole) is unusually weak because of repulsion between lone pairs on the small F atoms.
3. So \( \ce{Cl2} \) is highest.

Teacher's Note:
1. F2 is an exception to the trend.
2. Cl2 has the strongest X-X bond among halogens.