JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 13

Answer: (b) the ionisation energy of A is low and the electron affinity of B is high

Explanation:
1. A must lose an electron easily, so its ionisation energy must be low.
2. B must gain the electron readily, so its electron affinity must be high (a large negative value).
3. Both conditions together favour transfer of an electron from A to B.

Teacher's Note:
1. Ionic bond needs an easy electron donor and a good electron acceptor.
2. Also a high lattice energy helps.

Answer: (a) \( \ce{PbCl2} \)

Explanation:
1. By Fajans' rules a larger, lower-charged cation polarises the anion less and gives a more ionic bond.
2. \( \ce{Pb^2+} \) is large and carries a low charge (inert pair effect makes +2 stable for lead).
3. So \( \ce{PbCl2} \) is the most ionic of the four.

Teacher's Note:
1. Low charge and large size of the cation increase ionic character.
2. The inert pair effect makes +2 common in heavy group 14.

Answer: (d) all of these

Explanation:
1. Water molecules surround the ions, and the hydration energy is released as heat.
2. The polar water molecules weaken the electrostatic attraction between ions.
3. The ions then separate (dissociate) into solution.

Teacher's Note:
1. Hydration energy offsets the lattice energy.
2. Ionic compounds dissolve well in polar solvents like water.

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

Explanation:
1. Ionic bonds arise from electrostatic attraction, which acts equally in all directions, so they are non-directional.
2. \( \ce{RbCl} \) is ionic.
3. The other three are covalent, and covalent bonds are directional.

Teacher's Note:
1. Ionic bonds are non-directional and covalent bonds are directional.
2. This is why ionic solids form crystal lattices.

Answer: (d) CsF

Explanation:
1. Approximate radii in pm: Li+ 76, Na+ 102, Cs+ 167, F- 133, I- 220.
2. Ratios: LiF 1.75, NaF 1.30, CsI 1.32, CsF 0.80.
3. CsF has the lowest ratio.

Teacher's Note:
1. A large cation and small anion give a low anion-to-cation ratio.
2. Compare the ratios by estimating radii.

Answer: (d) all four

Explanation:
1. \( \ce{NH4Cl} \) has \( \ce{NH4+} \) and \( \ce{Cl-} \) ions, with covalent N-H bonds inside the cation.
2. KCN, NaOH and \( \ce{CuSO4.5H2O} \) all contain polyatomic ions bonded covalently inside and ionically to the other ion.
3. So all four contain both types of bond.

Teacher's Note:
1. A polyatomic ion always has covalent bonds within it.
2. Its link to the counter-ion is ionic.

Answer: (d) MgO

Explanation:
1. Lattice energy increases with the product of the ionic charges and decreases with ionic size.
2. MgO has charges 2+ and 2-, while the others are 1+ and 1-.
3. So MgO has by far the highest lattice energy.

Teacher's Note:
1. Charge has a stronger effect than size on lattice energy.
2. Hence MgO has a very high melting point.

Answer: (a) ionic

Explanation:
1. Two identical atoms have the same tendency to gain or lose electrons, so no electron transfer occurs.
2. Hence an ionic bond cannot form.
3. A coordinate bond can still occur between like atoms (for example in ozone), and covalent and metallic bonds do occur.

Teacher's Note:
1. Ionic bonding needs a difference in electronegativity.
2. Like atoms share electrons or pool them in a metal.

Answer: (c) alcohol

Explanation:
1. Green vitriol (\( \ce{FeSO4.7H2O} \)), Indian saltpetre (\( \ce{KNO3} \)) and potash alum are ionic and give ions in water.
2. Alcohol is a covalent non-electrolyte and gives no ions.

Teacher's Note:
1. Only electrolytes conduct in solution.
2. Alcohol and sugar are non-electrolytes.

Answer: (c) the non-directional nature of the bond

Explanation:
1. Space (geometric or optical) isomerism needs bonds fixed in particular directions.
2. Ionic bonds are non-directional, so no definite spatial arrangement exists for isomers.

Teacher's Note:
1. Directional bonds make isomerism possible.
2. Ionic bonds have none.

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

Explanation:
1. Covalent bonds are directional because they form by overlap of orbitals, and ionic bonds are non-directional.
2. A polar bond needs a difference in electronegativity, so (iii) is false.
3. Polar bonds can give a non-zero dipole moment, so (iv) is false.

Teacher's Note:
1. Polar bonds need unequal electronegativity.
2. Zero dipole moment needs a symmetric shape, not just the presence of polar bonds.

Answer: (a) \( \ce{C2H4} \) (C=C bond)

Explanation:
1. The C=C double bond has a pi bond that locks the two CH2 groups in one plane.
2. Rotation would break the pi bond, so it is restricted.
3. Single bonds (O-O and C-C) allow rotation.

Teacher's Note:
1. Free rotation is possible only about single sigma bonds.
2. This is the reason for cis-trans isomerism in alkenes.

Answer: (b) (ii) and (iii)

Explanation:
1. In \( \ce{BCl3} \) boron has only six electrons (electron deficient).
2. In \( \ce{PCl5} \) phosphorus has ten electrons (expanded octet).
3. \( \ce{CO2} \) and \( \ce{SiF4} \) obey the octet rule.

Teacher's Note:
1. Octet violations: electron-deficient (BCl3), expanded octet (PCl5, SF6) and odd-electron species.
2. Check the count around the central atom.

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

Explanation:
1. \( \ce{Be^2+} \) is very small with a high charge density, and \( \ce{Sn^4+} \) has a high charge.
2. Both strongly polarise the chloride ion, giving high covalent character (Fajans' rules).
3. NaCl and KF are typical ionic compounds.

Teacher's Note:
1. Small, highly charged cations give covalent compounds.
2. BeCl2 is the classic example in the s block.

Answer: (b) \( \ce{BrF5} \), \( \ce{SF6} \) and \( \ce{IF7} \)

Explanation:
1. In CO the C and O atoms both have an octet (with a triple bond and a coordinate bond).
2. Br in \( \ce{BrF5} \) has 12 electrons, S in \( \ce{SF6} \) has 12 and I in \( \ce{IF7} \) has 14.
3. So the octet rule is not applicable to the three hypervalent molecules.

Teacher's Note:
1. Hypervalent molecules have more than eight electrons on the central atom.
2. They occur in period 3 and below.

Answer: (b) 1 and 3 only

Explanation:
1. In the usual Lewis structures with double bonds, Cl in \( \ce{ClO4-} \) has 14 electrons and S in \( \ce{SO4^2-} \) has 12.
2. \( \ce{BF3} \) is electron deficient and \( \ce{CO3^2-} \) obeys the octet.
3. So only 1 and 3 are hypervalent.

Teacher's Note:
1. Hypervalent means more than eight electrons around the central atom.
2. Electron-deficient species are the opposite case.

Answer: (d) covalent and coordinate

Explanation:
1. \( \ce{N2O5} \) is O2N-O-NO2.
2. The N=O and N-O links are covalent.
3. Each nitrogen forms a coordinate bond to one oxygen atom.

Teacher's Note:
1. Oxides of non-metals have covalent and coordinate bonds.
2. Drawing the Lewis structure shows the coordinate links.

Answer: (c) a coordinate bond

Explanation:
1. \( \ce{NH3} \) has a lone pair (Lewis base) and \( \ce{BF3} \) has an empty p orbital (Lewis acid).
2. The lone pair is donated to boron, forming \( \ce{H3N->BF3} \), a coordinate bond.

Teacher's Note:
1. A coordinate bond forms between a lone pair donor and an electron-deficient acceptor.
2. The resulting adduct has tetrahedral boron.

Answer: (d) \( \ce{CO3^2-} \)

Explanation:
1. \( \ce{CH3NC} \) and CO have coordinate bonds in their usual structures, and ozone has a coordinate bond from the central oxygen.
2. \( \ce{CO3^2-} \) has one C=O, two C-O(-) bonds, and all are ordinary covalent bonds.

Teacher's Note:
1. A negative charge on an oxygen does not mean a coordinate bond.
2. Draw the Lewis structure to check for lone pair donation.

Answer: (d) 6

Explanation:
1. Sulphur forms six bonds in \( \ce{H2SO4} \) (two S=O and two S-OH links).
2. So its covalency is 6.

Teacher's Note:
1. Sulphur shows covalency 2, 4 and 6.
2. Maximum covalency of S is 6.

Answer: (c) 1.6 N

Explanation:
1. Molar mass of \( \ce{N3-} \) = 42 g, so 4.2 g is 0.1 mol, i.e. 0.1 N ions.
2. Valence electrons per ion = 3 x 5 + 1 = 16.
3. Total = 0.1 N x 16 = 1.6 N.

Teacher's Note:
1. Add one valence electron for each unit of negative charge.
2. Convert mass to moles first.

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

Explanation:
1. \( \ce{B2H6} \) and \( \ce{Al2H6} \) are electron deficient and have two bridging hydrogen atoms (three-centre two-electron bonds) with no direct B-B or Al-Al bond.
2. Ethane and disilane have a direct C-C and Si-Si bond.

Teacher's Note:
1. Diborane has banana bonds.
2. Its bridging hydrogens hold the two boron atoms together.

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

Explanation:
1. \( \ce{CO2} \) has 6 + 16 = 22 electrons.
2. \( \ce{N3-} \) has 21 + 1 = 22, \( \ce{CNO-} \) has 6 + 7 + 8 + 1 = 22 and \( \ce{NCN^2-} \) has 7 + 6 + 7 + 2 = 22.
3. \( \ce{NO2-} \) has 7 + 16 + 1 = 24.

Teacher's Note:
1. Count total electrons, not just valence electrons, for isoelectronic comparison.
2. These species are also linear.

Answer: (b) s-s < p-p < s-p

Explanation:
1. The extent of overlap decides the bond strength.
2. The s-s overlap is smallest, p-p is greater, and s-p overlap is the most effective of the three.
3. So the order is s-s < p-p < s-p.

Teacher's Note:
1. Learn this order as it is standard in NCERT.
2. More overlap means a stronger bond.

Answer: (d) sp2-sp3

Explanation:
1. C2 is part of a double bond, so it is sp2.
2. C3 is the CH2 group with four single bonds, so it is sp3.
3. The C2-C3 bond is therefore sp2-sp3.

Teacher's Note:
1. Hybridisation is decided by the number of sigma bonds plus lone pairs.
2. Double bond carbon is sp2, triple bond carbon is sp.