CBSE Class 11 Chemistry Chemical Bonding and Molecular Structure MCQs Set 06

Answer: (d) CO₂

Explanation:
1. In CO₂, carbon shares two electron pairs with each oxygen atom (two double bonds), so C has 8 electrons in its valence shell and each O also has 8 electrons.
2. Lewis structure of CO₂ :
3. LiCl (Li has only 2 electrons around it), BeH₂ (Be has 4 electrons) and BCl₃ (B has 6 electrons) are molecules with incomplete octet of the central atom.
4. Hence, CO₂ has a complete octet around the central atom.

Teacher's Note:
1. Count only the electrons around the central atom: shared pairs plus its own lone pairs.
2. Be, B and Al compounds such as BeH₂, BCl₃ and AlCl₃ are the standard electron-deficient exceptions.

Answer: (a) Only I

Explanation:
1. In structure I, oxygen is shown with three lone pairs along with the triple bond, i.e. 12 electrons around O. This breaks the octet rule, so structure I is wrong.
2. The correct Lewis structure of CO has one lone pair on C and one lone pair on O :
3. Structure II (NO₂⁻) and structure III (CO₃²⁻) give every atom a complete octet and the correct total number of electrons (18 and 24 valence electrons respectively), so they are correct.
4. Hence, only structure I is wrong.

Teacher's Note:
1. CO has 10 valence electrons: 6 are used in the triple bond, leaving one lone pair each on C and O.
2. Always check the total valence electrons and the octet of each atom when judging a Lewis structure.

Answer: (c) CCl₄

Explanation:
1. Carbon tetrachloride (CCl₄) has an octet around the central atom: C forms four single bonds (4 × 2 = 8 electrons).
2. The number of bonded electrons around the central atom in the other compounds are : PF₅ (10e⁻), SF₆ (12e⁻), BF₃ (6e⁻).
3. Structural representation of CCl₄ is as follows :

Teacher's Note:
1. PF₅ and SF₆ are expanded-octet (hypervalent) molecules, while BF₃ is electron deficient.
2. Multiply the number of bonds by 2 to count shared electrons around the central atom.

Answer: (a) 10

Explanation:
1. Krypton (Kr) forms two covalent bonds with two fluorine atoms (one with each fluorine).
2. Structure of KrF₂ :
3. Number of electrons in valence shell of Kr = 8
4. Number of electrons it shares with two F-atoms (one from each F) = 2
5. ∴ Total number of electrons found around Kr in KrF₂ = 8 + 2 = 10

Teacher's Note:
1. Noble gas compounds such as KrF₂, XeF₂ and XeF₄ always have more than 8 electrons around the central atom.
2. Kr in KrF₂ has 3 lone pairs and 2 bond pairs, i.e. 5 electron pairs = 10 electrons.

Answer: (b) XeO₄

Explanation:
1. XeO₄ contains equal number of σ and π-bonds.
2. Number of σ and π-bonds present in each species are as follows :
3. HCO₃⁻ : 4σ and 1π; XeO₄ : 4σ and 4π; (CN)₂ : 3σ and 4π; CH₂(CN)₂ : 6σ and 4π.
4. Only XeO₄ has the same number (4) of σ and π-bonds.

Teacher's Note:
1. Every single bond is one σ bond; a double bond is 1σ + 1π and a triple bond is 1σ + 2π.
2. In XeO₄, xenon forms four Xe=O double bonds, so σ and π counts are both 4.

Answer: (c) −0.75

Explanation:
1. In PO₄³⁻, the total charge of −3 is shared equally by the four oxygen atoms (resonance).
2. Formal charge on each O-atom of P—O bond = \( \frac{\text{total charge}}{\text{number of O-atoms}} \) = \( -\frac{3}{4} \) = −0.75

Teacher's Note:
1. For resonating oxoanions, the average formal charge on each O = total charge ÷ number of equivalent O atoms.
2. The same idea gives −2/3 = −0.67 for each O in CO₃²⁻ and −1/3 for each O in NO₃⁻.

Answer: (a) No definite geometry

Explanation:
1. Covalent bonds are directional. Therefore, covalent compounds have a definite shape (geometry), e.g. acetylene is linear and methane is tetrahedral in shape.
2. Covalent compounds are generally insoluble in polar solvents, are formed between atoms with small electronegativity difference and have low melting points.
3. Hence, ‘no definite geometry’ is not a characteristic of a covalent compound.

Teacher's Note:
1. Directional covalent bonds are the reason VSEPR and hybridisation can predict molecular shapes.
2. Ionic bonds, not covalent bonds, are non-directional.

Answer: (b) high electron affinity, low ionisation energy

Explanation:
1. Formation of ionic bond is favoured by
2. (i) low ionisation potential (energy) of the metal,
3. (ii) greater value of electron affinity of the non-metal,
4. (iii) higher value of lattice energy of the resulting ionic compound.
5. Hence, high electron affinity and low ionisation energy favour ionic compounds.

Teacher's Note:
1. The metal must lose electrons easily (low IE) and the non-metal must accept them readily (high EA).
2. That is why alkali metals and halogens form typical ionic compounds.

Answer: (a) CaCl₂

Explanation:
1. In CaCl₂, Ca—Cl bonds are electrovalent or ionic bonds because of the high electronegativity difference between Ca and Cl-atoms.
2. AlCl₃, SiCl₂ and PCl₃ are covalent because the electronegativity difference is small and the small, highly charged cation (Al³⁺) polarises Cl⁻ strongly.

Teacher's Note:
1. Ca is an s-block metal with low ionisation enthalpy, so it readily forms Ca²⁺ ions.
2. AlCl₃ is a common trap: by Fajans’ rule, small Al³⁺ makes it largely covalent.

Answer: (c) Al₂O₃ > CaO > MgBr₂ > NaCl

Explanation:
1. Lattice energy depends upon (i) the product of ionic charges, (ii) the interionic distance and (iii) the nature of the crystal structure.
2. Hence, more is the ionic charge, more will be the lattice energy.
3. Product of ionic charges : Al₂O₃ (Al³⁺, O²⁻) = 6; CaO (Ca²⁺, O²⁻) = 4; MgBr₂ (Mg²⁺, Br⁻) = 2; NaCl (Na⁺, Cl⁻) = 1.
4. Approximate lattice enthalpies : Al₂O₃ ≈ 15900, CaO ≈ 3400, MgBr₂ ≈ 2400 and NaCl ≈ 790 kJ mol⁻¹.
5. So, the correct order of lattice energy is Al₂O₃ > CaO > MgBr₂ > NaCl.

Teacher's Note:
1. Charge has a much bigger effect on lattice energy than size, so start by multiplying the ionic charges.
2. Oxides with 2− ions usually have far higher lattice energies than halides with 1− ions.

Answer: (a) MgO

Explanation:
1. Lattice energy ∝ \( \frac{\text{charge of ion}}{\text{size of ion}} \)
2. All four are oxides of M²⁺ ions, so the charges are the same. On moving down the group of alkaline earth metals, the size of the ion increases due to increase in number of shells. Thus, Mg²⁺ has the smallest size.
3. Hence, lattice energy is highest for MgO.

Teacher's Note:
1. When charges are equal, the smallest ions give the highest lattice energy.
2. Lattice energy order here is MgO > CaO > SrO > BaO.

Answer: (b) C—H < C=C < C—O < C—C

Explanation:
1. The C—H bond is the shortest due to the small size of H.
2. Double bonds are shorter than single bonds. Hence, C=C bond is shorter than C—O and C—C.
3. Due to the electronegativity difference between C and O and the smaller size of O, the C—O bond length is shorter than the C—C bond.
4. Hence, the correct order is C—H (109 pm) < C=C (134 pm) < C—O (143 pm) < C—C (154 pm).

Teacher's Note:
1. Bond length increases with atomic size and decreases with bond multiplicity.
2. Remember the typical values: C—H 109 pm, C=C 134 pm, C—O 143 pm, C—C 154 pm.

Answer: (a) H₂S (2lp)

Explanation:
1. H₂S has the lowest bond angle because two lone pairs are present on S and S has low electronegativity, so the bond pairs lie farther from S and repel each other less.
2. H₂S (92°) < H₂O (104.5°) < NH₃ (107°) < SO₂ (119.5°)

Teacher's Note:
1. Lone pairs squeeze bond angles; a larger, less electronegative central atom squeezes them further.
2. That is why H₂S (92°) has a much smaller angle than H₂O (104.5°).

Answer: (c) opposite polarity of N in the two molecules

Explanation:
1. In NF₃, N is less electronegative than F, so N carries partial positive charge; in NH₃, N is more electronegative than H, so N carries partial negative charge. Thus, the polarity of N is opposite in the two molecules.
2. In NF₃ the bonding electron pairs are pulled away from N towards F, so bond pair-bond pair repulsion near N is smaller and the bond angle decreases. In NH₃, the bond pairs lie closer to N and repel more.
3. Thus, the bond angle in NF₃ is smaller than that in NH₃ because of the opposite polarity of N in these molecules.

Teacher's Note:
1. With the same central atom, the bond angle decreases as the electronegativity of the attached atoms increases.
2. Compare also: PF₃ < PCl₃ and OF₂ < H₂O follow the same reasoning.

Answer: (c) 464.5 kJ mol⁻¹

Explanation:
1. The mean or average bond enthalpy of O—H bonds in H₂O is obtained by dividing the total bond dissociation enthalpy by the number of bonds broken.
2. Average bond enthalpy = \( \frac{502 + 427}{2} \) = \( \frac{929}{2} \) = 464.5 kJ mol⁻¹

Teacher's Note:
1. Mean bond enthalpy = total enthalpy to break all identical bonds ÷ number of bonds.
2. 929 kJ mol⁻¹ is the total for both bonds, and 75 is only their difference, which are common traps.

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

Explanation:
1. H₂, Cl₂ and Br₂ each have a single bond, so their bond order is 1.
2. In CO (three shared electron pairs between C and O) the bond order is 3. For N₂ (N≡N), bond order is 3.
3. Hence, both (a) and (b) options are correct.

Teacher's Note:
1. Isoelectronic species such as CO, N₂, CN⁻ and NO⁺ (14 electrons) all have bond order 3.
2. Bond order = number of shared electron pairs between the two atoms in a Lewis structure.

Answer: (b) 1 D = 3.33564 × 10⁻³⁰ C m

Explanation:
1. 1 D = 3.33564 × 10⁻³⁰ C m
2. In CGS units, 1 D = 1 × 10⁻¹⁸ esu cm.

Teacher's Note:
1. Dipole moment = charge × distance, so its SI unit is coulomb metre (C m).
2. Keep both conversions ready: 1 D = 3.33564 × 10⁻³⁰ C m = 10⁻¹⁸ esu cm.

Answer: (b) is vector sum of the dipole moments of various bonds

Explanation:
1. In case of a polyatomic molecule, the dipole moment is the vector sum of the dipole moments of the various bonds.
2. Therefore, it depends on both the bond dipoles and the geometry (spatial arrangement) of the bonds.

Teacher's Note:
1. Because dipole moment is a vector, symmetrical molecules like CO₂, BF₃ and CH₄ have zero net dipole moment.
2. Lone pairs also contribute to the resultant dipole moment.

Answer: (c) NH₃

Explanation:
1. CO₂ and CH₄ have zero dipole moment as these are symmetrical in nature.
2. CO₂ (μ_net = 0) and CH₄ (μ_net = 0) :
3. Between NH₃ and NF₃, NH₃ has the greater dipole moment though N possesses one lone pair of electrons in both.
4. In NH₃, the net N—H bond dipole is in the same direction as the dipole of the lone pair, but in NF₃, the resultant of the three N—F bond dipoles is opposite to the dipole moment of the lone pair.
5. Resultant of 3 N—H bonds lies in the same direction as μ₄ (lone pair) in NH₃; resultant of 3 N—F bonds lies opposite to μ₄ in NF₃ (μ_net = 0.2 D).
6. Hence, NH₃ (1.47 D) has the maximum dipole moment.

Teacher's Note:
1. Always add the lone pair dipole as a vector along with the bond dipoles.
2. μ(NH₃) = 1.47 D is much larger than μ(NF₃) = 0.23 D even though F is more electronegative than H.

Answer: (d) Stability ∝ bond order ∝ bond enthalpy ∝ \( \frac{1}{\text{bond length}} \)

Explanation:
1. A general correlation useful for understanding the stabilities of molecules is that, with increase in bond order, bond enthalpy increases and bond length decreases.
2. Stability increases with increase in bond order and bond enthalpy.
3. Therefore, stability ∝ bond order ∝ bond enthalpy, and stability increases with decrease in bond length.
4. Therefore, stability ∝ bond order ∝ bond enthalpy ∝ \( \frac{1}{\text{bond length}} \)

Teacher's Note:
1. Higher bond order means more shared electrons, a shorter and stronger bond.
2. Only bond length is inversely related; stability, bond order and bond enthalpy rise together.

Answer: (b) MnO₂

Explanation:
1. According to Fajans’ rule, a cation with a higher charge (higher oxidation state) has greater polarising power, so its compound is more covalent and less ionic.
2. Oxidation states : Mn⁺⁴ in MnO₂, Mn⁺⁷ in Mn₂O₇, P⁺⁵ in P₂O₅ and P⁺³ in P₂O₃.
3. P₂O₅ and P₂O₃ are oxides of a non-metal and are covalent (molecular) oxides.
4. Between the two manganese (metal) oxides, Mn in MnO₂ has the lower oxidation state (+4), so MnO₂ is more ionic than Mn₂O₇ (+7), which is a covalent, acidic oxide.
5. So, MnO₂ will be the most ionic.

Teacher's Note:
1. Metal oxides in low oxidation states are more ionic; very high oxidation states (Mn₂O₇, CrO₃) give covalent oxides.
2. Non-metal oxides such as P₂O₃, P₂O₅, SO₂ and CO₂ are covalent.

Answer: (c) NaCl < LiCl < BeCl₂

Explanation:
1. On the basis of Fajans’ rule, the smaller the size of the cation, the higher will be its polarising power and the higher will be the covalent character.
2. ∴ Polarising power ∝ \( \frac{1}{\text{size of cation}} \) and covalent character ∝ polarising power
3. The order of size of cations is Na⁺ > Li⁺ > Be²⁺ (Be²⁺ also has a higher charge).
4. So, the correct order of increasing covalent character is NaCl < LiCl < BeCl₂.

Teacher's Note:
1. Small size and high charge of the cation both increase polarisation and covalent character.
2. BeCl₂ is largely covalent, LiCl is partly covalent and NaCl is typically ionic.

Answer: (b) AlCl₃

Explanation:
1. Al³⁺ cation is smaller than Na⁺ and Mg²⁺ (because of greater nuclear charge) and carries a high charge.
2. According to Fajans’ rule, a small, highly charged cation polarises the anion to a greater extent.
3. Hence, Al³⁺ polarises Cl⁻ ions to a great extent, therefore AlCl₃ has covalent bonds between Al and Cl-atoms.
4. Na₂S, NaH and MgCl₂ are ionic.

Teacher's Note:
1. AlCl₃ exists as covalent Al₂Cl₆ dimers in the vapour state, which shows its covalent nature.
2. NaH is an ionic (saline) hydride containing the H⁻ ion.

Answer: (d) All of these

Explanation:
1. All the given molecules can be represented in terms of resonance. These are shown below :
2. (a) O₃, (b) CO₃²⁻ and (c) CO₂ :

Teacher's Note:
1. Resonance is needed whenever a single Lewis structure cannot show the real, equal bond lengths.
2. Canonical structures differ only in the positions of electrons, never of nuclei.

Answer: (c) 1.33

Explanation:
1. The resonance structures of NO₃⁻ ion are as :
2. The N—O bond order = \( \frac{\text{number of bonds between two atoms}}{\text{total number of resonating structures}} \)
3. = \( \frac{4}{3} \) = 1.33

Teacher's Note:
1. In NO₃⁻, four bonds (one double + two single) are shared by three equivalent N—O links.
2. Similarly, CO₃²⁻ has C—O bond order 4/3 = 1.33 and PO₄³⁻ has P—O bond order 5/4 = 1.25.

Answer: (b) sp, sp² and sp³

Explanation:
1. The type of hybrid orbitals of nitrogen can be decided by using VSEPR theory, counting bp and lp in
2. NO₂⁺ = 2 bp + 0 lp = linear ⇒ sp-hybridised
3. NO₃⁻ = 3 bp + 0 lp ⇒ sp²⁻hybridised
4. NH₄⁺ = 4 bp + 0 lp ⇒ sp³⁻hybridised

Teacher's Note:
1. Count σ bond pairs plus lone pairs on the central atom: 2 → sp, 3 → sp², 4 → sp³.
2. π bonds do not count in deciding hybridisation.

Answer: (c) more positive and less negative

Explanation:
1. The two sp-hybrid orbitals point in the opposite direction along the z-axis with projecting more positive lobes and less negative lobes.
2. This provides more effective overlapping, resulting in the formation of stronger bonds.

Teacher's Note:
1. A hybrid orbital has one big lobe and one small lobe; the big lobe is used for overlapping.
2. Unequal lobes are why bonds formed by hybrid orbitals are stronger than those by pure s or p orbitals.

Answer: (d) ClO₃⁻, SO₃²⁻

Explanation:
1. Structures of ClO₃⁻, SO₃²⁻, CO₃²⁻ and NO₃⁻ :
2. Number of electrons :
3. CO₃²⁻ = 6 + 24 + 2 = 32
4. SO₃²⁻ = 16 + 24 + 2 = 42
5. ClO₃⁻ = 17 + 24 + 1 = 42
6. NO₃⁻ = 7 + 24 + 1 = 32
7. Hence, ClO₃⁻ and SO₃²⁻ are isoelectronic (42 electrons) and both are pyramidal in shape (3 bp + 1 lp).

Teacher's Note:
1. Isoelectronic means equal total electrons; isostructural means the same shape.
2. CO₃²⁻ and NO₃⁻ are also an isoelectronic, isostructural (trigonal planar) pair.

Answer: (d) A-1, B-3, C-4, D-2

Explanation:
1. The correct match is A → 1, B → 3, C → 4, D → 2.
2. The structures of the xenon compounds are represented below :
3. XeF₆ (6 bp + 1 lp) : distorted octahedral; XeO₃ (3 bp + 1 lp) : pyramidal; XeOF₄ (5 bp + 1 lp) : square pyramidal; XeF₄ (4 bp + 2 lp) : square planar.

Teacher's Note:
1. For xenon compounds, count bond pairs and lone pairs around Xe (Xe has 8 valence electrons).
2. XeF₄ is square planar because its two lone pairs sit opposite each other.

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

Explanation:
1. The valence bond theory explains the shape, the formation and the directional properties of bonds in polyatomic molecules like CH₄, NH₃ and H₂O etc. in terms of overlap and hybridisation of atomic orbitals.

Teacher's Note:
1. Simple overlap of pure orbitals cannot explain the 109.5° angles of CH₄; hybridisation is needed.
2. VBT = overlap of half-filled orbitals + hybridisation to explain geometry.

Answer: (c) A-2, B-3, C-1

Explanation:
1. The correct match is (A) → (2), (B) → (3), (C) → (1).
2. A : Zero overlap (out of phase due to different orientation or direction of approach).
3. B : Negative overlap (orbitals forming bond have different sign and orientation).
4. C : Positive overlap (orbitals forming bond have same sign and orientation in space).
5.

Teacher's Note:
1. Overlap is positive when lobes of the same sign overlap, negative when opposite signs overlap.
2. A p_x orbital and an s orbital approaching along z give zero net overlap.

Answer: (c) C

Explanation:
1. At point C in the curve, H₂ is found in the most stable state due to minimum energy.
2. Energy is released when the bond is formed between two hydrogen atoms; the energy is minimum at the equilibrium bond length of 74 pm, and 435.8 kJ mol⁻¹ is released.

Teacher's Note:
1. The minimum of a potential energy curve gives the bond length and the bond enthalpy.
2. At distances shorter than C, nuclear repulsion makes the energy rise sharply.

Answer: (c) sp

Explanation:
1. Number of hybrid orbitals = \( \frac{\text{number of σ-electrons (bond pair + lone pair)}}{2} \) = \( \frac{2 + 2\ (\text{negative ion})}{2} \) = 2 ⇒ sp
2. The negatively charged carbon has one σ bond pair (to the other carbon) and one lone pair.
3. Hence, in the carbanion CH₃C≡C⁻, the pair of electrons (negative charge) is present in an sp-hybridised orbital.

Teacher's Note:
1. A triple-bonded carbon is always sp hybridised; its lone pair occupies the remaining sp orbital.
2. Higher s-character makes the acetylide anion relatively stable.

Answer: (b) A-3, B-1, C-2

Explanation:
1. A → (3); B → (1); C → (2)
2. A. Tetrahedral shape – sp³⁻hybridisation
3. B. Trigonal shape – sp²⁻hybridisation
4. C. Linear shape – sp-hybridisation

Teacher's Note:
1. Bond angles help: 109.5° → sp³, 120° → sp², 180° → sp.
2. Number of hybrid orbitals equals the number of electron pairs around the central atom.

Answer: (c) The average formal charge on each oxygen atom is 0.67 units

Explanation:
1. The hybridisation of the central atom in CO₃²⁻ is sp². Hence, (a) is wrong.
2. Each resonance structure has two C—O single bonds and one C=O double bond, so (b) is wrong :
3. Due to resonance all C—O bond lengths are equal, so (d) is wrong.
4. Formal charge on each O-atom = \( \frac{\text{total charge}}{\text{number of O-atoms}} \) = \( \frac{-2}{3} \) = −0.67 units
5. Hence, statement (c) is correct (the magnitude of the average formal charge is 0.67 units, with a negative sign).

Teacher's Note:
1. CO₃²⁻ is trigonal planar (sp²) with three equal C—O bonds of bond order 1.33.
2. Average formal charge = total ionic charge ÷ number of equivalent O atoms.

Answer: (a) A-3, B-1, C-5, D-4

Explanation:
1. The correct match is (A) → (3); (B) → (1); (C) → (5); (D) → (4).
2. (A) SF₄ = number of bp (4) + number of lp (1) = 5 ⇒ sp³d-hybridisation
3. (B) IF₅ = number of bp (5) + number of lp (1) = 6 ⇒ sp³d²⁻hybridisation
4. (C) NO₂⁺ = number of bp (2) + number of lp (0) = 2 ⇒ sp-hybridisation
5. (D) NH₄⁺ = number of bp (4) + number of lp (0) = 4 ⇒ sp³⁻hybridisation

Teacher's Note:
1. d²sp³ (inner d-orbitals) is used for complexes, not for main-group molecules like IF₅.
2. 5 electron pairs → sp³d, 6 electron pairs → sp³d².

Answer: (b) molecular orbital

Explanation:
1. The electron probability distribution around a group of nuclei in a molecule is given by a molecular orbital.
2. An atomic orbital is monocentric, whereas a molecular orbital is polycentric (spread over all the nuclei).

Teacher's Note:
1. Atomic orbitals belong to one nucleus; molecular orbitals belong to the whole molecule.
2. Both bonding and antibonding MOs describe electron distribution, so ‘only’ options are wrong.

Answer: (a) 2 unpaired electrons in π_b MO

Explanation:
1. B₂ = 5 + 5 = 10 electrons
2. B₂ : σ1s², σ*1s², σ2s², σ*2s², π2p_x¹ = π2p_y¹
3. Due to the presence of 2 unpaired electrons in the bonding π (π_b) MOs, B₂ shows paramagnetic behaviour.

Teacher's Note:
1. For B₂, C₂ and N₂, π2p orbitals are filled before σ2p_z; this order makes B₂ paramagnetic.
2. Hund’s rule puts one electron each in the degenerate π2p_x and π2p_y orbitals.

Answer: (c) Be₂

Explanation:
1. (a) F₂ = σ1s², σ*1s², σ2s², σ*2s², σ2p_z², π2p_x² = π2p_y², π*2p_x² = π*2p_y²; Bond order = \( \frac{1}{2}(N_b - N_a) = \frac{1}{2}(10 - 8) = 1 \)
2. (b) O₂ = σ1s², σ*1s², σ2s², σ*2s², σ2p_z², π2p_x² = π2p_y², π*2p_x¹ = π*2p_y¹; Bond order = \( \frac{1}{2}(10 - 6) = 2 \)
3. (c) Be₂ = σ1s², σ*1s², σ2s², σ*2s²; Bond order = \( \frac{1}{2}(4 - 4) = 0 \)
4. (d) Li₂ = σ1s², σ*1s², σ2s²; Bond order = \( \frac{1}{2}(4 - 2) = 1 \)
5. Therefore, Be₂ molecule has zero bond order.

Teacher's Note:
1. Zero bond order means the molecule is not expected to exist (Be₂, He₂, Ne₂).
2. Bond order = ½ (bonding electrons − antibonding electrons).

Answer: (b) C₂

Explanation:
1. According to MO theory, C₂ is the diatomic molecular species that has only π-bonds.
2. C₂ (Z = 12) ⇒ σ1s², σ*1s², σ2s², σ*2s², π2p_x² = π2p_y²
3. The double bond in C₂ consists of both π-bonds because the last (valence) four electrons are in the two π-molecular orbitals.
4. N₂ (Z = 14) : σ1s², σ*1s², σ2s², σ*2s², π2p_x² = π2p_y², σ2p_z² (1σ and 2π-bonds)
5. Be₂ (Z = 8) : σ1s², σ*1s², σ2s², σ*2s² (σ electrons only, bond order 0)
6. O₂ (Z = 16) : σ1s², σ*1s², σ2s², σ*2s², σ2p_z², π2p_x² = π2p_y², π*2p_x¹ = π*2p_y¹ (σ and π bonds)

Teacher's Note:
1. C₂ is the classic example of a double bond made of two π bonds and no σ bond.
2. This happens because σ2p_z lies above π2p in energy for molecules up to N₂.

Answer: (c) H₂⁺ is more stable than H₂⁻

Explanation:
1. H₂⁺ : (σ1s¹) ∴ BO = \( \frac{1}{2}(1 - 0) = \frac{1}{2} \)
2. H₂⁻ : (σ1s)²(σ*1s¹) ∴ BO = \( \frac{1}{2}(2 - 1) = \frac{1}{2} \)
3. Even though the bond orders of H₂⁺ and H₂⁻ are equal, H₂⁺ is more stable than H₂⁻ as in the latter, an electron is present in the higher energy antibonding (σ*1s) orbital.

Teacher's Note:
1. When bond orders are equal, the species with fewer antibonding electrons is more stable.
2. Both ions exist because their bond order (½) is positive.

Answer: (c) A-3, B-1, C-2, D-2

Explanation:
1. He₂ does not exist (bond order 0).
2. O₂ is paramagnetic (two unpaired electrons in π*2p_x and π*2p_y).
3. Li₂ and C₂ are diamagnetic (all electrons paired).
4. Hence, correct match is (A) → (3), (B) → (1), (C) → (2), (D) → (2).

Teacher's Note:
1. O₂ is the classic case where MO theory explains paramagnetism that the Lewis structure cannot.
2. A column II entry may be used more than once in such matching questions.

Answer: (d) H₂S

Explanation:
1. H₂S does not show hydrogen bonding because S is not electronegative enough and is large in size; that is why H₂S exists as a gas at room temperature.
2. The other given molecules such as HF, H₂O and NH₃ show hydrogen bonding (H attached to F, O or N).

Teacher's Note:
1. Hydrogen bonding needs H bonded to a small, highly electronegative atom: F, O or N.
2. That is why H₂O is a liquid but H₂S is a gas at room temperature.

Answer: (d) Hydrogen bonding

Explanation:
1. The higher boiling point of water is due to hydrogen bonding.
2. The O-atom of each H₂O molecule is covalently linked with two H-atoms of its own molecule and with other H-atoms of adjacent H₂O molecules by H-bonding. Extra energy is needed to break these H-bonds.

Teacher's Note:
1. Intermolecular hydrogen bonds cause molecular association and raise boiling points.
2. Covalent O—H bonds are not broken on boiling; only intermolecular forces are overcome.

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

Explanation:
1. The boiling point of a substance increases with increase in intermolecular hydrogen bonding and molecular mass.
2. Association of molecules takes place due to intermolecular hydrogen bonding, which increases the boiling point of a substance. Larger molecular mass means stronger van der Waals forces.

Teacher's Note:
1. Intramolecular H-bonding (as in o-nitrophenol) lowers boiling point, because molecules do not associate.
2. Intermolecular H-bonding (as in p-nitrophenol) raises the boiling point.