JEE Chemistry Chemical Bonding And Molecular Structure MCQs Set 47

Answer: (c) [ICl₄]⁻

Explanation:
1. Hybrid orbitals H = ½(V + M − C + A), where V = valence electrons of the central atom, M = monovalent atoms, C = cationic charge and A = anionic charge.
2. [ICl₂]⁻ and [BrF₂]⁻: ½(7 + 2 + 1) = 5 (sp³d); [ICl₄]⁻: ½(7 + 4 + 1) = 6 (sp³d²); [IF₆]⁻: ½(7 + 6 + 1) = 7 (sp³d³).
3. So [ICl₄]⁻ has sp³d² hybridisation.

Teacher's Note:
1. Add one for each negative charge and subtract one for each positive charge.
2. H = 6 corresponds to sp³d².

Answer: (a) C₂²⁻

Explanation:
1. C₂²⁻ (14 e⁻): all paired, bond order 3. O₂²⁻ (18 e⁻): all paired, bond order 1. O₂ and N₂²⁻ (16 e⁻) each have two unpaired electrons.
2. Diamagnetic species: C₂²⁻ and O₂²⁻. Bond length is inversely related to bond order, so C₂²⁻ (BO = 3) has the shorter bond.

Teacher's Note:
1. Write the MO configuration to check for unpaired electrons.
2. Higher bond order means shorter bond length.

Answer: (a) CO

Explanation:
1. CO has 14 electrons with all of them paired (isoelectronic with N₂).
2. B₂ (two unpaired electrons in π orbitals), NO (one unpaired electron) and O₂ (two unpaired electrons in π*) are paramagnetic.

Teacher's Note:
1. A diamagnetic species has no unpaired electron.
2. Count electrons and fill the MOs.

Answer: (d) I and III only

Explanation:
1. VBT cannot explain colour (which needs d-orbital splitting) and cannot classify ligands as weak or strong field, so I and III are correct.
2. VBT gives only a qualitative idea of magnetic behaviour, so II is incorrect.

Teacher's Note:
1. Crystal field theory is needed for colour and for the spectrochemical series.
2. Option (d) combines I and III.

Answer: (d) N³⁻, O²⁻, F⁻ and Na⁺

Explanation:
1. Electrons: N³⁻ = 10, O²⁻ = 10, F⁻ = 10, Na⁺ = 10, Li⁺ = 2, Mg²⁺ = 10.
2. Only the set N³⁻, O²⁻, F⁻ and Na⁺ has the same number of electrons (10).

Teacher's Note:
1. Isoelectronic species have the same number of electrons.
2. Compare each ion's electron count.

Answer: (b) π*2pₓ

Explanation:
1. O₂ (16 e⁻) has the configuration ... π2pₓ² π2p_y² π*2pₓ¹ π*2p_y¹.
2. The extra electron is added to one of the half-filled antibonding π* orbitals (π*2pₓ or π*2p_y), giving O₂⁻ with one unpaired electron.

Teacher's Note:
1. Electrons enter the lowest-energy vacant or half-filled orbital.
2. π* orbitals lie below σ*2p_z in O₂.

Answer: (c) N₂⁺

Explanation:
1. N₂⁺ (13 e⁻): σ1s² σ*1s² σ2s² σ*2s² π2p_x² π2p_y² σ2p_z¹.
2. The bonding MOs hold four π electrons (two π bonds) and one σ electron (half a σ bond).
3. N₂ has two π and one σ bond, O₂⁺ has bond order 2.5 with a different distribution.

Teacher's Note:
1. Count the electrons in bonding MOs after cancelling those in antibonding ones.
2. The σ2p_z orbital holds a single electron in N₂⁺.

Answer: (d) H₂²⁻

Explanation:
1. Bond orders: He₂²⁺ = 1, He₂⁺ = 0.5, H₂⁻ = 0.5, H₂²⁻ = (2 − 2)/2 = 0.
2. A species with zero bond order is not viable, so H₂²⁻ does not exist.

Teacher's Note:
1. Viability requires a positive bond order.
2. H₂²⁻ has two electrons in σ1s and two in σ*1s.

Answer: (c) KCl

Explanation:
1. PH₃, O₂, B₂H₆ and H₂SO₄ all contain covalent bonds.
2. KCl is purely ionic, so only KCl has no covalent bond.

Teacher's Note:
1. B₂H₆ has two three-centre two-electron bonds besides normal covalent bonds.
2. Metal halides of alkali metals are ionic.

Answer: (b) CO

Explanation:
1. NO has one unpaired electron, O₂ has two, B₂ has two (π orbitals half-filled).
2. CO (14 e⁻) has all electrons paired, so it is diamagnetic.

Teacher's Note:
1. CO is isoelectronic with N₂.
2. Count unpaired electrons from the MO diagram.

Answer: (b) It contains Cs⁺ and I₃⁻ ions

Explanation:
1. Cs⁺ is a very large cation that stabilises the large polyhalide anion I₃⁻.
2. Cs³⁺ is not possible, so CsI₃ is an ionic compound of Cs⁺ and I₃⁻.

Teacher's Note:
1. Large cations stabilise large polyhalide anions.
2. Cs shows only the +1 oxidation state.

Answer: (a) C₂ and N₂

Explanation:
1. C₂ (12 e⁻): all paired in σ and π bonding MOs, diamagnetic. N₂ (14 e⁻): all paired, diamagnetic.
2. O₂ and S₂ have two unpaired electrons each in π* orbitals, so both are paramagnetic.

Teacher's Note:
1. O₂ and S₂ are isoelectronic in their valence shells.
2. Fill MOs using the 2p order for C₂ and N₂ with π below σ.

Answer: (c) H₂²⁺, He₂

Explanation:
1. H₂²⁺ has no electron, so BO = 0. He₂ has 4 electrons, BO = (2 − 2)/2 = 0.
2. Both have zero bond order and cannot exist; the other pairs contain at least one species with positive bond order (H₂⁺, H₂⁻, He₂²⁺).

Teacher's Note:
1. A species exists only if its bond order is positive.
2. Check each member of each pair.

Answer: (b) Li₂⁻ < Li₂⁺ < Li₂

Explanation:
1. Bond orders: Li₂ = (4 − 2)/2 = 1, Li₂⁺ = (3 − 2)/2 = 0.5, Li₂⁻ = (4 − 3)/2 = 0.5.
2. Li₂ is the most stable. Li₂⁺ and Li₂⁻ have equal bond order but Li₂⁻ has more electrons in antibonding orbitals, so it is less stable.
3. Order: Li₂⁻ < Li₂⁺ < Li₂.

Teacher's Note:
1. Compare bond orders first, then the number of antibonding electrons.
2. Extra antibonding electrons destabilise a species.

Answer: (c) PF₅ and BrF₅

Explanation:
1. CO₃²⁻ and NO₃⁻ are both trigonal planar; PCl₄⁺ and SiCl₄ are both tetrahedral; AlF₆³⁻ and SF₆ are both octahedral.
2. PF₅ is trigonal bipyramidal, while BrF₅ is square pyramidal (one lone pair on Br), so they are not isostructural.

Teacher's Note:
1. Isostructural species have the same shape and hybridisation.
2. Count lone pairs for BrF₅.

Answer: (c) AlCl₃

Explanation:
1. By Fajans' rules, covalent character increases with higher cationic charge and smaller cationic size.
2. Al³⁺ has the highest charge and smallest size among these cations, so AlCl₃ is the most covalent.

Teacher's Note:
1. Polarising power ∝ charge/size of the cation.
2. The anion (Cl⁻) is the same in all four.

Answer: (b) sp², sp, sp³

Explanation:
1. Count σ bonds + lone pairs on N: NO₃⁻ has 3 σ bonds and no lone pair, giving sp².
2. NO₂⁺ has 2 σ bonds and no lone pair, giving sp.
3. NH₄⁺ has 4 σ bonds, giving sp³.

Teacher's Note:
1. Steric number 2, 3, 4 corresponds to sp, sp², sp³.
2. Lone pairs are included in the count.

Answer: (d) pentagonal bipyramid

Explanation:
1. Iodine is excited to give 7 unpaired electrons: sp³d³ hybridisation, seven bond pairs and no lone pair.
2. The shape is a pentagonal bipyramid.

Teacher's Note:
1. Seven electron pairs with sp³d³ hybridisation give a pentagonal bipyramid.
2. Five F atoms lie in a plane at 72° and two lie axial.

Answer: (a) one sigma, two pi

Explanation:
1. CaC₂ contains the carbide ion C₂²⁻, isoelectronic with N₂.
2. The two carbon atoms are joined by a triple bond: one σ and two π bonds.

Teacher's Note:
1. Carbide ion, −C≡C−, has a triple bond.
2. Count one σ bond plus the remaining bonds as π.

Answer: (a) O₂²⁺

Explanation:
1. Bond orders: O₂²⁺ = 3, O₂⁺ = 2.5, O₂⁻ = 1.5, O₂²⁻ = 1.
2. Bond length is inversely proportional to bond order, so O₂²⁺ is shortest.

Teacher's Note:
1. Removing antibonding electrons raises the bond order.
2. O₂²⁺ is isoelectronic with N₂.