Chemical Bonding And Molecular Structure

Let’s understand the structure and resonance in each compound:

1. Ethyl propanoate (I): CH₃CH₂COOCH₂CH₃
   → No conjugation with the carbonyl (C=O) group. So, it has a typical ester C=O bond length.
2. Ethyl propenoate (II): CH₂=CH–COOCH₂CH₃
   → The carbonyl group is conjugated with the adjacent C=C double bond (alkene), which causes resonance.
   → This partial delocalization lengthens the C=O bond slightly, making it longer than in (I).
3. Ethenyl propanoate (III): CH₃CH₂COOCH=CH₂
   → The vinyl group is attached to the –OR part (not conjugated with the C=O group).
   → But –OCH=CH₂ can still donate some electron density through resonance, increasing delocalization over the ester group.

Now consider the resonance effects:

• Conjugation with the C=O group increases delocalization and results in longer C=O bonds.
• More conjugation → more delocalization → longer C=O bond.

So, the bond length order is:

II (most conjugation, longest C=O) > I (normal ester) > III (least conjugation affecting C=O)

Correct order: II > I > III

Correct answer: Option (D)

To determine if all atoms in a molecule lie on the same plane, we need to look for:

• Aromatic ring systems (which are planar due to conjugation)
• Substituents that are sp² hybridized or conjugated with the ring

Option (A): 4-nitrobenzaldehyde

• Benzene ring: planar
• –NO₂: the nitrogen is sp² hybridized and conjugated with the ring
• –CHO (aldehyde): carbon is sp² hybridized and also conjugated
• All atoms lie in the same plane → correct

Other options:

• 4-methoxybenzaldehyde: –OCH₃ group has an sp³ carbon (methyl), so not all atoms are planar
• 4-methyl nitrobenzene: –CH₃ is sp³ → non-planar group
• 4-nitroacetophenone: –COCH₃ group has a methyl (sp³) → not completely planar

Correct answer: Option (A): 4-nitrobenzaldehyde

Self-ionisation of BrF₃:

Just like water (H₂O) and sulfur dioxide (SO₂), bromine trifluoride (BrF₃) undergoes self-ionisation:

2BrF₃ ⇌ BrF₂⁺ + BrF₄⁻

This is the correct reaction for the self-ionisation of BrF₃.

Geometry and bonding:

• BrF₂⁺ is linear
• BrF₄⁻ is square planar

In BrF₄⁻, there are four F−Br−F bond angles of 90° due to the square planar geometry, and this matches the description in the question.

Therefore, the correct answer is option (B): 2BrF₃ ⇌ BrF₂⁺ + BrF₄⁻

To determine the number of lone pairs and molecular shapes for SF₄, CF₄, and XeF₄, let's examine each molecule:

1) SF₄:

• Central atom = Sulfur (S), atomic number = 16
• Valence electrons = 6
• SF₄ has 4 bonding pairs (from 4 F atoms) and 1 lone pair
• Electron pair geometry = Trigonal bipyramidal
• Molecular geometry = See-saw

2) CF₄:

• Central atom = Carbon (C), atomic number = 6
• Valence electrons = 4
• CF₄ has 4 bonding pairs and 0 lone pairs
• Electron pair geometry = Tetrahedral
• Molecular geometry = Tetrahedral

3) XeF₄:

• Central atom = Xenon (Xe), atomic number = 54
• Valence electrons = 8
• XeF₄ has 4 bonding pairs and 2 lone pairs
• Electron pair geometry = Octahedral
• Molecular geometry = Square planar

Conclusion:

• SF₄ → 1 lone pair
• CF₄ → 0 lone pairs
• XeF₄ → 2 lone pairs

Correct answer: (D) the same with 1, 0 and 2 lone pairs of electrons on the central atoms, respectively.

To find the hybridisation of nitrogen in different species, we use this rule:

• Count the regions of electron density (i.e., number of sigma bonds + lone pairs).
• Use the following chart:

1) NO₃⁻ (Nitrate ion):

• N is bonded to 3 O atoms.
• It has no lone pair.
• Total regions = 3 → sp² hybridisation.

2) NO₂ (Nitrogen dioxide):

• N is bonded to 2 O atoms.
• It has 1 lone pair.
• Total regions = 3 → sp² hybridisation.

3) NO₂⁻ (Nitrite ion):

• N is bonded to 2 O atoms.
• It has 1 lone pair.
• Total regions = 3 → sp² hybridisation.

4) NO₂⁺ (Nitronium ion):

• N is bonded to 2 O atoms.
• It has no lone pair.
• Total regions = 2 → sp hybridisation.

Conclusion: Only NO₂⁺ has nitrogen in sp hybridisation.

Correct answer: (C) NO₂⁺

The correct answer is option (B): Kr⁺

Explanation:

When a gaseous mixture of He, Ne, Ar, and Kr is irradiated with photons that have just enough energy to ionize argon (Ar), the following points are important:

• Ionization energy is the energy required to remove an electron from an atom.
• If a photon has energy equal to the ionization energy of Ar, then it has enough energy to ionize atoms with lower ionization energy than Ar, but not those with higher ionization energy.

Ionization energies (approximate values):

• He: 24.6 eV
• Ne: 21.6 eV
• Ar: 15.8 eV
• Kr: 14.0 eV

Since the photon energy is just enough to ionize Ar (15.8 eV), it cannot ionize He or Ne (which require more energy), but it can ionize Kr (which requires less).

Hence, only Kr will be ionized, forming Kr⁺.

We use the VSEPR (Valence Shell Electron Pair Repulsion) theory to determine the number of lone pairs and hybridization of the central atom, Xenon (Xe), in XeOF₂.
(A) Valence Electrons of Central Atom (Xe): Xenon is a noble gas (Group 18), so it has 8 valence electrons.
(B) Electrons Used for Bonding:Total electrons used by Xe in bonding = 2 (for O) + 2 (for F) = 4 electrons.
(E) Non-bonding Electrons on Xe: Number of non-bonding electrons = (Total valence electrons of Xe) - (Electrons used in bonding) Number of non-bonding electrons = 8 - 4 = 4 electrons.
(F) Number of Lone Pairs on Xe:
Number of lone pairs = (Number of non-bonding electrons) / 2
Number of lone pairs = 4 / 2 = 2 lone pairs.
(G) Steric Number (SN):
SN = (Number of atoms bonded to Xe) + (Number of lone pairs on Xe)
SN = (1 O + 2 F) + 2 = 3 + 2 = 5.
(H) Hybridization: The hybridization corresponding to a steric number of 5 is sp³d.
Thus, Xenon in XeOF₂ has 2 lone pairs of electrons and its hybridization is sp³d. This corresponds to option (D). The electron geometry is trigonal bipyramidal, and the molecular geometry is T-shaped.