Given below are two statements: Statement (I) : for C F₃$\mathrm{C}\ell \mathrm{F}_{3}$ , all three possible structures may be drawn as follows.

Statement (II) : Structure III is most stable, as the orbitals having the lone pairs are axial, where the p-$\ell \mathfrak{p}-$ bp repulsion is minimum. In the light of the above statements, choose the most appropriate answer from the options given below:

Related Formula

Paramagnetism $\implies$ Presence of at least one unpaired electron in the molecular orbitals.

Core Logic

According to Molecular Orbital Theory (MOT):

• O₂$O_2$ has 16 electrons. Its outer configuration contains two unpaired electrons in the anti-bonding orbitals: π2pₓ = π2py$\pi^{}_{2p_x} = \pi^{}_{2p_y}$. Thus, it is paramagnetic.
• S₂$S_2$ belongs to the same oxygen family group and shares an analogous valence configuration with two unpaired electrons in its anti-bonding π^*$\pi^*$ orbitals. Hence, it is also paramagnetic.
• N₂$N_2$ (14e-), F₂$F_2$ (18e-), and Cl₂$Cl_2$ (34e-) have completely paired electronic systems and behave diamagnetically.

Pattern Recognition

Both O₂$O_2$ and S₂$S_2$ contain 2 unpaired electrons in their highest occupied molecular orbitals, making them classic examples of paramagnetic diatomic species.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Related Formula

Core Logic

Let's perform steric calculations for each species:

• SO₂$\text{SO}_2$: Central sulfur atom has 6 valence electrons, forms 2 σ$2\,\sigma$-bonds (and 2 π$2\,\pi$-bonds) with oxygen, leaving 1 lone pair. Steric number = 2 + 1 = 3 sp²$= 2 + 1 = 3 \implies sp^2$.
• NO₂^-$\text{NO}_2^-$: Central nitrogen atom has 5 valence electrons + 1$+ 1$ from negative charge = 6$= 6$. It forms 2 σ$2\,\sigma$-bonds, leaving 1 lone pair. Steric number = 2 + 1 = 3 sp²$= 2 + 1 = 3 \implies sp^2$.
• N₃^-$\text{N}_3^-$ (Azide ion): Linear configuration structure can be drawn as:

The central nitrogen has 2 σ$2\,\sigma$-bonds and 0 lone pairs. Steric number = 2 + 0 = 2 sp$= 2 + 0 = 2 \implies sp$.

Step 1: Geometry Outlines

The individual orbital fields are represented visually:

Hence, hybridizations follow the order: sp²$sp^2$, sp²$sp^2$, and sp$sp$.

Pattern Recognition

Steric short tracking: Species with linear structures like CO₂, N₂O, N₃^-$\text{CO}_2, \text{N}_2O, \text{N}_3^-$ possess central atoms that are always sp$sp$ hybridized due to the requirement of two opposing σ$\sigma$-bonds.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Core Logic

Analysis of Statement I: Ozone (O₃$\mathrm{O}_3$) exhibits resonance. The two major canonical forms contribute equally to the resonance hybrid, meaning both oxygen-oxygen bonds are identical. Their bond order is 1.5$1.5$, making the bond length intermediate between a true single bond and a true double bond. Thus, Statement I is completely true.

Analysis of Statement II: Statement II claims that lone pair-lone pair repulsion is solely responsible for this intermediate bond length. This is incorrect. The intermediate bond parameter is fundamentally a direct consequence of resonance delocalization, not lone-pair repulsions. Thus, Statement II is false.

Pattern Recognition

Whenever a molecule has identical intermediate bond lengths instead of distinct single and double bonds, resonance delocalization is almost always the core underlying reason.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Core Logic

For atomic orbitals to successfully combine into molecular orbitals, they must share appropriate spatial symmetry relative to the internuclear axis (z$z$-axis).

• Combination A, B, C, and E involve orbitals with mismatching symmetry planes, resulting in a net zero overlap integral.
• Combination D (2s$2s$ and 2pz$2p_z$) preserves continuous spatial alignment along the z$z$-axis, allowing effective frontal overlap to synthesize a stable sigma molecular orbital.

Visual symmetry breakdowns:

• 

  (Symmetry mismatch for A)

• 

  (Symmetry mismatch for C)

• 

  (Valid overlapping leading to Sigma molecular orbital for D)

• 

  (Symmetry mismatch for E)

Pattern Recognition

Verify orbital symmetry signs across the designated internuclear reference line. Mismatched symmetries cancel out completely (Ioverlap = 0$I_{\text{overlap}} = 0$).

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure