The correct formal charges on oxygen atoms numbered 2, 1 and 3 respectively are :

Core Logic

Central atom: Cl (Group 17, 7 valence electrons). Atoms attached: 3 Fluorine atoms. Bonds: 3 single bonds $\implies$ 3 bond pairs (bp). Remaining electrons = 7 - 3 = 4 2$7 - 3 = 4 \implies 2$ lone pairs (lp). Total electron pairs = 3 (bp) + 2 (lp) = 5$3 \text{ (bp)} + 2 \text{ (lp)} = 5$. Hybridisation = sp³d$sp^3d$. Electron geometry = Trigonal bipyramidal. Molecular geometry (shape) = T-shape, with lone pairs occupying the equatorial positions to minimize repulsion.

Pattern Recognition

AX₃E₂$AX_3E_2$ systems always adopt a T-shaped molecular geometry because the two lone pairs prefer the equatorial plane (120° apart) to minimize 90° repulsions.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Core Logic

Let's analyze the bonding structure for each molecule: (A) C₂H₄$C_2H_4$ (Ethene): H₂C=CH₂$H_2C=CH_2$. The C=C$C=C$ double bond has 1 σ$\sigma$ and 1 π$\pi$ bond. Four C-H bonds are 4 σ$\sigma$ bonds. Total: 5 σ$\sigma$, 1 π$\pi$. (Matches IV) (B) C₂H₂$C_2H_2$ (Ethyne): HC ≡ CH$HC \equiv CH$. The C ≡ C$C \equiv C$ triple bond has 1 σ$\sigma$ and 2 π$\pi$ bonds. Two C-H bonds are 2 σ$\sigma$ bonds. Total: 3 σ$\sigma$, 2 π$\pi$. (Matches I) (C) CH₄$CH_4$ (Methane): Four single C-H bonds. Total: 4 σ$\sigma$. (Matches III) (D) NH₃$NH_3$ (Ammonia): Central N has 3 single N-H bonds and 1 lone pair. Total: 3 σ$\sigma$, 1 lone pair. (Matches II)

Step 1: Final Match

A arrow$\rightarrow$ IV B arrow$\rightarrow$ I C arrow$\rightarrow$ III D arrow$\rightarrow$ II

Pattern Recognition

Every single bond is a sigma (σ$\sigma$). Every double bond adds one pi (π$\pi$). Every triple bond adds two pi (π$\pi$) bonds. Ammonia's distinguishing feature is the explicit lone pair presence.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure