Identify the correct statement about CIF3 from the following options:

Year	2024	Total
Questions	3	3

Year

2024

Total

Questions

Solution & Explanation

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

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Q69 neet_2026_03_may_morning Sigma and Pi Bonds

Match List I with List II :

List I	List II
A. $C₂H₄$ $C_{2}H_{4}$	I. 3 $σ$ $\sigma$ bonds, 2 $π$ $\pi$ bonds
B. $C₂H₂$ $C_{2}H_{2}$	II. 3 $σ$ $\sigma$ bonds, one lone pair
C. $CH₄$ $CH_{4}$	III. 4 $σ$ $\sigma$ bonds
D. $NH₃$ $NH_{3}$	IV. 5 $σ$ $\sigma$ bonds, 1 $π$ $\pi$ bond

Choose the correct answer from the options given below :

A. A-IV, B-I, C-III, D-II
B. A-III, B-IV, C-II, D-I
C. A-I, B-II, C-IV, D-III
D. A-II, B-III, C-I, D-IV

Solution

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): $HC \equiv CH$ $HC \equiv CH$. The $C \equiv C$ $C \equiv C$ 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

Q86 neet_2026_03_may_morning Lewis Structures and Formal Charge

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

Lewis Structures and Formal Charge diagram for Q86 - NEET 2026 Code 12 — Displays the Lewis structure of an ozone (O3) molecule with numbered oxygen atoms.

A. -1, 0, +1
B. 0, +1, -1
C. 0, 0, 0
D. +1, 0, -1

Solution

Related Formula

Formal charge = V - L - (S)/(2)

Where $V$ $V$ = Valence electrons, $L$ $L$ = non-bonding (lone pair) electrons, and $S$ $S$ = shared (bonding) electrons.

Core Logic

Evaluating the $O₃$ $O_3$ molecule based on standard Lewis structure numbering (assuming Atom 1 is the central oxygen, Atom 2 is the double-bonded terminal, and Atom 3 is the single-bonded terminal, though we must trace the question's specific numbering carefully). Based on the source's solution image:

Atom 2 (Terminal double-bonded O): $V=6, S=4, L=4$ $V=6, S=4, L=4$. $FC = 6 - 4 - (4)/(2) = 0$ $FC = 6 - 4 - \frac{4}{2} = 0$.
Atom 1 (Central O): $V=6, S=6, L=2$ $V=6, S=6, L=2$. $FC = 6 - 2 - (6)/(2) = +1$ $FC = 6 - 2 - \frac{6}{2} = +1$.
Atom 3 (Terminal single-bonded O): $V=6, S=2, L=6$ $V=6, S=2, L=6$. $FC = 6 - 6 - (2)/(2) = -1$ $FC = 6 - 6 - \frac{2}{2} = -1$.

The sequence requested is for atoms numbered 2, 1, and 3 respectively. So the order is: 0, +1, -1.

Solution for Q86 - Ozone Formal Charges — Displays the Lewis structure of an ozone (O3) molecule with numbered oxygen atoms.

Step 1: Final Conclusion

The formal charges are 0, +1, and -1 respectively.

Pattern Recognition

In ozone ( $O₃$ $O_3$), the central oxygen always carries a +1 charge, the single-bonded terminal carries -1, and the double-bonded terminal is neutral (0).

Chapter Mix

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Identify the correct statement about $CIF₃$ $\mathrm{CIF}_3$ from the following options:

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

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Solution

Core Logic

Step 1: Final Match

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

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Pattern Recognition

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Identify the correct statement about CIF₃$\mathrm{CIF}_3$ from the following options:

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Chemical Bonding and Molecular Structure Previous-Year Questions

Solution

Core Logic

Step 1: Final Match

Pattern Recognition

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Core Logic

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Pattern Recognition

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Identify the correct statement about $CIF₃$ $\mathrm{CIF}_3$ from the following options: