Which of the following attain the linear structure: (Note: more than one of the given options may be correct.)

NCERT Exemplar · Q24

Q. Which of the following attain the linear structure: (Note: more than one of the given options may be correct.)

(i) BeCl2

(ii) NCO^+

(iii) NO2

(iv) CS2

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Concept understanding — Resonance Structures

The Problem: One Picture Isn't Enough

Imagine you're trying to draw a photograph of a friend who is laughing. A single still frame captures one expression, but it misses the movement, the energy, the in-between of the laugh. A single Lewis structure does the same thing to certain molecules — it freezes them into one arrangement of electrons, but the real molecule is more like a short video clip, with electrons moving smoothly between positions.

Take ozone, $O_3$ . If you try to draw a Lewis structure, you get a dilemma. You can put the double bond on the left:

O = O - O

Or on the right:

O - O = O

Both satisfy the octet rule. Both have the same atoms. But which one is correct? Neither, alone. The real ozone molecule has two identical $O-O$ bonds — each is halfway between a single and a double bond. No single Lewis picture can show that.

The Solution: Resonance Structures

Resonance structures are a set of two or more Lewis structures that collectively describe the actual electronic structure of a molecule where a single Lewis structure is inadequate. They are connected by a double-headed arrow ( $\leftrightarrow$ ) to show they are not different molecules, but different ways of drawing the same molecule.

Important

Resonance structures are not real, separate molecules that flip back and forth. They are imaginary "snapshots" that we average together to get the true structure. The real molecule is a resonance hybrid — a blend of all contributing structures.

The Rules (Precise Statement)

Same atomic positions. Only electrons (pi bonds and lone pairs) move; atoms never move.
Same total number of electrons. You are redistributing, not adding or removing.
Valid Lewis structures. Each resonance form must obey the octet rule (for second-period elements) and have correct formal charges.
Curved arrows show electron movement. An arrow from a lone pair or a pi bond points to where those electrons go next.

How to Draw Them: The Curved Arrow Method

Take the nitrate ion, $NO_3^-$ . Start with one valid Lewis structure:

\begin{bmatrix} O \\ || \\ O - N = O \end{bmatrix}^-

Now, push electrons:

Take the lone pair on the top oxygen (the one with the negative charge) and push it down to form a double bond with nitrogen.
Simultaneously, push the existing double bond on the right up to become a lone pair on that oxygen.

You get a second structure:

\begin{bmatrix} O = N - O \\ | \\ O^- \end{bmatrix}^-

Repeat the process from this new structure, and you get a third. All three are resonance structures of $NO_3^-$ .

Tip

A quick way to spot resonance: look for a pi bond next to an atom with a lone pair (or a pi bond next to a positive charge). That's the classic "conjugated system" that allows electrons to delocalize.

The Hybrid: What the Molecule Actually Looks Like

The resonance hybrid is not an average of the bond lengths — it is the actual molecule. In $NO_3^-$ , all three $N-O$ bonds are identical, with a bond order of $1\frac{1}{3}$ (one and one-third). The negative charge is spread equally over all three oxygens, not stuck on one.

You represent the hybrid by drawing dashed lines for partial bonds and placing the charge in a circle (or using fractional charges) to show delocalization.

Common Mistakes to Avoid …

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Linearity here follows the classic electron-count rule: 16-valence-electron triatomics with no lone pair on the central atom (BeCl₂, CS₂) are linear; NO₂ (17 electrons, odd electron on N) is bent, and NCO⁺ (14 valence electrons) does not adopt the linear 16-electron geometry. The answer is (i) and (iv).

Species by species

$\text{BeCl}_2$ — beryllium contributes two bond pairs and keeps no lone pair; the two Be–Cl bonds spread to $180^\circ$ . Linear ✓
$\text{NCO}^+$ — valence electrons: $5 + 4 + 6 - 1 = 14$ . The familiar linear species of this family (CO₂, NCO⁻, N₂O) all have 16 valence electrons; removing two electrons from cyanate changes the electronic structure so that the 16-electron linear picture no longer applies. Not grouped with the linear pair. ✗ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$\mathrm{CO_3^{2-}}$
›Reveal solutionSolution
Tests resonance in $\mathrm{CO_3^{2-}}$ : the central C is $sp^2$ (not $sp^3$ ), and the resonance hybrid has all C–O bonds equal with average formal charge $-0.67$ per O.

Concept and Intuition

$\mathrm{CO_3^{2-}}$ has carbon at the centre bonded to three oxygen atoms, no lone pair on carbon, giving a trigonal planar shape ( $sp^2$ hybridisation, bond angle $120°$ ). The ion is best described not by one Lewis structure but by resonance — three equivalent structures each with one C=O double bond and two C–O single bonds, differing only in which oxygen carries the double bond. The true structure is the resonance HYBRID: an average of all three contributors, not any single one of them.

Step-by-Step Solution

Hybridisation of C: 3 σ-bonds + 0 lone pairs on C ⇒ $sp^2$ . Statement I (claims $sp^3$ ) is false.

Average formal charge per O: Total charge on the ion is $-2$ , shared over 3 equivalent oxygens by resonance ⇒ average $= -2/3 = -0.67$ per O. Statement II is true. …
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AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$\ddot{S}=C=\ddot{N}$
›Reveal solutionSolution
Applying the formal-charge formula to sulphur in each of the three thiocyanate-type resonance structures gives $0, -1, +1$ respectively.

Concept and Intuition

Formal charge is a book-keeping tool: $FC = V - N - \dfrac{B}{2}$ , where $V$ is the atom's own valence electron count, $N$ is the number of nonbonding (lone-pair) electrons drawn on it, and $B$ is the total bonding electrons around it (each bond contributing 2, shared equally). As a bond changes from single to double to triple around an atom while its total octet is preserved, the number of lone pairs it retains must shrink accordingly, and this trade-off is exactly what shifts the formal charge.

Step-by-Step Solution

Sulphur's valence electron count is $V=6$ throughout.

Structure I ( $\ddot{S}=C=\ddot{N}$ ): S carries 2 lone pairs ( $N=4$ ) and one double bond to C ( $B=4$ ). $FC_S = 6-4-\frac{4}{2} = 6-4-2 = 0$ .

Structure II ( $:\!S-C\equiv N$ ): S carries 3 lone pairs ( $N=6$ ) and one single bond to C ( $B=2$ ). $FC_S = 6-6-\frac{2}{2} = 6-6-1 = -1$ . …
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AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
$Q. From the following identify the groups that exhibit negative resonance (-R) effect when attached to conjugated system formyl (A), amino (B), alkoxy (C), cyano (D), nitro (E) (A) A, C, E only (B) B, C, D only (C) A, D, E only (D) B, D, E only$
›Reveal solutionSolution
−R (electron-withdrawing resonance) groups pull electron density away via a conjugated multiple bond to an electronegative atom; formyl, cyano and nitro fit this, while amino and alkoxy are +R donors — option (C).

Concept and Intuition

A group shows +R (positive resonance/electron-donating) effect if it has a lone pair that can be delocalised INTO the conjugated system (like –NH $_2$ or –OR, both with a lone pair on an atom directly attached to the ring/chain). A group shows −R (negative resonance/electron-withdrawing) effect if it has a $\pi$ bond to a more electronegative atom that can accept electron density FROM the conjugated system by extending the conjugation (like –CHO, –CN, –NO $_2$ , all built around a C=O, C≡N, or N=O type multiple bond).

Step-by-Step Solution

Formyl (–CHO): has a C=O; the carbonyl can accept electron density via resonance from the attached conjugated system → −R.

Amino (–NH $_2$ ): nitrogen's lone pair donates into the ring/chain → +R (electron-donating), NOT −R.

Alkoxy (–OR): oxygen's lone pair donates into the system → +R, NOT −R.

Cyano (–CN): the C≡N triple bond can accept electron density by resonance → −R. …
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AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
$Q. For ozone molecule consider the following (A) It is a linear molecule with bond angle 180° (B) It is an angular molecule with bond angle 117° (C) The bond lengths of both O-O bonds are same (D) With respect to oxygen it is thermodynamically more stable The correct options are (A) (B), (C) only (B) (A), (B) only (C) (B), (D) only (D) (A), (D) only$
›Reveal solutionSolution
Ozone is bent (~117°) with two equal, resonance-averaged O–O bonds, and is actually less stable than O₂ — so only statements (B) and (C) are correct.

Concept and Intuition

Ozone's structure is a textbook resonance example: it's a bent, angular molecule (not linear, since the central oxygen has a lone pair that occupies space and bends the molecule, similar to $\mathrm{SO_2}$ ), and its two O–O bonds are equivalent because the true structure is a resonance hybrid of two equal contributing Lewis structures (double bond on one side, single on the other, averaging to identical bond lengths in reality).

Step-by-Step Solution

(A) Linear with 180°: false — ozone is bent due to a lone pair on the central O, giving it an angular ( $sp^2$ -like) geometry.

(B) Angular with ~117°: true — the experimentally measured bond angle of ozone is about 116.8°, commonly rounded to 117°.

(C) Both O–O bond lengths equal: true — resonance between the two equivalent Lewis structures (double bond alternating sides) means the actual structure is an average, giving two identical, intermediate-length O–O bonds (~128 pm each). …
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AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
$1\ \sigma$
›Reveal solutionSolution
Ozone's resonance hybrid structure has 2 $\sigma$ -bonds and 1 delocalised $\pi$ -bond over the bent O–O–O framework, with a bond angle of about $116.8° \approx 117°$ .

Concept and Intuition

Ozone is best described not by a single Lewis structure but by a resonance hybrid of two equivalent canonical forms, each showing one O=O double bond and one O–O single bond, with a lone pair and a formal positive charge on the central oxygen and a negative charge on one terminal oxygen. The true structure is an average: each terminal oxygen is connected to the central oxygen by a bond of order 1.5 — i.e., one full $\sigma$ -bond plus a $\pi$ -bond that is delocalised over the whole three-atom system rather than localised on just one side. The central oxygen has one lone pair (which, along with the bent shape from VSEPR-like reasoning on the delocalised system), gives the observed bent geometry with a bond angle slightly less than the ideal trigonal planar 120°.

Step-by-Step Solution

Draw the resonance structures of ozone: central O bonded to two terminal O atoms, with one O=O double bond and one O–O single bond in each canonical form, the double bond alternating between the two sides in the two resonance structures.

Each canonical structure contributes 2 $\sigma$ -bonds (framework bonds) and 1 $\pi$ -bond (in one of the two O–O linkages).

Because the two resonance structures are equivalent and interconvert, the real molecule has the $\pi$ -electron density delocalised (shared) over both O–O linkages, but there is still only one $\pi$ -bond's worth of electron density distributed across the system (not two full $\pi$ -bonds) — hence "2 $\sigma$ and 1 $\pi$ " bonds overall (with fractional/delocalised character rather than one $\pi$ localized on one side). …
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Q.Which of the following attain the linear structure: (Note: more than one of the given options may be correct.)

Species by species

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