Q.Explain the important aspects of resonance with reference to the $CO_3^{2-}$ ion.

Showing the 12 most recent of 13 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.In which of the following, the number of valence electrons is maximum? (A) $\mathrm{PO_4^{3-}}$ (B) $\mathrm{SO_3^{2-}}$ (C) $\mathrm{CO_3^{2-}}$ (D) $\mathrm{CNO^-}$

   ›Reveal solutionSolution

   Simple valence-electron counting (atoms' group valence electrons plus extra electrons equal to the ionic negative charge) shows $\mathrm{PO_4^{3-}}$ has the most, at 32.

   Concept and Intuition

   For a polyatomic ion, total valence electrons = (sum of each atom's own valence electrons) + (1 electron for each unit of negative charge), or $-$ 1 electron for each unit of positive charge. This total is what you'd distribute when drawing the Lewis structure.

   Step-by-Step Solution

   1. $\mathrm{PO_4^{3-}}$: P (group 15, 5 e⁻) + 4×O (group 16, 6 e⁻ each = 24) + 3 (extra for $3-$ charge) = $5+24+3 = 32$.
   2. $\mathrm{SO_3^{2-}}$: S (6 e⁻) + 3×O (18) + 2 (charge) = $6+18+2 = 26$.
   3. $\mathrm{CO_3^{2-}}$: C (4 e⁻) + 3×O (18) + 2 (charge) = $4+18+2 = 24$. …

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2. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.Observe the following molecules $\mathrm{ClF_3, SF_6, CH_4, NH_3, SF_4, XeF_4, PCl_5}$. The number of molecules in which central atom has expanded octet is (A) 6 (B) 4 (C) 5 (D) 3

   ›Reveal solutionSolution

   Counting the electrons around each central atom shows five of the seven molecules (all except CH₄ and NH₃) have an expanded octet.

   Concept and Intuition

   An expanded octet occurs when the central atom is surrounded by more than 8 electrons — possible for elements in period 3 or beyond that have accessible d-orbitals (or, in modern treatment, enough valence orbitals) to accommodate extra electron pairs.

   Step-by-Step Solution

   1. ClF₃: Cl has 3 bond pairs + 2 lone pairs = 10 electrons → expanded.
   2. SF₆: S has 6 bond pairs = 12 electrons → expanded.
   3. CH₄: C has 4 bond pairs = 8 electrons → normal octet.
   4. NH₃: N has 3 bond pairs + 1 lone pair = 8 electrons → normal octet.
   5. SF₄: S has 4 bond pairs + 1 lone pair = 10 electrons → expanded.
   6. XeF₄: Xe has 4 bond pairs + 2 lone pairs = 12 electrons → expanded. …

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3. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
   Q.Two statements are given below Statement I: Octet theory accounts for the shape of the molecules Statement II: Octet theory does not explain the relative stability of the molecules The correct answer is (A) Both statement-I and statement-II are correct (B) Both statement-I and statement-II are not correct (C) Statement-I is correct but statement-II is not correct (D) Statement-I is not correct but statement-II is correct

   ›Reveal solutionSolution

   The octet theory (Lewis theory) helps predict molecular shapes only in a very limited sense, but it fundamentally fails to explain the relative stability of molecules. Therefore Statement I is not correct, and Statement II is correct.

   The key here is to understand what the octet theory (also called the Lewis octet rule) actually does and does not do. It was a pioneering idea: atoms tend to gain, lose, or share electrons to achieve a stable configuration of eight valence electrons (like a noble gas). But it is a qualitative model with serious limitations.

   Why this approach works:

   We need to test each statement against the known capabilities and failures of the octet theory. Statement I claims it "accounts for the shape of molecules." Statement II claims it "does not explain relative stability." If we recall that molecular shape is determined by electron-pair repulsion (VSEPR theory), which is a separate idea built on Lewis structures, and that stability depends on bond energies and resonance—things the octet rule cannot handle—we can judge each statement.

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   1. Examine Statement I: "Octet theory accounts for the shape of molecules."

      The octet theory itself only tells us how many bonds an atom typically forms (e.g., carbon forms 4 bonds to get 8 electrons). It does not predict the three-dimensional arrangement of atoms. That job belongs to VSEPR theory (Valence Shell Electron Pair Repulsion), which uses the number of electron domains around a central atom—not just the octet rule. For example, both water (bent) and carbon dioxide (linear) satisfy the octet rule, but the octet theory alone cannot tell you why one is bent and the other linear. So Statement I is incorrect.

   2. Examine Statement II: "Octet theory does not explain the relative stability of molecules." …

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4. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.The correct formula used to determine the formal charge ($Q_f$) on an atom in the given Lewis structure of a molecule or ion is (V = number of valence electrons in free atom, U = number of unshared electrons on the atom, B = number of bonds around the atom) (A) $Q_f = V - \left(\dfrac{U}{B}\right)$ (B) $Q_f = V + (U - B)$ (C) $Q_f = V - (U + B)$ (D) $Q_f = V - \left(\dfrac{B}{U}\right)$

   ›Reveal solutionSolution

   Formal charge is the free-atom valence electron count minus the electrons "owned" by the atom in the structure — its lone-pair electrons plus one electron per bond.

   Concept and Intuition

   The standard formal charge formula is $FC = V - N_{nonbonding} - \frac12 N_{bonding}$, where $N_{bonding}$ is the total number of bonding electrons (2 per bond). Since the question defines $B$ as the number of bonds (not bonding electrons), $\frac12 N_{bonding} = B$ directly, giving $Q_f = V - U - B = V-(U+B)$.

   Step-by-Step Solution

   1. Start from $FC = V - (\text{lone-pair electrons}) - \frac12(\text{bonding electrons})$. …

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5. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.Observe the following structure: $\mathrm{\overset{\displaystyle ..}{\underset{(1)}{O}}=\underset{(2)}{N}-\overset{\displaystyle ..}{\underset{(3)}{O}}}\!:$ — atom

   (1) is the doubly-bonded oxygen (with two lone pairs shown), atom

   (2) is the central nitrogen, and atom

   (3) is the singly-bonded oxygen (with three lone pairs shown, i.e. bearing the negative charge). The formal charges on the atoms 1, 2, 3 respectively are (A) +1, 0, -1 (B) 0, 0, -1 (C) -1, 0, +1 (D) 0, 0, 0

   ›Reveal solutionSolution

   Applying the formal-charge formula to each atom in this O=N-O⁻ resonance structure gives 0 on

   the double-bonded oxygen, 0 on nitrogen, and −1 on the singly-bonded oxygen.

   Concept and Intuition

   Formal charge $= (\text{valence electrons}) - (\text{non-bonding electrons}) - \tfrac12(\text{bonding electrons})$. It's computed atom by atom from the Lewis structure exactly

   as drawn (this is the classic nitrite ion, $NO_2^-$, resonance form).

   Step-by-Step Solution

   1. Atom 1 (O, double-bonded to N, 2 lone pairs = 4 non-bonding e⁻): $FC = 6 - 4 - \tfrac12(4) = 6-4-2 = 0$.
   2. Atom 2 (N, central, one double bond = 4 bonding e⁻ + one single bond = 2 bonding e⁻, total 6 bonding e⁻; 1 lone pair = 2 non-bonding e⁻): $FC = 5 - 2 - \tfrac12(6) = 5-2-3 = 0$.
   3. Atom 3 (O, single-bonded to N, 3 lone pairs = 6 non-bonding e⁻): $FC = 6 - 6 - \tfrac12(2) = 6-6-1 = -1$. …

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6. AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
   Q.In $\mathrm{OF_2}$ number of bond pairs and lone pairs of electrons are respectively (A) 2, 6 (B) 2, 8 (C) 2, 9 (D) 2, 10

   ›Reveal solutionSolution

   OF$_2$ has 2 O–F bond pairs; counting the lone pairs on oxygen (2) and both fluorines (3 each, so 6 total) gives 8 lone pairs overall.

   Concept and Intuition

   Oxygen has 6 valence electrons; in OF$_2$ it uses 2 electrons to form 2 single bonds to fluorine, leaving 4 electrons as 2 lone pairs. Each fluorine has 7 valence electrons, uses 1 in the O-F bond, leaving 6 electrons as 3 lone pairs per fluorine atom.

   Step-by-Step Solution

   1. Structure: F–O–F (bent, like water, due to O's 2 lone pairs).
   2. Bond pairs: 2 (one O-F bond to each fluorine).
   3. Lone pairs on O: 2. …

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7. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.Identify incorrectly matched set from the following (A) Molecules with incomplete octet - $\mathrm{BeH_2, BCl_3}$ (B) Polar molecules - $\mathrm{BF_3, CCl_4}$ (C) Molecules with expanded octet - $\mathrm{PCl_5, SF_6}$ (D) Odd electron molecules - $\mathrm{NO, NO_2}$

   ›Reveal solutionSolution

   Check each pairing against molecular geometry/electron count; $\mathrm{BF_3}$ and $\mathrm{CCl_4}$ are actually nonpolar due to symmetry, so option (B) is the incorrect match.

   Concept and Intuition

   Molecular polarity depends on both bond polarity and molecular geometry — even polar bonds can give a nonpolar molecule if the geometry is symmetric enough that individual bond dipoles cancel vectorially.

   Step-by-Step Solution

   1. (A) $\mathrm{BeH_2}$ (2 bond pairs around Be) and $\mathrm{BCl_3}$ (3 bond pairs around B) both leave the central atom with fewer than 8 electrons — genuinely incomplete octet. Correctly matched.
   2. (B) $\mathrm{BF_3}$ is trigonal planar (symmetric, dipoles cancel) and $\mathrm{CCl_4}$ is tetrahedral (symmetric, dipoles cancel) — both are nonpolar, not polar. Incorrectly matched. …

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8. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.What are the formal charges on terminal oxygens of ozone molecule? (A) $+1, -1$ (B) $+1, +1$ (C) $-1, -1$ (D) $0, -1$

   ›Reveal solutionSolution

   In ozone's resonance structure, one terminal O is double-bonded (formal charge 0) and the other is singly bonded (formal charge $-1$); the central O carries $+1$.

   Concept and Intuition

   Formal charge $= (\text{valence electrons}) - (\text{non-bonding electrons}) - \tfrac12(\text{bonding electrons})$. Ozone's Lewis structure is $\mathrm{O=O^+-O^-}$ (with resonance delocalizing which terminal O is double-bonded), giving the central oxygen a formal +1 charge (3 bonds, one lone pair) and the two terminal oxygens different formal charges depending on their bonding.

   Step-by-Step Solution

   1. Central O: 2 lone electrons is not right — actually it has one lone pair (2 electrons) and forms 3 bonds (1 double + 1 single) = 4 bond pairs total; formal charge $=6-2-\tfrac12(8)=6-2-4=0$... but the standard, textbook-assigned value for the central O in ozone is $+1$ (it has one lone pair, and 3 sigma+pi bonding interactions counted per the standard resonance Lewis structure: 2 lone e$^-$ + 6 bonding e$^-$ shared $\Rightarrow$ FC $=6-2-3=+1$).
   2. Terminal O double-bonded to the central atom: 2 lone pairs (4 electrons) + 1 double bond (4 shared electrons): FC $=6-4-2=0$. …

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9. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.The set of molecules in which the central atom is not obeying the octet rule is (A) $\mathrm{CO_2}$, $\mathrm{SiH_4}$, $\mathrm{BeCl_2}$ (B) $\mathrm{H_2O}$, $\mathrm{Cl_2O}$, $\mathrm{CO_2}$ (C) $\mathrm{CH_4}$, $\mathrm{NH_3}$, $\mathrm{OF_2}$ (D) $\mathrm{SF_6}$, $\mathrm{PCl_5}$, $\mathrm{XeF_2}$

   ›Reveal solutionSolution

   The octet rule is violated by expansion (more than 8 electrons) in elements from period 3 onward that have accessible d-orbitals; SF$_6$, PCl$_5$, and XeF$_2$ are the classic textbook examples of expanded octets.

   Concept and Intuition

   The octet rule works well for period-2 elements (C, N, O, F) but elements in period 3 and beyond (S, P, Xe, Cl, etc.) can use empty low-lying d-orbitals to accommodate more than four electron pairs, forming hypervalent species. SF$_6$ has 6 bond pairs (12 electrons) on S, PCl$_5$ has 5 bond pairs (10 electrons) on P, and XeF$_2$ has 2 bond pairs plus 3 lone pairs (10 electrons total) on Xe — none of the three obeys the strict octet.

   Step-by-Step Solution

   1. Check set (A): CO$_2$ (C obeys octet via two double bonds = 8 e), SiH$_4$ (Si obeys octet, 4 single bonds = 8 e), BeCl$_2$ (Be has only 4 electrons — an incomplete octet, not expanded) — mixed/incomplete, not "expanded" set.
   2. Check set (B): H$_2$O, Cl$_2$O, CO$_2$ — central O/C atoms all obey the normal octet (8 electrons each).
   3. Check set (C): CH$_4$, NH$_3$, OF$_2$ — central C, N, O atoms all obey the octet normally (8 electrons each, with lone pairs where applicable). …

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10. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.The formal charges of atoms (1),

    (2) and

    (3) in the ion $[\mathrm{O=N=O}]^{+}$ (atoms labelled (1), (2),

    (3) respectively, left to right) is (A) 0, +2, -1 (B) 0, +1, 0 (C) +2, 0, -1 (D) +1, 0, 0

    ›Reveal solutionSolution

    The nitronium-like ion $[\mathrm{O=N=O}]^{+}$ is linear with N centrally double-bonded to both O atoms; formal charge bookkeeping gives 0 on each oxygen and +1 on nitrogen.

    Concept and Intuition

    Formal charge $= (\text{valence electrons}) - (\text{non-bonding electrons}) - \tfrac12(\text{bonding electrons})$. This ion is isoelectronic with CO$_2$: nitrogen is $sp$ hybridized, forms two double bonds (one to each O), and has no lone pair, while each oxygen retains two lone pairs after forming one double bond to N.

    Step-by-Step Solution

    1. Oxygen (atom 1, leftmost): valence electrons = 6; it has 2 lone pairs (4 non-bonding electrons) and one N=O double bond (4 bonding electrons). FC $= 6 - 4 - \tfrac12(4) = 6 - 4 - 2 = 0$.
    2. Nitrogen (atom 2, centre): valence electrons = 5; it has no lone pairs (0 non-bonding electrons) and two double bonds, i.e. 8 bonding electrons total. FC $= 5 - 0 - \tfrac12(8) = 5 - 4 = +1$.
    3. Oxygen (atom 3, rightmost): by symmetry, same as atom 1: FC $= 0$. …

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11. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.In the Lewis dot structure of carbonate ion shown under the formal charges on the oxygen atoms 1,2 & 3 are respectively: [FIGURE] (Lewis dot structure of the carbonate ion $\mathrm{CO_3^{2-}}$ in brackets with a 2- charge: a central C is double-bonded to an O labelled "2" at the top, and singly bonded to an O labelled "1" at the lower left and an O labelled "3" at the lower right) (A) -2, 0, 0 (B) -1, 0, -1 (C) 0, -1, -1 (D) -3, 0, +1

    ›Reveal solutionSolution

    This tests formal-charge calculation on a Lewis structure: formal charge = valence electrons − nonbonding electrons − (bonding electrons)/2.

    Concept and Intuition

    Formal charge lets us assign charges within a single resonance structure so the atoms' total charge matches the ion's overall charge ($-2$ here across 3 O atoms averaged, but in any one resonance form the charges are localized).

    Step-by-Step Solution

    1. O2 (double bonded to C): 2 bonding pairs (4 electrons) shared + 2 lone pairs (4 electrons) = FC $= 6 - 4 - \tfrac{4}{2} = 6-4-2=0$.
    2. O1 (single bonded to C): 1 bonding pair (2 electrons) + 3 lone pairs (6 electrons) = FC $=6-6-\tfrac{2}{2}=6-6-1=-1$.
    3. O3 (single bonded to C), symmetric with O1: FC $=-1$.
    4. So charges on O1, O2, O3 respectively are $-1, 0, -1$. Sum $=-2$, matching the ion's overall charge (with C's formal charge $0$).

    Common Mistakes …

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12. AP EAPCET 2021Set ap-2021-09-03-AN1 markMCQ
    Q.In the given electron dot structure, the formal charge on each nitrogen atom (respectively) from left to right is ______ $\mathrm{\overset{..}{N}=N=\overset{..}{N}}$ (each terminal N carries two lone pairs of dots) (A) $+1, 0, +1$ (B) $-1, +1, -1$ (C) $0, -1, 0$ (D) $+1, -1, +1$

    ›Reveal solutionSolution

    This tests formal-charge calculation on a cumulated double-bond (azide-like) structure; the terminal nitrogens each come out $-1$ and the central nitrogen $+1$.

    Concept and Intuition

    Formal charge lets us check how the drawn Lewis structure distributes the 'ownership' of electrons compared to the free atom. The formula is

    $FC = V - L - \frac{B}{2}$

    where $V$ = valence electrons of the free atom, $L$ = electrons in lone pairs on that atom, and $B$ = total electrons shared in bonds around that atom (i.e. $2\times$ number of bonds). Nitrogen has $V=5$.

    Step-by-Step Solution

    1. Terminal nitrogens: Each terminal N is joined to the central N by one double bond only, and each carries two lone pairs (given). So $B = 4$ (one double bond = 4 bonding electrons) and $L = 4$ (two lone pairs).

    $FC = 5 - 4 - \frac{4}{2} = 5-4-2 = -1$

    2. Central nitrogen: The central N has a double bond on each side, so it is involved in two double bonds: $B = 8$. It carries no lone pair (none shown/possible, since it is already using all orbitals for the two $\pi$ and $\sigma$ bonds — this matches the linear, sp-hybridized azide-type central atom).

    $FC = 5 - 0 - \frac{8}{2} = 5-0-4 = +1$

    3. Left-to-right sequence: terminal N $(-1)$, central N $(+1)$, terminal N $(-1)$. …

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