Q.Distinguish between a sigma and a pi bond.

Showing the 12 most recent of 52 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The increasing order of N–N bond length of $\mathrm{N_2O(A)}$, $\mathrm{N_2O_3(B)}$, $\mathrm{N_2O_4(C)}$ is (A) A, C, B (B) A, B, C (C) B, C, A (D) C, B, A

   ›Reveal solutionSolution

   This tests how N–N bond length varies across nitrogen oxides depending on bond order and the structural environment. Answer: A, C, B (increasing order).

   Concept and Intuition

   Bond length is inversely related to bond order/strength: a higher-order (more multiple-bond character) N–N linkage is shorter and stronger, while a weak single bond between two units pulling electron density away (as through delocalization into N=O groups) is unusually long. $N_2O$ has significant N≡N triple-bond character (it's isoelectronic with $CO_2$, structure $N\equiv N-O$), giving the shortest N–N distance. $N_2O_4$ is two planar $NO_2$ units joined by a single N–N bond, and $N_2O_3$ is an $NO$ unit joined to an $NO_2$ unit by an even weaker, longer N–N single bond — a well-known "unusually long" bond because of extensive delocalization pulling electron density away from the N–N linkage.

   Step-by-Step Solution

   1. $N_2O$: structure is $N\equiv N-O$ (like $CO_2$), giving the N–N bond significant triple-bond character, bond length $\approx1.12$–$1.13$ Å — shortest.
   2. $N_2O_4$: two $NO_2$ groups joined by a single N–N bond, bond length $\approx1.75$ Å. …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The number of species having bond order $\geq 2$ from the following is $N_2,\ O_2^{2-},\ NO^-,\ O_2^{+},\ NO^{+},\ NO$ (A) 4 (B) 6 (C) 5 (D) 3

   ›Reveal solutionSolution

   This tests MO-theory bond orders for a set of diatomics/ions; only $O_2^{2-}$ falls below bond order 2, so 5 of the 6 species qualify.

   Concept and Intuition

   Bond order (BO) from molecular orbital theory is $\text{BO}=\dfrac{1}{2}(N_b-N_a)$, where $N_b$/$N_a$ are electrons in bonding/antibonding MOs. Removing an electron from an antibonding MO (forming a cation) raises the bond order; adding an electron to an antibonding MO (forming an anion) lowers it. Isoelectronic species (same total electron count) have identical MO configurations and hence the same bond order — this is the fast way to solve such problems.

   Step-by-Step Solution

   1. $N_2$ (14 e⁻): triple bond, BO = 3.
   2. $O_2$ (16 e⁻) has BO = 2. $O_2^{2-}$ (18 e⁻, two extra electrons go into the antibonding $\pi^*$ orbitals) has BO $=2-1=1$. Not $\ge 2$.
   3. $NO$ (15 e⁻, isoelectronic with $O_2^{+}$) has BO = 2.5.
   4. $NO^-$ (16 e⁻, isoelectronic with $O_2$) has BO = 2.
   5. $O_2^{+}$ (15 e⁻, isoelectronic with $NO$) has BO = 2.5.
   6. $NO^{+}$ (14 e⁻, isoelectronic with $N_2$) has BO = 3. …

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3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Which of the following orders is not correct for the property mentioned against them? (A) $\mathrm{H_2O > HF > NH_3 > H_2S}$ — boiling point (B) $\mathrm{H_2O > NH_3 > NF_3 > CF_4}$ — dipole moment (C) $\mathrm{C-C > N-O > C-O > C-H}$ — bond length (D) $\mathrm{O_2^+ > O_2 > O_2^- > O_2^{2-}}$ — bond order

   ›Reveal solutionSolution

   Testing several classic periodic/bonding-trend orders at once; three check out against known values, but the bond-length order in (C) has N–O and C–O swapped — that's the incorrect one.

   Concept and Intuition

   Bond length shortens as bond order increases and as the atoms involved get smaller/more electronegative (stronger orbital overlap), while it lengthens with atomic size. For single (sigma) bonds among small second-period atoms, typical average bond lengths increase roughly as: C–H (shortest, small H) $<$ N–O $<$ C–O $<$ C–C (largest single bond among these, both atoms carbon-sized without extra electronegativity-driven shortening). Recognizing the correct relative order between C–O and N–O bonds is the crux of this question.

   Step-by-Step Solution

   1. (A) Boiling points: $\mathrm{H_2O}$ and HF both hydrogen-bond strongly, with water's extensive 3-D H-bond network giving it the highest boiling point (100°C) even over HF (19.5°C, which H-bonds only linearly and has a low molar mass). NH$_3$ (-33°C) H-bonds more weakly than HF, and H$_2$S (-60°C) barely hydrogen-bonds at all. Order $\mathrm{H_2O>HF>NH_3>H_2S}$ is correct.
   2. (B) Dipole moments: Water's bent shape with two lone pairs gives the largest net dipole (1.85 D); NH$_3$'s pyramidal shape gives 1.47 D; NF$_3$'s dipole is much reduced (0.24 D) because the lone pair and the three highly electronegative N–F bond dipoles largely oppose each other; CF$_4$ is perfectly tetrahedral and symmetric, so its bond dipoles cancel exactly (0 D). Order correct.
   3. (C) Bond lengths: using standard average bond lengths (pm): C–C $\approx154$, C–O $\approx143$, N–O $\approx136$, C–H $\approx109$. The true descending order is C–C $>$ C–O $>$ N–O $>$ C–H. The option states C–C $>$ N–O $>$ C–O $>$ C–H, which incorrectly places N–O ahead of C–O. This order is wrong. …

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4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.Match the following List – I (molecule / Ion) and List – II (Bond order and magnetic property): A. $O_2^-$ ... I. 1.5 – Paramagnetic B. $O_2$ ... II. 2.0 – Paramagnetic C. $C_2^{2-}$ ... III. 3.0 – Diamagnetic D. $B_2$ ... IV. 1.0 – Paramagnetic The correct answer is (A) A-I, B-III, C-II, D-IV (B) A-I, B-II, C-III, D-IV (C) A-IV, B-II, C-I, D-III (D) A-IV, B-I, C-II, D-III

   ›Reveal solutionSolution

   This tests MO-theory bond order and magnetic behaviour for four diatomic species. Matching gives $O_2^-\to$1.5-paramagnetic, $O_2\to$2.0-paramagnetic, $C_2^{2-}\to$3.0-diamagnetic, $B_2\to$1.0-paramagnetic, i.e. A-I, B-II, C-III, D-IV.

   Concept and Intuition

   Molecular orbital theory gives bond order $= \frac{1}{2}(N_b - N_a)$ where $N_b$/$N_a$ are bonding/antibonding electrons, and a species is paramagnetic if it has unpaired electrons in its MO configuration (following Hund's rule in degenerate orbitals like $\pi_{2p}$).

   Step-by-Step Solution

   1. $O_2^-$ (superoxide ion): total electrons = 8+8+1 = 17. Filling MOs beyond $O_2$'s 16, the extra electron goes into one of the degenerate $\pi^*_{2p}$ orbitals, giving bond order $= \frac{1}{2}(10-7) = 1.5$ and one unpaired electron → paramagnetic, matches I (1.5–Paramagnetic).
   2. $O_2$: 16 electrons; classic MO configuration gives bond order $= \frac12(10-6) = 2.0$ with two unpaired electrons in $\pi^*_{2p}$ → paramagnetic, matches II (2.0–Paramagnetic).
   3. $C_2^{2-}$: 6+6+2 = 14 electrons, isoelectronic with $N_2$. Bond order $=\frac12(10-4)=3.0$, all electrons paired → diamagnetic, matches III (3.0–Diamagnetic). …

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5. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.Identify the correct set in which sum of bond orders of all species is maximum (A) $N_2^+, O_2, O_2^{2+}$ (B) $O_2^-, O_2^+, B_2$ (C) $C_2, F_2, N_2$ (D) $O_2^{2-}, C_2^{2-}, O_2^{2+}$

   ›Reveal solutionSolution

   Computing MO bond orders species-by-species, Set A ($N_2^+,O_2,O_2^{2+}$) gives the largest total bond order sum, $7.5$.

   Concept and Intuition

   Molecular orbital theory gives bond order as $\tfrac12(\text{bonding e}^- - \text{antibonding e}^-)$. Removing an electron from an antibonding orbital (or adding one to a bonding orbital) raises bond order; removing a bonding electron or adding an antibonding one lowers it. Systematically working out each species' electron configuration lets us compare sums across the four sets.

   Step-by-Step Solution

   1. Set A: $N_2^+$: BO $=2.5$ (one electron removed from a bonding $\sigma2p$ MO of $N_2$, BO$=3$). $O_2$: BO$=2.0$ (standard). $O_2^{2+}$: BO$=3.0$ (both $\pi^*2p$ electrons removed from $O_2$). Sum $=2.5+2.0+3.0=7.5$.
   2. Set B: $O_2^-$: BO$=1.5$ (one electron added to $\pi^*2p$). $O_2^+$: BO$=2.5$ (one electron removed from $\pi^*2p$). $B_2$: BO$=1.0$. Sum $=1.5+2.5+1.0=5.0$.
   3. Set C: $C_2$: BO$=2.0$. $F_2$: BO$=1.0$. $N_2$: BO$=3.0$. Sum $=2.0+1.0+3.0=6.0$.
   4. Set D: $O_2^{2-}$: BO$=1.0$ ($\pi^*2p$ fully filled). $C_2^{2-}$: BO$=3.0$ (isoelectronic with $N_2$ after adding 2 electrons to $\sigma2p$). $O_2^{2+}$: BO$=3.0$. Sum $=1.0+3.0+3.0=7.0$. …

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6. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.The bond order of a homodiatomic molecule is 3. If the number of bonding electrons in it is 10, the number of antibonding electrons will be (A) 4 (B) 5 (C) 6 (D) 3

   ›Reveal solutionSolution

   Solving the standard bond-order formula $BO=(N_b-N_a)/2$ for $N_a$ with $N_b=10$ and $BO=3$ gives $N_a=4$.

   Concept and Intuition

   Molecular orbital theory defines bond order as half the difference between the number of electrons in bonding orbitals ($N_b$) and antibonding orbitals ($N_a$): $BO=\dfrac{N_b-N_a}{2}$. This is just algebra once two of the three quantities are known.

   Step-by-Step Solution

   1. Formula: $BO=\dfrac{N_b-N_a}2$.
   2. Substitute $BO=3$, $N_b=10$: $3=\dfrac{10-N_a}{2}$. …

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7. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.Match the following. List - I (Molecule / ion): A. $C_2$; B. $O_2^{2+}$; C. $O_2$; D. $O_2^-$. List - II (Bond order and magnetic nature): I. 3, Diamagnetic; II. 2, Paramagnetic; III. 2, Diamagnetic; IV. 1.5, Paramagnetic. The correct answer is (A) A-II, B-IV, C-I, D-III (B) A-III, B-I, C-IV, D-II (C) A-II, B-III, C-I, D-IV (D) A-III, B-I, C-II, D-IV

   ›Reveal solutionSolution

   Working out the MO electron configuration (and hence bond order and unpaired-electron count) for $C_2$, $O_2^{2+}$, $O_2$, and $O_2^{-}$ gives the match A-III, B-I, C-II, D-IV.

   Concept and Intuition

   For homonuclear diatomics of second-period elements, bond order $=\dfrac{N_b-N_a}{2}$, and whether the species is paramagnetic depends on whether any orbital (usually the degenerate $\pi^{*}2p$ pair) holds unpaired electrons. Removing or adding electrons from/to $O_2$ changes both its bond order and its magnetic character, since electrons are added/removed preferentially from the highest, and here degenerate, $\pi^{*}2p$ level.

   Step-by-Step Solution

   1. $C_2$ (6+6 = 12 valence e$^-$): configuration $\sigma2s^2\sigma^{*}2s^2\pi2p_x^2\pi2p_y^2$ (no $\sigma2p$ occupied at this electron count). Bonding e$^-=2+4=6$, antibonding $=2$. $BO=(6-2)/2=2$. Both $\pi$ orbitals are completely filled (2 each), so all electrons are paired $\Rightarrow$ diamagnetic. Matches item III (2, Diamagnetic).
   2. $O_2$ (16 e$^-$): standard configuration gives $BO=2$ with two unpaired electrons in the degenerate $\pi^{*}2p$ orbitals (Hund's rule) $\Rightarrow$ paramagnetic. Matches item II (2, Paramagnetic).
   3. $O_2^{2+}$ (16$-$2 = 14 e$^-$): the two electrons removed from $O_2$ come from the (highest-energy, half-filled) $\pi^{*}2p$ orbitals, removing both unpaired electrons entirely. New antibonding count is reduced by 2, giving $BO=(8-2)/2=3$, and no unpaired electrons remain $\Rightarrow$ diamagnetic. Matches item I (3, Diamagnetic). …

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8. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.Consider the following changes I and II $\mathrm{O_2^-} \xleftarrow{\text{II}} \mathrm{O_2} \xrightarrow{\text{I}} \mathrm{O_2^+}$ The correct statements about these changes (I) and (II) in accordance with MO theory are A) In (I) bond order increases by 0.5 from the existing value B) In (II) bond order decreases by 1.0 from the existing value C) In both (I) and (II) magnetic property is not changed D) In both (I) and (II) magnetic property is changed (A) A, B & C only (B) A & C only (C) A & D only (D) B & C only

   ›Reveal solutionSolution

   This tests MO-theory bond order and unpaired-electron counting for $O_2$, $O_2^+$, and $O_2^-$. Removing an electron raises bond order by 0.5; adding one lowers it by 0.5 (not 1.0); both ions stay paramagnetic like $O_2$ itself, so statements A and C are correct.

   Concept and Intuition

   In $O_2$'s MO configuration, the highest occupied orbitals are the degenerate antibonding $\pi_{2p}^*$ pair, each holding one electron (Hund's rule) — this is why $O_2$ is famously paramagnetic even though its Lewis structure suggests no unpaired electrons. Removing or adding a single electron from/to these antibonding orbitals changes the bond order by exactly $\tfrac{1}{2}$ per electron (since antibonding electrons subtract $\tfrac12$ from bond order), and — crucially — going from 2 unpaired electrons to 1 unpaired electron (either by removing one or by pairing one with an added electron) still leaves the species with an unpaired electron, so all three species remain paramagnetic.

   Step-by-Step Solution

   1. $O_2$ configuration: $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p})^2(\pi_{2p})^4(\pi_{2p}^*)^2$, with the two $\pi^*$ electrons unpaired in separate degenerate orbitals. Bond order $= \tfrac{1}{2}(8-4) = 2$. Paramagnetic (2 unpaired e$^-$).
   2. Change I: $O_2 \to O_2^+$ (remove one electron from $\pi_{2p}^*$): now $(\pi_{2p}^*)^1$. Bond order $= \tfrac12(8-3) = 2.5$, an increase of 0.5 from 2. One unpaired electron remains — still paramagnetic. …

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9. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.The pair of molecules / ions with the same bond order value is (A) $\mathrm{B_2, C_2}$ (B) $\mathrm{O_2, C_2}$ (C) $\mathrm{O_2^+, O_2^-}$ (D) $\mathrm{H_2^+, Li_2}$

   ›Reveal solutionSolution

   This tests molecular orbital bond-order calculation; $O_2$ and $C_2$ are the pair that share the same bond order (2).

   Concept and Intuition

   Bond order $= \tfrac{1}{2}(N_b - N_a)$ where $N_b$, $N_a$ are the number of electrons in bonding and antibonding MOs. Applying the standard MO filling order for homonuclear diatomics (with $\pi$ below $\sigma_{2p}$ for $B_2, C_2, N_2$; $\sigma_{2p}$ below $\pi$ for $O_2, F_2$) gives each species a definite bond order that also predicts bond strength/length trends.

   Step-by-Step Solution

   1. $\mathrm{B_2}$ (10 e⁻): configuration $\sigma_{2s}^2\sigma_{2s}^{*2}\pi_{2p_x}^1\pi_{2p_y}^1$; bonding = 4, antibonding = 2 (of core cancel); bond order $= (4-2)/2 = 1$.
   2. $\mathrm{C_2}$ (12 e⁻): configuration $\sigma_{2s}^2\sigma_{2s}^{*2}\pi_{2p_x}^2\pi_{2p_y}^2$; bonding = 6, antibonding = 2; bond order $= (6-2)/2 = 2$.
   3. $\mathrm{O_2}$ (16 e⁻): configuration up to $\pi^{*2p_x}{}^1\pi^{*2p_y}{}^1$; bonding = 8, antibonding = 4; bond order $= (8-4)/2 = 2$.
   4. $\mathrm{O_2^+}$ (15 e⁻, one antibonding electron removed from $O_2$): bond order $= 2.5$.
   5. $\mathrm{O_2^-}$ (17 e⁻, one extra antibonding electron): bond order $= 1.5$.
   6. $\mathrm{H_2^+}$ (1 e⁻ in $\sigma_{1s}$): bond order $= 0.5$. …

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10. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.The sum of bond order values of $\mathrm{C_2}$ and $\mathrm{O_2^{2+}}$ is x, which is equal to sum of bond order values of a, b and c. What are a, b and c ? (A) $\mathrm{O_2^-, O_2^+, O_2}$ (B) $\mathrm{B_2, N_2, F_2}$ (C) $\mathrm{He_2^+, F_2, N_2}$ (D) $\mathrm{O_2^{2-}, N_2, Be_2}$

    ›Reveal solutionSolution

    This tests MO-theory bond-order calculation for diatomics/ions and matching a target sum; the matching set is B₂, N₂, F₂, option (B).

    Concept and Intuition

    Molecular orbital bond order is $BO = \dfrac{N_b - N_a}{2}$, where $N_b$/$N_a$ are electrons in bonding/antibonding MOs. For second-period diatomics with $Z \le 7$ (B₂, C₂, N₂), the $\pi_{2p}$ orbitals fill before $\sigma_{2p_z}$; for O₂, F₂ the order reverses. Removing electrons from an antibonding orbital (as in cations) INCREASES bond order.

    Step-by-Step Solution

    1. $C_2$ (12 e⁻): $\sigma1s^2\sigma^*1s^2\sigma2s^2\sigma^*2s^2\pi2p_x^2\pi2p_y^2$. Bonding = 8, antibonding = 4, so $BO = (8-4)/2 = 2$.
    2. $O_2^{2+}$ (14 e⁻, removed from the antibonding $\pi^*$): $O_2$ (16 e⁻) has $BO=2$; removing 2 electrons from $\pi^*2p$ leaves antibonding = 4 (was 6), bonding stays 10, so $BO = (10-4)/2 = 3$.
    3. So $x = BO(C_2) + BO(O_2^{2+}) = 2 + 3 = 5$. …

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11. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
    Q.The sum of number of antibonding electrons present in $\mathrm{O_2, O_2^-, O_2^{2-}}$ is (A) 15 (B) 9 (C) 21 (D) 6

    ›Reveal solutionSolution

    Counting antibonding electrons in the MO diagrams of $O_2$, $O_2^-$ and $O_2^{2-}$ gives 6, 7 and 8 respectively; their sum is 21.

    Concept and Intuition

    Molecular orbital theory places $O_2$'s 16 electrons in the sequence

    $\sigma1s^2\ \sigma^*1s^2\ \sigma2s^2\ \sigma^*2s^2\ \sigma2p_z^2\ (\pi2p_x^2=\pi2p_y^2)\ (\pi^*2p_x^1=\pi^*2p_y^1)$

    The two unpaired electrons in the degenerate $\pi^*2p$ orbitals give $O_2$ its paramagnetism. Adding electrons (to form $O_2^-$, then $O_2^{2-}$) simply adds them one at a time into the still-incomplete $\pi^*2p$ set, since that is the highest occupied (partially filled) level.

    Step-by-Step Solution

    1. $O_2$ (16 electrons): antibonding electrons are in $\sigma^*1s$ (2), $\sigma^*2s$ (2), and $\pi^*2p$ (2, one each in $\pi^*_x,\pi^*_y$) $= 2+2+2 = 6$.
    2. $O_2^-$ (17 electrons, superoxide): the extra electron pairs up in one of the $\pi^*2p$ orbitals, making $\pi^*2p^3$: antibonding total $= 2+2+3 = 7$. …

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12. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.The correct order of increasing bond lengths of $\mathrm{C-H}$, $\mathrm{O-H}$, $\mathrm{C-C}$ and $\mathrm{H-H}$ is (A) $\mathrm{O-H < H-H < C-C < C-H}$ (B) $\mathrm{C-C < C-H < H-H < O-H}$ (C) $\mathrm{C-C < O-H < H-H < C-H}$ (D) $\mathrm{H-H < O-H < C-H < C-C}$

    ›Reveal solutionSolution

    Comparing the standard bond lengths of these four common single bonds gives the increasing order $\mathrm{H-H < O-H < C-H < C-C}$, matching option (D).

    Concept and Intuition

    Bond length depends chiefly on the sum of the covalent radii of the two bonded atoms (with some shortening from bond polarity/multiple-bond character, though all four bonds here are single bonds). Smaller atoms with smaller covalent radii form shorter bonds. Hydrogen has the smallest covalent radius, so H-H is the shortest bond among ordinary single bonds. Oxygen is small but larger than H, so O-H is next. Carbon is somewhat larger than oxygen in effective single-bond radius contribution when paired with H, giving C-H a bit longer than O-H. Two carbon atoms bonded together (C-C) sum two of the largest radii in this set, making it the longest.

    Step-by-Step Solution

    1. Recall standard/reference bond lengths: $\mathrm{H-H} \approx 0.74\,\text{Å}$, $\mathrm{O-H} \approx 0.96\,\text{Å}$, $\mathrm{C-H} \approx 1.09\,\text{Å}$, $\mathrm{C-C} \approx 1.54\,\text{Å}$.
    2. Arrange these in increasing numerical order: $0.74 < 0.96 < 1.09 < 1.54$. …

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