Q.How do you express the bond strength in terms of bond order?

Showing the 12 most recent of 13 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.According to Molecular orbital theory, which of the following is correct with respect to bond order? (A) Bond order of $\mathrm{N}_2^{+}$is less than $\mathrm{O}_2$ while that of $\mathrm{O}_2^{+}$is more than $\mathrm{O}_2$. (B) Bond order of $\mathrm{N}_2^{+}$is less than $\mathrm{N}_2$ while that of $\mathrm{O}_2^{+}$is more than $\mathrm{O}_2$. (C) Bond order of $\mathrm{N}_2^{+}$and $\mathrm{O}_2^{+}$is less than $\mathrm{O}_2$ (D) Bond order of $\mathrm{N}_2^{+}$and $\mathrm{O}_2^{+}$is more than $\mathrm{N}_2$

   ›Reveal solutionSolution

   Bond order is half the difference between bonding and antibonding electrons. For $\mathrm{N}_2^+$ (bond order 2.5) it is less than $\mathrm{N}_2$ (3.0), and for $\mathrm{O}_2^+$ (bond order 2.5) it is more than $\mathrm{O}_2$ (2.0). So option (B) is correct.

   Concept & Intuition

   Molecular orbital (MO) theory lets us calculate bond order as:

   $\text{Bond order} = \frac{\text{number of bonding electrons} - \text{number of antibonding electrons}}{2}$

   Removing an electron from a molecule changes its bond order depending on which orbital the electron came from. If you remove a bonding electron, bond order decreases; if you remove an antibonding electron, bond order increases. The trick is knowing the MO electron configurations for $\mathrm{N}_2$ and $\mathrm{O}_2$, which differ because of the relative energies of the $2p$ orbitals.

   Step-by-step reasoning

   1. MO configuration of $\mathrm{N}_2$ (14 electrons)

      For $\mathrm{N}_2$, the order is: \sigma_{1s}^2, \sigma_{1s}^*^2, \sigma_{2s}^2, \sigma_{2s}^*^2, \pi_{2p_x}^2 = \pi_{2p_y}^2, \sigma_{2p_z}^2.

      Bonding electrons: $\sigma_{1s}(2) + \sigma_{2s}(2) + \pi_{2p}(4) + \sigma_{2p_z}(2) = 10$

      Antibonding electrons: $\sigma_{1s}^*(2) + \sigma_{2s}^*(2) = 4$

      Bond order = $(10 - 4)/2 = 3$.

   2. MO configuration of $\mathrm{N}_2^+$ (13 electrons)

      Removing one electron from $\mathrm{N}_2$ takes it from the highest occupied molecular orbital (HOMO), which is $\sigma_{2p_z}$ (bonding).

      Bonding electrons become $9$, antibonding remain $4$.

      Bond order = $(9 - 4)/2 = 2.5$.

      So $\mathrm{N}_2^+$ has a lower bond order than $\mathrm{N}_2$.

   3. MO configuration of $\mathrm{O}_2$ (16 electrons)

      For $\mathrm{O}_2$, the order shifts: \sigma_{1s}^2, \sigma_{1s}^*^2, \sigma_{2s}^2, \sigma_{2s}^*^2, \sigma_{2p_z}^2, \pi_{2p_x}^2 = \pi_{2p_y}^2, \pi_{2p_x}^*^1 = \pi_{2p_y}^*^1.

      Bonding electrons: $\sigma_{1s}(2) + \sigma_{2s}(2) + \sigma_{2p_z}(2) + \pi_{2p}(4) = 10$

      Antibonding electrons: $\sigma_{1s}^*(2) + \sigma_{2s}^*(2) + \pi_{2p}^*(2) = 6$

      Bond order = $(10 - 6)/2 = 2$.

   4. MO configuration of $\mathrm{O}_2^+$ (15 electrons)

      Removing one electron from $\mathrm{O}_2$ takes it from a $\pi_{2p}^*$ orbital (antibonding).

      Antibonding electrons become $5$, bonding remain $10$.

      Bond order = $(10 - 5)/2 = 2.5$. …

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.Which of the following options represent the correct bond order? (A) $O_2^{+}<O_2^{-}<O_2$ (B) $O_2<O_2^{-}<O_2^{+}$ (C) $O_2^{-}<O_2<O_2^{+}$ (D) $O_2^{-}<O_2^{+}<O_2$

   ›Reveal solutionSolution

   [!TLDR]

   Using MO theory, bond orders are O₂⁻ = 1.5, O₂ = 2, O₂⁺ = 2.5, so O₂⁻ < O₂ < O₂⁺.

   Concept

   Bond order $= \tfrac{1}{2}(N_b - N_a)$, where $N_b$ and $N_a$ are the electrons in bonding and antibonding molecular orbitals — a standard CBSE/NCERT Class 11 MO-theory result. For O₂, the highest occupied orbitals are the antibonding $\pi^*$ set.

   Solution

   Neutral $O_2$ has 16 electrons with $N_b = 10$, $N_a = 6$:

   $\text{BO}(O_2) = \tfrac{1}{2}(10-6) = 2.$

   Removing one electron ($O_2^{+}$) takes it from an antibonding $\pi^*$ orbital, so $N_a = 5$: …

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3. COMEDK 2025Set 2025-A1 markMCQ
   Q.Which one of the following molecules attains greater stability on formation of its diatomic monovalent anion? (A) $\mathrm{C}_2$ (B) $\mathrm{O}_2$ (C) $\mathrm{N}_2$ (D) $\mathrm{F_2}$

   ›Reveal solutionSolution

   The stability of a diatomic monovalent anion depends on the bond order change upon gaining an electron. Adding an electron to $\mathrm{N}_2$ increases its bond order from 3 to 3.5, making it the most stable anion among the options.

   The key concept here is bond order and how it relates to molecular stability. In molecular orbital theory, a higher bond order means a stronger, more stable bond. When a neutral diatomic molecule gains an electron to form a monovalent anion ($\mathrm{X}_2^-$), the electron enters either a bonding or an antibonding orbital. If it enters a bonding orbital, the bond order increases and the anion is more stable than the neutral molecule; if it enters an antibonding orbital, the bond order decreases and the anion is less stable. We need to find which molecule gains the most stability — i.e., which one experiences the largest increase in bond order upon forming its anion.

   Let’s work through each molecule step by step.

   1. Recall the molecular orbital configurations for the homonuclear diatomic molecules of the second period.

      For $\mathrm{C}_2$, $\mathrm{N}_2$, $\mathrm{O}_2$, and $\mathrm{F}_2$, the order of molecular orbitals (for $\mathrm{Li}_2$ to $\mathrm{N}_2$) is:

      $\sigma_{1s}$, $\sigma_{1s}^*$, $\sigma_{2s}$, $\sigma_{2s}^*$, $\pi_{2p_x} = \pi_{2p_y}$, $\sigma_{2p_z}$, $\pi_{2p_x}^* = \pi_{2p_y}^*$, $\sigma_{2p_z}^*$.

      For $\mathrm{O}_2$ and $\mathrm{F}_2$, the $\sigma_{2p_z}$ orbital is lower in energy than the $\pi_{2p}$ orbitals, but the key is the filling order of valence electrons.

   2. Determine the bond order of each neutral molecule.

      Bond order = $\frac{(\text{number of bonding electrons}) - (\text{number of antibonding electrons})}{2}$.

      • $\mathrm{C}_2$: Total valence electrons = 8. Configuration: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\pi_{2p})^4$. Bonding = 2 (from $\sigma_{2s}$) + 4 (from $\pi_{2p}$) = 6; Antibonding = 2 (from $\sigma_{2s}^*$). Bond order = $(6-2)/2 = 2$.
      • $\mathrm{N}_2$: Total valence electrons = 10. Configuration: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\pi_{2p})^4 (\sigma_{2p_z})^2$. Bonding = 2 + 4 + 2 = 8; Antibonding = 2. Bond order = $(8-2)/2 = 3$.
      • $\mathrm{O}_2$: Total valence electrons = 12. Configuration: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p})^4 (\pi_{2p}^*)^2$. Bonding = 2 + 2 + 4 = 8; Antibonding = 2 + 2 = 4. Bond order = $(8-4)/2 = 2$.
      • $\mathrm{F}_2$: Total valence electrons = 14. Configuration: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p})^4 (\pi_{2p}^*)^4$. Bonding = 2 + 2 + 4 = 8; Antibonding = 2 + 4 = 6. Bond order = $(8-6)/2 = 1$.

   3. Now add one electron to form the monovalent anion $\mathrm{X}_2^-$ and see where it goes.

      The added electron will occupy the next available molecular orbital (the lowest unoccupied molecular orbital, LUMO, of the neutral molecule).

      • $\mathrm{C}_2^-$: Neutral $\mathrm{C}_2$ has the $\pi_{2p}$ orbitals fully filled (4 electrons). The next orbital is $\sigma_{2p_z}$ (bonding). Adding one electron gives: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\pi_{2p})^4 (\sigma_{2p_z})^1$. Bonding = 2 + 4 + 1 = 7; Antibonding = 2. Bond order = $(7-2)/2 = 2.5$.

        Change: from 2.0 to 2.5 → increase of 0.5.

      • $\mathrm{N}_2^-$: Neutral $\mathrm{N}_2$ has the $\sigma_{2p_z}$ filled. The next orbital is $\pi_{2p}^*$ (antibonding). Adding one electron gives: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\pi_{2p})^4 (\sigma_{2p_z})^2 (\pi_{2p}^*)^1$. Bonding = 2 + 4 + 2 = 8; Antibonding = 2 + 1 = 3. Bond order = $(8-3)/2 = 2.5$.

        Change: from 3.0 to 2.5 → decrease of 0.5. …

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4. COMEDK 2025Set 2025-M1 markMCQ
   Q.In which one of the following pairs does the stability of the monovalent ion increase with respect to the molecule, while the magnetic character remains the same for both? (A) $\mathrm{C}_2 / \mathrm{C}_2^{+}$ (B) $\mathrm{O}_2 / \mathrm{O}_2{ }^{+}$ (C) $\mathrm{B}_2 / \mathrm{B}_2{ }^{+}$ (D) $\mathrm{N}_2 / \mathrm{N}_2^{+}$

   ›Reveal solutionSolution

   The key is to compare bond orders and magnetic character (number of unpaired electrons) for each neutral molecule and its monovalent cation. Only in $\mathrm{O}_2 / \mathrm{O}_2^{+}$ does the cation have a higher bond order (greater stability) while both remain paramagnetic with two unpaired electrons.

   Concept & Intuition

   Stability of a diatomic molecule or ion is directly related to its bond order: higher bond order means stronger, shorter bonds and greater stability. Magnetic character depends on the number of unpaired electrons in the molecular orbital (MO) configuration. We need a pair where:

   • The cation has a higher bond order than the neutral molecule (so stability increases upon ionization).
   • Both the neutral and the cation have the same number of unpaired electrons (same magnetic character).

   We’ll use the standard MO diagrams for homonuclear diatomic molecules of the second period. For $\mathrm{B}_2$, $\mathrm{C}_2$, and $\mathrm{N}_2$, the ordering is $\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \pi_{2p_x} = \pi_{2p_y} < \sigma_{2p_z} < \pi^*_{2p_x} = \pi^*_{2p_y} < \sigma^*_{2p_z}$. For $\mathrm{O}_2$ and beyond, the $\sigma_{2p_z}$ drops below the $\pi_{2p}$ orbitals, but that doesn’t affect our reasoning here.

   Let’s work through each option.

   1. Option (A): $\mathrm{C}_2 / \mathrm{C}_2^{+}$

      • $\mathrm{C}_2$ has 12 electrons. MO configuration: $(\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p})^4$ (no electrons in $\sigma_{2p_z}$). Bond order = $\frac{8-4}{2}=2$. All electrons paired → diamagnetic.
      • $\mathrm{C}_2^{+}$ has 11 electrons. Remove one electron from the highest occupied MO, which is the $\pi_{2p}$ orbital. Configuration: $(\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p})^3$. Bond order = $\frac{7-4}{2}=1.5$ (lower than 2). It has one unpaired electron → paramagnetic.
      • Stability decreases (bond order drops), and magnetic character changes (diamagnetic → paramagnetic). Not our pair.

   2. Option (B): $\mathrm{O}_2 / \mathrm{O}_2^{+}$

      • $\mathrm{O}_2$ has 16 electrons. MO configuration: $(\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p})^4 (\pi^*_{2p})^2$. The two $\pi^*$ electrons are in different orbitals (Hund’s rule) → two unpaired electrons → paramagnetic. Bond order = $\frac{10-6}{2}=2$.
      • $\mathrm{O}_2^{+}$ has 15 electrons. Remove one electron from the highest occupied MO, which is a $\pi^*$ orbital. Configuration: $(\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p})^4 (\pi^*_{2p})^1$. Still one unpaired electron in $\pi^*$ → paramagnetic. Bond order = $\frac{10-5}{2}=2.5$ (higher than 2).
      • Stability increases (bond order rises), and both are paramagnetic (both have unpaired electrons). This matches perfectly.

   3. Option (C): $\mathrm{B}_2 / \mathrm{B}_2^{+}$

      • $\mathrm{B}_2$ has 10 electrons. MO configuration: $(\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p})^2$ (two unpaired electrons in degenerate $\pi$ orbitals) → paramagnetic. Bond order = $\frac{6-4}{2}=1$. …

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5. COMEDK 2024Set 2024-A1 markMCQ
   Q.The bond dissociation enthalpy of the species in their correct order is: (A) $\mathrm{O}_2^{2-}>\mathrm{O}_2>\mathrm{O}_2^{+}>\mathrm{O}_2^{-}$ (B) $\mathrm{O}_2>\mathrm{O}_2^{+}>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}$ (C) $\mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}$ (D) $\mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{2-}>\mathrm{O}_2^{-}$

   ›Reveal solutionSolution

   Higher bond order means a stronger, harder-to-break bond. Bond orders are O$_2^+$ = 2.5, O$_2$ = 2.0, O$_2^-$ = 1.5, O$_2^{2-}$ = 1.0, giving the BDE order O$_2^+$ > O$_2$ > O$_2^-$ > O$_2^{2-}$.

   Using MO theory, bond order $= \tfrac12(N_b - N_a)$ with total electrons:

   • O$_2^+$ (15 e$^-$): bond order 2.5
   • O$_2$ (16 e$^-$): bond order 2.0
   • O$_2^-$ (superoxide, 17 e$^-$): bond order 1.5 …

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6. COMEDK 2024Set 2024-M1 markMCQ
   Q.Given below are 4 molecular species/ions. Identify the species/ion which exhibits diamagnetic nature. $[\mathrm{A}]=\mathrm{N}_2^{+} \quad[\mathrm{B}]=\mathrm{O}_2^{-} \quad[\mathrm{C}]=\mathrm{B}_2 \quad[\mathrm{D}]=\mathrm{C}_2$ (A) [C] (B) [A] (C) [D] (D) [B]

   ›Reveal solutionSolution

   Diamagnetic species have all electrons paired. Using molecular orbital theory, we find that only $\mathrm{C}_2$ has a zero net spin (all electrons paired), so the correct option is (C).

   Concept & Intuition

   Diamagnetism arises when a molecule has no unpaired electrons — all spins are paired, so there is no permanent magnetic moment. To decide which of $\mathrm{N}_2^{+}$, $\mathrm{O}_2^{-}$, $\mathrm{B}_2$, or $\mathrm{C}_2$ is diamagnetic, we need their molecular orbital (MO) electron configurations. The key is to count total valence electrons, fill the MOs in order of increasing energy (remembering that for $\mathrm{B}_2$ and $\mathrm{C}_2$ the $\pi_{2p}$ orbitals lie below the $\sigma_{2p}$ orbital, unlike for $\mathrm{N}_2$ and $\mathrm{O}_2$), and then check for unpaired electrons.

   Step-by-step reasoning

   1. Determine total valence electrons for each species

      • $\mathrm{N}_2^{+}$: Nitrogen has 5 valence electrons each, so $\mathrm{N}_2$ has 10; removing one electron gives $10 - 1 = 9$ valence electrons.
      • $\mathrm{O}_2^{-}$: Oxygen has 6 valence electrons each, so $\mathrm{O}_2$ has 12; adding one electron gives $12 + 1 = 13$ valence electrons.
      • $\mathrm{B}_2$: Boron has 3 valence electrons each, so total = $3 + 3 = 6$ valence electrons.
      • $\mathrm{C}_2$: Carbon has 4 valence electrons each, so total = $4 + 4 = 8$ valence electrons.

   2. Recall the MO energy ordering for these homonuclear diatomic molecules

      • For $\mathrm{B}_2$ and $\mathrm{C}_2$ (elements before nitrogen):

   $\sigma_{1s} < \sigma_{1s}^* < \sigma_{2s} < \sigma_{2s}^* < \pi_{2p_x} = \pi_{2p_y} < \sigma_{2p_z} < \pi_{2p_x}^* = \pi_{2p_y}^* < \sigma_{2p_z}^*$

   • For $\mathrm{N}_2^{+}$ and $\mathrm{O}_2^{-}$ (elements from nitrogen onward):

   $\sigma_{1s} < \sigma_{1s}^* < \sigma_{2s} < \sigma_{2s}^* < \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi_{2p_x}^* = \pi_{2p_y}^* < \sigma_{2p_z}^*$

   (We ignore core $1s$ orbitals as they are fully filled and not involved in bonding.)

   3. Fill MOs for each species and check for unpaired electrons

      • $\mathrm{N}_2^{+}$ (9 valence electrons)

        Filling order: $\sigma_{2s}^2$, $\sigma_{2s}^{*2}$, $\sigma_{2p_z}^2$, $\pi_{2p_x}^2$, $\pi_{2p_y}^1$ (since $\pi$ orbitals are degenerate, Hund’s rule applies).

        Configuration: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^1$

        One unpaired electron → paramagnetic.

      • $\mathrm{O}_2^{-}$ (13 valence electrons)

        Filling order: $\sigma_{2s}^2$, $\sigma_{2s}^{*2}$, $\sigma_{2p_z}^2$, $\pi_{2p_x}^2$, $\pi_{2p_y}^2$, $\pi_{2p_x}^{*2}$, $\pi_{2p_y}^{*1}$ (the extra electron goes into a $\pi^*$ orbital). …

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7. COMEDK 2023Set 2023-E1 markMCQ
   Q.Identify the correct statement describing the characteristics of $\mathrm{C}_2$ molecule. (A) Bond order $=2.0 ; \quad$ Paramagnetic in nature; One sigma and one pi bond formed. (B) Bond order $=0 \quad \mathrm{C}_2$ molecule is non-existent. (C) Bond order $=2.0 ; \quad$ Diamagnetic in nature; Both bonds formed are pi bonds. (D) Bond order $=1.5 ; \quad$ Paramagnetic in nature; Both bonds formed are sigma bonds $(\sigma)$.

   ›Reveal solutionSolution

   $\mathrm{C_2}$ has 12 electrons; below oxygen the $\pi_{2p}$ orbitals lie below $\sigma_{2p}$, giving the configuration $\sigma_{2s}^2\sigma_{2s}^{*2}\pi_{2p_x}^2\pi_{2p_y}^2$. Bond order $=2$, all electrons paired (diamagnetic), and the double bond is made of two $\pi$ bonds.

   MO configuration of $\mathrm{C_2}$ ($\pi_{2p}$ below $\sigma_{2p}$ for $Z<8$):

   $\sigma_{1s}^2\,\sigma_{1s}^{*2}\,\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\pi_{2p_x}^2\,\pi_{2p_y}^2$

   Bond order (valence electrons: 8 bonding, 4 antibonding):

   $\text{B.O.} = \frac{N_b - N_a}{2} = \frac{8-4}{2} = 2.0$

   Magnetism: every orbital is doubly filled — no unpaired electrons — so $\mathrm{C_2}$ is diamagnetic. …

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8. COMEDK 2023Set 2023-M1 markMCQ
   Q.Which of these represents the correct order of decreasing bond order? (A) $\mathrm{C}_2^{2-}>\mathrm{O}_2^{2+}>\mathrm{O}_2^{-}>\mathrm{He}_2^{2+}$ (B) $\mathrm{O}_2^{-}>\mathrm{O}_2^{+}>\mathrm{He}_2^{+}>\mathrm{C}_2^{2-}$ (C) $\mathrm{He}_2^{+}>\mathrm{O}_2^{-}>\mathrm{C}_2^{2-}>\mathrm{O}_2^{+}$ (D) $\mathrm{He}_2^{+}>\mathrm{O}_2^{2-}>\mathrm{O}_2^{+}>\mathrm{O}_2^{-}$

   ›Reveal solutionSolution

   Using MO theory, $\mathrm{C_2^{2-}}$ and $\mathrm{O_2^{2+}}$ both have bond order 3, $\mathrm{O_2^{-}}$ has 1.5 and $\mathrm{He_2^{2+}}$ has 1. The sequence $3 \ge 3 > 1.5 > 1$ matches only option (A).

   Compute bond order (BO) $= \tfrac12(N_b - N_a)$ from the number of electrons:

   • $\mathrm{C_2^{2-}}$: 14 electrons $\Rightarrow$ configuration fills up to $\sigma_{2p}$, BO $= 3$.
   • $\mathrm{O_2^{2+}}$: 14 electrons $\Rightarrow$ BO $= 3$.
   • $\mathrm{O_2^{-}}$: 17 electrons $\Rightarrow$ BO $= 1.5$.
   • $\mathrm{He_2^{2+}}$: 2 electrons ($\sigma_{1s}^2$) $\Rightarrow$ BO $= 1$.

   So the decreasing order is $\mathrm{C_2^{2-}}(3) \approx \mathrm{O_2^{2+}}(3) > \mathrm{O_2^{-}}(1.5) > \mathrm{He_2^{2+}}(1)$. …

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9. COMEDK 2022Set 20221 markMCQ
   Q.Increasing order of bond order of oxygen and its ions is (A) ${O_2} < O_2^ + < O_2^{2 - } < O_2^ -$ (B) $O_2^ - < O_2^{2 - } < O_2^ + < {O_2}$ (C) $O_2^ + < {O_2} < O_2^ - < O_2^{2 - }$ (D) $O_2^{2 - } < O_2^ - < {O_2} < O_2^ +$

   ›Reveal solutionSolution

   Increasing order of bond order: O2^2- < O2^- < O2 < O2^+

   Concept: Molecular orbital theory. Bond order = (Nb - Na)/2. O2 has 16 electrons with configuration ... (pi2p)^2, giving bond order 2. Adding electrons goes into the antibonding pi orbitals (lowering bond order); removing an electron from pi* raises it.

   O2^+ (15 e): bond order = (10 - 5)/2 = 2.5

   O2 (16 e): bond order = (10 - 6)/2 = 2.0

   O2^- (17 e): bond order = (10 - 7)/2 = 1.5 …

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10. COMEDK 2022Set 20221 markMCQ
    Q.The P-O bond order in PO$_4^{3-}$ is (A) 1 (B) 1.5 (C) 1.45 (D) 1.25

    ›Reveal solutionSolution

    Bond order = total bonds / number of linkages = 5/4 = 1.25

    Concept: Fractional bond order from resonance. In PO4^3- the phosphorus is bonded to 4 oxygens by 4 sigma bonds, and there is one additional pi bond (one P=O) that is delocalised over the four P-O linkages by resonance.

    Total bonds (bond pairs) between P and O = 5 …

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11. KCET 2021Set B-21 markMCQ
    Q.Which of the following is an incorrect statement? (A) Hydrogen bonding is stronger than dispersion forces (B) Sigma bonds are stronger than $\pi$-bonds (C) Ionic bonding is non-directional (D) $\sigma$-electrons are referred to as mobile electrons

    ›Reveal solutionSolution

    Test each statement — the false one is (D): mobility belongs to $\pi$-electrons, not $\sigma$-electrons.

    The question asks for the incorrect statement, so each option must be checked in turn.

    Step 1 — (A) "Hydrogen bonding is stronger than dispersion forces." — TRUE.

    Typical bond-energy scale:

    • London dispersion forces: $\sim 0.05$–$40\ \mathrm{kJ\,mol^{-1}}$ (very weak; induced-dipole–induced-dipole)
    • Hydrogen bonds: $\sim 10$–$40\ \mathrm{kJ\,mol^{-1}}$, and up to $\sim 100$ in strong cases

    Hydrogen bonding is a dipole–dipole interaction of special strength, arising when H is bonded to the small, highly electronegative N, O or F. It comfortably outranks dispersion forces — this is why $\mathrm{H_2O}$ boils at $100^\circ\mathrm{C}$ while $\mathrm{H_2S}$, held only by weaker forces, is a gas.

    Step 2 — (B) "Sigma bonds are stronger than $\pi$-bonds." — TRUE.

    A $\sigma$ bond forms by head-on (axial) overlap of orbitals, giving maximum, concentrated overlap along the internuclear axis. A $\pi$ bond forms by sideways (lateral) overlap of parallel p orbitals, which is less effective. Less overlap → weaker bond. Hence $\pi$ bonds are the ones that break first in addition reactions of alkenes.

    Step 3 — (C) "Ionic bonding is non-directional." — TRUE.

    An ionic bond is simply the electrostatic attraction between oppositely charged ions, and the Coulomb field of an ion is spherically symmetric. An ion therefore attracts counter-ions equally in all directions — which is exactly why ionic compounds form extended 3-D lattices rather than discrete directional molecules. …

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12. COMEDK 2021Set 20211 markMCQ
    Q.Find the correct order of C$-$O bond length among CO, CO$_3^{2-}$, CO$_2$. (A) CO$_2$ < CO$_3^{2-}$ < CO (B) CO < CO$_3^{2-}$ < CO$_2$ (C) CO$_3^{2-}$ < CO$_2$ < CO (D) CO < CO$_2$ < CO$_3^{2-}$

    ›Reveal solutionSolution

    So increasing C-O bond length: CO < CO2 < CO3^2-.

    Concept: Bond order vs bond length - higher bond order gives shorter bond.

    • CO: carbon monoxide, C(triple)O, bond order 3, bond length ~ 1.13 A (shortest)
    • CO2: O=C=O, bond order 2, bond length ~ 1.16 A …

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