Skip to content
NCERT Exemplar · Q29

Q.Species having same bond order are : (Note: more than one of the given options may be correct.)

(i) N2
(ii) N2^-
(iii) F2^+
(iv) O2^-
Andhra Pradesh BieapMCQ· 1mImportance★★★★★est
65% · 73/112 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

To find species with the same bond order, we calculate the bond order for each using molecular orbital (MO) theory. This involves determining the total number of electrons, writing the MO configuration, and applying the formula: Bond Order = 1/2×(bonding electrons−anti-bonding electrons)1/2 \times (\text{bonding electrons} - \text{anti-bonding electrons}). We find that F2+_2^+ and O2−_2^- both have a bond order of 1.5.

In chemistry, bond order is a fundamental concept that describes the number of chemical bonds between a pair of atoms. It's a direct indicator of bond strength and stability: a higher bond order generally means a stronger, shorter, and more stable bond. We determine bond order using Molecular Orbital (MO) theory, which describes how atomic orbitals combine to form molecular orbitals.

The key idea is that electrons occupy these molecular orbitals, some of which are bonding (stabilizing the molecule) and some are anti-bonding (destabilizing the molecule). The net effect of these electrons determines the bond order.

The bond order (BO) is calculated as:

BO=12(Nb−Na)\text{BO} = \frac{1}{2} (N_b - N_a)

where NbN_b is the number of electrons in bonding molecular orbitals and NaN_a is the number of electrons in anti-bonding molecular orbitals.

A crucial aspect of MO theory for diatomic molecules is the energy ordering of the molecular orbitals, which changes depending on the total number of electrons in the molecule. This change is due to the extent of s-p mixing.

Important

The energy order of molecular orbitals for diatomic molecules:

  • For molecules with up to 14 electrons (e.g., N2_2, C2_2, B2_2): σ1s<σ1s∗<σ2s<σ2s∗<π2px=π2py<σ2pz<π2px∗=π2py∗<σ2pz∗\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 molecules with more than 14 electrons (e.g., O2_2, F2_2, Ne2_2): σ1s<σ1s∗<σ2s<σ2s∗<σ2pz<π2px=π2py<π2px∗=π2py∗<σ2pz∗\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}

Let's calculate the bond order for each given species:

  1. N2_2

    • Total number of electrons = 7(from N)+7(from N)=147 (\text{from N}) + 7 (\text{from N}) = 14 electrons.
    • Since it has 14 electrons, we use the MO energy order for ≤14\le 14 electrons.
    • MO configuration: \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2
    • Number of bonding electrons (NbN_b) = 2(σ1s)+2(σ2s)+4(π2p)+2(σ2pz)=102 (\sigma_{1s}) + 2 (\sigma_{2s}) + 4 (\pi_{2p}) + 2 (\sigma_{2p_z}) = 10
    • Number of anti-bonding electrons (NaN_a) = 2(σ1s∗)+2(σ2s∗)=42 (\sigma^*_{1s}) + 2 (\sigma^*_{2s}) = 4
    • Bond Order = 12(10−4)=12(6)=3\frac{1}{2} (10 - 4) = \frac{1}{2} (6) = \mathbf{3}

  2. N2−_2^-

    • Total number of electrons = 7(from N)+7(from N)+1(for negative charge)=157 (\text{from N}) + 7 (\text{from N}) + 1 (\text{for negative charge}) = 15 electrons.
    • Even though it has 15 electrons, the s-p mixing effect is still significant for nitrogen-based molecules. Therefore, we still use the MO energy order for ≤14\le 14 electrons.
    • MO configuration: \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi^*_{2p_x}^1
    • Number of bonding electrons (NbN_b) = 2(σ1s)+2(σ2s)+4(π2p)+2(σ2pz)=102 (\sigma_{1s}) + 2 (\sigma_{2s}) + 4 (\pi_{2p}) + 2 (\sigma_{2p_z}) = 10
    • Number of anti-bonding electrons (NaN_a) = 2(σ1s∗)+2(σ2s∗)+1(π2px∗)=52 (\sigma^*_{1s}) + 2 (\sigma^*_{2s}) + 1 (\pi^*_{2p_x}) = 5
    • Bond Order = 12(10−5)=12(5)=2.5\frac{1}{2} (10 - 5) = \frac{1}{2} (5) = \mathbf{2.5}

  3. F2+_2^+

    • Total number of electrons = 9(from F)+9(from F)−1(for positive charge)=179 (\text{from F}) + 9 (\text{from F}) - 1 (\text{for positive charge}) = 17 electrons. …
Start your 14-day free trial to unlock the full solution →

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests
Start 14-day free trial →See Plans & Pricing →

Full access for 14 days. No credit card required.

←Previous questionQ28 — Dimagnetic species are those which contain no unpaired electrons. Whic…→Next questionQ30 — Which of the following statements are not correct? (Note: more than on…