JEE Challenger
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ChemEasyAdv 2007 • Paper 1

Identify Species with Bond Order Different from Carbon Monoxide

The species having bond order different from that in CO\text{CO} is

Options

A

NO−\text{NO}^-

Correct
B

NO+\text{NO}^+

C

CN−\text{CN}^-

D

N2\text{N}_2

Topics & Concepts

Chemical Bonding and Molecular StructureMolecular Orbital Theory

Step-by-Step Solution

To determine which species has a bond order different from that of carbon monoxide (CO\text{CO}), we can use Molecular Orbital Theory (MOT) and the total number of electrons in each diatomic species.

1. Bond Order of CO\text{CO}:

  • Number of electrons in C=6\text{C} = 6
  • Number of electrons in O=8\text{O} = 8
  • Total number of electrons in CO=6+8=14\text{CO} = 6 + 8 = 14

The molecular orbital electronic configuration for a 1414-electron system is:
σ1s2 σ1s∗2 σ2s2 σ2s∗2 (π2px2=π2py2) σ2pz2\sigma_{1s}^2\ \sigma_{1s}^{*2}\ \sigma_{2s}^2\ \sigma_{2s}^{*2}\ (\pi_{2p_x}^2 = \pi_{2p_y}^2)\ \sigma_{2p_z}^2

The number of bonding electrons is Nb=10N_b = 10, and the number of antibonding electrons is Na=4N_a = 4.

The bond order is calculated as:
Bond Order=Nb−Na2=10−42=3\text{Bond Order} = \frac{N_b - N_a}{2} = \frac{10 - 4}{2} = 3

2. Analysis of the Options:

  • Option (A): NO−\text{NO}^-

    • Total electrons =7(from N)+8(from O)+1(negative charge)=16= 7 (\text{from N}) + 8 (\text{from O}) + 1 (\text{negative charge}) = 16
    • The molecular orbital electronic configuration for a 1616-electron system is:
      σ1s2 σ1s∗2 σ2s2 σ2s∗2 σ2pz2 (π2px2=π2py2) (π2px∗1=π2py∗1)\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})
    • Here, Nb=10N_b = 10 and Na=6N_a = 6.
    • Bond order:
      Bond Order=10−62=2\text{Bond Order} = \frac{10 - 6}{2} = 2
    • Since the bond order of NO−\text{NO}^- is 22, it is different from that of CO\text{CO} (which is 33).

  • Option (B): NO+\text{NO}^+

    • Total electrons =7+8−1=14= 7 + 8 - 1 = 14
    • Being isoelectronic with CO\text{CO}, its bond order is:
      Bond Order=10−42=3\text{Bond Order} = \frac{10 - 4}{2} = 3

  • Option (C): CN−\text{CN}^-

    • Total electrons =6+7+1=14= 6 + 7 + 1 = 14
    • Being isoelectronic with CO\text{CO}, its bond order is:
      Bond Order=10−42=3\text{Bond Order} = \frac{10 - 4}{2} = 3

  • Option (D): N2\text{N}_2

    • Total electrons =7+7=14= 7 + 7 = 14
    • Being isoelectronic with CO\text{CO}, its bond order is:
      Bond Order=10−42=3\text{Bond Order} = \frac{10 - 4}{2} = 3

Conclusion:

The species with a bond order different from that in CO\text{CO} is NO−\text{NO}^-.

Correct Answer: (A)

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