JEE Main · 2022 · Shift-IImediumBOND-106

The correct order of bond orders of C22-, N22- and O22- is

Chemical Bonding · Class 11 · JEE Main Previous Year Question

Question

The correct order of bond orders of C22−\mathrm{C_2^{2-}}, N22−\mathrm{N_2^{2-}} and O22−\mathrm{O_2^{2-}} is

Options
  1. a

    C22−>N22−>O22−\mathrm{C_2^{2-} > N_2^{2-} > O_2^{2-}}

    ✓
  2. b

    O22−>N22−>C22−\mathrm{O_2^{2-} > N_2^{2-} > C_2^{2-}}

  3. c

    N22−>O22−>C22−\mathrm{N_2^{2-} > O_2^{2-} > C_2^{2-}}

  4. d

    C22−>O22−>N22−\mathrm{C_2^{2-} > O_2^{2-} > N_2^{2-}}

Correct Answera

C22−>N22−>O22−\mathrm{C_2^{2-} > N_2^{2-} > O_2^{2-}}

Detailed Solution

🧠 Calculate bond orders for the dianion series using MO electron count.

| Species | Total e | Bonding e | Antibonding e | BO | |---|---|---|---|---| | \ceC22−\ce{C2^{2-}} | 14 | 10 | 4 | (10−4)/2 = 3 | | \ceN22−\ce{N2^{2-}} | 16 | 10 | 6 | (10−6)/2 = 2 | | \ceO22−\ce{O2^{2-}} | 18 | 10 | 8 | (10−8)/2 = 1 |

🗺️ \ceC22−\ce{C2^{2-}} is isoelectronic with \ceN2\ce{N2} (14 electrons, BO = 3). Each pair of electrons added from \ceC22−\ce{C2^{2-}} onwards goes into an antibonding orbital, reducing BO by 1.

Answer: (a) \ceC22−>N22−>O22−\boxed{\text{Answer: (a) } \ce{C2^{2-} > N2^{2-} > O2^{2-}}}

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