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 $\mathrm{C_2^{2-}}$ , $\mathrm{N_2^{2-}}$ and $\mathrm{O_2^{2-}}$ is

Options

a
$\mathrm{C_2^{2-} > N_2^{2-} > O_2^{2-}}$
✓
b
$\mathrm{O_2^{2-} > N_2^{2-} > C_2^{2-}}$
c
$\mathrm{N_2^{2-} > O_2^{2-} > C_2^{2-}}$
d
$\mathrm{C_2^{2-} > O_2^{2-} > N_2^{2-}}$

Correct Answera

$\mathrm{C_2^{2-} > N_2^{2-} > O_2^{2-}}$

Detailed Solution

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

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

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

$\boxed{\text{Answer: (a) } \ce{C2^{2-} > N2^{2-} > O2^{2-}}}$

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