Chemical Bonding Result Question 5
5. Among the following molecules/ions, $\mathrm{C}_2^{2-}, \mathrm{N}_2^{2-}, \mathrm{O}_2^{2-}, \mathrm{O}_2$ Which one is diamagnetic and has the shortest bond length?
(2019 Main, 8 April II)
(a) $\mathrm{C}_2^{2-}$
(b) $\mathrm{O}_2$
(c) $\mathrm{O}_2^{2-}$
(d) $\mathrm{N}_2^{2-}$
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Answer:
Correct Answer: 5. ( a )
Solution:
| Species | MO energy order | Bond order $(\mathrm{BO})$ | $\boldsymbol{n}$, number of unpaired $e^{-}$ | Magnetic character |
|---|---|---|---|---|
| $\mathrm{C}_2^{2-}\left(14 e^{-}\right)$ | $[8 \bar{e}] \pi_{2 p_x^2}=\pi_{2 p_y^2} \sigma_{2 p_x^2}$ | $\frac{6-0}{2}=3$ | 0 | Diamagnetic |
| $\mathrm{O}_2\left(16 e^{-}\right)$ | $[8 \bar{e}] \sigma_{2 p_z^2} \pi_{2 p_x^2}=\pi_{2 p_y^2} \stackrel{*}{\pi}{2 p_x^1}= \stackrel{*}{\pi}{2 p_y^1}$ | $\frac{6-2}{2}=2$ | 2 | Paramagnetic |
| $\mathrm{O}_2^{2-}\left(18 e^{-}\right)$ | $[8 \bar{e}] \sigma_{2 p_z^2} \pi_{2 p_x^2}=\pi_{2 p_y^2} \stackrel{*}{\pi}{2 p_x^2}= \stackrel{*}{\pi}{2 p_y^2}$ | $\frac{6-4}{2}=1$ | 0 | Diamagnetic |
| $\mathrm{N}_2^{2-}\left(16 e^{-}\right)$ | $[8 \bar{e}] \pi_{2 p_x^2}=\pi_{2 p_y^2} \sigma_{2 p_x^2} \stackrel{*}{\pi}{2 p_x^1}= \stackrel{*}{\pi}{2 p_y^1}$ | $\frac{6-2}{2}=2$ | 2 | Paramagnetic |
Bond length $\propto \frac{1}{\text { BO (Bond order) }}$. So order of bond length
$\underset{BO= 3}{C_2^ {2-}}$ < $\underset{BO= 2} {O_2}$ = ${N}_2^{2-}$ < $\underset{BO= 1}{O_2^ {2-}}$
The diamagnetic species with shortest bond length is $\mathrm{C}_2^{2-}$ (option-a).
BETA
