Which one of the following species will have maximum bond dissociation energy?

# Which one of the following species will have maximum bond dissociation energy?

1. A

${\mathrm{C}}_{2}$

2. B

${\mathrm{C}}_{2}^{+}$

3. C

${\mathrm{C}}_{2}^{-}$

4. D

${\mathrm{C}}_{2}^{{2}^{-}}$

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### Solution:

${\mathrm{C}}_{2}$  (8 electrons); $\mathrm{KK}\left(\sigma 2\mathrm{s}{\right)}^{2}\left(\sigma \ast 2\mathrm{s}{\right)}^{2}{\left(\pi \ast 2{\mathrm{p}}_{x}\right)}^{2}{\left(\pi 2{\mathrm{p}}_{y}\right)}^{2}$ Bond order $=\left(6-2\right)/2=2$

${\mathrm{C}}_{2}^{+}$ (7 electrons);  Bond order $=\left(5-2\right)/2=11/2$

${\mathrm{C}}_{2}^{-}$ (9 electrons);  $\mathrm{KK}\left(\sigma 2\mathrm{s}{\right)}^{2}\left(\sigma \ast 2\mathrm{s}{\right)}^{2}{\left(\pi 2{\mathrm{p}}_{x}\right)}^{2}{\left(\pi 2{\mathrm{p}}_{y}\right)}^{2}{\left(\sigma 2{\mathrm{p}}_{y}\right)}^{1}$ Bond order $=\left(7-2\right)/2=2\frac{1}{2}$

${\mathrm{C}}_{2}^{2-}$ -(10 electrons);   Bond order $=\left(8-2\right)/2=3$

Larger the bond order, larger the dissociation energy.

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