Chemical Bonding Result Question 4
4. Among the following, the molecule expected to be stabilised by anion formation is $\mathrm{C} _2, \mathrm{O} _2, \mathrm{NO}, \mathrm{F} _2$.
(2019 Main, 9 April I)
(a) $\mathrm{C} _2$
(b) $\mathrm{F} _2$
(c) $\mathrm{NO}$
(d) $\mathrm{O} _2$
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Answer:
Correct Answer: 4. ( a )
Solution:
- $\mathrm{C}_2$ will be stabilised after forming anion. The electronic configuration of carbon is $1 s^2 2 s^2 2 p^2$. There are twelve electrons in $\mathrm{C}_2$. After forming anion (i.e. $\mathrm{C}_2^{-}$ ), the electronic configuration is
- $\mathrm{C}_2^{-}:(\sigma 1 s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\pi 2 p_x^2=\pi 2 p_y^2\right)\left(\sigma 2 p_z^1\right)$ or $K K(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\pi 2 p_x^2=\pi 2 p_y^2\right), \sigma 2 p_z^1$
Bond order $=\frac{1}{2}\left(N_b-N_a\right)=\frac{1}{2}(9-4)=2.5$
For other options such as $\mathrm{F}_2^{-}, \mathrm{O}_2^{-}, \mathrm{NO}^{-}$, the electronic configurations are as follows :
$\mathrm{F}_2^{-}:(\sigma 1 s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\sigma 2 p_z\right)^2\left(\pi 2 p_x^2=\pi 2 p_y^2\right)\left(\pi^* 2 p_x^2=\pi^* 2 p_y^2\right)\left(\sigma^* 2 p_z^1\right)$
Bond order $=1 / 2\left(N_b-N_a\right)=1 / 2(10-9)=0.5$
- $\mathrm{O}_2^{-}:(\sigma 1 s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\sigma 2 p_z\right)^2\left(\pi 2 p_x^2=\pi 2 p_y^2\right)\left(\pi^* 2 p_x^2=\pi^* 2 p_y^1\right)$
Bond order $=\frac{1}{2}\left(N_b-N_a\right)=\frac{1}{2}(10-7)=1.5$
- $\mathrm{NO}^{-}:(\sigma \mid s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\sigma 2 p_z\right)^2\left(\pi 2 p_x^2=\pi 2 p_y^2\right)\left(\pi^* 2 p_x^1=\pi^* 2 p_y^1\right)$
Bond order $=\frac{1}{2}\left(N_b-N_a\right)=\frac{1}{2}(10-6)=2$
The value of bond order of $\mathrm{C}_2^{-}$ is highest among the given options. Bond order between two atoms in a molecule may be taken as an approximate measure of the bond length.
The bond length decreases as bond order increases. As a result, stability of a molecule increases.
BETA
