1. Tardigrade
  2. Question
  3. Chemistry
  4. Bond order of which among the following molecules is zero ?

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Q. Bond order of which among the following molecules is zero ?

MHT CETMHT CET 2014

Solution:

(a) $F _{2}$ molecule

$=\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{z}^{2}$

$\pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}, \pi^{*} 2 p_{x}^{2} \approx \pi^{*} 2 p_{y}^{2}$

Bond order $=\frac{1}{2}\left(N_{b}-N_{a}\right)=\frac{1}{2}(10-8)=1$

(b) $O _{2}$ molecule

$=\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{z}^{2}$

$\pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}, \pi^{*} 2 p_{x}^{1} \approx \pi^{*} 2 p_{y'}$

Bond order $=\frac{1}{2}(10-6)=2$

(c) $Be_{2}$ molecule

$=\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}$

Bond order $=\frac{1}{2}(4-4)=0$

(d) $Li _{2}$ molecule $=\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}$

Bond order $=\frac{1}{2}(4-2)=1$

Therefore, $Be _{2}$ molecule has zero bond order.