Question

As per molecular orbital theory, bond order of $L{i}_2$ is $x$ and bond order of $L{i}_2^{-}$is $y$. The value of $(x-y)$ is____.

Answer 1

Answer: 0.5

$L{i}_2\Rightarrow \sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2$
Bond order $=\frac{N_b-N_a}{2}$
$=\frac{4-2}{2}=1$
$\therefore x=1$
$L{i}_2^{-}\Rightarrow \sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^1$
Bond order $=\frac{4-3}{2}=\frac{1}{2}=0.5$
$\therefore y=0.5$
$\therefore (x-y)=(1-0.5)=0.5$