Question
What is meant by the term bond order? Calculate the bond order of : $\text{N}_2,\text{O}_2,\text{O}_2^+$ and $\text{O}_2^-$
✓
Answer
Bond order of a molecule or an ion is a measure of the strength or stability of the bond. Numerically, it is half of the difference between number of electrons in bonding $(N_b)$ and in anti-bonding $(N_a)$ molecular orbitals. Bond order $=\frac{1}{2}(\text{N}_\text{b}-\text{N}_\text{a})$ If B.O. is zero or negative, the bond is highly unstable, i.e. the molecule does not exist. Higher bond order means high bond dissociation energy and greater stability. Greater the bond order, lesser is the bond length. M.O. configuration of $N_2$: $\text{KK}\sigma(2\text{s})^2\sigma^*(2\text{s})^2\pi(2\text{p}_\text{x})^2=\pi(2\text{p}_\text{y})\sigma(2\text{p}_\text{z})$ Bond order $=\frac{1}{2}(8-2)=3$ Thus, nitrogen molecule has three bonds, one sigma and two pi. the molecule is diamagnetic.M.O. comfiguration $O_2$:
$\text{KK}\sigma (2\text{s})^2\sigma^*(2s)^2\sigma(2\text{p}_\text{x})^2=\pi(2\text{p}_\text{y})^2$
thus, nitrogen molecule has three bonds, one sigma and two pi. the molecule is diamagnetic.$\text{KK}\sigma (2\text{s})^2\sigma^*(2\text{s})^2\sigma(2\text{p}_\text{z})^2\pi(2\text{p}_\text{x})^2$ $=\pi(2\text{p}_\text{y})^2\pi^*(2\text{p}_\text{x})^1=\pi^*(2\text{p}_\text{y})^1$
[Note : $\sigma(2\text{p}_\text{z})$ is filled first and then $\pi(2\text{p}_\text{x})$and $\pi(2\text{p}_\text{y})$molecular orbitals are filled.] Bond order$=\frac{(8-4)}{2}=2$The oxygen molrcule has two bonds (one sigma and one pi). The molecule is paramagnetic.
$\text{KK}\sigma(2\text{s})^2\sigma^*(2\text{s})^2\sigma(2\text{p}_\text{z})^2\pi(2\text{p}_\text{x})^2=\pi(2\text{p}_\text{y})^2\pi^*(2\text{p}_\text{x})^1$ Bond order $=\frac{8-3}{2}=2.5$ M.O. configuration of $\text{O}_2^-$: $\text{KK}\sigma(2\text{s})^2\sigma^*(2\text{s})^2\sigma(2\text{p}_\text{z})^2\pi(2\text{p}_\text{x})^2$ $=\pi(2\text{p}_\text{y})^2\pi^*(2\text{p}_\text{x})^2=\pi^*(2\text{p}_\text{y})^1$ Bond order $\frac{(8-5)}{2}=1.5$
$\text{KK}\sigma (2\text{s})^2\sigma^*(2s)^2\sigma(2\text{p}_\text{x})^2=\pi(2\text{p}_\text{y})^2$
thus, nitrogen molecule has three bonds, one sigma and two pi. the molecule is diamagnetic.$\text{KK}\sigma (2\text{s})^2\sigma^*(2\text{s})^2\sigma(2\text{p}_\text{z})^2\pi(2\text{p}_\text{x})^2$ $=\pi(2\text{p}_\text{y})^2\pi^*(2\text{p}_\text{x})^1=\pi^*(2\text{p}_\text{y})^1$
[Note : $\sigma(2\text{p}_\text{z})$ is filled first and then $\pi(2\text{p}_\text{x})$and $\pi(2\text{p}_\text{y})$molecular orbitals are filled.] Bond order$=\frac{(8-4)}{2}=2$The oxygen molrcule has two bonds (one sigma and one pi). The molecule is paramagnetic.
$\text{KK}\sigma(2\text{s})^2\sigma^*(2\text{s})^2\sigma(2\text{p}_\text{z})^2\pi(2\text{p}_\text{x})^2=\pi(2\text{p}_\text{y})^2\pi^*(2\text{p}_\text{x})^1$ Bond order $=\frac{8-3}{2}=2.5$ M.O. configuration of $\text{O}_2^-$: $\text{KK}\sigma(2\text{s})^2\sigma^*(2\text{s})^2\sigma(2\text{p}_\text{z})^2\pi(2\text{p}_\text{x})^2$ $=\pi(2\text{p}_\text{y})^2\pi^*(2\text{p}_\text{x})^2=\pi^*(2\text{p}_\text{y})^1$ Bond order $\frac{(8-5)}{2}=1.5$
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