What is bond order ? Explain by example.
Bond order according to Lewis : The bond order is given by the number of bonds (shared electron pair) between the two atoms in a molecule.
e.g. : Bond order in $\mathrm{H}_{2}, \mathrm{~F}_{2}, \mathrm{Cl}_{2}, \mathrm{HCl}$ is one. In $\mathrm{O}_{2}$ bond order two and in $\mathrm{N}_{2}$ three because shared electron pairs between two atoms is respectively 2 and $3 .$
According to MO theory bond order: $\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)$
Where, $\mathrm{N}_{\mathrm{b}}=$ Total no. of bonding $\mathrm{e}^{-}$of $\mathrm{BMO}$
$\mathrm{N}_{\mathrm{a}}=$ Total no. of anti bonding $\mathrm{e}^{-}$of $\mathrm{ABMO}$
Bond order : The difference of electron in MO and ABMO orbitals divided by 2 is called Bond order.
e.g. : Bond order in $\mathrm{H}_{2}=\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)$
$=\frac{1}{2}(2-0)=1$
Bond order in $\mathrm{F}_{2}, \mathrm{Cl}_{2}, \mathrm{Br}_{2}=\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)$
$=\frac{1}{2}(10-8)=1$
Bond order $\mathrm{BO}$ in $\mathrm{O}_{2}=\frac{1}{2}(10-6)=2$
Bond order $\mathrm{BO}$ in $\mathrm{N}_{2}=\frac{1}{2}(10-4)=3$
When two atomic orbitals combine they form
Which of the following would have same magnetic nature whether $sp$ mixing is operative or not?
Given below are two statements:
Statement $(I)$ : A $\pi$ bonding $MO$has lower electron density above and below the inter-nuclear asix.
Statement $(II)$ : The $\pi^*$ antibonding $MO$ has a node between the nucles.In the light of the above statements, choose the most appropriate answer from the options given below:
Which one is paramagnetic and has the bond order $1/2$
Incorrect statement when $C_2$ change to $C_2^{-2}$