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Firstly draw lewis structure, then count the total number of bonds which is equal to $4$ here. Finally, count the number of bond groups between individual atoms, which is $3$. The bond order will be equal to $\frac{4}{3}$.
Thus, the overall structure can be obtained by looking at those two structure and averaging the bond order between them. For benzene one resonance form for each bond will be single and the other double $(2 + 1)/2$. The "$2$" on the bottom of the equation is the number of resonance structure being considered. Thus, for benzene the bond order is ...
A further example is given using the NH3 MO diagram. Here, they calculate the bond order as 3, ignoring the fact that the NH3 a1 orbital is weakly bonding. If we were to in theory calculate the MO diagram for NH3 in a trigonal planar geometry (same as for BH3), we would also get the bond order as 3.
In both cases, the observed bond order is probably closer to 2.5, while experiments suggest that the bond is stronger in $\ce{CO+}$. An orbital with bonding character has no node perpendicular to the bond axis; an orbital with anti-bonding character has at least one node perpendicular to the bond axis (electron density is zero).
If you now return hovering from the second atom in question to a position between the two atoms in question (and still on top of the bond in question), and left-click, the bond order may be recalculated, and its display will change if deemed suitable -- as shown in the example of the oxygen atoms of the sulfonamide.
The bond order of the molecule dilithium, Li2, is 1. Bond order can be calculated by using the equation below: 0.5 x (number of bonding electrons -... Explanation: plzz mark it as brainliest if it really hrlps u..!!!
thus Bond order = 9-4/2=5/2= 2.5 according the molecular orbital theory, N2 has more antibonding electrons than N2+. Also, more antibonding electrons lead to instability.
In that case, $\ce{SO_3}$ contains one double bond and two single bonds, which is why people tend to list the overall bond-order as 1.33. The actual bonding structure of $\ce{SO_3}$ is a little more complicated than that, as J. LS points out, so you might need to brush up on molecular-orbital theory to get into the nitty-gritty of its bonding ...
The reason for the special order for molecules with 14 or less electrons is due to the s-p mixing that arises due to the s and p atomic orbitals being close in energy. It is not actually possible to definitely answer your question without looking at the actual orbital energies.
A positive bond order means a stable molecule .but negative or zero bond order means an unstable molecule. stability of a molecule is directly proportional to bond order . Calculation of the bond order of N2, O2, O2+ and O2- :-