Question

What is the bond order of $NO$ and $N{O^ + }$?

Answer 1

Hint: To solve this question first we must know what bond order is. Bond order is the difference between the number of bonding and antibonding electrons divided by $2$. So, first find out the number of electrons present in $NO$ and $N{O^ + }$ write down its electronic configuration. From this we find out the number of bonding and antibonding electrons and use them in formula.

Complete answer:
Bond order is the number of chemical bonds between a pair of atoms and indicates the stability of a bond. Bond order and bond length indicate the type and strength of covalent bonds between atoms. Bond order and length are inversely proportional to each other: when bond order is increased, bond length is decreased.
We use bond orders to predict the stability of molecules. If the bond order for a molecule is equal to zero, the molecule is unstable. A bond order of greater than zero suggests a stable molecule. The higher the bond order is, the more stable the bond.
Bond order $ = \dfrac{{a - b}}{2}$
Where $a = $Number of electrons in bonding molecular orbitals.
And, $b = $Number of electrons in antibonding molecular orbitals.
Let’s first calculate Bond order for $NO$
Molecular configuration of $NO$ can be written as
\[{\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\pi 2px} \right)^2}{\left( {\pi 2py} \right)^2}{\left( {\sigma 2pz} \right)^2}{\left( {{\pi ^*}2px} \right)^1}\]
The number of bonding electrons are $10$
The number of antibonding electrons are $5$
Bond order $ = \dfrac{{10 - 5}}{2}$
Bond order $ = \dfrac{5}{2}$
Bond order = $2.5$
Bond order for $N{O^ + }$
Total number of electrons in $N{O^ + }$$ = 14$
Molecular configuration can be written as
\[{\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( \pi \right)^4}{(2{p_z})^2}\]
$a = 10$
$b = 4$
Bond order $ = \dfrac{{10 - 4}}{2}$
Bond order $ = \dfrac{6}{2}$
Bond order $ = 3$

Note:
The length of the bond is determined by the number of bonded electrons. The higher the bond order, the stronger the pull between the two atoms and the shorter the bond length. Generally, the length of the bond between two atoms is approximately the sum of the covalent radii of the two atoms.