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Question: What is the bond order of \(NO\) and \(N{O^ + }\)?...

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

Explanation

Solution

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 22. So, first find out the number of electrons present in NONO and NO+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 =a−b2 = \dfrac{{a - b}}{2}
Where a=a = Number of electrons in bonding molecular orbitals.
And, b=b = Number of electrons in antibonding molecular orbitals.
Let’s first calculate Bond order for NONO
Molecular configuration of NONO can be written as
(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px)2(π2py)2(σ2pz)2(π∗2px)1{\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 1010
The number of antibonding electrons are 55
Bond order =10−52 = \dfrac{{10 - 5}}{2}
Bond order =52 = \dfrac{5}{2}
Bond order = 2.52.5
Bond order for NO+N{O^ + }
Total number of electrons in NO+N{O^ + } =14 = 14
Molecular configuration can be written as
(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π)4(2pz)2{\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=10a = 10
b=4b = 4
Bond order =10−42 = \dfrac{{10 - 4}}{2}
Bond order =62 = \dfrac{6}{2}
Bond order =3 = 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.