Construct the molecular orbital diagram for N2 and then identify the bond order. What is the…
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Question “Construct the molecular orbital diagram for N2 and then identify the bond order. What is the…”
Answer
This question can be solved using a concept that calculates bond order from molecular orbital diagram.
The bond order of a molecular is equal to half the difference in electrons in antibonding and bonding orbitals of the molecular orbital chart.
In the molecular orbital diagram for a specific molecule, the sum of the electrons in the outermost shells of the atoms involved with bonding is filled.
The expression below calculates the bond order:
{\rm{Bond}}\;{\rm{order}} = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{bonding}}\;{\rm{electrons}} - {\rm{Number}}\;{\rm{of}}\;{\rm{antibonding}}\;{\rm{electrons}}}}{2}
(1.1)
The molecular orbital diagram for {{\rm{N}}_2}
(1.2)
The bond order for {{\rm{N}}_2} \begin{array}{c}\\{\rm{Bond}}\;{\rm{order}} = \frac{{8 - 2}}{2}\\\\ = 3\\\end{array}
(2.1)
The number of electrons in {\rm{C}}_2^ +
The electronic ground state configuration of {\rm{C}}_2^ + {\rm{Electronic}}\;{\rm{configuration}}\;{\rm{of}}\;{\rm{\;C}}_2^ + = {\left( {{\sigma _{1{\rm{s}}}}} \right)^2}{\left( {\sigma _{{\rm{1s}}}^*} \right)^2}{\left( {{\sigma _{{\rm{2s}}}}} \right)^2}{\left( {\sigma _{2{\rm{s}}}^*} \right)^2}{\left( {{\pi _{{\rm{2}}{{\rm{p}}_x}}} - {\pi _{{\rm{2}}{{\rm{p}}_y}}}} \right)^3}
(2.2)
The bond order for {\rm{C}}_2^ +
\begin{array}{c}\\{\rm{Bond}}\;{\rm{order}} = \frac{{7 - 4}}{2}\\\\ = 1.5\\\end{array}Ans Part 1.1
The molecular orbital diagram for $math_tag_9
Part 1.2
Part 2.1: The bond order for {{\bf{N}}_{\bf{2}}}
Part 2.2: The ground state electronic configuration for {\bf{C}}_{\bf{2}}^{\bf{ + }}
The bond order for $math_tag_11
Conclusion
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