Question 1) By drawing molecular orbital diagrams for B2, C2, N2, O2, and F2, predict which…
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Question “Question 1) By drawing molecular orbital diagrams for B2, C2, N2, O2, and F2, predict which…”
Question 1) By drawing molecular orbital diagrams for B2, C2,
N2, O2, and F2, predict which of these homonuclear diatomic
molecules are magnetic.
Question 2) Based on the molecular orbital diagram for NO, which
of the following electronic configurations and statements are most
correct?
Answer
This concept is based upon the molecular orbital theory.
First, draw the molecular orbital diagram. Then calculate the bond order using the antibonding and bonding electrons. Then, verify its magnetic behavior.
Bond Order:
It indicates the number of chemical bonds that exist between atoms.
You can calculate the bond order using this formula.
{\rm{Bond order }} = \frac{{{\rm{ }}\left[ {{N_{\rm{b}}}{\rm{ }} - {N_{\rm{a}}}{\rm{ }}} \right]}}{{\rm{2}}}Here, {N_{\rm{b}}}
(A)
The total number electrons in {{\rm{B}}_{\rm{2}}} {\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}},{{\rm{\sigma }}^*}{\rm{1}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{,}}{{\rm{\sigma }}^*}{\rm{2}}{{\rm{s}}^{\rm{2}}},{\rm{\pi 2}}{{\rm{p}}_X}^1 = {\rm{\pi 2}}{{\rm{p}}_X}^1
The total number electrons in {{\rm{O}}_{\rm{2}}} {\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}},{{\rm{\sigma }}^*}{\rm{1}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{,}}{{\rm{\sigma }}^*}{\rm{2}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{p}}_z}^{\rm{2}},{\rm{\pi 2}}{{\rm{p}}_X}^2 = {\rm{\pi 2}}{{\rm{p}}_X}^2,{{\rm{\pi }}^*}{\rm{2}}{{\rm{p}}_X}^1 = {{\rm{\pi }}^*}{\rm{2}}{{\rm{p}}_X}^1
The total number electrons in {{\rm{C}}_{\rm{2}}} {\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}},{{\rm{\sigma }}^*}{\rm{1}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{,}}{{\rm{\sigma }}^*}{\rm{2}}{{\rm{s}}^{\rm{2}}},{\rm{\pi 2}}{{\rm{p}}_X}^2 = {\rm{\pi 2}}{{\rm{p}}_X}^2
$codeMathTag8’s total electron count
{\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}},{{\rm{\sigma }}^*}{\rm{1}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{,}}{{\rm{\sigma }}^*}{\rm{2}}{{\rm{s}}^{\rm{2}}},{\rm{\pi 2}}{{\rm{p}}_X}^2 = {\rm{\pi 2}}{{\rm{p}}_X}^2{\rm{,\sigma 2}}{{\rm{p}}_z}^{\rm{2}}
The total number electrons in {{\rm{F}}_{\rm{2}}} {\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}},{{\rm{\sigma }}^*}{\rm{1}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{,}}{{\rm{\sigma }}^*}{\rm{2}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{p}}_z}^{\rm{2}},{\rm{\pi 2}}{{\rm{p}}_X}^2 = {\rm{\pi 2}}{{\rm{p}}_X}^2,{{\rm{\pi }}^*}{\rm{2}}{{\rm{p}}_X}^2 = {{\rm{\pi }}^*}{\rm{2}}{{\rm{p}}_X}^2
(B)
The total number electrons in {\rm{NO}} {\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}},{{\rm{\sigma }}^*}{\rm{1}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{,}}{{\rm{\sigma }}^*}{\rm{2}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{p}}_z}^{\rm{2}},{\rm{\pi 2}}{{\rm{p}}_X}^2 = {\rm{\pi 2}}{{\rm{p}}_X}^2,{{\rm{\pi }}^*}{\rm{2}}{{\rm{p}}_X}^1
In the orbital {\rm{NO}}Ans, there is an unpaired electron. Part A
Part B: The homonuclear diatommic mole {{\bf{B}}_{\bf{2}}}
The configuration $codeMathTag16
Conclusion
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