To find:
The Bond order of N2+, O2+, C2+, and Br22- from the MO theory and predict the bond orders.
Answer to Problem 5.97QA
Solution:
1) Bond order of N2+ = 2.5
2) Bond order of O2+ = 2.5
3) Bond order of C2+ = 1.5
4) Bond order of Br22- = 0
All species with nonzero bond order may exist, therefore, N2+, O2+ and C2+ may exist.
Explanation of Solution
The Molecular orbital diagram of N2+, O2+, C2+, and Br22- are below
1) Bond order of N2+
:
Electronic configuration of N2+ is
2) Bond order of O2+:
Electronic configuration of O2+ is
3) Bond order of C2+:
Electronic configuration of C2+ is
4) Bond order of Br22- :
In case of Br22- after 4s there is 3d subshell which makes the orbital diagram complicated and as it is completely filled, it is going to cancel out with bonding and antibonding electrons;hence, it is neglected in this case. Valence electronic configuration of Br22- is
The bond order of Br22- is 0 because bonding and antibonding electrons are equal.
All species with nonzero bond order may exist, therefore, N2+, O2+ and C2+ may exist.
Conclusion:
The bond order of the given molecular ions is calculated from the MO diagram. It is used to check the existence of species from the bond order.
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