A) \[B{{e}_{2}}\]
B) \[O_{2}^{+}\]
C) \[N_{2}^{+}\]
D) \[N_{2}^{{}}\]
Correct Answer: D
Solution :
Bond order of CO is 3 and bond order of \[{{N}_{2}}\] is also 3. It can be easily calculated using formula \[\operatorname{Bond} order =\,\,\frac{Nb-Na}{2}\] where, Nb = number of electrons in BMO. Na = number of electrons in ABMO.You need to login to perform this action.
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