Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT

Question

Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is(are)\

Professor PC Thomas Classes · Answer

(b,c): $\mathrm{Ne}_{2}$ : Bond order (B.O.) $$ =\frac{1}{2}\left[N_{b}-N_{a}ight]=\frac{1}{2}[10-10]=0 $$ For $F_{2}$, highest occupied molecular orbital (HOMO) is $\pi^{*}{ }_{2 p}$.\ Bond order of $\mathrm{O}_{2}{ }^{+}=2.5$\ Bond order of $\mathrm{O}_{2}=2.0$\ Bond order $\propto$ Bond energy $$ \propto \frac{1}{	ext { Bond length }} $$ So, $\mathrm{O}_{2}{ }^{+}$has higher bond energy than $\mathrm{O}_{2}$. Though $\mathrm{Li}_{2}$ and $\mathrm{B}_{2}$ both have same bond order i.e., 1 , but larger atom forms larger bond. So, bond length of $\mathrm{Li}_{2}$ is larger than that of $B_{2}$.\