(a) For $\mathrm{B}_{2}^{+}$: Each boron atom has 3 valence electrons, so for $\mathrm{B}_{2}$, there are $3 + 3 = 6$ electrons. Since it is $\mathrm{B}_{2}^{+}$, we remove one electron, giving us a total of 5 valence electrons.
(b) For $\mathrm{Li}_{2}^{+}$:
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