Chemical Bonding and Molecular Structure Part 1

Note:(1) Bond length $\propto \frac{1}{\;\text{ Bond order }\;}$(2) Bond order $= \frac{1}{2} [ N_b - N_a ]$$N_b =$ No of electrons in bonding molecular orbital$N_a =$ No of electrons in anti bonding molecular orbital(4) upto 14 electrons, molecular orbital configuration isHere $N_a =$ Anti bonding electron $=4$ and $N_b = 10$(5) After 14 electrons to 20 electrons molecular orbital configuration isHere $N_a = 10$and $N_b = 10$In 0 atom 8 electrons present, so in $\mathrm{O}_2, 8 \times 2 = 16$ electrons present.Then in $\mathrm{O}_2^{+}$no of electrons $=15$in $\mathrm{O}_2^{-}$no of electrons $=17$in $\mathrm{O}_2^{2-}$ no of electrons $=18$ $\therefore$ Molecular orbital configuration of $\mathrm{O}_2$ (16 electrons) is $\therefore N_a = 6$$N_b = 10$$\therefore \;\; BO = \frac{1}{2}[10-6]=2$ Molecular orbital configuration of $\mathrm{O}_2^{+}$(15 electrons) is$\therefore \;\; N_b = 10$$N_a = 5$$\therefore \;\; BO = \frac{1}{2}[10-5]=2.5$ Molecular orbital configuration of $\mathrm{O}_2^{2+}$ (14 electrons) is$\therefore \;\; N_b = 10$$N_a = 4$$\therefore \;\; BO = \frac{1}{2}[10-4]=3$ Molecular orbital configuration of $\mathrm{O}_2^{-}$(17 electrons) is$\therefore \;\; N_b = 10$$N_a = 7$$\therefore \;\; BO = \frac{1}{2}[10-7]=1.5$ Molecular orbital configuration of $\mathrm{O}_2^{2-}$ (18 electrons) is$\therefore \;\; N_b = 10$$N_a = 8$$\therefore \;\; B 0 = \frac{1}{2}[10-8]=1$ As Bond length $\propto \frac{1}{\;\text{ Bond order }\;}$So $\mathrm{O}_2^{2+}$ has the shortest bond length.