Question

Find the bond order of the peroxide ions.

Answer 1

Step-1 Calculating the Electronic configuration and bond order formula

1. Total electron count in peroxide ion $\left({{\mathrm{O}}_2}^{2-}\right)$= $$\begin{gathered} 2\times No.ofelectronofOxygen+2\left(from-vecharge\right) \\ =2\times 8+2 \\ =18 \end{gathered}$$
2. The Electronic configuration of peroxide ion $\left({{\mathrm{O}}_2}^{2-}\right)$${{\mathrm{O}}_2}^{2-}=\mathrm{\sigma }1{\mathrm{s}}^2\cdot {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2\cdot \mathrm{\sigma }2{\mathrm{s}}^2\cdot \mathrm{\sigma }*2{\mathrm{s}}^2\cdot \mathrm{\sigma }2{\mathrm{pz}}^2\cdot \mathrm{\pi }2{\mathrm{px}}^2=\mathrm{\pi }2{\mathrm{py}}^2\cdot \mathrm{\pi }{*}^{}2{\mathrm{px}}^2={\mathrm{\pi }}^{*}2{\mathrm{py}}^2\cdot \mathrm{\sigma }{*}^{}2{\mathrm{pz}}^0$
3. The anti-bonding molecular orbitals are denoted with a $\left(\ast \right)$and bonding molecular orbitals are simply denoted
4. From the electronic configuration, we can calculate the bond order by calculating the number of electrons in the bonding (BMO) and anti-bonding (ABMO) orbitals of the ion.
5. The formula used to calculate the bond order is:

$Bondorder=1/2\times (No.of{e}^{-}inBMO-No.of{e}^{-}inABMO)$

Step-2 Calculating the Bond order

• No. of electrons in $\mathrm{BMO}$ = $10$
• No. of electrons in $\mathrm{ABMO}$ = $8$
  $$\text{Bond order =}{\displaystyle \frac{1}{2}}\times (10-8)$$

Bond order =$1$

Hence, Bond Order for Peroxide ion $\left({{\mathrm{O}}_2}^{2-}\right)$= $1$