Decreasing order of stability of O2, \text{O}_2^ -\text{O}_2^ + and \text{O}_2^{2 - } is :

  • Option 1)

    \text{O}_2^ + > \text{O}_2 > \text{O}_2^ - > \text{O}_2^{2 - }

  • Option 2)

    \text{O}_2^{2 - } > \text{O}_\text{2}^ - > \text{O}_\text{2} > \text{O}_2^ +

  • Option 3)

    \text{O}_\text{2} > \text{O}_\text{2}^ + > \text{O}_\text{2}^{\text{2} - } > \text{O}_2^ -

  • Option 4)

    \text{O}_\text{2}^ - > \text{O}_\text{2}^{2 - } > \text{O}_\text{2}^\text{ + } > \text{O}_\text{2}

 

Answers (1)

As learnt

Bond Order -

Bond order is defined as one half the difference between the number of electrons present in the bonding and the antibonding orbitals.

- wherein

Bond \:Order=\frac{N_{b}-{N_{a}}}{2}

 

Bonder \: order=\frac{Nb-Na}{2}

For \: O_{2}^{+} \: : \: Bond \: order \: = \frac{10-5}{2}=2.5

For \: O_{2}^{-} \: : \: Bond \: order \: = \frac{10-7}{2}=1.5

For \: O_{2}^{2-} \: : \: Bond \: order \: = \frac{10-8}{2}=1.0

For \: O_{2} \: : \: Bond \: order \: = \frac{10-6}{2}=2

 

Since \: bond \: order \: of \: O_{2}^{+}> B.O. \: of \: O_{2}>B.O. \: of \ O_{2}^{-}>B.O \: of \: O_{2}^{2-}

Hence order of stability

O_{2}^{+}> O_{2}>O_{2}^{-}>O_{2}^{2-}

 


Option 1)

\text{O}_2^ + > \text{O}_2 > \text{O}_2^ - > \text{O}_2^{2 - }

This option is correct

Option 2)

\text{O}_2^{2 - } > \text{O}_\text{2}^ - > \text{O}_\text{2} > \text{O}_2^ +

This option is incorrect

Option 3)

\text{O}_\text{2} > \text{O}_\text{2}^ + > \text{O}_\text{2}^{\text{2} - } > \text{O}_2^ -

This option is incorrect

Option 4)

\text{O}_\text{2}^ - > \text{O}_\text{2}^{2 - } > \text{O}_\text{2}^\text{ + } > \text{O}_\text{2}

This option is incorrect

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