Question

Decreasing order of stability of $O_2$, $O_2^{-}$,$O_2^{+}$, and $O_2^{2-}$ is:

• A. $O_2>O_2^{+}>O_2^{2-}>O_2^{-}$
• B. $O_2^{-}>O_2^{2-}>O_2^{+}>O_2$
• C. $O_2^{+}>O_2>O_2^{-}>O_2^{2-}$
• D. $O_2^{2-}>O_2^{-}>O_2>O_2^{+}$

Answer 1

Answer: (C)

$O_2^{+}>O_2>O_2^{-}>O_2^{2-}$

Correct option- C.

Reason:-

$O_2\to$ ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}({\sigma }_{2pz}{)}^2{\pi }_{2px}^2{\pi }_{2py}^2{\pi }_{2px}^{\ast 1}{\pi }_{2py}^{\ast 1}$

Bond Order =$\frac{10-6}{2}$=$\frac{4}{2}$=2

$O_2^{-}\to$ ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2pz}^2{\pi }_{2px}^2{\pi }_{2py}^2{\pi }_{2px}^{\ast 2}{\pi }_{2py}^{\ast 1}$

Bond Order =$\frac{10-7}{2}$=$\frac{3}{2}$=1.5

$O_2^{+}\to$ ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2pz}^2{\pi }_{2px}^2{\pi }_{2py}^2{\pi }_{2px}^{\ast 1}$

Bond Order =$\frac{10-5}{2}$=2.5

$O_2^{2-}\to$ ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2pz}^2{\pi }_{2px}^2{\pi }_{2py}^2{\pi }_{2px}^{\ast 2}{\pi }_{2py}^{\ast 2}$

Bond Order =$\frac{10-8}{2}$=1

A higher bond order also means that the atoms are held together more tightly. Bond order also indicates the stability of the bond. The higher the bond order, the more electrons holding the atoms together, and therefore the greater the stability.

$O_2^{+}$>$O_2$>$O_2^{-}$>$O_2^{2-}$