Question

The correct order of the $\mathrm{O}-\mathrm{O}$ bond length in ${\mathrm{O}}_2,{\mathrm{H}}_2{\mathrm{O}}_2$ and ${\mathrm{O}}_3$ is

Answer 1

Bond length:

• The bond length of a molecule is the distance between the nuclei of two chemically linked atoms.
• It is close to the total of the covalent radii of the two linked atoms.

Bond order:

• The bond order shows the number of chemical bonds present between a pair of atoms
• The more the bond order, the larger the pull between the two atoms and the shorter the bond length.
• Bond order can be found out by using the formula:

$\mathrm{Bond}\mathrm{order}=\frac{(\mathrm{Bonding}\mathrm{molecules}’\mathrm{number}\mathrm{of}\mathrm{electrons})–(\mathrm{Antibonding}\mathrm{molecules}’\mathrm{number}\mathrm{of}\mathrm{electrons})}{2}$

Calculation of Bond order of ${\mathrm{O}}_2$:

• Electronic configuration of ${\mathrm{O}}_2$ $(\mathrm{Z}=16)$

${\left[\mathrm{\sigma }\right(1\mathrm{s}\left)\right]}^2{\left[{\mathrm{\sigma }}^{\ast }\right(1\mathrm{s}\left)\right]}^2{\left[\mathrm{\sigma }\right(2\mathrm{s}\left)\right]}^2{\left[{\mathrm{\sigma }}^{\ast }\right(2\mathrm{s}\left)\right]}^2{\left[\mathrm{\sigma }\right(1\mathrm{pz}\left)\right]}^2{\left[\mathrm{\pi }\right(2\mathrm{px}\left)\right]}^2{\left[\mathrm{\pi }\right(2\mathrm{py}\left)\right]}^2{\left[{\mathrm{\pi }}^{\ast }\right(2\mathrm{px}\left)\right]}^1{\left[{\mathrm{\pi }}^{\ast }\right(2\mathrm{py}\left)\right]}^1$

• So, Bond order $=\frac{10-6}{2}=2$
• Similarly the bond order of ${\mathrm{H}}_2{\mathrm{O}}_2$ and ${\mathrm{O}}_3$ is $1$and $1.5$.

So, the order of bond length is ${\mathrm{O}}_2<{\mathrm{O}}_3<{\mathrm{H}}_2{\mathrm{O}}_2$.