Question

The bond orders of \[O_{2}^{2-},O_{2}^{-},{{O}_{2}}\]are respectively x, y and z. then what is the value of 2(x+y+z)?

Answer 1

Hint:For finding the bond order of given molecules and ions we have to find electrons in bonding and non bonding orbitals, pi and sigma bonds.

Complete step by step solution:
We know that bond order is a measurement of the number of electrons involved in bonds between two atoms in a molecule. Most of the time, bond order is equal to the number of bonds between two atoms.

Bond order = $\dfrac{(Bonding\ orbital\ electrons – Antibonding\ orbital\ electrons)}{2}$


Here we see the number of electrons in every molecule and ion:
\[{{O}_{2}}\]​has 16 electrons, in which 8 electrons of every oxygen atom:

And if we fill these electrons in bonding and antibonding orbitals: \[{{(\sigma 1s)}^{2}},{{(\sigma *1s)}^{2}},{{(\sigma 2s)}^{2}},{{(\sigma *2s)}^{2}},{{(2{{p}_{z}})}^{2}},{{(\pi )}^{4}},{{(\pi *)}^{2}}\]

Bond Order (x) = \[\dfrac{(10-6)}{2}\]= 2

\[{{O}_{2}}^{-}\]has 17= (8+8+1) electrons:

\[{{(\sigma 1s)}^{2}},{{(\sigma *1s)}^{2}},{{(\sigma 2s)}^{2}},{{(\sigma *2s)}^{2}},{{(2{{p}_{z}})}^{2}},{{(\pi )}^{4}},{{(\pi *)}^{3}}\]

Bond Order (y) = \[\dfrac{(10-7)}{2}\]=1.5

For bond order of \[{{O}_{2}}^{2-}\]: it has (8+8+2 = 18 electrons):
\[{{(\sigma 1s)}^{2}},{{(\sigma *1s)}^{2}},{{(\sigma 2s)}^{2}},{{(\sigma *2s)}^{2}},{{(2{{p}_{z}})}^{2}},{{(\pi )}^{4}},{{(\pi *)}^{4}}\]

Bond Order (z) = \[\dfrac{(10-8)}{2}\] = 1

So, the value of 2 x (x+y+z) = 2 x (2 + 1.5 + 1) = 2 x 4.5 = 9
Then the answer is = 9.

Note: Bond order decreases and bond distance increases as bond distance is inversely proportional to bond order. It is used as an indicator of the stability of a chemical bond, the higher the bond order, the stronger the chemical bond.