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Question: The bond orders of \[O_{2}^{2-},O_{2}^{-},{{O}_{2}}\]are respectively x, y and z. then what is the v...

The bond orders of O22−,O2−,O2O_{2}^{2-},O_{2}^{-},{{O}_{2}}are respectively x, y and z. then what is the value of 2(x+y+z)?

Explanation

Solution

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

Here we see the number of electrons in every molecule and ion:
O2{{O}_{2}}​has 16 electrons, in which 8 electrons of every oxygen atom:

And if we fill these electrons in bonding and antibonding orbitals: (σ1s)2,(σ∗1s)2,(σ2s)2,(σ∗2s)2,(2pz)2,(π)4,(π∗)2{{(\sigma 1s)}^{2}},{{(\sigma *1s)}^{2}},{{(\sigma 2s)}^{2}},{{(\sigma *2s)}^{2}},{{(2{{p}_{z}})}^{2}},{{(\pi )}^{4}},{{(\pi *)}^{2}}

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

O2−{{O}_{2}}^{-}has 17= (8+8+1) electrons:

(σ1s)2,(σ∗1s)2,(σ2s)2,(σ∗2s)2,(2pz)2,(π)4,(π∗)3{{(\sigma 1s)}^{2}},{{(\sigma *1s)}^{2}},{{(\sigma 2s)}^{2}},{{(\sigma *2s)}^{2}},{{(2{{p}_{z}})}^{2}},{{(\pi )}^{4}},{{(\pi *)}^{3}}

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

For bond order of O22−{{O}_{2}}^{2-}: it has (8+8+2 = 18 electrons):
(σ1s)2,(σ∗1s)2,(σ2s)2,(σ∗2s)2,(2pz)2,(π)4,(π∗)4{{(\sigma 1s)}^{2}},{{(\sigma *1s)}^{2}},{{(\sigma 2s)}^{2}},{{(\sigma *2s)}^{2}},{{(2{{p}_{z}})}^{2}},{{(\pi )}^{4}},{{(\pi *)}^{4}}

Bond Order (z) = (10−8)2\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.