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Question: In the species \({O_2}{\text{ , O}}_2^ + {\text{ O}}_2^ - \) and \({\text{O}}_2^{2 - }\) the correct...

In the species O2 , O2+ O2−{O_2}{\text{ , O}}_2^ + {\text{ O}}_2^ - and O22−{\text{O}}_2^{2 - } the correct decreasing order of bond strength is:
A. O2>O2+>O2−>O22−{O_2} > O_2^ + > O_2^ - > O_2^{2 - }
B. O2+>O2>O2−>O22−O_2^ + > {O_2} > O_2^ - > O_2^{2 - }
C. O22−>O2−>O2+>O2O_2^{2 - } > O_2^ - > O_2^ + > {O_2}
D. O2−>O22−>O2>O2+O_2^ - > O_2^{2 - } > {O_2} > O_2^ +

Explanation

Solution

Bond strength depends upon the Bond order. Bond orders of the given species are determined by using MOT. Bond order is the measurement of bond strength and bond energy.

Complete step by step answer:
Let us find out the bond order of the given species:
Bond strength is measured by bond energy and bond energy is directly proportional to bond order in the given species. Bond order of diatomic molecules can be determined by using molecular orbital theory (MOT). Bond orders of the given species are as:
For O2{O_2}
Total number of electrons = 16
Molecular electronic configuration of O2{O_2} = (σ1s)2{(\sigma _{1s})^2} (σ1s∗)2{(\sigma _{1s}^*)^2} (σ2s)2{(\sigma _{2s})^2} (σ2s∗)2{(\sigma _{2s}^*)^2} (σ2pz)2{(\sigma _{2{p_z}})^2} [(π2px)2=(π2py)2][{(\pi _{2{p_x}})^2} = {(\pi _{2{p_y}})^2}] [(π2px∗)1=(π2py∗)1][{(\pi _{2{p_x}}^*)^1} = {(\pi _{2{p_y}}^*)^1}]
Number of electrons present in bonding molecular orbitals (BMO) = Nb{N_b}
Nb{N_b}= 2 + 2 + 2 + 2 + 2 = 10
Number of electrons present in antibonding molecular orbitals (ABMO) = Na{N_a}
Na{N_a}= 2 + 2 + 1 + 1 = 6

Formula for Bond order = Nb−Na2\dfrac{{{N_b} - {N_a}}}{2}

Bond order in O2{O_2} molecule = 10−62 = 2\dfrac{{10 - 6}}{2}{\text{ = 2}}
For O2+O_2^ +
Total number of electrons = 15
One electron is removed from (π2py∗){(\pi _{2{p_y}}^*)}
Molecular electronic configuration of O2+O_2^ + = (σ1s)2{(\sigma _{1s})^2} (σ1s∗)2{(\sigma _{1s}^*)^2} (σ2s)2{(\sigma _{2s})^2} (σ2s∗)2{(\sigma _{2s}^*)^2} (σ2pz)2{(\sigma _{2{p_z}})^2} [(π2px)2=(π2py)2][{(\pi _{2{p_x}})^2} = {(\pi _{2{p_y}})^2}] [(π2px∗)1=(π2py∗)0][{(\pi _{2{p_x}}^*)^1} = {(\pi _{2{p_y}}^*)^0}]
Number of electrons present in bonding molecular orbitals (BMO) = Nb{N_b}
Nb{N_b}= 2 + 2 + 2 + 2 + 2 = 10
Number of electrons present in antibonding molecular orbitals (ABMO) = Na{N_a}
Na{N_a}= 2 + 2 + 1 + 0 = 5

Formula for Bond order = Nb−Na2\dfrac{{{N_b} - {N_a}}}{2}

Bond order in O2+O_2^ + = 10−52 = 2.5\dfrac{{10 - 5}}{2}{\text{ = 2}}{\text{.5}}
For O2−O_2^ -
Total number of electrons = 17
One electron is added to (π2px∗){(\pi _{2{p_x}}^*)}.
Molecular electronic configuration of O2−O_2^ - = (σ1s)2{(\sigma _{1s})^2} (σ1s∗)2{(\sigma _{1s}^*)^2} (σ2s)2{(\sigma _{2s})^2} (σ2s∗)2{(\sigma _{2s}^*)^2} (σ2pz)2{(\sigma _{2{p_z}})^2} [(π2px)2=(π2py)2][{(\pi _{2{p_x}})^2} = {(\pi _{2{p_y}})^2}] [(π2px∗)2=(π2py∗)1][{(\pi _{2{p_x}}^*)^2} = {(\pi _{2{p_y}}^*)^1}]

Number of electrons present in bonding molecular orbitals (BMO) = Nb{N_b}
Nb{N_b}= 2 + 2 + 2 + 2 + 2 = 10
Number of electrons present in antibonding molecular orbitals (ABMO) = Na{N_a}
Na{N_a}= 2 + 2 + 2 + 1 = 7

Formula for Bond order = Nb−Na2\dfrac{{{N_b} - {N_a}}}{2}

Bond order in O2−O_2^ - = 10−72 = 1.5\dfrac{{10 - 7}}{2}{\text{ = 1}}{\text{.5}}
For O22−O_2^{2 - }
Total number of electrons = 18
Two electrons are added to ABMO, one electron to each(π2px∗){(\pi _{2{p_x}}^*)} and (π2py∗){(\pi _{2{p_y}}^*)}
Molecular electronic configuration of O22−O_2^{2 - } = (σ1s)2{(\sigma _{1s})^2} (σ1s∗)2{(\sigma _{1s}^*)^2} (σ2s)2{(\sigma _{2s})^2} (σ2s∗)2{(\sigma _{2s}^*)^2} (σ2pz)2{(\sigma _{2{p_z}})^2} [(π2px)2=(π2py)2][{(\pi _{2{p_x}})^2} = {(\pi _{2{p_y}})^2}] [(π2px∗)2=(π2py∗)2][{(\pi _{2{p_x}}^*)^2} = {(\pi _{2{p_y}}^*)^2}]

Number of electrons present in bonding molecular orbitals (BMO) = Nb{N_b}
Nb{N_b} = 2 + 2 + 2 + 2 + 2 = 10
Number of electrons present in antibonding molecular orbitals (ABMO) = Na{N_a}
Na{N_a} = 2 + 2 + 2 + 2 = 8

Formula for Bond order = Nb−Na2\dfrac{{{N_b} - {N_a}}}{2}

Bond order in O22−O_2^{2 - } = 10−82 = 1\dfrac{{10 - 8}}{2}{\text{ = 1}}

Decreasing bond order is O2+>O2>O2−>O22−O_2^ + > {O_2} > O_2^ - > O_2^{2 - }
Bond strength is directly proportional to bond order, so decreasing order of bond strength is: O2+>O2>O2−>O22−O_2^ + > {O_2} > O_2^ - > O_2^{2 - }.

Hence, the correct answer is (B).

Note: Bond energy and bond strength are determined with the help of bond order. Bond order is the number of bonds between two atoms of a molecule. Bond strength is directly proportional to bond order.