Question

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

Answer 1

Hint: 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 ${O_2}$
Total number of electrons = 16
Molecular electronic configuration of ${O_2}$ = \[{(\sigma _{1s})^2}\] ${(\sigma _{1s}^*)^2}$ ${(\sigma _{2s})^2}$ ${(\sigma _{2s}^*)^2}$ ${(\sigma _{2{p_z}})^2}$ $[{(\pi _{2{p_x}})^2} = {(\pi _{2{p_y}})^2}]$ $[{(\pi _{2{p_x}}^*)^1} = {(\pi _{2{p_y}}^*)^1}]$
Number of electrons present in bonding molecular orbitals (BMO) = ${N_b}$
${N_b}$= 2 + 2 + 2 + 2 + 2 = 10
Number of electrons present in antibonding molecular orbitals (ABMO) = ${N_a}$
${N_a}$= 2 + 2 + 1 + 1 = 6

Formula for Bond order = $\dfrac{{{N_b} - {N_a}}}{2}$

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

Formula for Bond order = $\dfrac{{{N_b} - {N_a}}}{2}$

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

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

Formula for Bond order = $\dfrac{{{N_b} - {N_a}}}{2}$

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

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

Formula for Bond order = $\dfrac{{{N_b} - {N_a}}}{2}$

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

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

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.