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# In ${\text{O}}_2^ - {\text{, }}{{\text{O}}_{\text{2}}}$ and ${\text{O}}_2^{2 - }$ molecular species, the total number of antibonding electrons respectively are:(A) 7,6,8(B) 1,0,2(C) 6,6,6(D) 8,6,8

Last updated date: 13th Jun 2024
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Hint: Determine total number of electrons present in each species. Then write the molecular orbital electronic configuration of each species. From the molecular orbital electronic configuration of each species, determine the number of antibonding electrons present.

Complete step by step answer:
The atomic number of oxygen is 8. Thus, an oxygen atom has 8 electrons. If a negative charge is present, there is an additional electron. If two negative charges are present, then two additional electrons are present.
a) Calculate total number of electrons present in ${\text{O}}_2^ -$ ion:
8 + 8 + 1 = 17
Each oxygen atom brings 8 valence electrons. There is an additional electron due to negative charge. So ${\text{O}}_2^ -$ ion has 17 electrons. Write MO electronic configuration of ${\text{O}}_2^ -$ ion.
$\sigma _{1s}^2\sigma _{1s}^{*2}\sigma _{2s}^2\sigma _{2s}^{*2}\sigma _{2p_z}^2\pi _{2p_x}^2\pi _{2p_y}^2\pi _{2p_x}^{*2}\pi _{2p_y}^{*1}$
Count the number of antibonding electrons present:
There are 7 antibonding electrons present in ${\text{O}}_2^ -$ ion. These are present in $\sigma _{1s}^*,\sigma _{2s}^*,\pi _{2p_x}^*,\pi _{2p_y}^*$ orbitals.
b) Calculate total number of electrons present in ${{\text{O}}_{\text{2}}}$ molecule:
8 + 8 = 16
Each oxygen atom brings 8 valence electrons. So ${{\text{O}}_{\text{2}}}$ molecule has 16 electrons. Write MO electronic configuration of ${{\text{O}}_{\text{2}}}$ molecule.
$\sigma _{1s}^2\sigma _{1s}^{*2}\sigma _{2s}^2\sigma _{2s}^{*2}\sigma _{2p_z}^2\pi _{2p_x}^2\pi _{2p_y}^2\pi _{2p_x}^{*2}$
Count the number of antibonding electrons present:
There are 6 antibonding electrons present in ${{\text{O}}_{\text{2}}}$ molecules. These are present in $\sigma _{1s}^*,\sigma _{2s}^*,\pi _{2p_x}^*$ orbitals.
c) Calculate total number of electrons present in ${\text{O}}_2^{2 - }$ ion:
8 + 8 + 2 = 18
Each oxygen atom brings 8 valence electrons. There are two additional electrons due to two negative charges. So ${\text{O}}_2^{2 - }$ ion has 18 electrons. Write MO electronic configuration of ${\text{O}}_2^{2 - }$ ion.
$\sigma _{1s}^2\sigma _{1s}^{*2}\sigma _{2s}^2\sigma _{2s}^{*2}\sigma _{2p_z}^2\pi _{2p_x}^2\pi _{2p_y}^2\pi _{2p_x}^{*2}\pi _{2p_y}^{*2}$
Count the number of antibonding electrons present:
There are 8 antibonding electrons present in ${\text{O}}_2^{2 - }$ ion. These are present in $\sigma _{1s}^*,\sigma _{2s}^*,\pi _{2px}^*,\pi _{2py}^*$ orbitals.
Thus, in ${\text{O}}_2^ - {\text{, }}{{\text{O}}_{\text{2}}}$ and ${\text{O}}_2^{2 - }$ molecular species, the total number of antibonding electrons are 7,6,8 respectively.

Hence, the correct option is the option (A).

Note: From the molecular orbital electronic configuration of a species, you can determine the number of bonding electrons and the number of antibonding electrons. From this, you can calculate the bond order. For example, for ${{\text{O}}_{\text{2}}}$ molecule, $B.O = \dfrac{{{N_b} - {N_a}}}{2} = \dfrac{{10 - 6}}{2} = \dfrac{4}{2} = 2$