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Question: For diatomic species are listed below, identify the correct order in which bond order is increasing ...

For diatomic species are listed below, identify the correct order in which bond order is increasing them.
A. O2−<NO<C22−<He2+{{O}_{2}}^{-} < NO < {{C}_{2}}^{2-} < H{{e}_{2}}^{+}
B. C22−<He2+<O2−<NO{{C}_{2}}^{2-} < H{{e}_{2}}^{+} < {{O}_{2}}^{-} < NO
C. He2+<O2−<NO<C22−H{{e}_{2}}^{+} < {{O}_{2}}^{-} < NO < {{C}_{2}}^{2-}
D. NO<O2−<C22−<He2+NO < {{O}_{2}}^{-} < {{C}_{2}}^{2-} < H{{e}_{2}}^{+}

Explanation

Solution

Bond order is defined as the difference between the number of bonds and antibonds. Bond-dissociation energy measure of the strength of a chemical bond A–B. It can be defined as the standard enthalpy change when A–B is cleaved by homolysis to give fragments A and B, which are usually radical species.

Complete answer:
In molecular orbital theory, bond order is defined as half the difference between the number of bonding electrons and the number of antibonding electrons which can be shown as:
B.O.=number of bonding electrons−number of antibonding electrons2B.O.=\dfrac{number\ of\ bonding\ electrons-number\ of\ antibonding\ electrons}{2}
Bond order is also an index of bond strength and is also used extensively in valence bond theory. Generally higher the bond order stronger is the bond.
NO it contains 15 electrons and electronic configuration can be written as:
σ1s2σ∗1s2σ2s2σ∗2s2π2px2π2py2π∗2px1\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}\pi {{2}_{px}}^{2}\pi {{2}_{py}}^{2}{{\pi }^{*}}{{2}_{px}}^{1};
Bonding electrons = 8 ; Antibonding electrons = 3
B.O. = 8−32=2.5\dfrac{8-3}{2}=2.5
O2−{{O}_{2}}^{-}contain 17 electrons having E.C. σ1s2σ∗1s2σ2s2σ∗2s2π2px2π2py2π∗2px2π∗2py1\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}\pi {{2}_{px}}^{2}\pi {{2}_{py}}^{2}{{\pi }^{*}}{{2}_{px}}^{2}{{\pi }^{*}}{{2}_{py}}^{1}
Bonding electrons = 8 ; Antibonding electrons = 5
B.O. = 8−52=1.5\dfrac{8-5}{2}=1.5
C22−{{C}_{2}}^{2-}contain 14 electrons having E.C. σ1s2σ∗1s2σ2s2σ∗2s2π2px2π2py2\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}\pi {{2}_{px}}^{2}\pi {{2}_{py}}^{2}
Bonding electrons = 8 ; Antibonding electrons = 2
B.O. = 8−22=3\dfrac{8-2}{2}=3
He22+H{{e}_{2}}^{2+}contain 3 electrons having E.C. σ1s2σ∗1s1\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{1}}
Bonding electrons = 2 ; Antibonding electrons = 1
B.O. = 2−12=0.5\dfrac{2-1}{2}=0.5

**Hence the bond order follows the order mentioned in the option C He2+<O2−<NO<C22−H{{e}_{2}}^{+} < {{O}_{2}}^{-} < NO < {{C}_{2}}^{2-} **

Note:
Except for diatomic molecules, the bond-dissociation energy differs from the bond energy. While the bond-dissociation energy is the energy of a single chemical bond, the bond energy is the average of all the bond-dissociation energies of the bonds of the same type for a given molecule.