Question

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

Answer 1

Hint: 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:
$B.O.=\dfrac{number\ of\ bonding\ electrons-number\ of\ antibonding\ electrons}{2}$
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:
\[\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}\pi {{2}_{px}}^{2}\pi {{2}_{py}}^{2}{{\pi }^{*}}{{2}_{px}}^{1}\];
Bonding electrons = 8 ; Antibonding electrons = 3
B.O. = \[\dfrac{8-3}{2}=2.5\]
${{O}_{2}}^{-}$contain 17 electrons having E.C. $\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}\pi {{2}_{px}}^{2}\pi {{2}_{py}}^{2}{{\pi }^{*}}{{2}_{px}}^{2}{{\pi }^{*}}{{2}_{py}}^{1}$
Bonding electrons = 8 ; Antibonding electrons = 5
B.O. = $\dfrac{8-5}{2}=1.5$
${{C}_{2}}^{2-}$contain 14 electrons having E.C. $\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}\pi {{2}_{px}}^{2}\pi {{2}_{py}}^{2}$
Bonding electrons = 8 ; Antibonding electrons = 2
B.O. = $\dfrac{8-2}{2}=3$
$H{{e}_{2}}^{2+}$contain 3 electrons having E.C. $\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{1}}$
Bonding electrons = 2 ; Antibonding electrons = 1
B.O. = $\dfrac{2-1}{2}=0.5$

Hence the bond order follows the order mentioned in the option C $H{{e}_{2}}^{+} < {{O}_{2}}^{-} < NO < {{C}_{2}}^{2-}$

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.