Solveeit Logo
Login

Question

Question: The correct order of bond dissociation energy among \({{N}_{2}}\),\({{O}_{2}}\),\[{{O}_{2}}^{-}\] is...

The correct order of bond dissociation energy among N2{{N}_{2}},O2{{O}_{2}},O2−{{O}_{2}}^{-} is shown in which of the following arrangements?
A. N2>O2−>O2{{N}_{2}}>{{O}_{2}}^{-}>{{O}_{2}}
B. O2>O2−>N2{{O}_{2}}>{{O}_{2}}^{-}>{{N}_{2}}
C. N2>O2>O2−{{N}_{2}}>{{O}_{2}}>{{O}_{2}}^{-}
D. O2−>O2>N2{{O}_{2}}^{-}>{{O}_{2}}>{{N}_{2}}

Explanation

Solution

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

Complete Solution :
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 antibondin⁡g electrons2B.O.=\dfrac{number\ of\ bonding\ electrons-number\ of\ antibond\operatorname{in}g\ electrons}{2}
Bond order also tells us about the bond strength and is also used comprehensively in valence bond theory. Generally higher the bond order stronger is the bond.
Hence we can say that Bond order∝Bond energyBond~order\propto Bond~energy

A. N2{{N}_{2}} discuss the electronic configuration according to MO theory
σ1s2σ∗1s2σ2s2σ∗2s2π2px2π2py2π2pz2\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}\pi 2{{p}_{x}}^{2}\pi 2{{p}_{y}}^{2}\pi 2{{p}_{z}}^{2};
Bonding electron = 10 Antibonding electron = 4
B.0. = 10−42=3\dfrac{10-4}{2}=3

B. O2{{O}_{2}}; σ1s2σ∗1s2σ2s2σ∗2s2π2px2π2py2π2pz2π∗2px2\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}\pi 2{{p}_{x}}^{2}\pi 2{{p}_{y}}^{2}\pi 2{{p}_{z}}^{2}{{\pi }^{*}}2{{p}_{x}}^{2}
Bonding electron = 10 Antibonding electron = 6
B.O. = 10−62=2\dfrac{10-6}{2}=2

C. O2−{{O}_{2}}^{-}; σ1s2σ∗1s2σ2s2σ∗2s2π2px2π2py2π2pz2π∗2px2π∗2py1\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}\pi 2{{p}_{x}}^{2}\pi 2{{p}_{y}}^{2}\pi 2{{p}_{z}}^{2}{{\pi }^{*}}2{{p}_{x}}^{2}{{\pi }^{*}}2{{p}_{y}}^{1}
Bonding electron = 10 Antibonding electron = 7
B.O. = 10−72=1.5\dfrac{10-7}{2} = 1.5
Hence we discuss higher the bond order, higher is the energy so it follows the order of option C; N2>O2>O2−{{N}_{2}}>{{O}_{2}}>{{O}_{2}}^{-}
So, the correct answer is “Option C”.

Note: The bond-dissociation energy is always different from the bond energy except for diatomic molecules. The bond-dissociation energy is the energy of a single chemical bond while the bond energy is the average of all the bond-dissociation energies of the bonds of the same type for a given molecule.