(a) `H_(2)^(+)and H_(2)^(-)` ions have same bond order but `H_(2)^(-)` ion is more stable. Explain.
(b) `N_(2)` has a greater bond dissociation enthalpy than `N_(2)^(+)` ion hals more bond dissociation enthalpy than `O_(2).` Why ?
(c ) Can we have ahomonuclear diatomic molecule with all its ground state molecular orbitals full of electrons ?
(d) When a magnet is dipped in a jar of liquid oxygen, some oxygen clings to it. Assign reason :

Explanation

(a) Both the ions have the same bond other (0.5) but they differ in the configurations.
`H_(2)^(+)ion : [sigma_(1s)]^(1),H_(2)^(-)ion :[sigma_(1s)]^(2)[sigma_(1s)^(**)]^(1)`
Since `H_(2)^(-)` ion has an electron in the anitibonding molecular orbital, it is therefore, less stable.
(b) When `N_(2)` is converted ot `N_(2)^(+)` ion, an electron is removed from the bonding M.O. Therefore, the bond order of `N_(2)^(+)(2.5)` is less than that of `N_(2)(3.0)` and its bond dissociation enthalpy is also less. On the other hand when `O_(2)` changes to `O_(2)^(+)` ion, the electron is removed from antibonding M.O. As a result, the bond order of `O_(2)^(+)` ion (2.5) is more than that of `O_(2)(2.0).` This means that bond dissociation enthalpy of `O_(2)^(+)` ion is more than that of `O_(2).`
(c ) No such a molecule cannot exist an it will be only hypothetical in nature (e.g. `He_(2)` molecule). If the ground state is full of electrons, then `N_(b)=N_(a).` This means that the bond order will be zero. Such a molecule can not exist.
(d) Molecular orbital electronic configuration of oxygen `(O_(2))` shows the presence of two unpaired electrons. This means that it is of paramagnetic nature. Thus, when a magnet is placed in a jar of liquid oxygen, some molecules will cling to it.