Chemistry - CHEMICAL BONDING Question with Solution | TestHub

To determine the existence of a species, we calculate its bond order using Molecular Orbital Theory. A bond order of zero indicates that the molecule/ion is unlikely to exist.

Let's analyze each option:

A. $H_2^{+}$ and $H{e}_2^{2-}$

For $H_2^{+}$: Total electrons = 1. Configuration: $({\sigma }_{1s}{)}^1$. Bond order = $(1-0)/2=0.5$. Exists.

For $H{e}_2^{2-}$: Total electrons = 6. Configuration: $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2$. Bond order = $(4-2)/2=1$. Exists.

B. $H_2^{-}$ and $H{e}_2^{2-}$

For $H_2^{-}$: Total electrons = 3. Configuration: $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^1$. Bond order = $(2-1)/2=0.5$. Exists.

For $H{e}_2^{2-}$: Bond order = 1. Exists.

C. $H_2^{2+}$ and $H{e}_2$

For $H_2^{2+}$: Total electrons = 0. Configuration: None. Bond order = $(0-0)/2=0$. Does not exist.

For $H{e}_2$: Total electrons = 4. Configuration: $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2$. Bond order = $(2-2)/2=0$. Does not exist.

D. $H_2^{-}$ and $H{e}_2^{2+}$

For $H_2^{-}$: Bond order = 0.5. Exists.

For $H{e}_2^{2+}$: Total electrons = 2. Configuration: $({\sigma }_{1s}{)}^2$. Bond order = $(2-0)/2=1$. Exists.

Therefore, both $H_2^{2+}$ and $H{e}_2$ have a bond order of zero, meaning they are not likely to exist.

The final answer is $\boxed{C}$.