Question

Order of bond energy of ${{H}_{2}}^{+}$, ${{H}_{2}}$ and ${{H}_{2}}^{-}$ is:
A. ${{H}_{2}}<{{H}_{2}}^{+}<{{H}_{2}}^{-}$
B. ${{H}_{2}}^{+}<{{H}_{2}}<{{H}_{2}}^{-}$
C. ${{H}_{2}}^{-}<{{H}_{2}}^{+}<{{H}_{2}}$
D. ${{H}_{2}}^{+}<{{H}_{2}}^{-}<{{H}_{2}}$

Answer 1

Hint: To find out the correct order of bond energy for the given elements, first of all calculate the bond order of each element. Remember, bond order is directly proportional to the bond energy and stability.

Complete step by step answer:
Given that,
We have to find out the order of bond energy of ${{H}_{2}}^{+}$, ${{H}_{2}}$ and ${{H}_{2}}^{-}$.
In order to see which one has the highest bond energy and which one has the lowest, we can simply figure this out by calculating the bond orders of each element that is given in the question.
Bond order is defined as half the difference between the number of electrons present in bonding molecular orbital (B) and the number of electrons present in the antibonding molecular orbitals (NB).
And the formula od bond order is written as:
$Bond\text{ }Order=\dfrac{1}{2}(B-NB)$ where,
B is the bonding electrons and NB is the non-bonding electrons.

Let us now look into each element given.
In ${{H}_{2}}^{+}$, the total number of electrons is $1$ and it is present in the bonding molecular orbital. So, it has one bonding electron and zero non-bonding electrons.
Thus, the bond order of ${{H}_{2}}^{+}$ will be $=\dfrac{1}{2}(1-0)=\dfrac{1}{2}=0.5$.
In ${{H}_{2}}$, the total number of electrons is $2$ and it is present in the bonding molecular orbital. So, it has two bonding electrons and zero non-bonding electrons.
Thus, the bond order of ${{H}_{2}}$ will be $=\dfrac{1}{2}(2-0)=1$

While in ${{H}_{2}}^{-}$, the total number of electrons is $3$ and where two electrons are present in the bonding molecular orbital and one electron is present in the antibonding molecular orbital. So, it has two bonding electrons and one nonbonding electron.
Therefore, the bond order of ${{H}_{2}}^{-}$ will be $=\dfrac{1}{2}(3-1)=1$
We know bond order is directly proportional to that of bond energy. Thus, an element having highest bond order will have highest bond energy.

We can see that ${{H}_{2}}$ has the highest bond order while ${{H}_{2}}^{+}$ and ${{H}_{2}}^{-}$ are having same bond order. ${{H}_{2}}^{+}$ will be more stable and will have more bond energy than ${{H}_{2}}^{-}$, as in ${{H}_{2}}^{+}$ the electrons are present in only bonding molecular orbital but in ${{H}_{2}}^{-}$, two electrons are present in bonding orbital and one electron in antibonding.
Therefore, the correct order of bond energy will be ${{H}_{2}}^{-}<{{H}_{2}}^{+}<{{H}_{2}}$
So, the correct answer is “Option C”.

Note: It is important to note that when the bond order of an element is high its bond length will be shorter and we know that shorter the bond length, greater will be its bond energy and less likely will be ease of its bond breaking.