To determine which species has the highest bond energy, we need to analyze their bond orders. Bond energy is directly proportional to bond order; higher bond order generally means higher bond energy.

Step 1: Determine the number of electrons for each species.

• ${\mathrm{H}}_2$: 2 electrons (1 from each H atom)
• ${\mathrm{H}}_2^{+}$: 1 electron (2 - 1 for the positive charge)
• ${\mathrm{H}}_2^{-}$: 3 electrons (2 + 1 for the negative charge)

Step 2: Write the molecular orbital (MO) configuration and calculate the bond order.

Bond Order (B.O.) = $\frac{1}{2}$ (Number of electrons in bonding MOs - Number of electrons in antibonding MOs)

• For ${\mathrm{H}}_2$ (2 electrons):
  MO configuration: $({\sigma }_{1s}{)}^2$
  Bond Order = $\frac{1}{2}(2-0)=1$
• For ${\mathrm{H}}_2^{+}$ (1 electron):
  MO configuration: $({\sigma }_{1s}{)}^1$
  Bond Order = $\frac{1}{2}(1-0)=0.5$
• For ${\mathrm{H}}_2^{-}$ (3 electrons):
  MO configuration: $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^1$
  Bond Order = $\frac{1}{2}(2-1)=0.5$

Step 3: Compare bond orders and determine bond energy.

Since bond energy is directly proportional to bond order, the species with the highest bond order will have the highest bond energy.

• ${\mathrm{H}}_2$: Bond Order = 1
• ${\mathrm{H}}_2^{+}$: Bond Order = 0.5
• ${\mathrm{H}}_2^{-}$: Bond Order = 0.5

Thus, ${\mathrm{H}}_2$ has the highest bond order (1) and therefore the highest bond energy.

Option Analysis:

• A) ${\mathrm{H}}_2$: Has a bond order of 1, which is the highest among the given options. Therefore, it has the highest bond energy.
• B) ${\mathrm{H}}_2^{+}$: Has a bond order of 0.5, which is lower than that of ${\mathrm{H}}_2$.
• C) ${\mathrm{H}}_2^{-}$: Has a bond order of 0.5, which is lower than that of ${\mathrm{H}}_2$.
• D) All have same: Incorrect, as their bond orders are different.

Correct Answer: (A)