Question

Arrange the following in the correct order of their bond orders.
\[ \mathrm{N_2}\quad (I), \qquad \mathrm{O_2}\quad (II), \qquad \mathrm{O_2^{+}}\quad (III), \qquad \mathrm{O_2^{-}}\quad (IV) \]

• A. \(II \gt III \gt I \gt IV\)
• B. \(III \gt II \gt IV \gt I\)
• C. \(I \gt III \gt II \gt IV\)
• D. \(I \gt II \gt III \gt IV\)

Hint:

Removing an electron from an antibonding orbital increases bond order, while adding an electron to an antibonding orbital decreases bond order. Therefore, \[ \mathrm{O_2^{+}} \gt \mathrm{O_2} \gt \mathrm{O_2^{-}} \] in bond order.

Answer 1

The Correct Option is C

Step 1: Recall the formula for bond order.
According to Molecular Orbital Theory, \[ \text{Bond Order} = \frac{N_b-N_a}{2} \] where \[ N_b=\text{number of bonding electrons} \] and \[ N_a=\text{number of antibonding electrons}. \]

Step 2: Determine the bond order of \(\mathrm{N_2}\).
The molecular orbital configuration of \(\mathrm{N_2}\) contains \[ N_b=10,\qquad N_a=4 \] Therefore, \[ \text{Bond Order of } \mathrm{N_2} = \frac{10-4}{2} = 3 \] Thus, \[ \mathrm{N_2} \rightarrow 3 \]

Step 3: Determine the bond order of \(\mathrm{O_2}\).
For \(\mathrm{O_2}\), \[ N_b=10,\qquad N_a=6 \] Hence, \[ \text{Bond Order of } \mathrm{O_2} = \frac{10-6}{2} = 2 \] Thus, \[ \mathrm{O_2} \rightarrow 2 \]

Step 4: Determine the bond order of \(\mathrm{O_2^{+}}\).
Formation of \(\mathrm{O_2^{+}}\) involves removal of one electron from the highest occupied antibonding orbital.
Therefore, antibonding electrons decrease by one. \[ N_b=10,\qquad N_a=5 \] Hence, \[ \text{Bond Order of } \mathrm{O_2^{+}} = \frac{10-5}{2} = 2.5 \] Thus, \[ \mathrm{O_2^{+}} \rightarrow 2.5 \]

Step 5: Determine the bond order of \(\mathrm{O_2^{-}}\).
Formation of \(\mathrm{O_2^{-}}\) involves addition of one electron to an antibonding orbital.
Therefore, antibonding electrons increase by one. \[ N_b=10,\qquad N_a=7 \] Hence, \[ \text{Bond Order of } \mathrm{O_2^{-}} = \frac{10-7}{2} = 1.5 \] Thus, \[ \mathrm{O_2^{-}} \rightarrow 1.5 \]

Step 6: Arrange the species in decreasing order of bond order.
The bond orders are \[ \mathrm{N_2}=3 \] \[ \mathrm{O_2^{+}}=2.5 \] \[ \mathrm{O_2}=2 \] \[ \mathrm{O_2^{-}}=1.5 \] Therefore, \[ \mathrm{N_2} \gt \mathrm{O_2^{+}} \gt \mathrm{O_2} \gt \mathrm{O_2^{-}} \] or \[ I \gt III \gt II \gt IV \]

Step 7: Final conclusion.
Hence, the correct order is \[ \boxed{I \gt III \gt II \gt IV} \]