Question

The sum of bond order of \(O_2\), \(O_2^-\), \(O_2^{2-}\) and \(O_2^+\) is equal to

• A. 5
• B. 4
• C. 6
• D. 9

Hint:

A useful shortcut for bond orders of O₂ and its ions: O₂ has a bond order of 2. Adding an electron (to an antibonding orbital) decreases the bond order by 0.5. Removing an electron (from an antibonding orbital) increases it by 0.5. O₂⁺: 2 + 0.5 = 2.5 O₂: 2.0 O₂⁻: 2 - 0.5 = 1.5 O₂²⁻: 2 - 1.0 = 1.0

Answer 1

The Correct Option is D

Step 1: Understanding Bond Order
The bond order of a molecule is given by the formula: \[ \text{Bond Order} = \frac{1}{2} \left( \text{Number of bonding electrons} - \text{Number of antibonding electrons} \right) \] We will calculate the bond order for \( O_2 \), \( O_2^- \), \( O_2^{2-} \), and \( O_2^+ \) using this formula. Step 2: Bond Order of \( O_2 \)
For the molecule \( O_2 \), the molecular orbital configuration is: \[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \pi_{2p_y}^*^1 \] The number of bonding electrons is \( 10 \), and the number of antibonding electrons is \( 6 \). Thus, the bond order is: \[ \text{Bond Order of } O_2 = \frac{1}{2} (10 - 6) = 2 \] Step 3: Bond Order of \( O_2^- \)
For the \( O_2^- \) ion, the molecular orbital configuration is: \[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^2 \pi_{2p_y}^*^1 \] The number of bonding electrons is \( 11 \), and the number of antibonding electrons is \( 6 \). Thus, the bond order is: \[ \text{Bond Order of } O_2^- = \frac{1}{2} (11 - 6) = 2.5 \] Step 4: Bond Order of \( O_2^{2-} \)
For the \( O_2^{2-} \) ion, the molecular orbital configuration is: \[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^2 \pi_{2p_y}^*^2 \] The number of bonding electrons is \( 12 \), and the number of antibonding electrons is \( 6 \). Thus, the bond order is: \[ \text{Bond Order of } O_2^{2-} = \frac{1}{2} (12 - 6) = 3 \] Step 5: Bond Order of \( O_2^+ \)
For the \( O_2^+ \) ion, the molecular orbital configuration is: \[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \pi_{2p_y}^*^0 \] The number of bonding electrons is \( 9 \), and the number of antibonding electrons is \( 5 \). Thus, the bond order is: \[ \text{Bond Order of } O_2^+ = \frac{1}{2} (9 - 5) = 2 \] Step 6: Total Sum of Bond Orders
Now, we sum the bond orders of \( O_2 \), \( O_2^- \), \( O_2^{2-} \), and \( O_2^+ \): \[ \text{Total Sum of Bond Orders} = 2 + 2.5 + 3 + 2 = 9.5 \] Thus, the sum of the bond orders is \( \boxed{9} \).