Question

The bond order values of anions of sodium peroxide and potassium superoxide are respectively:

• A. \(1,\;1.5\)
• B. \(1,\;0.5\)
• C. \(0.5,\;1\)
• D. \(2,\;0.5\)

Hint:

Important bond orders of oxygen species: \[ O_2 \rightarrow 2 \] \[ O_2^- \rightarrow 1.5 \] \[ O_2^{2-} \rightarrow 1 \] As electrons are added to antibonding orbitals, bond order decreases.

Answer 1

The Correct Option is A

Step 1: Identify the anions present in the compounds.
Sodium peroxide contains the peroxide ion \[ O_2^{2-} \] Potassium superoxide contains the superoxide ion \[ O_2^{-} \] Therefore, we need to determine the bond orders of \[ O_2^{2-} \] and \[ O_2^{-} \] respectively.

Step 2: Calculate the bond order of peroxide ion \((O_2^{2-})\).
For molecular oxygen, \[ O_2 \] the bond order is \[ 2 \] When two electrons are added to form peroxide ion, \[ O_2^{2-} \] these electrons enter the antibonding \(\pi^{*}\) molecular orbitals.
Thus, \[ N_b=10 \] \[ N_a=8 \] Using the bond order formula, \[ \text{Bond Order} = \frac{N_b-N_a}{2} \] \[ = \frac{10-8}{2} \] \[ =1 \] Hence, the bond order of \[ O_2^{2-} \] is \[ 1 \]

Step 3: Calculate the bond order of superoxide ion \((O_2^-)\).
Superoxide ion is formed by adding one electron to \(O_2\), \[ O_2 + e^- \rightarrow O_2^- \] One electron enters an antibonding orbital.
Therefore, \[ N_b=10 \] \[ N_a=7 \] Hence, \[ \text{Bond Order} = \frac{10-7}{2} \] \[ = \frac{3}{2} \] \[ =1.5 \] Thus, the bond order of \[ O_2^- \] is \[ 1.5 \]

Step 4: Final conclusion.
Therefore, the bond order values of the anions of sodium peroxide and potassium superoxide are \[ \boxed{1,\;1.5} \] Hence, the correct option is \[ \boxed{(1)} \]