Step 1: Use the Bond Order Formula Bond order is given by: \[ \text{Bond order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2} \]
Step 2: Compute Bond Orders \( C_2 \): Electronic configuration in molecular orbitals gives bond order = 2. \( O_2 \): Electronic configuration in molecular orbitals also gives bond order = 2. Since both species have the same bond order, they are the correct answer.
Identify the correct orders against the property mentioned:
A. H$_2$O $>$ NH$_3$ $>$ CHCl$_3$ - dipole moment
B. XeF$_4$ $>$ XeO$_3$ $>$ XeF$_2$ - number of lone pairs on central atom
C. O–H $>$ C–H $>$ N–O - bond length
D. N$_2$>O$_2$>H$_2$ - bond enthalpy
Choose the correct answer from the options given below: