Question

Number of molecules having bond order 2 from the following molecule is ...................
$C_2, O_2, Be_2, Li_2, Ne_2, N_2, He_2$

Answer 1

Correct Answer: 2

The bond order (B.O.) is calculated using the formula:
\[ \text{B.O.} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2}. \]
{C$_2$:}
\[ (12e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2, \sigma_{2s}^{{\ast}2}, [\pi_{2px} = \pi_{2py}]^4. \]
\[ \text{B.O.} = \frac{8 - 4}{2} = 2. \]
{O$_2$:} \[ (16e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2, \sigma_{2s}^{{\ast}2}, \sigma_{2p_z}^2, [\pi_{2px} = \pi_{2py}]^4, [\pi_{2px}^{{\ast}} = \pi_{2py}^{{\ast}}]^2. \]
\[ \text{B.O.} = \frac{10 - 6}{2} = 2. \]
{Be$_2$:} \[ (8e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2, \sigma_{2s}^{{\ast}2}. \]
\[ \text{B.O.} = \frac{4 - 4}{2} = 0. \]
{Li$_2$:} \[ (6e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2. \]
\[ \text{B.O.} = \frac{4 - 2}{2} = 1. \]
{Ne$_2$:} \[ (20e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2, \sigma_{2s}^{{\ast}2}, \sigma_{2p_z}^2, [\pi_{2px} = \pi_{2py}]^4, [\pi_{2px}^{{\ast}} = \pi_{2py}^{{\ast}}]^4, \sigma_{2p_z}^{{\ast}2}. \]
\[ \text{B.O.} = \frac{10 - 10}{2} = 0. \]
{N$_2$:} \[ (14e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2, \sigma_{2s}^{{\ast}2}, \pi_{2px}^2, \pi_{2py}^2, \sigma_{2p_z}^2. \]
\[ \text{B.O.} = \frac{10 - 4}{2} = 3. \]
{H$_2$:} \[ (2e^-) : \sigma_{1s}^2. \]
\[ \text{B.O.} = \frac{2 - 0}{2} = 1. \]
Molecules with bond order 2: C$_2$ and O$_2$.