To determine the number of molecules with a bond order of 2, we first need to calculate the bond order for each given molecule. Bond order is given by the formula:
Bond Order = (Number of bonding electrons - Number of antibonding electrons) / 2
Let's apply this formula to each molecule:
| $C_2$ | Bond Order = (6 bonding electrons - 2 antibonding electrons) / 2 = 2 |
| $O_2$ | Bond Order = (8 bonding electrons - 4 antibonding electrons) / 2 = 2 |
| $Be_2$ | Bond Order = (2 bonding electrons - 2 antibonding electrons) / 2 = 0 |
| $Li_2$ | Bond Order = (2 bonding electrons - 0 antibonding electrons) / 2 = 1 |
| $Ne_2$ | Bond Order = (8 bonding electrons - 8 antibonding electrons) / 2 = 0 |
| $N_2$ | Bond Order = (10 bonding electrons - 4 antibonding electrons) / 2 = 3 |
| $He_2$ | Bond Order = (2 bonding electrons - 2 antibonding electrons) / 2 = 0 |
Checking our results, $C_2$ and $O_2$ each have a bond order of 2. So, there are 2 molecules with a bond order of 2.
This matches the expected range of 2 to 2, confirming the calculated result is correct.
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$_2$.
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