Question

Consider the following statements
Statement I: The electronic configuration of $N_2$ in terms of molecular orbital theory is $KK (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi 2p_z)^2$.
Statement II: According to molecular orbital theory, the double bond in $C_2$ molecule consists of two $\pi$ bonds.
The correct answer is

• A. Both statements I and II are correct
• B. Both statements I and II are not correct
• C. Statement I is correct, but statement II is not correct
• D. Statement I is not correct, but statement II is correct

Hint:

{MOT energy crossover:} For $Li_2$ to $N_2$, the $\pi 2p$ orbitals are lower in energy than $\sigma 2p_z$. For $O_2$ and $F_2$, $\sigma 2p_z$ is lower than $\pi 2p$.

Answer 1

The Correct Option is D

Step 1: Analyze Statement I ($N_2$ Configuration):
Nitrogen ($N_2$) has 14 electrons. For diatomic molecules with $Z \le 7$, the energy order is: $\sigma 1s<\sigma^* 1s<\sigma 2s<\sigma^* 2s<(\pi 2p_x = \pi 2p_y)<\sigma 2p_z<\dots$ The configuration given in Statement I lists $(\pi 2p_z)^2$ at the end. However, the $z$-axis orbital forms a sigma bond ($\sigma 2p_z$), not a pi bond. The term $(\pi 2p_z)$ is notationally incorrect for standard diatomic convention where $z$ is the internuclear axis. Furthermore, if the statement meant to show equality or incorrect filling, it violates the standard order where $\sigma 2p_z$ is the HOMO for $N_2$. Thus, the statement is Incorrect.
Step 2: Analyze Statement II ($C_2$ Bonding):
Carbon ($C_2$) has 12 electrons. Configuration: $KK (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2$. Bond Order = $\frac{1}{2}(8 - 4) = 2$. The last 4 electrons occupy the degenerate $\pi 2p_x$ and $\pi 2p_y$ orbitals. The $\sigma 2p_z$ orbital is empty. This implies that both bonds contributing to the double bond are $\pi$-bonds. There is no $\sigma$-bond in $C_2$. This is a unique feature predicted by MOT. Thus, Statement II is Correct.
Step 3: Conclusion:
Statement I is false, Statement II is true.