Question 5

Analyze each statement: (A) $Ne_2$ has 20 electrons. Configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^2, \pi^* 2p_y^2, \sigma^* 2p_z^2$. Bonding electrons = 10, Antibonding = 10. Bond order = $(10-10)/2 = 0$. This is a correct statement. (B) $F_2$ has 18 electrons. The configuration ends in $... \pi^* 2p_x^2, \pi^* 2p_y^2$. The HOMO is a $\pi^*$ orbital, not a $\sigma$ orbital. This statement is INCORRECT. (C) $O_2$ has a bond order of 2. $O_2^+$ is formed by removing an electron from an antibonding $\pi^*$ orbital, increasing the bond order to 2.5. Higher bond order implies higher bond energy. Thus, bond energy of $O_2^+$ is greater than $O_2$. This statement is INCORRECT. (D) $Li_2$ and $B_2$ both have a bond order of 1. However, $Li$ (atomic number 3) has a significantly larger atomic radius than $B$ (atomic number 5) due to periodic trends. Consequently, the bond length of $Li_2$ (~267 pm) is larger than $B_2$ (~159 pm). This is a correct statement. The question asks for INCORRECT statements, which are (B) and (C).