Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Statement-1 is correct. The electronic configuration of ${\text{N}}_2$ is $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2\pi 2p_y^2\sigma 2p_z^2$. Bond order of ${\text{N}}_2=\frac{1}{2}(10-4)=3$. For ${\text{N}}_2^{+}$, an electron is removed from the highest occupied molecular orbital, which is $\sigma 2p_z$ (a bonding orbital). So, the configuration of ${\text{N}}_2^{+}$ is $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2\pi 2p_y^2\sigma 2p_z^1$. Bond order of ${\text{N}}_2^{+}=\frac{1}{2}(9-4)=2.5$. Thus, Statement-1 is correct. Statement-2 is also correct and explains why the bond order decreases, as an electron is removed from a bonding molecular orbital.