The correct order of energies of molecular orbitals of $N _2$ molecule, is
$\sigma 1 s < \sigma^* 1 s < \sigma 2 s < \sigma^* 2 s < \left(\pi 2 p_{ x }=\pi 2 p_{ y }\right) < \left(\pi^* 2 p_{ x }=\pi^* 2 p_{ y }\right) < \sigma 2 p_{ z } < \sigma^* 2 p_{ z }$
$\sigma 1 s < \sigma^* 1 s < \sigma 2 s < \sigma^* 2 s < \left(\pi 2 p_{ x }=\pi 2 p_{ y }\right) < \sigma 2 p_{ z } < \left(\pi^* 2 p_{ x }=\pi^* 2 p_{ y }\right) < \sigma^* 2 p_{ z }$
$\sigma 1 s < \sigma^* 1 s < \sigma 2 s < \sigma^* 2 s < \sigma 2 p_{ z } < \left(\pi 2 p_{ x }=\pi 2 p_{ y }\right) < \left(\pi^* 2 p_{ x }=\pi^* 2 p_{ y }\right) < \sigma^* 2 p_{ z }$
$\sigma 1 s < \sigma^* 1 s < \sigma 2 s < \sigma^* 2 s < \sigma 2 p_{ z } < \sigma^* 2 p_{ z } < \left(\pi 2 p_{ x }=\pi 2 p_{ y }\right) < \left(\pi^* 2 p_{ x }=\pi^* 2 p_{ y }\right)$
Which of the following contains $(2C -1e^-)$ bond
Molecular orbital theory was developed mainly by
Given below are two statements:
Statement $(I)$ : A $\pi$ bonding $MO$has lower electron density above and below the inter-nuclear asix.
Statement $(II)$ : The $\pi^*$ antibonding $MO$ has a node between the nucles.In the light of the above statements, choose the most appropriate answer from the options given below:
In which of the following transformations, the bond order has increased and the magnetic behaviour has changed ?
Which has the highest bond energy