The correct order of energies of molecular orbitals of $N_2$ molecule is

  • A
    $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma 2p_z < \sigma^* 2p_z$
  • B
    $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$
  • C
    $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$
  • D
    $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < \sigma^* 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y)$

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Similar Questions

For a stable molecule,the value of bond order must be

What is the number of unpaired electron$(s)$ in the highest occupied molecular orbital $(HOMO)$ of the following species: $N_2$,$N_2^{+}$,$O_2$,$O_2^{+}$?

Which one of the following pairs of species have the same bond order?

The increasing order of bond order of $O_2, O_2^+, O_2^-$,and $O_2^{2-}$ is:

Match the following:
Pair of species Identical property
$A$. $B_2 \& O_2$ $P$. Bond order $= 2.5$
$B$. $Be_2 \& H_2^{2-}$ $Q$. Paramagnetic nature
$C$. $N_2^{+} \& N_2^{-}$ $R$. Diamagnetic nature
$D$. $O_2^{+} \& O_2^{-}$ $S$. Doesn't exist

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