What is the bond order of $N_2$?

Which of the following pictorial diagrams most correctly represents the $\pi^*$ ($\pi$-antibonding) molecular orbital between two atoms if the internuclear axis is taken to be in the $z$-direction?

For diatomic molecules,the correct statement$(s)$ about the molecular orbitals formed by the overlap of two $2p_z$ orbitals is(are):
$(A)$ $\sigma$ orbital has a total of two nodal planes.
$(B)$ $\sigma^*$ orbital has one node in the $xz$-plane containing the molecular axis.
$(C)$ $\pi$ orbital has one node in the plane which is perpendicular to the molecular axis and goes through the center of the molecule.
$(D)$ $\pi^*$ orbital has one node in the $xy$-plane containing the molecular axis.

Which of the following molecular orbitals has a nodal plane and is relatively higher in energy compared to bonding orbitals?

What is the bond order of the $O_2$ molecule?

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