ANSWER 1:  i) 2p_y & 2p_y

Reason: The shape of the s-orbital is Sperical and thus it can only overlap any orbital (maybe s maybe p or any other) AXIALLY. But p-orbital can overlap in two ways: Axially as well as Laterally. Thus answer is (i). See this figures:

The Overlaps in Figure 2 shown in Navy blue and Red are LATERAL overlaps while that of the Black is Axial Overlap.

ANSWER 2: (i) Pi-orbital

Reason: This is becaise, if the axis of the molecule is Z, then the p-orbitals lying in this axis will overlap  AXIALLY and the others will overlap LATERALLY, as they both lie Perpendicular to the Z-axis, and thus will form pi-bonds with their corresponding orbitals of the other molecule. take a look at the following figure:

I hope my answers are correct and u understand it clrearly............ though i am quite sure of both of them but still do inform me if i am wrong!!!  .........

The configuration Diagrams for increasing energy are as follows..........

I Believe u can write the configuraitions urself, with the help of these diagrams, THE ELECTRONS ARE FILLED FROM THE MOST STABLE ORBITAL TO THE LEAST STABLE ORBITAL.............

AND....

I Hope it Helps.............. All the Best........

look u have to find the no of VALANCE electrons of Both Oxygen and Nitrogen................ only the VALANCE electron.............  now the total number of VALANCE electrons are to be arranged in the molecular orbitals.......... not ALL the electrons...... ok???

The relation is to applied for the TOTAL no. of valance eledctrons........... that means

VALANCE e^- of Oxygen + VALANCE e^- of Nitrogen

Is it clear now??? After all its a Orbital of a COMPOUND!!! :) :) so the valance electrons of both the atoms are to be considered!!!

The correct Order of stability is

O₂⁺ > O₂ > O₂^- > O₂^2-

Reason: The electronic configuration of O₂ is 1s² 2s² 2p⁴, thus the valance shell contains 6 electrons. and since there are 2 oxygen atoms in a molecule, thus, 12 electrons are to be filled in the molecular orbitals as :

( 2s)² ( * 2s)²  (2p_x)²   (pi 2p_y)² (pi 2p_z)² (pi * 2p_y)² (pi 2p_z)²

Thus the bond Order = (No. of Bonding e^- - No. Of anti-bonding e^-) / 2

We get, Bond Order = (8 - 4)/2 = 2.

Similarly, repeating the steps for O₂⁺,  O₂^- and O₂^2- we find the above result.

Increasing bond order = increasing stability.... Hope this Helps......