1. Interpretation of the question

The species referred to is an AX$_3$E$_2$ type species (such as $\mathrm{ClF_3}$-type electron arrangement), where the electron-pair geometry is trigonal bipyramidal.

In a trigonal bipyramidal arrangement:

• there are 2 axial positions,
• there are 3 equatorial positions.

The important repulsion order is: $\text{lp-lp} > \text{lp-bp} > \text{bp-bp}$ And among positions:

• an axial position has 3 equatorial neighbors at $90^\circ$,
• an equatorial position has only 2 axial neighbors at $90^\circ$.

So lone pairs prefer equatorial positions because that minimizes $90^\circ$ repulsions.

2. Check Statement I

Statement I says that all three possible structures may be drawn.

For a trigonal bipyramidal arrangement with two lone pairs, the three positional possibilities are:

1. both lone pairs axial,
2. one axial and one equatorial,
3. both equatorial.

These are indeed the three possible ways of placing two lone pairs in a trigonal bipyramidal set of five positions.

So Statement I is correct.

3. Check Statement II

Statement II says structure III is most stable because the orbitals having lone pairs are axial, where lp-bp repulsion is minimum.

This is incorrect.

Why?

In trigonal bipyramidal geometry:

• lone pairs are most stable at equatorial positions, not axial.
• equatorial positions experience fewer $90^\circ$ interactions.

Thus, the most stable structure is the one with lone pairs in equatorial positions.

Therefore, the reason given in Statement II is wrong.

So Statement II is incorrect.

4. Conclusion

• Statement I: Correct
• Statement II: Incorrect

Hence, the correct option is: $\boxed{\text{C}}$

5. Comparison with stored answer

Stored correct answer: C

My derived answer: C

They match.