The compound which does not exist is

Solution

Chemistry ma’am · 3 steps

Burn sodium in air and you get the peroxide Na₂O₂, not the superoxide NaO₂.

1. Check each compound. Do not reject a compound just because its octet looks unusual. Ask whether it is actually known and stable under normal conditions.

   Compound	What it is	Known?
   (NH₄)₂BeF₄	contains [BeF₄]²⁻; Be uses one 2s and three 2p orbitals, so covalency 4, its maximum	yes
   PbEt₄	tetraethyllead, a covalent Pb(IV) compound, once used as an antiknock agent in petrol	yes
   BeH₂	only 4 electrons on Be in the monomer, so solid BeH₂ forms a polymer through Be-H-Be bridges	yes
   NaO₂	superoxide of sodium; sodium burnt in air gives Na₂O₂ instead	no

2. Why not NaO2. The superoxide ion O₂⁻ is a large anion. Only large cations like K⁺, Rb⁺ and Cs⁺ form stable superoxides with it. Na⁺ is too small, so sodium forms the peroxide Na₂O₂. NaO₂ can be made only by forcing it under high oxygen pressure, so it is not a normal compound.
3. So. The compound that does not exist under ordinary conditions is NaO₂, option (d).

The idea to keep. Big anions pair best with big cations. Burnt in air, Li gives the oxide, Na gives the peroxide, and K, Rb, Cs give superoxides. An unusual octet alone never proves a compound cannot exist.

Answer from the NTA final answer key. Solution by PrepWiser.