Doubt from aromaticity

Question 2 and 3 with reason

For 2nd i think it should be option D because in structure 5 as the lone pair orbitals are on same size hence conjugated pie bond can be formed
Do confirm the answer

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Yes what about 3rd

Is it C? All orbitals are out of phase with the orbitals they are bonding with, making it the most unstable structure, and therefore having the highest energy.

Answer is option a

Question 6,7 and 8

Option a is answer for 8th

For 8th
AgNo3 will take out I and ppt AgI will be formed so
1st compound will become aromatic
In 2nd compound there will be cross conjugation but there will be more resonating structures than C
That’s why A should be 1st and then out of B and C I’m a bit skeptical because B has more reso structures but there is cross conjugation in it

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For 6th
Option b seems correct to me , as 3H2 gets absorbed and hence randomness decreases
So Δ S< 0
And as reactant was more stable than product therefore
Δ H > 0

But answer key says e is it correct or wrong

I’m sorry ΔHo must be -ve as -T ΔSo is + ve therefore for reaction to be spontaneous ΔHo must be -ve

Option e must be correct

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In question 20 i think c should be the incorrect option but answer key mentions a as incorrect option

20 c is correct , because of resonance all the bond strength and length are same in benzene

Also option a is incorrect as all the C are Sp² hybridised

Question 27 and 28

28 b?

Yes but why what are the rules for dimerization

To attain stability some compounds undergo dimerization.
Here cyclobutadiene is antiaromatic it has 4pi electron . It is relatively less stable than aromatic compounds . It will undergo dimerization to become more stable (aromatic).
Common example of dimerization is NO2 (it has odd electrons , it is less stable). It undergoes dimerization to become N2O4

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Thanks

Question 32 and 33

32 d ?