Are carbocations stabilised by 2p-2p back bonding more stable than cycloproplylmethyl+ carbocations?

[Ayush_2020]

Basically pls tell whether the order of stability is correct or not. If not pls tell the correct answer.
It's taught to me that while comparing order of stability , maximum stability is brought by 2p-2p back bonding then resonance and other electronic displacement affect. Is it always true?

[ananda_krishnan]

Acc to me it should be : 3>1>2>4.
1.3rd structure(CPM-Cyclo propyl methyl cation)is exceptionally stable coz of Dancing resonance..if you havent studied the concept of dancing resonance,do check it in YouTube for better explanation .It is also called sigma bond resonance as the sigma bond is utilised here for completing the octet of carbonation.
2.1st structure comes next in stability...same reason as you have stated above...backbonding b/w 2p-2p orbitals.
3.2nd and 4th structures have simple resonance..and thus are second last and last in stability.2nd structure is more stable than the 4th structure coz of more number of resonating structures in 2nd structure.I hope there are 7 reso structures and 5 reso structures respectively in 2nd and 4th structure.
(I hope the 4th structure is that on the rightmost end of the page).

[Ayush_2020]

Sir, why 1 is more stable than 2 ? 2 is aromatic and 1 is non aromatic . And stability of aromatic will always be more than non aromatic.
Or in this question we have to think like 1 after formation of back bond , octate of all atom will complete whereas octate of all elements won't be complete in 2.
Pls explain sir.

[ananda_krishnan]

Structures in which all the atoms have complete octet are highly stable. In 1st structure this is applicable.But in case of structure 2 ,all the atoms do not have complete octet on proceeding with resonance.
As long as rings are not there..we don't have to compare structures using aromaticity principles.I mean 1st structure and 2nd structure can't be compared on aromaticity basis as structure 1 doesn't have a ring.

[Ayush_2020]

Sir , on the whole we apply aromatic>non aromatic>anti aromatic
Only when we are given cyclic compound.
Am I correct.

[ananda_krishnan]

Yes

[seshu_ram]

Bro for first one back bonding how??

[Ayush_2020]

CH2+ will have vacant p orbital , whereas oxygen has lone pair , so there will be back bonding.
Just remember that as nitrogen can form max of 4 bonds , similarly oxygen can form at max 3 bonds.