Question 1

π electron cloud kya hota hai aur wo kahan hota hai?

Accepted Answer

π bond do parallel p orbitals ke sideways (lateral) overlap se banta hai. Iska electron density molecular plane ke upar aur neeche do hisso mein baithta hai — plane par bilkul zero hota hai. Isi zero-density plane ko nodal plane kehte hain. Isliye π ko ek line nahi, ek cloud samajhna chahiye.

Question 2

Ethyne ka π cloud cylindrical kyun hota hai?

Accepted Answer

sp hybridisation mein har carbon ke paas do unhybridised p orbitals bachte hain, jo aapas mein 90° par hote hain. Dono alag-alag π bond banate hain — ek upar-neeche, doosra aage-peeche. Ye do perpendicular clouds mil kar C≡C axis ke chaaro taraf ek cylinder jaisa symmetric electron cloud bana dete hain.

Question 3

Benzene mein saare C–C bond lengths barabar kyun hote hain?

Accepted Answer

Har carbon ka bacha hua p orbital sirf ek padosi se nahi, dono padosiyon se barabar overlap karta hai. Isliye 6 π electrons kisi ek bond mein locked nahi rehte — wo poori ring par delocalised ho jaate hain. Result: har C–C identical, length 139 pm (single 154 aur double 134 ke beech) aur bond order 1.5.

Question 4

Resonance ka matlab kya hai — kya benzene do structures ke beech switch karta hai?

Accepted Answer

Nahi, ye sabse common NEET trap hai. Benzene kabhi Kekulé A, kabhi Kekulé B nahi hota. Wo hamesha ek hi cheez hai — resonance hybrid — jise hum ek single Lewis structure se theek se likh hi nahi paate. Kekulé structures sirf hamari kaagaz wali limitation hain, koi real oscillation nahi.

Question 5

σ aur π bond mein basic farak kya hai?

Accepted Answer

σ head-on (axial) overlap se banta hai, uska density nuclei ko jodne wali line par hota hai, wo strong hota hai aur us axis ke around rotation free hota hai. π sideways overlap se banta hai, density plane ke upar-neeche hoti hai, wo weaker hota hai aur rotation lock kar deta hai. Har bond ka pehla hamesha σ hota hai; double mein 1σ+1π, triple mein 1σ+2π.

Ethene, Ethyne & Benzene — the π Electron Cloud in 3D

Practice

Learned it? Now prove it under the clock.