How many pπ–pπ and pπ–dπ bond present in SO3(s):
How many pπ–pπ and pπ–dπ bond present in SO3(s):
Option 1 - <p>pπ–pπ = 5, pπ–dπ = 2</p>
Option 2 - <p>pπ–pπ = 10, pπ–dπ = 2</p>
Option 3 - <p>pπ–pπ = 0, pπ–dπ = 6</p>
Option 4 - <p>pπ–pπ = 12, pπ–dπ = 2</p>
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4 months ago
Correct Option - 3
Detailed Solution:
SO3 (s) exist as cyclic trimer ⇒ S3O9
→ sp3, so all bond π–bonds would be of pπ–dπ so pπ–pπ bond = 0
pπ–dπ bonds = 6
Similar Questions for you
CH3COOH + NaOH → CH3COONa + H2O
ΔH = –50.6 kJ/mol
NaOH + SA [HCl] → NaCl + H2O
ΔH = –55.9 kJ/mol
the value of ΔH for ionisation of CH3COOH
⇒ ΔH = +55.9 – 50.6
5.3 kJ/mol
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Chemistry Chemical Bonding and Molecular Structure 2025
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