See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1: Identify the molecular formula. The species is C2H3 with a positive charge, so it is C2H3+. Step 2: Analyze the hybridization clues given. The positively charged carbon is sp-hybridized and the uncharged carbon is sp2-hybridized. An empty p-orbital is perpendicular to the pi system. Step 3: Build the structure. An sp-hybridized carbon forms two sigma bonds and has two p-orbitals available for pi bonding. An sp2-hybridized carbon forms three sigma bonds and has one p-orbital. For C2H3+: the sp2 carbon bears two hydrogens (accounting for 2 H) and one sigma bond to the sp carbon; the sp carbon bears one hydrogen (accounting for the 3rd H) and has one p-orbital forming a pi bond with the sp2 carbon AND one empty p-orbital perpendicular to that pi system. Step 4: This gives the structure H2C=C+-H, i.e., a vinyl cation. In the vinyl cation (ethenyl cation), one carbon is sp2 (the =CH2 end) and the positively charged carbon is sp-hybridized (the =C+-H end), with an empty sp-hybridized orbital (or equivalently an empty p-orbital perpendicular to the pi bond) on the charged carbon. Step 5: Evaluate other options. Allenyl cation would be C3-based. Alkyl cation (sp3) does not match. Allyl cation involves sp2 carbons with resonance delocalization and is C3H5+. None of these match C2H3+ with the described hybridization pattern. Step 6: The description matches the vinyl cation (ethenyl cation), option (a). Therefore, the correct answer is A.