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Chapter 1: Problem 198

Match the elements of Column I to elements of Column II. There can be single or multiple matches. Column I (a) Carbocation (b) Free radicals (c) Singlet carbene (d) Triplet carbene Column II (p) \(\mathrm{sp}^{2}\) hybridization (q) paramagnetic (r) diamagnetic (s) linear

Short Answer

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Question: Match each chemical species in Column I with its corresponding properties in Column II. Column I: a) Carbocation b) Free radicals c) Singlet carbene d) Triplet carbene Column II: p) \(\mathrm{sp}^{2}\) hybridization q) paramagnetic r) diamagnetic s) linear geometry Answer: a) Carbocation: (p) \(\mathrm{sp}^{2}\) hybridization, (r) diamagnetic b) Free radicals: (p) \(\mathrm{sp}^{2}\) hybridization, (q) paramagnetic c) Singlet carbene: (p) \(\mathrm{sp}^{2}\) hybridization, (r) diamagnetic, (s) linear geometry d) Triplet carbene: (q) paramagnetic, (s) linear geometry

Step by step solution

01

Analyze Carbocation

Carbocation is a positively charged carbon species with an empty p orbital. It has three sigma bonds with neighboring atoms and no lone pairs, which means it has a trigonal planar geometry and is \(\mathrm{sp}^{2}\) hybridized. It's also diamagnetic as it has no unpaired electrons.
02

Free Radicals

Free radicals are species with unpaired electrons, which makes them paramagnetic. Their geometry is similar to carbocations as they have three sigma bonds with neighboring atoms, and no lone pairs, which makes them \(\mathrm{sp}^{2}\) hybridized as well.
03

Singlet Carbene

A singlet carbene has a divalent carbon atom with two single bonds, a lone pair of electrons, and a vacant p orbital. It has \(\mathrm{sp}^{2}\) hybridized carbon with a lone pair and a linear geometry. Singlet carbenes are diamagnetic, as all their electrons are paired.
04

Triplet Carbene

Triplet carbene has a divalent carbon atom with two single bonds and two unpaired electrons in different orbitals. The unpaired electrons make triplet carbene paramagnetic. It has linear geometry and \(\mathrm{sp}\) hybridized carbon, as the electrons occupy separate orbitals rather than a common p orbital. Now we can match the elements of Column I to elements of Column II based on our analysis: (a) Carbocation: (p) \(\mathrm{sp}^{2}\) hybridization, (r) diamagnetic (b) Free radicals: (p) \(\mathrm{sp}^{2}\) hybridization, (q) paramagnetic (c) Singlet carbene: (p) \(\mathrm{sp}^{2}\) hybridization, (r) diamagnetic, (s) linear (d) Triplet carbene: (q) paramagnetic, (s) linear

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