To determine the shape of a carbocation, we need to understand the electronic configuration and hybridization involved when a carbocation forms.
A carbocation is a positively charged carbon atom with only six electrons in its valence shell. Because it is electron-deficient, it lacks an octet and thus cannot exhibit tetrahedral geometry, which requires an electron pair for each bond in a 3D space.
The most stable configuration that a carbocation can achieve is trigonal planar. Here is the reasoning behind this:
Therefore, the correct answer, based on hybridization and geometry theory, is that a carbocation takes a trigonal planar shape.
Let's briefly consider why the other options are incorrect:
Thus, considering all the points, the shape of the carbocation is trigonal planar.
A carbocation has a central carbon atom with three bonded groups and one empty {p}-orbital. The geometry around the carbocation is trigonal planar because the three groups are arranged at 120$^\circ$ to minimize electron repulsion.
\[\textbf{Carbocation structure:} \quad \text{H-C$^+$-H}.\]
The trigonal planar shape is due to {sp}$^2$-hybridization of the central carbon atom.
The reaction represented by \( A \rightarrow B \) follows first-order kinetics. At a given temperature, 20% of the reaction is completed in 223 s. The time taken to complete 50% of the reaction at the same temperature is _________ s (rounded off to the nearest integer).
Statement-1: \( \text{ClF}_3 \) has 3 possible structures.
Statement-2: \( \text{III} \) is the most stable structure due to least lone pair-bond pair (lp-bp) repulsion.

Which of the following options is correct?