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.
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).

Match the LIST-I with LIST-II for an isothermal process of an ideal gas system. 
Choose the correct answer from the options given below: