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.
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
The expression given below shows the variation of velocity \( v \) with time \( t \): \[ v = \frac{At^2 + Bt}{C + t} \] The dimension of \( A \), \( B \), and \( C \) is:
The dimensions of a physical quantity \( \epsilon_0 \frac{d\Phi_E}{dt} \) are similar to [Symbols have their usual meanings]
