Question

Geometry of methyl free radical is:
A.Pyramidal
B.Planar
C.Tetragonal
D.Linear

Answer 1

To answer this question you must recall the VSEPR theory. The Valence shell electron pair repulsion theory proposes that the hybridized orbitals in an atom arrange themselves in such a way so as to minimize the repulsion between them, hence determining the geometry of a molecule on the basis of its hybridization.

Complete answer:
We know that carbon has the atomic number 6. The number of valence electrons is 4. The ground state electronic configuration of carbon can be written as
$C:\left[ {He} \right]2{s^2}2{p^2}$
One electron is excited from the $2s$ orbital to $2p$ orbital to increase the covalency. The electronic configuration in this excited state is given as $C:\left[ {He} \right]2{s^1}2{p^3}$
Now we know that there is one free electron in a free radical. A methyl free radical consists of a carbon atom bonded to three hydrogen atoms and with one extra unpaired electron. This unpaired electron is highly reactive and thus is not included in the hybridization. So the hybridization of the carbon atom is $s{p^2}$ in the methyl free radical.
The geometry of the molecule is planar with three hybridized orbitals forming three bonds and the unpaired electron in an unhybridized $p -$ orbital.

Thus, the correct answer is B.

Note:
The concept of mixing of atomic orbitals in order to form new hybrid orbitals that possess different shapes and energies as compared to the original parent atomic orbitals is known as hybridisation. Hybrid orbitals are suitable to form chemical bonds of equal energies. Also hybridization of orbitals leads to the formation of more stable compounds because hybrid orbitals have lower energy than the unhybrid orbitals.